Home Explore M1-Allens Made Maths Theory + Exercise [II]

# M1-Allens Made Maths Theory + Exercise [II]

## Description: M1-Allens Made Maths Theory + Exercise [II]

JEE-Mathematics FUNCTION 1. CARTESIAN PRODUCT OF TWO SETS : 2. 3. The cartesian product of two non-empty sets A & B is the set of all possible ordered pair of the form (a, b) where E the first entry comes from set A & second comes from set B. A × B = {(a, b) | a  A, b  B} e.g. A = {1, 2, 3} B = {p, q} A × B = {(1, p)(1, q), (2, p), (2, q), (3, p), (3, q)} Note : (i) If either A or B is the null set, then A × B will also be empty set, i.e. A × B =  (ii) If n(A) = p & n(B) = q, then n(A × B) = p × q, where n(X) denotes the number of elements in set X. RELATION : A relation R from a non-empty set A to non-empty set B is subset of cartesian product A × B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B. The second element is called the image of the first element. Note : (i) The set of all first elements of the ordered pairs in a relation R from a set A to a set B is called the domain of the relation R. (ii) The set of all second elements in a relation R from a set A to a set B is called the range of the relation R. The whole set B is called the co-domain of the relation R. (iii) Range subset of co-domain. FUNCTION : A relation R from a set A to a set B is called a function if each element of A has unique image in B. It is denoted by the symbol. ƒ : A  B or A ƒ  B which reads ‘ƒ ’ is a function from A to B ‘or’ ƒ maps A to B, If an element a  A is associated with an element b  B, then b is called ‘the ƒ image of a’ or ‘image of a under ƒ ‘or’ the value of the function ƒ at a’. Also a is called the pre-image of b or argument of b under the function ƒ. We write it as b = ƒ (a) or ƒ : a  b or ƒ : (a, b) Thus a function ‘ƒ ’ from a set A to a set B is a subset of A × B in which each 'a' belonging to A appears in one and only one ordered pair belonging to ƒ. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 Representation of Function : ( a ) Ordered pair : Every function from A  B satisfies the following conditions : (i) ƒ  A x B (ii)  a  A there exist b  B and (iii) (a, b)  ƒ & (a, c)  ƒ  b = c (b) Formula based (uniformly/nonuniformly) : e.g. (i) ƒ : R  R, y = ƒ (x) = 4x , ƒ (x) = x2 (uniformly defined) (ii) ƒ( x )  x  1 1  x  4 (non-uniformly defined) x 4 x7 (iii) x2 x 0 (non-uniformly defined) ƒ(x)   x 1 x  0 (c) Graphical representation : y1 y1 x1 x1 y2 Graph (1) Graph (2) Graph(1) represent a function but graph(2) does not represent a function. 1

JEE-Mathematics Note : (i) If a vertical line cuts a given graph at more than one point then it can not be the graph of a function. (ii) Every function is a relation but every relation is not necessarily a function. 4 . DOMAIN, CO-DOMAIN & RANGE OF A FUNCTION : Let ƒ : A  B, then the set A is known as the domain of ƒ & the set B is known as co-domain of ƒ . The set of ƒ images of all the elements of A is known as the range of ƒ . Thus : Domain of ƒ = {a  a  A, (a, ƒ (a))  ƒ } Range of ƒ = {ƒ (a)  a  A, ƒ (a)  B} Note : (i) It should be noted that range is a subset of co-domain. (ii) If only the rule of function is given then the domain of the function is the set of those real numbers, where function is defined. For a continuous function, the interval from minimum to maximum value of a function gives the range Illustration 1 : Find the Domain of the following function : (i) f(x) = sin x  16  x2 (ii) y = log(x–4) (x2 – 11x + 24)   1  (iii) f(x) = log2   log1 / 2 1  4 x   1 Solution : (i) sinx  0 and 16 – x2  0  2n  x  (2n + 1) and –4  x  4  Domain is [–4, –]  [0, ] (ii) y = log( (x–4) x2 – 11x + 24) Here 'y' would assume real value if, x – 4 > 0 and  1, x2 – 11x + 24 > 0  x > 4 and  5, (x – 3) (x – 8) > 0  x > 4 and  5, x < 3 or x > 8  x > 8  Domain (y) = (8, ) (iii) We have   1  f(x )  log2   log1 / 2 1  4 x   1  1 f(x) is defined if  log1 / 2 1  4 x   1  0 or  1 or if 1  1   (1 / 2 ) 1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 if log1 / 2 1  4 x   1  4x   or 1 if 1 or if x1/4 < 1 or if 0 < x < 1 if 1   2 or 1 4x 4x D(f) = (0, 1) Illustration 2 : Find the range of following functions : 1  sin x  cos x  3 2  (i) f(x) = 8  3 sin x (ii) f(x) = log2  2   (iii) f(x) = log 2 2  log2 (16 sin2 x  1) Solution : 1 (i) f(x) = 8  3 sin x –1  sinx  1  Range of f = 1 , 1  11 5  2 E

JEE-Mathematics  sin x  cos x  3 2  (ii) Let y = log2  2   2y = sin  x   + 3  –1  2y – 3  1  4   2  2y  4   y  [1, 2]  (iii) f(x) = log 2 2  log2 (16 sin2 x  1) 1  16 sin2 x + 1  17  0  log2 (16 sin2x +1)  log2 17  2 – log2 17  2 – log2 (16 sin2 x + 1)  2 Now consider 0 < 2 – log2 (16 sin2x + 1)  2  – < log 2 [2  log2 (16 sin2 x  1)]  log 22 2  the range is (–, 2] Do yourself - 1 : ( i ) Find the domain of following functions : (a) y = 1 – log x (b) y = 5 – 2x 1 1 10 (c) y = (d) y = x2 – 4x | x| – x ( i i ) Find the range of the following function : (a) f(x) = sin3x 1 (c) f(x) = e–3x (b) f(x) = 3  cos x (d) f(x) = cos  2 x    (e) f(x) = 2 sin  2 x    x2 4   4  (f) f(x) = x4  1 (i i i ) Given 'n' real numbers a , a .....a . Determine the value of x at which the function 12 n f(x) = (x – a )2 + (x – a )2 + .....+ (x – a )2 takes on the minimum value. 12 n 5.Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 IMPORTANT TYPES OF FUNCTION : E (a) Polynomial function : If a function ‘f’ is called by f(x) = a0xn + a1xn–1 + a2xn–2 + ...+ an–1 x + an where n is a non negative integer and a0, a1, a2,....an are real numbers and a0 0, then f is called a polynomial function of degree n. Note : (i) A polynomial of degree one with no constant term is called an odd linear function. i.e. f(x) = ax, a 0 (ii) There are two polynomial functions, satisfying the relation; f(x). f(1/x) = f(x) + f(1/x). They are (1) f(x) = xn + 1 & (2) f(x) = 1 – xn , where n is a positive integer. (iii) Domain of a polynomial function is R (iv) Range of odd degree polynomial is R whereas range of an even degree polynomial is never R. (b) Algebraic function : A function ‘f’ is called an algebraic function if it can be constructed using algebraic operations (such as addition, subtraction, multiplication, division, and taking radicals) starting with polynomials. x4  16x2 Examples : f(x) = x2  1 ; g(x) = x  x + (x – 2) 3 x  1 If y is an algebraic function of x, then it satisfies a polynomial equation of the form P0(x)yn + P1(x)yn–1 + ........ + Pn–1(x)y + Pn(x) = 0, where ‘n’ is a positive integer and P0(x), P1(x),....... are polynomial in x. Note that all polynomial functions are Algebraic but the converse in not true. A functions that is not algebraic is called TR ANSCEDENTAL function. 3

JEE-Mathematics Basic algebraic function : (i) y = x2 y Domain : R y = x2 R a n g e : R+  {0} or [0, ) ox D o m a i n : R – {0} or R0 R a n g e : R – {0} 1 y (ii) y = y = 1/x Domain : R0 R a n g e : R+ or (0, ) x ox 1 (iii) y = x2 y y = 1/x2 ox (iv) y = x3 y y = x3 Domain : R 1 Range : R o x 1 (c) Rational function : Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 g(x) A rational function is a function of the form y = f(x) = h(x) , where g(x) & h(x) are polynomials & h(x)  0, Domain : R–{x | h(x)=0} Any rational function is automatically an algebraic function. (d) Identity function : y The function f : A  A defined by f(x) = x  x  A is called the identity x function on A and is denoted by IA. It is easy to observe that identity function is a bijection. y y=C (e) Constant function : ox f : A   B is said to be constant function if every element of A has the same f image in B. Thus f : A   B ; f(x) = c,  x  A, c  B is E constant function. Note that the range of a constant function is a singleton set. Domain : R Range : {C} 4

JEE-Mathematics (f ) Trigonometric functions : (–3/2,1) Y (i) Sine function (/2,1) f(x) = sin x Domain : R o X Range : [–1, 1], period 2 (ii) Cosine function (–/2,–1) (3/2,–1) f(x) = cos x Domain : R Y Range : [–1, 1], period 2 (0,1) (iii) Tangent function f(x) = tan x o  /2 X (– ,–1) ( ,–1) Y   /2 –3/2 x| (2n  1)  O /2 3X  2  Domain : R  x  ,n  I  Range : R , period  (iv) Cosecant function Y f(x) = cosec x Domain : R – {x|x = n, n I} (–3/2,1) (/2,1) y=1 Range : R – (–1, 1), period 2 o X (v) Secant function (–/2,–1) y=–1 f(x) = sec x Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65Domain : R – {x|x = (2n + 1) /2 : n  I} x=–2 x=  x=  x=2 Range : R – (–1, 1), period 2 Y (vi) Cotangent function f(x) = cot x (–2,1) (0,1) y=1 Domain : R – {x|x = n, n I} (-,-1) Range : R, period  o X (, -1) y=-1 E5 x=–3 x=–  x=  x= 3 22 2 2 Y O x (–3 ,0) (–  ,0) (  ,0) ( 3 ,0) 2 22 2 x=–2 x=– x= x=2

JEE-Mathematics (g ) Exponential and Logarithmic Function : A function f(x) = ax(a > 0), a  1, x R is called an exponential function. The inverse of the exponential function is called the logarithmic function, i.e. g(x)= logax. Note that f(x) & g(x) are inverse of each other & their graphs are as shown. (If functions are mirror image of each other about the line y = x) Domain of ax is R Range R+ Domain of logax is R+ Range R y y + + f(x)=ax, a>1 (0,1) (0,1) f(x)=ax, a (0,1) (1,0) x 45° x (1,0) y=x y = x g(x)=logax g(x)=logax y y = log2x 1 01 y = log3x y = log5x y = log10x x y y=e–x y Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 y=ex x x O O Note-1 : f(x) = a1/x, a > 0 Domain : R – {0} Range : R+ – {1} 1 Domain : R+ – {1} Range : R – {0} N o t e - 2 : f(x) = lo gxa = loga x (a > 0) (a   1) 6 E

JEE-Mathematics (h) Absolute value function : y=–x y The absolute value (or modulus) of a real number x y=x (written |x|) is a non negative real number that satisfies the conditions. O x x if x  0 |x| = x if x  0 Domain : R The properties of absolute value function are Range : [0, ) (i) the inequality |x|   means that   x  ; if  > 0 (ii) the inequality |x|   means that x or x –if  > 0 (iii) |x ± y|  |x| + |y| (iv) |x ± y|  ||x| – |y|| (v) |xy| = |x|.|y| (vi) x | x| , (y  0) y | y| 1 Range : R+ Note : f(x)= , Domain : R – {0}, | x| Illustration 3 : Determine the values of x satisfying the equality. Solution : |(x2 + 4x + 9) + (2x – 3)| = |x2 + 4x + 9| + |2x – 3|. Illustration 4 : The equality |a + b| = |a| + |b| is valid if and only if both summands have the same sign,  x2 + 4x + 9 = (x + 2)2 + 5 > 0 at any values of x, the equality is satisfied at those values of x at which 2x – 3  0, i.e. at x  3 . 2 Determine the values of x satisfying the equality |x4 – x2 – 6| = |x4 – 4| – |x2 + 2|. Solution : The equality |a – b| = |a|–|b| holds true if and only if a and b have the same sign and |a||b|. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 In our case the equality will hold true for the value of x at which x4 – 4  x2 + 2. Hence x2 – 2  1; |x|  3 . Do yourself - 2 : ( i ) Determine the values of x satisfying the equality |x2 – 8| =|3x2 – 5| –|2x2 + 3|. (i) Signum function : y Signum function y = sgn (x) is defined as follows y= 1  x 1 for x  0 x  ,x  0 0 y = Sgn x y   x  for x  0 o 0, x  0 1 for x  0 Domain : R y = -1 R a n g e : {–1, 0, 1} 7 E

JEE-Mathematics y (j) Greatest integer or step up function : graph of y = [x] 3 The function y = f(x) = [x] is called the 2 greatest integer function where [x] 1 denotes the greatest integer less than or equal to x. Note that for : x [x] –3 –2 –1 01 2 3 x [–2,–1) –2 –1 [–1,0) –1 0 –2 [0,1) 1 [1,2) –3 Domain : R Range : I Properties of greatest integer function : (i) [x]  x < [x] + 1 and x – 1 < [x]  x, 0  x – [x] < 1 (ii) [x + m] = [x] + m if m is an integer. (iii) [x] + [–x] = 0, x  I 1, x  I 1 1 n Note : f(x) = [x] Domain : R – [0, 1) Range : {x| x  , 0 } n (k) Fractional part function : y It is defined as : g(x) = {x} = x – [x] e.g. the fractional part of the number 2.1 graph of y = {x} Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 is 2.1–2 = 0.1 and the fractional part 1 of –3.7 is 0.3 The period of this function is 1 and graph of this function is as shown. x {x} D o m a i n : R -2 0 1 23 x [–2,–1) x+2 -1 [–1,0) x+1 Range : [0,1) [0,1) x 1 [1,2) x–1 N o t e : f(x )= {x} D o m a i n : R – I R a n g e : ( 1,  ) Properties of Fractional part function : (i) 0  {x} < 1 (ii) {[x]} = [{x}] = 0 (iii) {{x}} = {x} 1, x  I (iv) {x+m} = {x}, m  I (v) {x} +{–x} = 0, x  I Illustration 5 : If y = 2[x] + 3 & y = 3[x – 2] + 5 then find [x + y] where [ . ] denotes greatest integer function. 8E

