JEE-Mathematics d2x [AIEEE-2011] 1 2 . dy2 equals :- d2 y dy 2 d2y dy 3 d2 y 1 d2 y 1 dy 3 (1) (2) dx2 dx (3) (4) dx2 dx dx2 dx dx2 dy [JEE-(Main)-2013] 1 3 . If y = sec(tan–1x), then dx at x = 1 is equal to : 1 1 (3) 1 (4) 2 (1) (2) 2 2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] Que. 1 2 3 456 7 8 9 10 424 2 343 Ans. 1 1 3 Que. 11 12 13 Ans. 2 2 1 E 105

JEE-Mathematics JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [B] 1 . (a) If ln(x + y) = 2xy, then y'(0) = [JEE 2004 (Scr.)] (D) 0 (A) 1 (B) –1 (C) 2 sin 1 x c , 1 x 0 b 2 2 1, at x 0 (b) f(x) = 2 eax / 2 1 , x 0x1 2 If f(x) is differentiable at x = 0 and c < 1/2 then find the value of 'a' and prove that 64b2 = 4 – c2. [JEE 2004, 4] 2 . (a) If y = y(x) and it follows the relation x cos y + y cos x = , then y\"(0) :- (A) 1 (B) –1 (C) (D) – (b) If P(x) is a polynomial of degree less than or equal to 2 and S is the set of all such polynomials so that P(1) =1, P(0) = 0 and P'(x) > 0 x [0, 1], then :- (A) S = (B) S = (1 – a)x2 + ax, 0 < a < 2 (C) (1 – a)x2 + ax, a (0, ) (D) S = (1 – a)x2 + ax, 0 < a < 1 (c) If f(x) is a continuous and differentiable function and f(1/n) = 0, n 1 and n I, then :- (A) f(x) = 0, x (0, 1] (B) f(0) = 0, f ' (0) = 0 (C) f ' (x) = 0 = f \" (x), x (0, 1] (D) f(0) = 0 and f ' (0) need not to be zero [JEE 2005 (Scr.)] (d) If f(x – y) = f(x) · g(y) – f(y) · g(x) and g(x – y) = g(x) · g(y) + f(x) · f(y) for all x, y R. If right hand derivative at x = 0 exists for f(x). Find derivative of g(x) at x = 0. [JEE 2004 (Scr.)] 3 . For x > 0, Lim((sin x )1 / x (1 / x)sin x ) is :- [JEE 2006, 3] x 0 (D) 2 (A) 0 (B) –1 (C) 1 d2 x [JEE 2007, 3] 4 . dy2 equals :- (A) d2 y 1 (B) d2y 1 dy 3 (C) d2 y dy 2 (D) d2y dy 3 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 dx2 dx – dx2 dx 2 dx2 dx dx 5 . (a) Let g(x) = n f(x) where f(x) is a twice differentiable positive function on (0, ) such that f(x + 1) = x f(x). Then for N = 1, 2, 3 g\" N 1 – g\" 1 = 2 2 (A) –4 1 1 1 1 (B) 4 1 1 1 1 9 25 ....... 9 25 ....... (2N 1)2 (2N 1)2 11 ....... 1 4 1 1 1 1 (C) –4 1 (2N (D) 9 25 ....... 9 25 1)2 (2N 1)2 106 E

JEE-Mathematics (b) Let f and g be real valued functions defined on interval (–1, 1) such that g\"(x) is continuous, g(0) 0, g'(0) = 0, g\"(0) 0, and f(x) = g(x) sin x. Statement-1 : Lim [g(x) cot x – g(0)cosecx] = f \"(0) x 0 and Statement-2 : f ' (0) = g(0) (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation of statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1 (C) Statement-1 is true, statement-2 is false. [JEE 2008, 3+3] (D) Statement-1 is false, statement-2 is true. x [JEE 2009, 4] 6 . If the function f(x) = x3 e2 and g(x) = f –1(x), then the value of g'(1) is 7. Let ƒ() sin tan 1 sin , where . Then the value of d (ƒ()) is 44 d(tan ) cos2 [JEE 2011, 4] Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\04.MOD\\MOD.p65 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] 1. (a) A ; (b) a = 1 2 . (a) C ; (b) B ; (c) B, (d) g' (0) = 0 3. C 4. D 5. (a) A ; (b) A 6. 2 7. 1 E 107