JEE-Mathematics Solution : y = 3[x – 2] + 5 = 3[x] – 1 so 3 [x] – 1 = 2 [x] + 3 [x] = 4  4  x < 5 then y = 11 so x + y will lie in the interval [15, 16) so [x + y] = 15 Illustration 6 : Find the value of 1   1  1  ...... 1  2946  where [ . ] denotes greatest integer 2  2 1000  2 1000  function ? Solution : 1   1  1  ...... 1  499   1  500   ......  1  1499   1  1500   ...... 2   2 1000   2 1000  2 1000   2 1000  2 1000  + 1  2499   1  2500   ......  1  2946  2 1000  2 1000   2 1000  = 0 + 1 × 1000 + 2 × 1000 + 3 × 447 = 3000 + 1341 = 4341 Ans. 1 Illustration 7 : Find the domain f(x) = [| x| 5]  11 where [.] denotes greatest integer function. Solution : | x| 5   11 so  x  5  11 or  x  5  11  x   16  x   6 x  17 or x  6 (Not Possible)  x  –17 or x  17 so x  (–, –17]  [17, ) Illustration 8 : Find the range of f(x) = x  [x] , where [.] denotes greatest integer function. 1  x  [x] Solution : y = x  [x]  {x} 1  x  [x] 1  {x} Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65  1  1 1  1 1y y y {x} {x} y  {x} = 1  y 0  {x} < 1  0  y  1 1y Range = [0, 1/2) Illustration 9 : Solve the equation |2x – 1| = 3[x] + 2{x} where [.] denotes greatest integer and { . } denotes Solution : fractional part function. We are given that, |2x – 1| = 3[x] + 2{x} Let, 2x – 1 i.e. x 1 The given equation yields.  0  . 2 1 – 2x = 3[x] + 2{x}  1 – 2[x] – 2{x} = 3[x] + 2{x}  1  5[x] 1 – 5[x] = 4{x}  {x} = 4 1  5[x] 31  0  4 < 1  0  1 – 5[x] < 4  – 5 < [x]  5 E9

JEE-Mathematics 31 Now, [x] = 0 as zero is the only integer lying between – and 55 11 11  {x} = 4  x = 4 which is less than 2 , Hence 4 is one solution. 1 2[x] + 2{x} – 1 = 3[x] + 2{x} Now, let 2x – 1 > 0 i.e x > 2  2x – 1 = 3[x] + 2{x}  1  [x] = –1  –1  x < 0 which is not a solution as x > 2 1  x = 4 is the only solution. Do yourself - 3 : ( i ) Let {x} & [x] denotes the fraction and integral part of a real number x respectively, then match the column. Column-I Column-II (A) [x2] > 3 (p) x  [2, 4) (B) [x]2 – 5[x] + 6 = 0 (q) x  (–, –2]  [2, ) (C) x = {x} (r) x  {0} (D) {x} = [x] (s) x  (–, –5) (E) [x] < –5.2 (t) x  {–2} (F) 1 + x = sgn(x) (u) x  [0, 1) 6 . ALGEBR AIC OPER ATIONS ON FUNCTIONS : If f & g are real valued functions of x with domain set A, B respectively, f + g, f – g, (f. g) & (f/g) as follows : ( a ) (f ± g)(x) = f(x) ± g(x) domain in each case is A  B (b ) (f.g)(x) = f(x).g(x) domain is A  B (c)  f  (x)  f(x) domain A  B – {x|g(x)  0}  g  g(x) 7 . EQUAL OR IDENTICAL FUNCTION : Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 Two function f & g are said to be equal if : ( a ) The domain of f = the domain of g (b) The range of f = range of g and ( c ) f(x) = g (x), for every x belonging to their common domain (i.e. should have the same graph) 1x e.g. f(x) = x & g(x) = x2 are identical functions.  x 1 Illustration 10 : The functions f(x) = log (x – 1) – log (x – 2) and g(x) = log  x  2  are identical when x lies in the interval (A) [1, 2] (B) [2, ) (C) (2, ) (D) (–, ) Solution : Since f(x) = log (x – 1) –log (x – 2). Domain of f(x) is x > 2 or x  (2, ) .....(i)  x 1 x 1  x  (–, 1)  (2, ) .....(ii) g(x) = log  x  2  is defined if x  2 > 0 From (i) and (ii), x  (2, ). Ans. (C) 10 E

JEE-Mathematics Do yourself - 4 : ( i ) Are the following functions identical ? x x2 (b) f(x) = x & (x) = x2 (c) f(x) = log x2 &  (x) = 2log |x| (a) f(x) = x2 & (x) = x 10 10 8 . CLASSIFICATION OF FUNCTIONS : (a) One-One function (Injective mapping) : A function f : A  B is said to be a one-one function or injective mapping if different elements of A have different f images in B. Thus for x1, x2  A & f(x1), f(x2)  B, f(x1) = f(x2)  x1 = x2 or x1  x2  f(x1)  f(x2). Note: (i) Any continuous function which is entirely increasing or decreasing in whole domain is one-one. (ii) If a function is one-one, any line parallel to x-axis cuts the graph of the function at atmost one point Diagramatically an injective mapping can be shown B A BA or (b) Many-one function : A function f : A B is said to be a many one function if two or more elements of A have the same f image in B. Thus f : A  B is many one if  x1, x2  A, f(x1) = f(x2) but x1  x2 Diagramatically a many one mapping can be shown B A x1 or x2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 Note : If a continuous function has local maximum or local minimum, then f(x) is many-one because atleast one line parallel to x-axis will intersect the graph of function atleast twice. (c) Onto function (Surjective mapping) : If the function f : A  B is such that each element in B (co-domain) is the ‘f’ image of atleast one element in A, then we say that f is a function of A 'onto' B. Thus f : A  B is surjective if  b  B,  some a  A such that f(a) = b Diagramatically surjective mapping can be shown B A or Note that : If range = co-domain, then f(x) is onto. (d) Into function : If f : A  B is such that there exists atleast one element in co-domain which is not the image of any element in domain, then f(x) is into. E 11

JEE-Mathematics Diagramatically into function can be shown B BA A or Thus a function can be one of these four types : (i) one-one onto (injective & surjective) (ii) one-one into (injective but not surjective) (iii) many-one onto (surjective but not injective) (iv) many-one into (neither surjective nor injective) Note : (i) If ‘f’ is both injective & surjective, then it is called a Bijective mapping. The bijective functions are also named as invertible, non singular or biuniform functions. (ii) If a set A contains n distinct elements then the number of different functions defined from A  A is nn & out of it n! are one one and rest are many one. (iii) f : R  R is a polynomial (a) Of even degree, then it will neither be injective nor surjective. (b) Of odd degree, then it will always be surjective, no general comment can be given on its injectivity. Illustration11 : Let A = {x : –1  x  1} = B be a mapping f : A  B. For each of the following functions from Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 A to B, find whether it is surjective or bijective. (a) f(x) = |x| (b) f(x) = x|x| (c) f(x) = x3 (d) f(x) = [x] x (a) f(x) = |x| (e) f(x) = sin Graphically ; 2 Solution : Y Which shows many one, as the straight line is –1 O +1 X parallel to x-axis and cuts at two points. Here range for f(x)  [0, 1] Which is clearly subset of co-domain i.e., [0, 1]  [–1,1] Thus, into. Hence, function is many-one-into  Neither injective nor surjective E 12

(b) f(x) = x|x|= x2 , 1  x  0  , JEE-Mathematics   x 2 , 0  x  1  Y 1 Graphically, –1 O 1 X The graph shows f(x) is one-one, as the straight –1 line parallel to x-axis cuts only at one point. Here, range f(x)  [–1, 1] Thus, range = co-domain Hence, onto. Therefore, f(x) is one-one onto or (Bijective). Y (c) f(x) = x3, 1 Graphically; Graph shows f(x) is one-one onto –1 O 1 X –1 (i.e. Bijective) [as explained in above example] 2Y 3X (d) f(x) = [x], –2 –1 1 Graphically; O1 Which shows f(x) is many-one, as the straight line –1 parallel to x-axis meets at more than one point. Here, range –2 f(x)  {–1, 0, 1} which shows into as range  co-domain Hence, many-one-into x Y (e) f(x) = sin 1 2 –1 Graphically; O1 Which shows f(x) is one-one and onto as range –1 X = co-domain. Therefore, f(x) is bijective. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 Illustration 12 : Let f : R  R be a function defined by f (x) = x + x2 , then f is (A) injective (B) surjective (C) bijective (D) None of these Solution : We have, f(x) = x + x2 = x + | x | Clearly, f is not one-one as f(–1) = f(–2) = 0 and –1  – 2 Also, f is not onto as f(x)  0  x  R  range of f = (0, )  R Ans.(D) x2  3x  a Illustration 13 : Let f(x) = x2  x  1 , where f : R  R. Find the value of parameter 'a' so that the given function is one-one. Solution : x2  3x  a f(x) = x2  x  1 f'(x) = (x2  x  1)(2x  3)  (x2  3x  a )(2x  1) 2x2  2x(1  a )  (3  a) = (x2  x  1)2 (x2  x  1)2 Let, g(x) = –2x2 + 2x (1 – a) + (3 – a) g(x) will be negative if 4 (1 – a)2 + 8 (3 – a) < 0  1 + a2 – 2a + 6 – 2a < 0  (a – 2)2 + 3 < 0 which is not possible. Therefore function is not monotonic. Hence, no value of a is possible. E 13

JEE-Mathematics Do yourself - 5 : ( i ) Is the function f : N  N (the set of natural numbers) defined by f(x) = 2x + 3 surjective ? ( i i ) Let A = R – {3}, B = R – {1} and let f : A  B defined by f(x )  x  2 . Check whether the function f(x) x 3 is bijective or not. (i i i ) The function f : R  R, where R is the set of all real numbers defined by f(x) = 2x +3 is - (A) f is both one-one and onto (B) f is one-one but not onto (C) f is onto but not one-one (D) f is neither one-one nor onto 9 . BASIC TR ANSFORMATIONS ON GR APHS : (i) Drawing the graph of y = f(x) + b, b  R, from the known graph of y = f(x) A1 y=f(x)+b,b > 0 x0 OA y=f(x) y=f(x)+b,b < 0 A2 It is obvious that domain of f(x) and f(x) + b are the same. Let us take any point x0 in the domain of f(x). y  f(x0 ) . x x0 The corresponding point on f(x) + b would be f(x0) + b. For b > 0    f(x0) + b > f(x0) it means that the corresponding point on f(x) + b would be lying at a distance 'b' units above the point on f(x). For b < 0    f(x0) + b < f(x0) it means that the corresponding point on f(x) + b would be lying at a distance 'b' units below the point on f(x). Accordingly the graph of f(x) + b can be obtained by translating the graph of f(x) either in the positive y-axis direction (if b > 0) or in the negative y-axis direction (if b < 0), through a distance |b| units. (ii) Drawing the graph of y = –f(x) from the known graph of y = f(x) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 To draw y = –f(x), take the image of the curve y = f(x) in the x-axis as plane mirror. y y=f(x) y  0x x 0 y=–f(x) (iii) Drawing the graph of y = f(–x) from the known graph of y = f(x) To draw y = f(–x), take the image of the curve y = f(x) in the y-axis as plane mirror. y y=f(x) y  y=f(–x) 0x 0x 14 E

JEE-Mathematics (iv) Drawing the graph of y = |f(x)| from the known graph of y = f(x) |f(x)| = f(x) if f(x)  0 and |f(x)| = –f(x) if f(x) < 0. It means that the graph of f(x) and |f(x)| would coincide if f(x)  0 and for the portions where f(x) < 0 graph of |f(x)| would be image of y = f(x) in x-axis. y x y x y=f(x) y=|f(x)| O O (v) Drawing the graph of y = f(|x|) from the known graph of y = f(x) f(x), x  0 It is clear that, f(|x|) = f(x), . Thus f(|x|) would be a even function, graph of f(|x|) and f(x) x 0 would be identical in the first and the fourth quadrants (ax x  0) and as such the graph of f(|x|) would be symmetric about the y-axis (as (|x|) is even). yy y=f(x)  y=f(|x|) O x Ox (vi) Drawing the graph of |y| = f(x) from the known graph of y = f(x) Clearly |y|  0. If f(x) < 0, graph of |y| = f(x) would not exist. And if f(x)  0, |y| = f(x) would give y = ±f(x). Hence graph of |y| = f(x) would exist only in the regions where f(x) is non-negative and will be reflected about the x-axis only in those regions. yy O  O |y|=f(x) y=f(x) x x (vii) Drawing the graph of y = f(x + a), a  R from the known graph of y = f(x) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 y=f(x+a),a>0 y=f(x) y=f(x–a) x0–|a| x0 a<0 x0+|a| (i) If a > 0, shift the graph of f(x) through 'a' units towards left of f(x). (ii) If a < 0, shift the graph of f(x) through 'a' units towards right of f(x). (viii) Drawing the graph of y = af(x) from the known graph of y = f(x) y=af(x), a >1 y=f(x) y=af(x), 0<a<1 x It is clear that the corresponding points (points with same x co-ordinates) would have their ordinates in the ratio of 1 : a. E 15

JEE-Mathematics (ix) Drawing the graph of y = f(ax) from the known graph of y = f(x). y y=f(x) y=f(ax), a >1 y=f(ax),0<a<1 Ox Let us take any point x0  domain of f(x). Let ax = x0 or x = x0 . a 1 Clearly if 0 < a < 1, then x > x0 and ƒ (x) will stretch by a units along the y-axis and if a > 1, x < x0, then f(x) will compress by 'a' units along the y-axis. Illustration 14 : Find f(x) = max {1 + x, 1 – x, 2}. y y=2 Solution : From the graph it is clear that 1  x ; x  1 (–1,0) 0,1 f(x) = 2 ; 1  x  1 (1,0) x x 1 1  x ; (0,0) Illustration 15 : Draw the graph of y = |2 –|x –1||. Solution : y=x–1  (0,1) y=|x–1|  O y=–|x–1| x O 1 (0,–1) (1,0) (1,0) 1 O (–1,0) 2 y=2–|x–1| 2 y=|2–|x–1| Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 (–1,0) O 1 (3,0)  O1 x Illustration 16 : Draw the graph of y = 2  4 | x  1| y y= 1 y y= 1 y x x–1 y=|x–11| x=1 Solution : O x O x  x (0,–1) O 16 E

JEE-Mathematics y x=1 y y=2 y y=2– 4 y=– 1 x=1 |x–1| y=|x––41| |x–1| O x O x (0,–1)  (–1,0) (3,0) x (0,–2) (0,–4) Illustration 17 : Draw the graph of y = |e|x| – 2| y=ex y=e|x| 1 Solution :  O y y y=|e|x|–2| O (0,1) y=e|x|–2 x  Ox (0,–1) Illustration 18 : Draw the graph of f(x) = cosx cos(x + 2) – cos2(x + 1). Solution : f(x) = cosx cos(x + 2) – cos2(x + 1) = 1 co s(2 x  2 )  cos 2  – 1 cos(2 x  2)  1 0 2 2 ½cos2 – ½  1 cos2  1  0 . 22 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 1 0 . COMPOSITE OF UNIFORMLY & NON-UNIFORMLY DEFINED FUNCTION: Let f : A B & g : B C be two functions. Then the function gof : A C defined by (gof ) (x) = g(f(x))  x A is called the composite of the two functions f & g. Diagramatically x f(x) g g (f(x)) f Thus the image of every x  A under the function gof is the g-image of f-image of x. Note that gof is defined only if  x  A, f(x) is an element of the domain of ‘g’ so that we can take its g-image. Hence in gof(x) the range of ‘f’ must be a subset of the domain of ‘g’. Properties of composite functions: ( a ) In general composite of functions is not commutative i.e. gof  fog. ( b ) The composite of functions is associative i.e. if f, g, h are three functions such that fo(goh) & (fog)oh are defined, then fo(goh) = (fog)oh. ( c ) The composite of two bijections is a bijection i.e. if f & g are two bijections such that gof is defined, then gof is also a bijection. E 17

JEE-Mathematics Illustration 19 : If f be the greatest integer function and g be the modulus function, then (gof)   5  – (fog)   5  =  3   3  (A) 1 (B) –1 (C) 2 (D) 4 Solution :  5  – (fog)  5  = g f  5    f g  5   = g(–2) – f 5 Ans.(A) Illustration 20 : Given (gof)  3   3    3     3    3  = 2 – 1 = 1   Find the domain and range of h(x) = g(f(x)), where f(x) =  [x], 2  x  1 and g(x) =  [x],   x  0  1  x  2 sin x, , [.] denotes the greatest integer function.  x  1, 0x Solution :  [f(x)],   f(x)  0 f(x)= h(x) = g(f(x)) = sin(f(x)), 0  f(x)   +3 From graph of f(x), we get +1 –2 –1 –1 2 h(x) =  [[x ]], 1), 2  x  1 –2 sin (x  1  x  2 f(x)=–  Domain of h(x) is [–2, 2] and Range of h(x) is {–2, 1}  [sin3, 1] Illustration 21 :  x  1, x 1 and g(x) =  x2, 1  x  2 Let f(x) = 2x  1, 1 x 2  2  x  3 , find (fog) x  2, Solution : f(g(x)) =  g(x )  1, g(x)  1 y 2g(x)  1, 1  g(x)  2 5 Here, g(x) becomes the variable that means 4 x+2 we should draw the graph. It is clear that g(x)  1 ;  x  [–1, 1] 3 x2 2 and 1 < g(x)  2 ;  x  (1, 2 ] 1  x2  1, 1  x  1 –2 –1 0 1 2 3 x 1x 2  f(g(x)) =  2 2 x  1, Do yourself - 6 : Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 ( i ) f(x) = x3 – x & g(x) = sin 2x, find (a) f(f(1)) (b) f(f(–1)) (c) f  g       2     (d) f  g     (e) g(f(1)) (f) g  g       4     2       x  1; 0  x  2 (ii) If f(x) = | x| ; , then find fof(x). 2  x 3 11. HOMOGENEOUS FUNCTIONS : A function is said to be homogeneous with respect to any set of variables when each of its terms is of the same degree with respect to those variables. For examples 5x2 + 3y2 – xy is homogenous in x & y. Symbolically if, f(tx, ty) = tn f(x, y) then f(x, y) is homogeneous function of degree n. 18 E

JEE-Mathematics Illustration 22 : Which of the following function is not homogeneous ? (A) x3 + 8x2y + 7 y3 (B) y2 + x2 + 5xy xy 2x  y 1 (C) x2  y2 (D) 2y  x  1 Solution : It is clear that (D) does not have the same degree in each term. Ans. (D) 12. BOUNDED FUNCTION : A function is said to be bounded if |f(x)|  M, where M is a finite quantity. Do yourself - 7 : x2 ( i ) Find the boundness of the function f(x) = x4  1 13. IMPLICIT & EXPLICIT FUNCTION : A function defined by an equation not solved for the dependent variable is called an implicit function. e.g. the equations x3 + y3 = 1 & xy = yx, defines y as an implicit function. If y has been expressed in terms of x alone then it is called an Explicit function. Illustration 23 : Which of the following function is implicit function ? (A) y  x2  ex  5 (B) y = x2 x2 log x 1  cos1 x (C) xy – sin(x + y) = 0 (D) y  sin x Solution : It is clear that in (C) y is not clearly expressed in x. Ans. (C) Do yourself - 8 : ( i ) Which of the following function is implicit function ? (A) xy – cos(x + y) = 0 (B) y = x3 (C) y = log(x2 + x + 1) (D) y = |x| Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 ( i i ) Convert the implicit form into the explicit function : (a) xy = 1 (b) x2y = 1. 14. INVERSE OF A FUNCTION : Let f : A  B be a one-one & onto function, then their exists a unique function g : B  A such that f(x) = y  g(y) = x,   x  A & y  B. Then g is said to be inverse of f. Thus g = f–1 : B  A = {(f(x), x))|(x, f (x))  f}. Properties of inverse function : ( a ) The inverse of a bijection is unique. ( b ) If f : A  B is a bijection & g : B  A is the inverse of f, then fog = IB and gof = IA, where IA & IB are identity functions on the sets A & B respectively. If fof = I, then f is inverse of itself. ( c ) The inverse of a bijection is also a bijection. ( d ) If f & g are two bijections f : A B, g : B C then the inverse of gof exists and (gof)–1 = f–1 o g–1. E 19

JEE-Mathematics ( e) Since f(a) = b if and only if f–1(b) = a, the point (a, b) is on the graph of ‘f’ if and only if the point (b, a) is on the graph of f–1. But we get the point (b, a) from (a, b) by reflecting about the line y = x. y (b, a) y y y= f(x) y= x 0 (a, b) f–1 x 0x y= x x 0 (–1,0) (0,–1) y= x f y= f–1(x) The graph of f1 is obtained by reflecting the graph of f about the line y  x. Drawing the graph of y = f–1(x) from the known graph of y = f(x) For drawing the graph of y = f–1(x) we have to first of all find the interval in which the function is bijective (invertible). Then take the reflection of y = f(x) (within the invertible region) about the line y = x. The reflected part would give us the graph of y = f–1(x). e.g. let us draw the graph of y = sin–1x. We know that y = f(x) = sinx is invertible if f :   ,   [1, 1] 2 2   the inverse mapping would be f–1 : [1, 1]   ,   . 2 2  y y=x (1,/2) y=sinx /2 1 (/2,1) –/2 –1 x O 1 /2 (–/2,1) (0,–1) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 1/2) Illustration 24 : Let f : R  R be defined by f(x) = (ex–e–x)/2. Is f(x) invertible ? If so, find its inverse. Solution : Let us check for invertibility of f(x) : (a) One-One : Let x1, x2  R and x1< x2  ex1  e x2 (Because base e > 1) .......... (i) Also x1 < x2  –x2 < –x1  ex2  ex1 (Because base e > 1) .......... (ii) (i) + (ii)   ex1  ex2  ex2  ex1    1 e x1  ex11   e x2  ex2  f(x1) < f(x2) i.e. f is one-one. 22 20 E

JEE-Mathematics (b) Onto : As x tends to larger and larger values so does f(x) and when x  f(x)  Similarly as x  , f(x)  i.e.  < f(x) <  so long as x  ( ) Hence the range of f is same as the set R. Therefore f(x) is onto. Since f(x) is both one-one and onto, f(x) is invertible. (c) To find f–1 : Let f–1 be the inverse function of f, then by rule of identity fof–1(x) = x e f1 ( x )  e f1 ( x )  x  e2 f1 ( x )  2 xe f1 ( x )  1  0 2  e f1 (x )  2 x  4 x2  4  e f1 (x )  x  1  x2 2 Since >ef1 (x) 0, hence negative sign is ruled out and Hence ef1 (x ) = x  1  x2 Taking logarithm, we have f–1(x) = n(x  1  x2 ) .  x; 1  Illustration 25 : Find the inverse of the function f(x) =  x 2 ;1  x  4 8 x; x  4  x; 1  Solution : Given f(x) =  x 2 ;1  x  4 8 x; x  4 Let f(x) = y  x = f–1(y)  y, y  1   x   y, 1 y 4 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65  y 2 , y2  6 4  4 64  y, y  1   f 1 (y)   y, 1  y  16  y2 , y  16  64   x;  1  Hence f–1(x) =  x;1  x  16 Ans.   x2  ;x  16  64 Do yourself - 9 : ( i ) Let f : [–1, 1]  [–1, 1] defined by f(x) = x|x|, find f–1(x). E 21

JEE-Mathematics 15. ODD & EVEN FUNCTIONS : If a function is such that whenever 'x' is in it's domain '–x' is also in it's domain & it satisfies f(–x) = f(x) it is an even function f(–x) = –f(x) it is an odd function Note : (i) A function may neither be odd nor even. (ii) Inverse of an even function is not defined, as it is many – one function. (iii) Every even function is symmetric about the y-axis & every odd function is symmetric about the origin. (iv) Every function which has '–x' in it's domain whenever 'x' is in it's domain, can be expressed as the sum of an even & an odd function . e.g. f(x) = f(x) + f(-x) + f(x) - f(-x) 2 2 EVEN ODD (v) The only function which is defined on the entire number line & even and odd at the same time is f(x) = 0 ƒ(x) g(x) ƒ (x) + g(x) ƒ(x) – g (x) ƒ (x) . g(x) ƒ (x)/ g(x) (goƒ) (x) (ƒog)(x) odd odd odd odd even even odd odd e ven even eve n even even even even e ven odd even neither odd nor even neither odd nor even odd odd even e ven e ven odd neither odd nor even neither odd nor even odd odd even e ven Illustration 26 : Which of the following functions is (are) even, odd or neither : (i) f(x) = x2sinx (ii) f(x) = 1  x  x2  1  x  x2 (iii) 1  x f(x) = log  1  x  (iv) f(x) = sinx – cosx ex  ex (v) f(x) = 2 Solution : (i) f(–x) = (–x)2 sin(–x) = –x2 sinx = –f(x). Hence f(x) is odd. (ii) f(–x) = 1  (x)  (x)2  1  (x)  (x)2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 = 1  x  x2  1  x  x2 = –f(x). Hence f(x) is odd. (iii) f(–x) = log  1  (x )  1  x = –f(x). Hence f(x) is odd  1  (x )  = log  1  x  (iv) f(–x) = sin(–x) – cos(–x) = –sinx – cosx. Hence f(x) is neither even nor odd. ex  e(x) ex  ex (v) f(–x) =  = f(x). Hence f(x) is even 22 Illustration 27 : Identify the given functions as odd, even or neither : xx (ii) f(x + y) = f(x) + f(y) for all x, y  R (i) f(x) = ex  1  2  1 22 E

Solution : xx JEE-Mathematics (i) f(x) = ex  1  2  1 f(0) = 0 Clearly domain of f(x) is R ~ {0}. We have, f(x) = –f(–x) f(–x) = x x ex .x x (ex 1  1)x x ex 1  2 1  1  ex  2 1   1 (ex 1) 2 xx xx = x  ex  1  2  1  ex  1  2  1 = f(x) Hence f(x) is an even function. (ii) f(x + y) = f(x) + f(y) for all x, y  R Replacing x, y by zero, we get f(0) = 2f(0)  Replacing y by –x, we get f(x) + f(–x) = f(0) = 0  Hence f(x) is an odd function. Do yourself - 10 : ( i ) Which of the following functions is (are) even, odd or neither : (a) f(x) = x3 sin 3x ex2  e–x2 (b) f(x) = 2x ex – e–x (d) f(x) = x2 + 2x (c) f(x) = ex  e –x Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 16. PERIODIC FUNCTION : A function f(x) is called periodic if there exists a least positive number T(T >0) called the period of the function such that f(x + T) = f(x), for all values of x within the domain of f(x). e.g. The function sinx & cosx both are periodic over 2 & tan x  is periodic over . Note : For periodic function : (i) f(T) = f(0) = f(–T), where 'T' is the period. (ii) Inverse of a periodic function does not exist. (iii) Every constant function is periodic, but its period is not defined. (iv) If f(x) has a period T & g(x) also has a period T then it does not mean that f(x) + g(x) must have a period T. e.g. f(x) = |sin x| + |cos x|. 1 (v) If f(x) has period p, then f(x) and f(x) also has a period p. (vi) If f(x) has period T then f(ax + b) has a period T/|a| (a  0). Illustration 28 : Find the periods (if periodic) of the following functions, where [.] denotes the greatest integer function (i) f(x) = en(sinx) + tan3x – cosec(3x – 5) (ii) f(x) = x – [x – b], b  R sin x  cos x  (iii) f(x) = (iv) f(x) = tan 2 [x] sin x  cos x (v) f(x) = cos(sinx) + cos(cosx) (1  sin x)(1  sec x) (vi) f(x) = (1  cos x)(1  cos ec x) Solution : (vii) f(x) = e x[x ] cos x  cos 2 x ........ cos n (i) f(x) = en(sinx) + tan3x – cosec(3x – 5) Period of ensinx = 2, tan3x =  2 cosec (3x – 5) = 3  Period = 2 E 23

JEE-Mathematics (ii) f(x) = x – [x – b] = b + {x – b}  Period = 1 sin x  cos x (iii) f(x) = sin x  cos x Since period of |sinx + cosx| =  and period of |sinx| + |cosx| is  . Hence f(x) is periodic 2 with  as its period  (iv) f(x) = tan 2 [x]    tan 2 [x + T]= tan 2 [x]  [x + T] = n + [x] 2 2  T=2  Period = 2 (v) Let f(x) is periodic then f(x + T) = f(x)  cos(sin(x + T))+ cos(cos(x + T)) = cos(sinx) + cos(cosx) If x = 0 then cos(sinT) + cos(cosT) = cos(0) + cos(1) = cos  cos   + cos  sin   2   2   On comparing T = 2 (1  sin x)(1  sec x) (1  sin x)(1  cos x) sin x (vi) f(x) = (1  cos x)(1  cos ec x) = cos x(1  sin x)(1  cos x)  f(x) = tanx Hence f(x) has period . (vii) f(x) = e x[x ] cos x  cos 2 x ........ cos n Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 Period of x – [x] = 1 Period of |cosx| = 1 1 Period of |cos2x| = 2 ...................................... 1 Period of |cosnx| = n So period of f(x) will be L.C.M. of all period = 1 Illustration 29 : Find the periods (if periodic) of the following functions, where [.] denotes the greatest integer function (i) f(x) = ex–[x] + sinx (ii) f(x) = sin x  cos x (iii) f(x) = sin x  cos x (i) Period of ex–[x] = 1 23 3 23 Solution : period of sinx = 2  L.C.M. of rational and an irrational number does not exist.  not periodic. (ii) Period of sin x = 2  2 2 2 / 2 Period of cos x = 2  2 3 E 3 / 3  L.C.M. of two different kinds of irrational number does not exist.  not periodic. 24

JEE-Mathematics (iii) Period of sin x  2 2 3 3 / 3 Period of cos x  2  4 3 2 3 /2 3  L.C.M. of two similar irrational number exist.  Periodic with period = 4 3 Ans. Do yourself - 11 : ( i ) Find the periods (if periodic) of the following functions. (a) f(x) = 2n(cosx) + tan3x. (b) f(x) = sin (sinx) + sin (cosx) (c) f(x) = ex–[x], [.] denotes greatest integer function (d) f(x) = sin x  cos x 22 17. GENERAL : If x, y are independent variables, then : ( a ) f(xy) = f(x) + f(y)  f(x) = kn x ( b ) f(xy) = f(x) . f(y)  f(x) = xn, n  R or f(x) = 0 (c) f(x + y) = f(x) . f(y)  f(x) = akx or f(x) = 0 (d ) f(x + y) = f(x) + f(y)  f(x) = kx, where k is a constant. Miscellaneous Illustration : Illustration 30 : ABCD is a square of side . A line parallel to the diagonal BD at a distance 'x' from the vertex A cuts two adjacent sides. Express the area of the segment of the square with A at a vertex, as a function of x. Find this area at x = 1 / 2 and at x = 2, when  = 2. Solution : There are two different situations D C  F O  Case-I : when x = AP  OA, i.e., x  x P B 2 A E Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 ar(AEF )  1 x.2x  x2 ( PE = PF = AP = x) 2  D FC Case-II : when x = AP > OA, i.e., x  2 but x  2 2–x ar(ABEFDA) = ar(ABCD) – ar(CFE) P     2  1 2  x .2 2  x [ CP  2  x ] OE 2   2  22  x2  2 2x  2 2x  x2  2 AB  the required function s(x) is as follows :  x2,  1 at x  1  0x 2 s(x) =  ; area of s(x) =  2 Ans. 2 2 2  2 x – x2 – 2 ,  x 8( 2 – 1) at x  2  2 Illustration 31 : If the function f(x) satisfies the functional rule, f(x + y) = f(x) + f(y)  x,y  R & f(1) = 5, then find m  f(n) and also prove that f(x) is odd function. n 1 E 25

JEE-Mathematics Solution : Here, f(x + y) = f(x) + f(y); put x = t – 1, y = 1 f(t) = f(t – 1) + f(1) ....(1)  f(t) = f(t – 1) + 5  f(t) = {f(t – 2) + 5} + 5  f(t) = f(t – 2) + 2(5)  f(t) = f(t – 3) + 3(5) ............................. .............................  f(t) = f{t – (t –1)} + (t – 1)5  f(t) = f(1)+ (t – 1)5  f(t) = 5 + (t – 1)5  f(t) = 5t  m  m  5[1 2 3  .....  m]  5m(m  1)  f(n)  (5 n ) 2 n 1 n 1 m 5m(m  1) Hence, f(n)  2 .....(i) n 1 Now putting x=0, y=0 in the given function, we have f(0 + 0) = f(0) + f(0)  f(0) = 0 Also putting (–x) for (y) in the given function. f(x – x) = f(x) + f(– x)  f(0) = f(x) + f(–x)  0 = f(x) + f(–x)  f(–x) = –f(x) ....(ii) Thus, m 5m(m  1) and f(x) is odd function. f(n)  n 1 2 ANSWERS FOR DO YOURSELF 1 : ( i ) ( a ) x  (0, ) ( b ) x  (– , 5/2] ( c ) x  (– , 0)  (4, ) ( d ) x  (– , 0) (i i ) ( a ) [–1, 1] (b) 1 , 1 ( c ) [0, ) (d) [–1, 1] ( e ) [–2, 2] ( f ) 0, 1   4 2  2  (iii) x = a1  a2  ...  an Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\01.FUNCTION\\01.THEORY.p65 n 2 : ( i ) x  (,  8 )  ( 8, ) 3 : ( i ) (A)  (q), (B)  (p), (C)  (u), (D)  (r), (E)  (s), (F)  (t) 4 : (i) ( a) no (b) no (c) yes 5 : ( i ) not onto ( i i ) yes (iii) A 6 : (i) ( a ) 0 (b ) 0 (c) 0 (d) 0 (e) 0 (f) 0 x  2, 0  x  1 ( i i ) x  1, 1  x  2 x, 2  x  3 7 : (i) 0, 1  2  8 : (i) 1 1 9 : (i) A (ii) (a) y = (b ) y = x2 10 : (i) 11 : (i) x f 1 (x )   x, 1  x  0   x, 0  x  1 ( a ) even (b ) odd (c) odd (d ) neither (c) 1 (d)  (a) 2 (b) 2 26 E

JEE-Mathematics HYPERB O L A The Hyperbola is a conic whose eccentricity is greater than unity. (e > 1). 1 . STANDARD EQUATION & DEFINITION(S) : y Standard equation of the hyperbola is x=– a x=+ a e e x2 y2 b2 = a2 (e2  1) B L x = (ae, b2 /a) – = 1 , where (0, b) a2 b2 b2 S' A' A S x or a2 e2 = a2 + b2 i.e. e2 = 1 + a2 (–ae, 0) (–a, 0) c (0, 0) (a, 0) (ae, 0) 2 (0,–b) B' (ae, –b2 /a) L'  Conjugate Axis  = 1 +  Transverse Axis  ( a ) Foci : & S  ( ae, 0). S  (ae, 0) (b) Equations of directrices : aa x= e & x= e. ( C ) Vertices : A  (a , 0) & A  ( a , 0). (d) Latus rectum : (i) Equation : x = ± ae (ii) Length 2b2 Conjugate Axis 2 2a (e2  1) = 2e (distance from focus to directrix) = = Transverse Axis  = a (iii) Ends :  b2  –b 2  ;  b2  –b 2   a e , a ,  a e , a   –a e , a ,  –a e , a  (e) (i) Transverse Axis : NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 The line segment AA of length 2a in which the foci S & S both lie is called the Transverse Axis of the Hyperbola. (ii) Conjugate Axis : The line segment BB between the two points B  (0,  b) & B  (0, b) is called as the Conjugate Axis of the Hyperbola. The Transverse Axis & the Conjugate Axis of the hyperbola are together called the Principal axes of the hyperbola. (f) Focal Property : The difference of the focal distances of any point on the hyperbola is constant and equal to transverse axis i.e. P S – P S  = 2 a . The distance SS = focal length. (g) Focal distance : Distance of any point P(x, y) on Hyperbola from foci PS = ex – a & PS = ex + a. E 55

JEE-Mathematics Illustration 1 : Find the equation of the hyperbola whose directrix is 2x + y = 1, focus (1, 2) and eccentricity 3 . Solution : Let P(x, y) be any point on the hyperbola and PM is perpendicular from P on the directrix. Then by definition SP = e PM 2x  y 1 2  (SP)2 = e2 (PM)2  (x – 1)2 + (y – 2)2 = 3    4 1   5(x2 + y2 – 2x – 4y + 5) = 3(4x2 + y2 + 1 + 4xy – 2y – 4x)  7x2 – 2y2 + 12 xy – 2x + 14y – 22 = 0 which is the required hyperbola. Illustration 2 : The eccentricity of the hyperbola 4x2 – 9y2 – 8x = 32 is - 5 13 13 3 (A) (B) (C) (D) 3 3 2 2 Solution : 4x2 – 9y2 – 8x = 32  4(x – 1)2 – 9y2 = 36  (x  1)2 y2  1 94 Here a2 = 9, b2 = 4 b2 4 13  eccentricity e = 1  a2  1  Ans.(B) 9 3 x2 y2 Illustration 3 : If foci of a hyperbola are foci of the ellipse   1 . If the eccentricity of the hyperbola be 2, 25 9 then its equation is - x2 y2 x2 y2 x2 y2 (D) none of these (A)   1 (B)   1 (C)   1 4 12 12 4 12 4 Solution : 4 For ellipse e = , so foci = (±4, 0) 5 ae 4  2, For hyperbola e = 2, so a=  b= e 2 2 4 1  2 3 x2 y2 Ans.(A) Hence equation of the hyperbola is   1 4 12 Illustration 4 : Find the coordinates of foci, the eccentricity and latus-rectum, equations of directrices for the NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 hyperbola 9x2 – 16y2 – 72x + 96y – 144 = 0. Solution : Equation can be rewritten as x  4 2 y  32 1 so a = 4, b = 3 42  32 b2 = a2(e2 – 1) given e = 5 4 Foci : X = ± ae, Y = 0 gives the foci as (9, 3), (–1, 3) Centre : X = 0, Y = 0 i.e. (4, 3) a 16  directrices are 5x – 36 = 0; 5x – 4 = 0 Directrices : X = ± e i.e. x – 4 = ± 5 2b2 9 9 Latus-rectum = = 2. = a 42 56 E

JEE-Mathematics Do yourself - 1 : (i) Find the eccentricity of the hyperbola x2 y2 1 which passes through (4, 0) & (3 2, 2)  a2 b2 ( i i ) Find the equation to the hyperbola, whose eccentricity is 5 , focus is (a, 0) and whose directrix is 4x – 4 3y = a. (i i i ) In the hyperbola 4x2 – 9y2 = 36, find length of the axes, the co-ordinates of the foci, the eccentricity, and the latus rectum. ( i v ) Find the equation to the hyperbola, the distance between whose foci is 16 and whose eccentricity is 2. 2 . CONJUGATE HYPERBOLA : Two hyperbolas such that transverse & conjugate axes of one hyperbola are respectively the conjugate & the x2 y2 transverse axes of the other are called Conjugate Hyperbolas of each other. eg. – =1 & a2 b2 x2 y2 – + =1 are conjugate hyperbolas of each other . a2 b2 Note that : (i) If e & e are the eccentricities of the hyperbola & its conjugate then e 2 + e 2 = 1. 12 12 (ii) The foci of a hyperbola and its conjugate are concyclic and form the vertices of a square. (iii) Two hyperbolas are said to be similar if they have the same eccentricity. Illustration 5 : The eccentricity of the conjugate hyperbola to the hyperbola x2 – 3y2 = 1 is - (A) 2 (B) 2 / 3 (C) 4 (D) 4/3 Solution : Equation of the conjugate hyperbola to the hyperbola x2 – 3y2 = 1 is –x2 + 3y2 =1  x2 y2   1 1 1/3 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 Here a2 = 1, b2 = 1/3  eccentricity e = 1  a2 / b2  1  3  2 Ans. (A) Do yourself - 2 : ( i ) Find eccentricity of conjugate hyperbola of hyperbola 4x2 – 16y2 = 64, also find area of quadrilateral formed by foci of hyperbola & its conjugate hyperbola 3. RECTANGULAR OR EQUILATER AL HYPERBOLA : The particular kind of hyperbola in which the lengths of the transverse & conjugate axis are equal is called an E Equilateral Hyperbola. Note that the eccentricity of the rectangular hyperbola is 2 and the length of it's latus rectum is equal to it's transverse or conjugate axis. 57

JEE-Mathematics 4 . AUXILIARY CIRCLE : y A circle drawn with centre C & T.A. as Q P(asec, btan) a diameter is called the Auxiliary Circle of the hyperbola. Equation of the auxiliary (–a, 0) (0, 0)  A x circle is x2 + y2 = a2 . A' C (a, 0) N Note from the figure that P & Q are called the \"Corresponding Points\" on the hyperbola & the auxiliary circle. '' is called the eccentric angle of the point 'P' on the hyperbola. (0    2  ). Parametric Equation : The equations x = a sec  & y = b tan  together represents the hyperbola x2 – y2 = 1 where   is a a2 b2 parameter. The parametric equations ; x = a cos h , y = b sin h  also represents the same hyperbola. General Note : Since the fundamental equation to the hyperbola only differs from that to the ellipse in having b2 instead of b2 it will be found that many propositions for the hyperbola are derived from those for the ellipse by simply changing the sign of b2 . 5 . POSITION OF A POINT 'P' w.r.t. A HYPERBOLA : The quantity x12 – y12 = 1 is positive , zero or negative according as the point (x , y ) lies within , a2 b2 11 upon or outside the curve. 6 . LINE AND A HYPERBOLA : x2 y2 The straight line y= mx + c is a secant , a tangent or passes outside the hyperbola – =1 a2 b2 according as : c2 > = < a2 m2  b2. Equation of a chord of the hyperbola x2 y2 joining its two points P( & Q() is – =1 a2 b2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 x – y   cos – sin  cos a 2b 2 2 Illustration 6 : Show that the line x cos  + y sin  = p touches the hyperbola x2 y2 if a2 cos2  – b2 sin2  = p2.  1 a2 b2 Solution : The given line is x cos  + y sin  = p  y sin  = – x cos  + p  y = – x cot  + p cosec  Comparing this line with y = mx + c m = – cot , c = p cosec  x2 y2 1 Since the given line touches the hyperbola  then a2 b2 c2 = a2m2 – b2  p2 cosec2  = a2 cot2  – b2 or p2 = a2 cos2  – b2 sin2  58 E

JEE-Mathematics Illustration 7 : If (a sec, b tan) and (asec, b tan) are the ends of a focal chord of x2 y2  1 , then tan  tan  Solution : a2  b2 22 equal to - e 1 1e 1e e 1 (A) (B) (C) (D) e 1 1e 1e e 1 Equation of chord connecting the points (asec, b tan) and (asec, b tan) is x cos      y sin      cos     ........ (i) a  2  b  2   2  If it passes through (ae, 0); we have, e    cos      cos  2   2  cos     1  tan  . tan   2  22 e   tan  . tan   1  e      1  tan  .tan   2 2 1e  2  22 co s Similarly if (i) passes through (–ae, 0), tan  . tan   1  e Ans. (B, C) 2 2 1e Do yourself - 3 : (i) Find the condition for the line x + my + n = 0 to touch the hyperbola x2 y2 1  a2 b2 (ii) If the line y = 5x +1 touch the hyperbola x2 y2 1 {b > 4}, then -  4 b2 (A) b2  1 (B) b2 = 99 (C) b2 = 4 (D) b2 = 100 5 x2 y2 7 . TANGENT TO THE HYPERBOLA a 2  b 2  1 : (a) Point form : Equation of the tangent to the given hyperbola at the point (x , y ) is x x1 – y y1 = 1. 11 a2 b2 Note : In general two tangents can be drawn from an external point (x y ) to the hyperbola and they are 11 y  y = m (x  x) & y  y = m (x  x ), where m& m are roots of the equation 1 1 1 1 2 1 1 2 (x 2  a2) m2  2 xy m + y 2 + b2 = 0 . If D < 0, then no tangent can be drawn from (x y ) 1 11 1 11 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 to the hyperbola. ( b ) Slope form : The equation of tangents of slope m to the given hyperbola is y = m x ± a 2 m 2 – b 2 .   a2m , b2   a 2 m 2 – b 2  Point of contact are a2m2 – b2 Note that there are two parallel tangents having the same slope m. ( c ) Parametric form : Equation of the tangent to the given hyperbola at the point (a sec  , b tan ) is x sec  – y tan  = 1 . a b cos  1 – 2   2   1 + 2  Note : Point of intersection of the tangents at 1 & 2 is x = a , y = b tan  2   1 + 2  c o s  2  Illustration 8 : Find the equation of the tangent to the hyperbola x2 – 4y2 = 36 which is perpendicular to the line x – y + 4 = 0. E 59

JEE-Mathematics Solution : Let m be the slope of the tangent. Since the tangent is perpendicular to the line x – y = 0  m × 1 = –1  m = –1 Since x2 – 4y2 = 36 or x2  y2  1 36 9 Comparing this with x2  y2 1 a2 b2  a2 = 36 and b2 = 9 So the equation of tangents are y = (–1)x ± 36  (1)2  9 y = – x ± 27  x + y ± 3 3 = 0 Ans. Illustration 9 : x2 y2 The locus of the point of intersection of two tangents of the hyperbola a2  b2  1 if the product of their slopes is c2, will be - (A) y2 – b2 = c2(x2 + a2) (B) y2 + b2 = c2(x2 – a2) (C) y2 + a2 = c2(x2 – b2) (D) y2 – a2 = c2(x2 + b2) Solution : Equation of any tangent of the hyperbola with slope m is y = mx ± a2m2  b2 If it passes through (x , y ) then 11 (y – mx )2 = a2m2 – b2  (x 2 – a2) m2 – 2x y m + (y 2 + b2) = 0 11 1 11 1 If m = m , m then as given m m = c2 y 2  b2  c2 1  12 12 x12  a2 Hence required locus will be : y2 + b2 = c2(x2 – a2) Ans.(B) Illustration 10 : A common tangent to 9x2 – 16y2 = 144 and x2 + y2 = 9 is - 2 15 2 15 (C) y  3 2 15 2 15 (A) y  3 x (B) y  3 x x (D) y  3 x 7 7 7 7 7 7 7 7 Solution : x2 y2 1, x2 + y2 = 9  16 9 Equation of tangent y = mx + 16m2  9 (for hyperbola) (circle) Equation of tangent y = m'x + 3 1  m' 2 For common tangent m = m' and 3 1  m' 2 = 16m2  9 or 9 + 9m2 = 16m2 – 9 or 7m2 = 18  m = 3 2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 7 2 18  required equation is y = 3 x  3 1  77 2 15 Ans. (A,B,C,D) or y = 3 x  77 Do yourself - 4 : ( i ) Find the equation of the tangent to the hyperbola 4x2 – 9y2 = 1, which is parallel to the line 4y = 5x + 7. (ii) Find the equation of the tangent to the hyperbola 16x2 – 9y2 = 144 at  5, 16  .  3  (iii) Find the common tangent to the hyperbola x2 y2 1 and an ellipse x2 y2   1. 16 9 43 60 E

JEE-Mathematics x2 y2 8 . NORMAL TO THE HYPERBOLA a 2  b2  1 : ( a ) Point form : The equation of the normal to the given hyperbola at the point P (x , y ) on it is 11 a 2 x + b 2 y = a 2 + b 2 = a2 e2. x1 y1 ( b ) Slope form : The equation of normal of slope m to the given hyperbola is y = m x  m ( a 2 + b 2 ) (a2 – m2b2 )  a2 , mb2  foot of normal are     ( a 2 - m 2 b 2 ) ( a 2 - m 2 b 2 )  (c) Parametric form : The equation of the normal at the point P (a sec , b tan) to the given hyperbola is a x + b y = a 2 + b 2 = a2 e2. sec tan x2 y2  1 , if - Illustration 11 : Line x cos  + y sin  = p is a normal to the hyperbola  a2 b2 (A) a2 sec2 – b2 cosec2 = (a2  b2 )2 (C) a2 sec2 + b2 cosec2 = (a2  b2 )2 p2 p2 (C) a2 cos2 – b2 sin2 = (a2  b2 )2 (D) a2 cos2 + b2 sin2 = (a2  b2 )2 p2 p2 Solution : Equation of a normal to the hyperbola is ax cos + by cot = a2 + b2 comparing it with the given line equation a cos  b cot  a2  b2 sec   ap bp  , tan   cos  sin  p  cos (a2  b2 ) sin (a2  b2 ) Eliminating , we get a2p2 b2p2 (a2  b2 )2  1  a2 sec2 – b2 cosec2 = Ans.(A) cos2(a2  b2 )2 sin2(a2  b2 )2 p2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 x2 y2 Illustration 12 : The normal to the hyperbola a2  b2  1 meets the axes in M and N, and lines MP and NP are drawn at right angles to the axes. Prove that the locus of P is hyperbola (a2x2 – b2y2) = (a2 + b2)2. Solution : Equation of normal at any point Q is ax cos   by cot   a2  b2  a2  b2   a2  b2   M   a sec , 0 , N   0, b tan   N P x  Let P  (h, k) M a2  b2 a2  b2 x' Q  h sec , k  tan  A' C A ab a2h2 b2k2 sec2  tan2  1  a2  b2   a2  b2 2   locus of P is (a2x2 – b2y2) = (a2 + b2). E 61

JEE-Mathematics Do yourself - 5 : x2 y2 ( i ) Find the equation of normal to the hyperbola   1 at (5, 0). 25 16 (ii) Find the equation of normal to the hyperbola x2 y2 1 at the point  6, 3 5  .   2  16 9 (iii) Find the condition for the line x + my + n = 0 is normal to the hyperbola x2 y2 1. a2  b2 9 . HIGHLIGHTS ON TANGENT AND NORMAL : x2 y2 ( a ) Locus of the feet of the perpendicular drawn from focus of the hyperbola – = 1 upon any a2 b2 tangent is its auxiliary circle i.e. x2 + y2 = a2 & the product of lengths to these perpendiculars is b2 (semi Conjugate Axis)2 ( b ) The portion of the tangent between the point of contact & the directrix subtends a right angle at the corresponding focus. y Light Ray Tangent ( c ) The tangent & normal at any point of a Q hyperbola bisect the angle between the P focal radii. This spells the reflection  property of the hyperbola as \"An incoming light ray \" aimed towards one focus is reflected from the outer S' x surface of the hyperbola towards S the other focus. It follows that if an ellipse and a hyperbola have the same foci, they cut at right angles at any of their common point. Note that the ellipse x2 y2 1 & the hyperbola x2 y2 =1 (a > k > b > 0) a2  b2 a2  k2  k2  b2 are confocal and therefore orthogonal. ( d ) The foci of the hyperbola and the points P and Q in which any tangent meets the tangents at the vertices are concyclic with PQ as diameter of the circle. 1 0 . DIRECTOR CIRCLE : NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 The locus of the intersection of tangents which are at right angles is known as the Director Circle of the hyperbola. The equation to the director circle is : x2 + y2 = a2  b2. If b2 < a2, this circle is real ; if b2 = a2 the radius of the circle is zero & it reduces to a point circle at the origin. In this case the centre is the only point from which the tangents at right angles can be drawn to the curve. If b2 > a2, the radius of the circle is imaginary, so that there is no such circle & so no tangents at right angle can be drawn to the curve. Note : Equations of chord of contact, chord with a given middle point, pair of tangents from an external point are to be interpreted in the similar way as in ellipse. 11. ASYMPTOTES : Definition : If the length of the perpendicular let fall from a point on a hyperbola to a straight line tends to zero as the point on the hyperbola moves to infinity along the hyperbola, then the straight line is called the Asymptote of the Hyperbola. 62 E

JEE-Mathematics To find the asymptote of the hyperbola : Let y = mx + c is the asymptote of the hyperbola x 2 – y 2 = 1 . .........(1) a2 b2 Solving these two we get the quadratic as (b2 a2m2) x2 2a2 mcx  a2 (b2 + c2) = 0 In order that y = mx + c be an asymptote, both y roots of equation (1) must approach infinity, the conditions for which are : coefficient of x – y =0 ab x2 = 0 & coefficient of x = 0. B (0, b) b (–a,0) A'  x  b2  a2m2 = 0 or m =  & C A (a, 0) x + y =0 a B' (0, –b) ab a2 mc = 0   c = 0.  equations of asymptote are x + y = 0 ab and x – y = 0 . ab combined equation to the asymptotes x2 – y2 = 0 . a2 b2 Particular Case : When b = a the asymptotes of the rectangular hyperbola. x2  y2 = a2 are y =   x which are at right angles. Note : (i) Equilateral hyperbola   rectangular hyperbola. (ii) If a hyperbola is equilateral then the conjugate hyperbola is also equilateral. (iii) A hyperbola and its conjugate have the same asymptote. (iv) The equation of the pair of asymptotes differ the hyperbola & the conjugate hyperbola by the same constant only. (v) The asymptotes pass through the centre of the hyperbola & the bisectors of the angles between the asymptotes are the axes of the hyperbola. (vi) The asymptotes of a hyperbola are the diagonals of the rectangle formed by the lines drawn through the extremities of each axis parallel to the other axis. (vii) Asymptotes are the tangent to the hyperbola from the centre. (viii) A simple method to find the coordinates of the centre of the hyperbola expressed as a general equation of degree 2 should be remembered as : Let f (x , y) = 0 represents a hyperbola. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 f f f f Find  x &  y . Then the point of intersection of  x = 0 &  y = 0 gives the centre of the hyperbola. Illustration 13 : Find the asymptotes of the hyperbola 2x2 + 5xy + 2y2 + 4x + 5y = 0. Find also the general equation of all the hyperbolas having the same set of asymptotes. Solution : Let 2x2 + 5xy + 2y2 + 4x + 5y +  = 0 be asymptotes. This will represent two straight line so 4  25  25  8  25   0 24  =2  2x2 + 5xy + 2y2 + 4x + 5y + 2 = 0 are asymptotes  (2x + y + 2) = 0 and (x + 2y + 1) = 0 are asymptotes and 2x2 + 5xy + 2y2 + 4x + 5y + c = 0 is general equation of hyperbola. E 63

JEE-Mathematics Illustration 14 : Find the hyperbola whose asymptotes are 2x – y = 3 and 3x + y – 7 = 0 and which passes through the point (1, 1). Solution : The equation of the hyperbola differs from the equation of the asymptotes by a constant  The equation of the hyperbola with asymptotes 3x + y – 7 = 0 and 2x – y = 3 is (3x + y – 7)(2x – y – 3) + k = 0 It passes through (1, 1)  k = –6. Hence the equation of the hyperbola is (2x – y – 3)(3x + y – 7) = 6. Do yourself - 6 : ( i ) Find the equation to the chords of the hyperbola x2 – y2 = 9 which is bisected at (5, –3) x2 y2 ( i i ) If m and m are the slopes of the tangents to the hyperbola   1 which pass through the point 12 25 16 (6, 2), then find the value of 11m m and 11(m + m ). 12 12 (iii) Find the locus of the mid points of the chords of the circle x2 + y2 = 16 which are tangents to the hyperbola 9x2 – 16y2 = 144. ( i v ) The asymptotes of a hyperbola are parallel to lines 2x + 3y = 0 and 3x + 2y = 0. The hyperbola has its centre at (1, 2) and it passes through (5, 3). Find its equation. 12. HIGHLIGHTS ON ASYMPTOTES ( a ) If from any point on the asymptote a straight line be drawn perpendicular to the transverse axis, the product of the segments of this line, intercepted between the point & the curve is always equal to the square of the semi conjugate axis. ( b ) Perpendicular from the foci on either asymptote meet it in the same points as the corresponding directrix & the common points of intersection lie on the auxiliary circle. x2 y2 (c) The tangent at any point P on a hyperbola – =1 with centre C, meets the asymptotes in Q a2 b2 and R and cuts off a  CQR of constant area equal to ab from the asymptotes & the portion of the tangent intercepted between the asymptote is bisected at the point of contact . This implies that locus of the centre of the circle circumscribing the   CQR in case of a rectangular hyperbola is the hyperbola itself. x2 y2 (d) If the angle between the asymptote of a hyperbola – =1 is 2  then the eccentricity of a2 b2 the hyperbola is sec . 1 3 . RECTANGULAR HYPERBOLA : y NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 Rectangular hyperbola referred to its asymptotes as axis of coordinates. ( a ) Equation is xy = c2 with parametric representation x = ct, y = c/t, t  R – {0}. ( b ) Equation of a chord joining the points P (t ) & Q(t ) is o x 12 E x + t t y = c (t + t ) with slope, m = –1 12 12 t1t2 xy ( c ) Equation of the tangent at P (x , y ) is + = 2 11 x1 y1 x & at P (t) is + t y = 2 c. t ( d ) Equation of normal is y – c = t 2 ( x – ct ) t ( e ) Chord with a given middle point as (h, k) is kx + hy = 2hk. 64

Note : JEE-Mathematics For the hyperbola, xy = c2 (i) Vertices : (c, c) & (–c, –c). (ii) Foci : ( 2 c, 2 c) & ( – 2 c, – 2 c) (iv) Latus rectum :  = 2 2 c = T . A = C . A (iii) Directrices : x + y =  2 c Illustration 15 : A triangle has its vertices on a rectangular hyperbola. Prove that the orthocentre of the triangle also lies on the same hyperbola. Solution : Let t , t and t are the vertices of the triangle ABC, described on the rectangular hyperbola xy = c2. 12 3  co-ordinates of A, B and C are  c   c  and  c respectively ct1, t1 , ct2 , t2   ct 3 ,       t3  c c y HG KJc Now slope of BC is t3 t2   1 A ct1, t1 ct3  ct2 t2 t3 E  Slope of AD is t t GH KJc 23 C ct3, t3 Equation of altitude AD is y c  t2 t3 (x  ct1 ) D x t1 o or t y – c = xt t t – ct12 t2 t3 ........ (i) HG KJ Bc 1 123 ct2, t2 Similarly equation of altitude BE is t2y – c = xt1t2t3 – ct1 t22 t3 ........ (ii) Solving (i) and (ii), we get the orthocentre   t1 c t ,  ct1 t 2 t3  which lies on xy = c2.  t2   3  Do yourself - 7 : (i) If equation ax2 + 2hxy + by2 + 2gx + 2ƒy + c = 0 represents a rectangular hyperbola then write required conditions. ( i i ) Find the equation of tangent at the point (1, 2) to the rectangular hyperbola xy = 2. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 (i i i ) Prove that the locus of point, tangents from where to hyperbola x2 – y2 = a2 inclined at an angle    with x-axis such that tan tan = 2 is also a hyperbola. Find the eccentricity of this hyperbola. Miscellaneous Illustrations : Illustration 16 : Chords of the circle x2 + y2 = a2 touch the hyperbola x2/a2 – y2/b2 = 1. Prove that locus of their Solution : middle point is the curve (x2 + y2)2 = a2x2 – b2y2. Let (h, k) be the mid-point of the chord of the circle x2 + y2 = a2, so that its equation by T = S1 is hx + ky = h2 + k2 h h2  k2 or y = – x  i.e. of the form y = mx + c kk It will touch the hyperbola if c2 = a2m2 – b2   h2  k2 2  a2   h 2  b2 or (h2 + k2)2 = a2h2 – b2k2  k  k       Generalising, the locus of mid-point (h, k) is (x2 + y2)2 = a2x2 – b2y2 E 65

JEE-Mathematics x2 y2  1 . The tangent at any point P on this hyperbola meets Illustration 17 : C is the centre of the hyperbola  Solution : a2 b2 the straight lines bx – ay = 0 and bx + ay = 0 in the points Q and R respectively. Show that CQ . CR = a2 + b2. P is (a sec, b tan) Tangent at P is x sec   y tan   1 ab It meets bx – ay = 0 i.e. xy in Q ab  Q is  sec  a tan  , b   sec   tan   xy It meets bx + ay = 0 i.e.  in R. ab  R is  sec  a ta n  , b    sec   tan    CQ.CR = (a2  b2 ) . (a  b2 ) ( sec2 – tan2 = 1) Ans. = a2 + b2 sec   tan  sec   tan  Illustration 18 : A circle of variable radius cuts the rectangular hyperbola x2 – y2 = 9a2 in points P, Q, R and S. Determine the equation of the locus of the centroid of triangle PQR. Solution : Let the circle be (x – h)2 + (y – k)2 = r2 where r is variable. Its intersection with x2 – y2 = 9a2 is obtained by putting y2 = x2 – 9a2. x2 + x2 – 9a2 – 2hx + h2 + k2 – r2 = 2k (x2  9a2 ) or [2x2 – 2hx + (h2 + k2 – r2 – 9a2)]2 = 4k2(x2 – 9a2) or 4x4 – 8hx3 +.....= 0  Above gives the abscissas of the four points of intersection. 8h  x1  4  2h x + x + x + x = 2h 1234 Similarly y + y + y + y = 2k. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 1234 Now if ( ) be the centroid of PQR, then 3 = x+ x + x, 3 = y + y + y 1 2 3 1 2 3  x = 2h – 3, y = 2k – 3 44 But (x , y ) lies on x2 – y2 = 9a2 44  (2h – 3)2 + (2k – 3)2 = 9a2 Hence the locus of centroid ( ) is (2h – 3x)2 + (2k – 3y)2 = 9a2 or  x  2h 2   y 2k 2  a2  3      3  Illustration 19 : If a circle cuts a rectangular hyperbola xy = c2 in A, B, C, D and the parameters of these four points be t , t , t and t respectively, then prove that : 123 4 (a) t t t t = 1 1234 (b) The centre of mean position of the four points bisects the distance between the centres of the two curves. 66 E

JEE-Mathematics Solution : (a) Let the equation of the hyperbola referred to rectangular asymptotes as axes be xy = c2 or its parametric equation be x = ct, y = c/t ........... (i) and that of the circle be x2 + y2 + 2gx + 2ƒy + k = 0 ........... (ii) Solving (i) and (ii), we get c2 c c2t2 +  2gct  2ƒ  k  0 t2 t or c2 t4  2gct3  kt2  2ƒct  c2  0 ........... (iii) Above equation being of fourth degree in t gives us the four parameters t , t , t , t of the 1234 points of intersection. 2gc 2g  t + t + t + t = –  ........... (iv) 1 2 3 4 c2 c ............ (v) ttt +ttt +ttt +ttt 123 124 341 342 2ƒc 2ƒ = – c2  c tttt = c2  1 . It proves (a) ............ (vi) 1234 c2 Dividing (v) by (vi), we get 1 1 1 1 2ƒ     t1 t2 t3 t4 c ............ (vii) (b) The centre of mean position of the four points of intersection is c c  1 1 1 1  = c   2g , c   2ƒ  , by (iv) and (vii) (t1  t2  t3  t4 ),    t4   c 4  c   4 4  t1   4     t2 t3  = (–g/2, –ƒ/2) Above is clearly the mid-point of (0, 0) and (–g, –ƒ ) i.e. the join of the centres of the two curves. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 1 : (i) 3 ANSWERS FOR DO YOURSELF ( i i ) 7y2 + 24xy – 24ax – 6ay + 15a2 = 0 (iii) 6, 4; ( 13, 0) ; 13 / 3 ; 8/3 ( i v ) x2 – y2 = 32 2 : ( i ) 5 & 40 sq. units (ii) B (iii) y = ± x ± 7 3 : ( i ) n2 = a22 – b2m2 (i i ) 5x – 3y = 9 4 : (i) 24y = 30x ± 161 a2 b2 (a2  b2 )2 5: (i) y = 0 ; (ii) 8 5x 18y  75 5 (iii)  6: 2 m 2 n2 7: (i) 5x + 3y = 16 (ii) 20 & 24 (iii) (x2 + y2)2 = 16x2 – 9y2 E (iv) (2x + 3y – 8)(3x + 2y – 7) = 154 ( i )   0, h2 > ab, a + b = 0 (i i ) 2x + y = 4 (iii) e = 3 67

JEE-Mathematics EXERCISE - 01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . The eccentricity of the hyperbola 4x2 – 9y2 – 8x = 32 is - 5 13 4 3 (A) (B) (C) (D) 3 3 3 2 2 . The locus of the point of intersection of the lines 3x  y  4 3k  0 and 3kx  ky  4 3  0 for different values of k is - (A) ellipse (B) parabola (C) circle (D) hyperbola 3 3 . If the latus rectum of an hyperbola be 8 and eccentricity be 5 then the equation of the hyperbola can be - (A) 4x2 – 5y2 = 100 (B) 5x2 – 4y2 = 100 (C) 4x2 + 5y2 = 100 (D) 5x2 + 4y2 = 100 4 . If the centre, vertex and focus of a hyperbola be (0,0), (4, 0) and (6,0) respectively, then the equation of the hyperbola is – (A) 4x2 – 5y2 = 8 (B) 4x2 – 5y2 = 80 (C) 5x2 – 4y2 = 80 (D) 5x2 – 4y2 = 8 5 . The equation of the hyperbola whose foci are (6,5), (–4, 5) and eccentricity 5/4 is– (x  1)2 (y  5)2 x2 y2 (x  1)2 (y  5)2 (D) (x  1)2  (y  5)2  1 (A)   1 (B)   1 (C)   1 16 9 16 9 16 9 49 6 . The vertices of a hyperbola are at (0, 0) and (10,0) and one of its foci is at (18,0). The possible equation of the hyperbola is – x2 y2 (x  5)2 y2 x2 (y  5)2 1 (x  5)2 (y  5)2 (A)   1 (B)   1 (C)  (D)   1 25 144 25 144 25 144 25 144 7 . The length of the transverse axis of a hyperbola is 7 and it passes through the point (5, –2). The equation of the hyperbola is – (A) 4 x2  196 y2  1 (B) 49 x2  51 y2  1 (C) 4 x2  51 y2  1 (D) none of these 49 51 4 196 49 196 x2 y2 8 . AB is a double ordinate of the hyperbola – = 1 such that AOB (where 'O' is the origin) is an equilateral a2 b2 triangle, then the eccentricity e of the hyperbola satisfies - (A) e > 3 2 2 2 (B) 1 < e < (C) e = (D) e > 3 3 3 9 . The equation of the tangent lines to the hyperbola x2 – 2y2 = 18 which are perpendicular to the line y = x are - NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 (A) y = x ± 3 (B) y = – x ± 3 (C) 2x + 3y + 4 = 0 (D) none of these x2 y2 y2 x2 1 10. The equations to the common tangents to the two hyperbolas  1 and  are - a2 b2 a2 b2 (A) y = ± x ± b2  a2 (B) y = ± x ± (a2 – b2) (C) y = ± x ± a2  b2 (D) y = ± x ± a2  b2 1 1 . Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola 16y2 – 9x2 = 1 is - (A) x2 + y2 = 9 (B) x2 + y2 = 1/9 (C) x2 + y2 = 7/144 (D) x2 + y2 = 1/16 1 2 . The ellipse 4x2 + 9y2 = 36 and the hyperbola 4x2 – y2 = 4 have the same foci and they intersect at right angles then the equation of the circle through the points of intersection of two conics is - (A) x2 + y2 = 5 (B) 5 (x2 + y2) – 3x – 4y = 0 (C) 5 (x2 + y2) + 3x + 4y = 0 (D) x2 + y2 = 25 1 3 . The equation of the common tangent to the parabola y2 = 8x and the hyperbola 3x2 – y2 = 3 is - (A) 2x ± y + 1 = 0 (B) x ± y + 1 = 0 (C) x ± 2y + 1 = 0 (D) x ± y + 2 = 0 68 E

JEE-Mathematics 1 4 . Equation of the chord of the hyperbola 25x2 – 16y2 = 400 which is bisected at the point (6, 2) is - (A) 16x–75y=418 (B) 75x–16y=418 (C) 25x–4y = 400 (D) none of these 1 5 . The asymptotes of the hyperbola xy – 3x – 2y = 0 are- (A) x – 2 = 0 and y – 3 = 0 (B) x – 3 = 0 and y – 2 = 0 (C) x + 2 = 0 and y + 3 = 0 (D) x + 3 = 0 and y + 2 = 0 x2 y2 1 6 . If the product of the perpendicular distances from any point on the hyperbola – = 1 of eccentricity a2 b2 e = 3 on its asymptotes is equal to 6, then the length of the transverse axis of the hyperbola is - (A) 3 (B) 6 (C) 8 (D) 12 1 7 . If the normal to the rectangular hyperbola xy = c2 at the point 't' meets the curve again at 't ' then t3t has 11 the value equal to - (A) 1 (B) –1 (C) 0 (D) none 1 8 . Area of triangle formed by tangent to the hyperbola xy = 16 at (16, 1) and co-ordinate axes equals - (A) 8 (B) 16 ( C) 32 (D) 64 19 . Locus of the middle points of the parallel chords with gradient m of the rectangular hyperbola xy = c2 is - (A) y + mx = 0 (B) y – mx = 0 (C) my – x = 0 (D) my + x = 0  x2 y2  1. 20. Let P (asec, btan) and Q (asec, btan), where     , be two points on the hyperbola  a2 b2 2 If (h,k) is the point of intersection of the normals at P & Q, then k is equal to - [JEE 99] a2  b2  a2  b2  a2  b2  a2  b2  (A) (B)   a  (C) (D)   b  a b 2 1 . If x = 9 is the chord of contact of the hyperbola x2–y2 = 9, then the equation of the corresponding pair of tangents, is - [JEE 99] (A) 9x2–8y2 + 18x –9 = 0 (B) 9x2–8y2 –18x +9 = 0 (C) 9x2–8y2 –18x –9 = 0 (C) 9x2–8y2 + 18x +9 = 0 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 2 2 . Consider the hyperbola 3x2 – y2 – 24x + 4y – 4 = 0 - (A) its centre is (4, 2) (B) its centre is (2, 4) (C) length of latus rectum = 24 (D) length of latus rectum = 12 x2 y2 23. Let an incident ray L1 = 0 gets reflected at point A(–2, 3) on hyperbola  1 & passes through focus a2 b2 S(2, 0), then - (A) equation of incident ray is x + 2 = 0 (B) equation of reflected ray is 3x + 4y = 6 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 (C) eccentricity, e = 2 (D) length of latus rectum = 6 2 4 . For the curve 5(x – 1)2 + 5(y – 2)2 = 3(2x + y – 1)2 which of the following is true - (A) a hyperbola with eccentricity 3 (B) a hyperbola with directrix 2x + y – 1 = 0 (C) a hyperbola with focus (1, 2) (D) a hyperbola with focus (2, 1) 2 5 . The equation of common tangent of hyperbola 9x2 – 9y2 = 8 and the parabola y2 = 32x is/are - (A) 9x + 3y – 8 = 0 (B) 9x – 3y + 8 = 0 (C) 9x + 3y + 8 = 0 (D) 9x – 3y – 8 = 0 CHECK YOUR GRASP ANSWER KEY EXERCISE-1 Que. 1 2 3 4 5 6 7 8 9 10 Ans. B D A C A B C D B C Que. 11 12 13 14 15 16 17 18 19 20 Ans. D A A B A B B C A D Que. 21 22 23 24 25 Ans. B A,C A,B,C,D A,B,C B,C E 69

JEE-Mathematics EXERCISE - 02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1 . Variable circles are drawn touching two fixed circles externally, then locus of centre of variable circle is - (A) parabola (B) ellipse ( C) hyperbola (D) circle x2 y2 1 2. The locus of the mid points of the chords passing through a fixed point ( ) of the hyperbola,  is - a2 b2 (A) a circle with centre   ,   (B) an ellipse with centre   ,    2 2   2 2  (C) a hyperbola with centre   ,   (D) straight line through   ,    2 2   2 2  3 . The locus of the foot of the perpendicular from the centre of the hyperbola xy = c2 on a variable tangent is : (A) (x2 – y2)2 = 4c2 xy (B)(x2 + y2)2 = 2c2 xy (C) (x2 – y2) = 4c2xy (D) (x2 + y2)2 = 4c2 xy 4 . The equation to the chord joining two points (x , y ) and (x , y ) on the rectangular hyperbola xy = c2 is - 11 22 xy xy 1  1  (A) x1  x2 y1  y2 (B) x1  x2 y1  y2 xy 1 xy 1   (C) y1  y2 x1  x2 (D) y1  y2 x1  x2 5 . The equation 9x2 – 16y2 – 18x + 32y – 151 = 0 represent a hyperbola - (A) The length of the transverse axes is 4 (B) Length of latus rectum is 9 21 11 (D) none of these (C) Equation of directrix is x = 5 and x = – 5 6 . From the points of the circle x2 + y2 = a2, tangents are drawn to the hyperbola x2 – y2 = a2; then the locus of the middle points of the chords of contact is - (A) (x2 – y2)2 = a2 (x2 + y2) (B) (x2 – y2)2 = 2a2 (x2 + y2) (C) (x2 + y2)2 = a2 (x2 – y2) (D) 2(x2 – y2)2 = 3a2 (x2 + y2) 7 . The tangent to the hyperbola xy = c2 at the point P intersects the x-axis at T and the y-axis at T'. The normal to the hyperbola at P intersects the x-axis at N and the y-axis at N'. The areas of the triangles PNT and 11 PN ' T ' are  and ' respectively, then + is -  ' (A) equal to 1 (B) depends on t (C) depends on c (D) equal to 2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 8 . The tangent to the hyperbola, x2 – 3y2 = 3 at the point ( 3, 0) when associated with two asymptotes constitutes - (B) an equilateral triangle (A) isosceles triangle (C) a triangles whose area is 3 sq. units (D) a right isosceles triangle. x2 y2 9 . The asymptote of the hyperbola   1 form with any tangent to the hyperbola a triangle whose area is a2 b2 a2 tan  in magnitude then its eccentricity is - (A) sec  (B) cosec  (C) sec2  (D) cosec2  1 0 . From any point on the hyperbola H : (x2/a2) – (y2/b2) = 1 tangents are drawn to the hyperbola 1 H : (x2/a2) – (y2/b2) = 2. The area cut-off by the chord of contact on the asymptotes of H is equal to - 22 (A) ab/2 (B) ab (C) 2 ab (D) 4 ab 70 E

JEE-Mathematics xy 1 1 . The tangent at P on the hyperbola (x2/a2) – (y2/b2) = 1 meets the asymptote   0 at Q. If the locus of the ab mid point of PQ has the equation (x2/a2) – (y2/b2) = k, then k has the value equal to - (A) 1/2 (B) 2 (C) 3/4 (D) 4/3 x2 y2 1 sec  can be - 12. If  is the angle between the asymptotes of the hyperbola  with eccentricity e, then 2 a2 b2 (A) e (B) e/2 (C) e/3 e (D) e2  1 1 3 . If (5, 12) and (24, 7) are the focii of a conic passing through the origin, then the eccentricity of conic is - (A) 386 / 12 (B) 386 / 13 (C) 386 / 25 (D) 386 / 38 1 4 . The point of contact of line 5x + 12y = 9 and hyperbola x2 – 9y2 = 9 will lie on (A) 4x + 15y = 0 (B) 7x + 12y = 19 (C) 4x + 15y + 1 = 0 (D) 7x – 12y = 19 1 5 . Equation (2 + )x2 – 2xy + ( – 1)y2 – 4x – 2 = 0 represents a hyperbola if - (A)  = 4 (B)  = 1 (C)  = 4/3 (D)  = –1 1 6 . If a real circle will pass through the points of intersection of hyperbola x2 – y2 = a2 & parabola y = x2, then - (A) a  (–1, 1) (B) a   1 , 1  – {0} 2 2  (C) area of circle =  – a2; a   1 , 1   {0} (D) area of circle =  – 4a2  2 2  17. If least numerical value of slope of line which is tangent to hyperbola x2 y2 is 3  1 4 , a  R0 a2 (a3  a2  a)2 is obtained at a = k. For this value of 'a', which of the following is/are true - 1 1 9 5 (A) a = – (B) a = (C) LR = 16 (D) e = 2 2 4 1 8 . If the normal at point P to the rectangular hyperbola x2 – y2 = 4 meets the transverse and conjugate axes at A and B respectively and C is the centre of the hyperbola, then - (A) PA = PC (B) PA = PB (C) PB = PC (D) AB = 2PC NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 1 9 . If the circle x2 + y2 = a2 intersects the hyperbola xy = c2 in four points P (x , y ), Q(x ,y ), R(x ,y ), S(x , y4), 11 22 33 4 then - [JEE 98] (A) x + x + x + x = 0 (B) y + y +y + y = 0 (C) x x x x = c4 (D) y y y y = c4 12 3 4 1 23 4 1234 1234 2 0 . The curve described parametrically by, x = t2 + t + 1, y = t2 – t +1 represents - [JEE 99] (A) a pair of straight lines (B) an ellipse (C) a parabola (D) a hyperbola BRAIN TEASERS ANSWER KEY EXERCISE-2 Que. 1 2 3 4 5 6 7 8 9 10 Ans. C Que. 11 C D A C A C A,B,C A D Ans. C 12 13 14 15 16 17 18 19 2 0 E A,D A,D A,B B,D B,C A,C,D A,B,C,D A,B,C,D C 71

JEE-Mathematics EXERCISE - 03 MISCELLANEOUS TYPE QUESTIONS MATCH THE COLUMN Following question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given statement in Column-I can have correct matching with ONE statement in Column-II. 1 . Consider the hyperbola 9x2 – 16y2 – 36x + 96y + 36 = 0. Column - I Column - II (A) If directrices of the hyperbola are y = k1 & y = k2 then (p) 16 k1 + k2 is equal to (q) 1 0 (r) 6 (B) If foci of hyperbola are (a, b) & (a, c) then a + b + c (s) 8 is equal to (C) Product of the perpendiculars drawn from the foci upon its any tangent is (D) Distance between foci of the hyperbola is Following question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II. 2 . Column - I Column - II (A) A tangent drawn to hyperbola x2 y2 at p    (p) 17  1  6  (q) 8 a2 b2 (r) 1 6 (s) 2 4 forms a triangle of area 3a2 square units, with coordinate axes, then the square of its eccentricity is equal to (B) If the eccentricity of the hyperbola x2 – y2 sec2  5 is 3 times the eccentricity of the ellipse x2sec2 + y2 = 25 then smallest positive value of  is 6 , value of 'p' is p (C) For the hyperbola x2  y2  3 , angle between its 3 asymptotes is  then value of '' is 24 (D) For the hyperbola xy = 8 any tangent of it at P meets co-ordinate axes at Q and R then area of triangle CQR where 'c' is centre of the hyperbola is NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 ASSERTION & REASON These questions contain, Statement-I (assertion) and Statement-II (reason). (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for Statement-I. (C) Statement-I is true, Statement-II is false. (D) Statement-I is false, Statement-II is true. 1. Statement-I : Ellipse x2 y2 and 12x2 – 4y2 = 27 intersect each other at right angle.  1 25 16 Because Statement-II : Whenever confocal ellipse & hyperbola intersect, they intersect each other orthogonally. (A) A (B) B (C) C (D) D 72 E

JEE-Mathematics 55 2 . Statement-I : 3 and 4 are the eccentricities of two hyperbola which are conjugate to each other. Because Statement-II : If e and e1 are the eccentricities of two conjugate hyperbolas than ee1 > 1. (A) A (B) B (C) C (D) D 3 . Statement-I : A bullet is fired and hit a target. An observer in the same plane heard two sounds the crack of the riffle and the thud of the ball striking the target at the same instant, then locus of the observer is hyperbola where velocity of sound is smaller than velocity of bullet. Because Statement-II : If difference of distances of a point 'P' from the two fixed points is constant and less than the distance between the fixed points then locus of 'P' is a hyperbola. (A) A (B) B (C) C (D) D 4 . Statement-I : If a circle S = 0 intersects a hyperbola xy = 4 at four points. Three of them are (2, 2)(4, 1) and (6, 2/3) then co-ordinates of the fourth point are (1/4, 16). Because Statement-II : If a circle S = 0 intersects a hyperbola xy = c2 at t1, t2, t3, t4 then t1. t2. t3. t4 = 1. (A) A (B) B (C) C (D) D COMPREHENSION BASED QUESTIONS Comprehension # 1 : If we rotate the axes of the rectangular hyperbola x2 – y2 = a2 through an angle /4 in the clockwise direction then the equation x2 – y2 = a2 reduces to xy = a2 =  a 2 = c2 (say). Since x = ct, y= c satisfies 2   t  2 xy = c2.  (x, y) =  ct, c  (t  c) is called a 't' point on the rectangular hyperbola.  t  On the basis of above information, answer the following questions : 1 . If t1 and t2 are the roots of the equation x2 – 4x + 2 = 0, then the point of intersection of tangents at 't1' and 't2' on xy = c2 is - (A)  c , 2c  (B)  2 c, c  (C)  c , c  (D)  c, c   2   2   2   2  2 . If e1 and e2 are the eccentricities of the hyperbolas xy = 9 and x2 – y2 = 25, then (e1, e2) lie on a circle C1 with centre origin then the (radius)2 of the director circle of C1 is - (A) 2 (B) 4 (C) 8 (D) 16 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 3 . If the normal at the point 't1' to the rectangular hyperbola xy = c2 meets it again at the point 't2' then the value of t1t2 is - (A) – t11 (B) – t12 (C) – t13 (D) – t14 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE-3  Match the Column 1 . (A)(r) ; (B) (s); (C)  (p); (D)  (q) 2 . (A)(p); (B) (s); (C)  (q,r); (D)  (r)  Assertion & Reason 1. A 2. B 3. A 4. D  Comprehension Based Questions Comprehension # 1 : 1. D 2. C 3. B E 73

JEE-Mathematics CONCEPTUAL SUBJECTIVE EXERCISE EXERCISE - 04 [A] 1 . The hyperbola x2/a2  y2/b2 = 1 (a,b > 0) passes through the point of intersection of the lines, 7x + 13y  87 = 0 & 5x  8y + 7 = 0 and the latus rectum is 32 2 / 5 . Find 'a' & 'b'. 2 . Find the eccentricity of the hyperbola whose latus rectum is half its transverse axis. 3 . Find the centre, the foci, the directrices, the length of the latus rectum, the length & the equations of the axes 16x2  9y2 + 32x + 36y 164 = 0. 4 . For the hyperbola x2/100  y2/25 = 1, prove that (a) eccentricity = 5 / 2 (b) SA . S'A = 25, where S & S' are the foci & A is the vertex. 5 . Find the eccentricity of the conic represented by x2 – y2 – 4x + 4y + 16 = 0. 6 . If C is the centre of a hyperbola x2/a2  y2/b2 = 1, S, S its foci and P a point on it. Prove that SP. SP = CP2  a2 + b2. 7 . For what value of  does the line y = 3x +  touch the hyperbola 9x2 – 5y2 = 45. 8 . Find the equation of the tangent to the hyperbola x²  4y² = 36 which is perpendicular to the line x  y + 4 = 0. 9 . Tangents are drawn to the hyperbola 3x²  2y² = 25 from the point (0, 5/2). Find their equations. 1 0 . If the normal at a point P to the hyperbola x2/a2  y2/b2 = 1 meets the x  axis at G, show that SG = e. SP, S being the focus of the hyperbola. x2 y2 1 1 . The tangents and normal at a point on a2 – b2 = 1 cut the y-axis at A & B. Prove that the circle on AB as diameter passes through the foci of the hyperbola. 1 2 . Show that the locus of the middle points of normal chords of the rectangular hyperbola x2  y2 = a2 is (y2  x2)3 = 4 a2 x2 y2. 1 3 . The variable chords of the hyperbola x2/a2  y2/b2 = 1 (b > a) whose equation is x cos  + y sin  = p subtends a right angle at the centre. Prove that it always touches a circle. 1 4 . Find the asymptotes of the hyperbola 2x2 – 3xy – 2y2 + 3x – y + 8 = 0. Also find the equation to the conjugate hyperbola & the equation of the principal axes of the curve. 1 5 . Find the equation of the standard hyperbola passing through the point ( 3,3) and having the asymptotes as straight lines x 5 ± y = 0. CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 1 . a2 = 25/2 ; b2 = 16 2. 3 3. (1, 2); (4, 2) & (6, 2); 5 x  4 = 0 & 5 x + 14 = 0; 32 2 ; 6; 8; 3 y  2 = 0; x + 1 = 0 5. 2 7.  = ± 6 8. x + y ± 3 3 = 0 9 . 3x + 2y  5 = 0 ; 3x  2y + 5 = 0 1 4 . x – 2y + 1 = 0; 2x + y + 1 = 0; 2x2 – 3xy – 2y2 + 3x – y – 6 = 0; 3x – y + 2 = 0; x + 3y = 0 1 5 . 5x2 – y2 = 6 74 E

EXERCISE - 04 [B] JEE-Mathematics BRAIN STORMING SUBJECTIVE EXERCISE 1 . If 1 & 2 are the parameters of the extremities of a chord through (ae, 0) of a hyperbola x2/a2  y2/b2 = 1, then show that tan 1 . tan 2  e 1 = 0 . 2 2 e 1 2 . If the tangent at the point (h, k) to the hyperbola x2/a2  y2/b2 = 1 cuts the auxiliary circle in points whose ordinates are y and y then prove that 1/y + 1/y = 2/k. 12 12 3 . Tangents are drawn from the point (, ) to the hyperbola 3x2  2y2 = 6 and are inclined at angles  &  to the x axis. If tan . tan  = 2, prove that 2 = 22  7. 4 . The perpendicular from the centre upon the normal on any point of the hyperbola x2/a2  y2/b2 = 1 meets at R . Find the locus of R. 5 . If the normal to the hyperbola x2/a2  y2/b2 = 1 at the point P meets the transverse axis in G & the conjugate axis in g & CF be perpendicular to the normal from the centre C , then prove that PF . PG = b² & PF . Pg = a² where a & b are the semi transverse & semi-conjugate axes of the hyperbola. 6 . If a rectangular hyperbola have the equation , x y = c2, prove that the locus of the middle points of the chords of constant length 2 d is (x2 + y2) (x y  c2) = d2 x y. 7 . Prove that the locus of the middle point of the chord of contact of tangents from any point of the circle x2 + y2 = r2 to the hyperbola x2/a2  y2/b2 = 1 is given by the equation (x2/a2  y2/b2)2 = (x2 + y2) / r2. 8 . Find the equations of the tangents to the hyperbola x2  9 y2 = 9 that are drawn from (3, 2) . Find the area of the triangle that these tangents form with their chord of contact. 9 . A tangent to the parabola x2 = 4 ay meets the hyperbola xy = k2 in two points P & Q . Prove that the middle point of PQ lies on a parabola . 1 0 . Given the base of a triangle and the ratio of the tangent of half the base angles. Show that the vertex moves on a hyperbola whose foci are the extremities of the base. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 53 4. (x2 + y2)2 (a2y2 – b2x2 ) = x2y2 (a2 + b2)2 8. y x ; x – 3 = 0 ; 8 sq. unit 12 4 E 75

JEE-Mathematics JEE-[MAIN] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [A] 1 . The latus rectum of the hyperbola 16x2 – 9y2 = 144 is- [AIEEE-2002] (1) 16/3 (2) 32/3 (3) 8/3 (4) 4/3 x2 y2 x2 y2 1 coincide. Then the value of b2 is- 2. The foci of the ellipse + b2 = 1 and the hyperbola 144 – 81 = 25 16 [AIEEE-2003] (1) 9 (2) 1 (3) 5 (4) 7 3 . The locus of a point P(, ) moving under the condition that the line y = x +  is a tangent to the hyperbola x2 y2 1 is- [AIEEE-2005]  a2 b2 (4) an ellipse (1) a hyperbola (2) a parabola (3) a circle x2 y2  4. For the hyperbola   1 , which of the following remains constant when varies ? cos2  sin2  [AIEEE-2007, IIT-2003] (1) Abscissae of vertices (2) Abscissae of foci (3) Eccentricity (4) Directrix 5 . The equation of the hyperbola whose foci are (–2,0) and (2, 0) and eccentricity is 2 is given by : (1) –3x2 + y2 = 3 (2) x2 – 3y2 = 3 (3) 3x2 – y2 = 3 [AIEEE-2011] (4) –x2 + 3y2 = 3 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 Qu e. 1 2 34 5 E Ans 24 1 2 3 76

EXERCISE - 05 [B] JEE-Mathematics JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS 1 . The equation of the common tangent to the curve y2 = 8x and xy = –1 is - [JEE 2002 Screening] (D) y = x + 2 (A) 3y = 9x + 2 (B) y = 2x + 1 (C) 2y = x + 8 x2 y2 1 2. For hyperbola  which of the following remains constant with change in  - cos2 sin2 [JEE 2003 Screening] (A) abscissae of vertices (B) abscissae of foci (C) eccentricity (D) directrix 3 . The point of contact of the line and the hyperbola x2  2y2  4 is - [JEE 2004 Screening] 2x  6 y  2 (A) (4 ,  6 ) (B) ( 6, 1) 1 1  (D) 1 , 3 (C)  2 , 6   6 2  4. Tangents are drawn from any point on the hyperbola x2 y2 1 to the circle x2 + y2 = 9. Find the locus of  94 mid-point of the chord of contact. [JEE 2005 Mains 4M out of 60] x2 y2 5 . If a hyperbola passes through the focus of the ellipse   1 and its transverse and conjugate axis coincides 25 16 with the major and minor axis of the ellipse and product of their eccentricities is 1, then - [JEE 2006, 5M)] x2 y2 (A) equation of hyperbola   1 9 16 x2 y2 (B) equation of hyperbola   1 9 25 (C) focus of hyperbola (5, 0) (D) focus of hyperbola (5 3, 3) 6 . A hyperbola, having the transverse axis of length 2sin, is confocal with the ellipse 3x2 + 4y2 = 12. Then its equation is - [JEE 2007, 3M] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 (A) x2cosec2 – y2sec2 = 1 (B) x2sec2 – y2cosec2 = 1 (C) x2sin2 – y2cos2 = 1 (D) x2cos2 – y2sin2 = 1 7 . Match the column - [2007, 6M] Column I Column II (A) Two intersecting circles (p) have a common tangent (B) Two mutually external circles (q) have a common normal (C) Two circles, one strictly inside the other (r) do not have a common tangent (D) Two branches of a hyperbola (s) do not have a common normal 8. Let a and b be non-zero real numbers. Then, the equation (ax2 + by2 + c) (x2 – 5xy + 6y2) = 0 represents - (A) four straight lines, when c = 0 and a, b are of the same sign E (B) two straight lines and a circle, when a = b, and c is of sign opposite to that of a (C) two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a (D) a circle and an ellipse, when a and b are of the same sign and c is of sign opposite to that of a [JEE 2008, 3M, –1M] 77

JEE-Mathematics 9 . Consider a branch of the hyperbola x2 – 2y2 – 2 2 x – 4 2 y – 6 = 0 with vertex at the point A. Let B be one of the end points of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is - [JEE 2008, 3M, –1M] 2 3 2 3 (A) 1 – (B)  1 (C) 1  (D)  1 3 2 3 2 1 0 . An ellipse intersects the hyperbola 2x2 – 2y2 =1 orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then :- [JEE 2009, 4M, –1M] (A) equation of ellipse is x2 + 2y2 =2 (B) the foci of ellipse are (±1, 0) (C) equation of ellipse is x2 + 2y2 =4  (D) the foci of ellipse are  2,0 Paragraph for Question 11 and 12 [JEE 10, (3M each), –1M] The circle x2 + y2 – 8x = 0 and hyperbola x2 y2 intersect at the points A and B.  1 94 1 1 . Equation of a common tangent with positive slope to the circle as well as to the hyperbola is - (A) 2x  5y  20  0 (B) 2x  5y  4  0 (C) 3x – 4y + 8 = 0 (D) 4x – 3y + 4 = 0 1 2 . Equation of the circle with AB as its diameter is - (A) x2 + y2 – 12x + 24 = 0 (B) x2 + y2 + 12x + 24 = 0 (C) x2 + y2 + 24x – 12 = 0 (D) x2 + y2 – 24x – 12 = 0 x2 y2  1 . If this line passes through the point of intersection 13. The line 2x + y =1 is tangent to the hyperbola  a2 b2 of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is [JEE 10, 3M] x2 y2 14. Let the eccentricity of the hyperbola  1 be reciprocal to that of the ellipse a2 b2 x2 + 4y2 = 4. If the hyperbola passes through a focus of the ellipse, then - [JEE 2011, 4M] x2 y2 (B) a focus of the hyperbola is (2,0) (A) the equation of the hyperbola is   1 32 5 (D) the equation of the hyperbola is x2–3y2=3 (C) the eccentricity of the hyperbola is 3 15. Let P(6, 3) be a point on the hyperbola x2 y2 1 . If the normal at the point P intersects the x-axis at  a2 b2 (9, 0), then the eccentricity of the hyperbola is - [JEE 2011, 3M] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\03 Hyperbola.p65 5 3 (C) 2 (D) 3 (A) (B) 2 2 16. Tangents are drawn to the hyperbola x2 y2 1 , parallel to the straight line 2x – y = 1. The points of  94 contact of the tangents on the hyperbola are [JEE 2012, 4M] (A)  9 , 1 (B)   9 ,  1  (C) 3 3,  2 2  (D) 3 3, 2 2  2   2   2 2   2 2  PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] F Ix2 y2 6. A x2  y2 2 10. A, B GH KJ4 . 1. D 2. B 3. A  9 5. A,C E 94 9. B 7 . (A)  (p, q) ; (B)  (p, q) ; (C)  (q, r) ; (D)  (q, r) 8. B 16 . A,B 11. B 12. A 13. 2 14. B,D 15. B 78