JEE-Mathematics MONOTONICITY 1 . MONOTONICITY AT A POINT : ( a ) A function f(x) is called an increasing function at point x = a, if in a sufficiently small neighbourhood of x = a ; f(a – h) < f(a) < f(a + h) f(a+ h) f(a) f(a+ h) f(a) f(a–h) f(a–h) a–h a a+ h a ( b ) A function f(x) is called a decreasing function at point x = a, if in a sufficiently small neighbourhood of x= a ; f(a – h) > f(a) > f(a + h) f(a–h) f(a–h) f(a) f(a+ h) f(a) f(a+ h) a–h a a+ h x=a Note : If x = a is a boundary point, then use the appropriate one sides inequality to test Monotonicity of f(x). f(a) f(a+ h) f(a–h) f(a) x=a x=a f(a) > f(a–h) f(a+ h) > f(a) increasing at x = a decreasing at x = a ( c ) Te st i ng of monotonicit y of different iable funct ion at a poi nt. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 (i) If f'(a) > 0, then f(x) is increasing at x = a. (ii) If f'(a) < 0, then f(x) is decreasing at x = a. (iii) If f'(a) = 0, then examine the sign of f'(a+) and f'(a–). (1) If f'(a+) > 0 and f'(a–) > 0, then increasing (2) If f'(a+) < 0 and f'(a–) < 0, then decreasing (3) otherwise neither increasing nor decreasing. Illustration 1 : Let f(x) = x3 – 3x + 2. Examine the nature of function at points x = 0, 1 & 2. Solution : f(x) = x3 – 3x + 2 +–+ f'(x) = 3(x2 – 1) = 0 –1 1 x = ±1 (i) f'(0) = –3 decreasing at x = 0 (ii) f'(1) = 0 also, f'(1+) = positive and f'(1–) = negative neither increasing nor decreasing at x = 0. (iii) f'(2) = 9 increasing at x = 2 52 E

JEE-Mathematics Do yourself - 1 : ( i ) If function f(x) = x3 + x2 –x + 1 is increasing at x = 0 & decreasing at x = 1, then find the greatest integral value of 2 . MONOTONICITY OVER AN INTERVAL : (a) A function f(x) is said to be monotonically increasing (MI) in (a, b) if f'(x) 0 where equality holds only for discrete values of x i.e. f'(x) does not identically become zero for x (a, b) or any sub interval. (b) f(x) is said to be monotonically decreasing (MD) in (a, b) if f'(x) 0 where equality holds only for discrete values of x i.e. f'(x) does not identically become zero for x (a, b) or any sub interval. By discrete points, we mean that points where f'(x) = 0 does not form an interval. Note : (i) A function is said to be monotonic if it's either increasing or decreasing. (ii) If a function is invertible it has to be either increasing or decreasing. Illustration 2 : Prove that the function f x log x3 x6 1 is entirely increasing. Solution : Now, f x log x3 x6 1 f 'x 1 3 x 2 6x5 3x2 Illustration 3 : x6 x6 1 0 f(x) is increasing. x3 1 2 x6 1 Find the intervals of monotonicity of the function y x2 loge x , x 0 . Solution : Let y f x x2 loge x Illustration 4 : x2 loge x , x 0 2 x 1 1, x 0 x2 loge x , x 0 f x f 'x x 2x 1, x 0 x f 'x 2x 1 ; for all x x 0 + –+ –1 x 01 2 2 f 'x 2x2 1 x f 'x 2x 1 2x 1 x NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 So f 'x 0 when x 1 , 0 1 , and f 'x 0 when x , 1 0, 1 2 2 2 2 f(x) is increasing when x 1 , 0 1 , 2 2 and decreasing when x , 1 0, 1 Ans. 2 2 x a2 ab ac is decreasing in x b2 bc If a, b, c are real, then f x ab x c2 bc ac (A) 2 , 0 (B) 2 3 a2 b2 c2 0, 3 a2 b2 c2 a2 b2 c2 (C) 0, 3 (D) no where E 53

JEE-Mathematics 10 0 x a2 ab ac x a2 ab ac Solution : f 'x ab x b2 bc 0 1 0 ab x b2 bc Illustration 5 : ac bc x c2 ac bc x c2 0 01 Solution : x b2 x c2 b2c2 x a2 x c2 a2c2 x a2 x b2 a2b2 3x2 2x a2 b2 c2 +– + – 2 (a2 + b2+c2) f(x) will be decreasing when f 'x 0 0 3 2 a2 b2 c2 , 0 3x2 2x a2 b2 c2 x 3 0 Ans. (A) Prove the following (i) y = ex + sinx is increasing in x R+ (ii) y = 2x – sin x – tan x is decreasing in x (0, /2) (i) f(x) = ex + sinx, x R+ f'(x) = ex + cosx Clearly f'(x) > 0 x R+ (as ex > 1, xR+ and –1 cosx 1, x R+) Hence f(x) is increasing. (ii) f(x) = 2x – sin x – tan x x (0, /2) f'(x) = 2 – cos x – sec2 x f'(x) = cos2x – cos x – (cos2 x + sec2 x – 2) = cos2 x – cos x – (cos x – sec x)2 f'(x) < 0, x (0, /2) cos2 x < cos x, x (0, /2) Hence f(x) is decreasing in (0, /2) Do yourself - 2 : (i) If ƒ (x) = sinx + n |sec x + tan x| – 2x for x , then check the monotonicity of ƒ (x) 2 2 ( i i ) The function ƒ (x) = 2 log(x – 2) – x2 + 4x + 1 increases in the interval. (A) (1,2) (B) (2,3) (C) 5 , 3 (D) (2,4) 2 3 . GREATEST AND LEAST VALUE OF A FUNCTION : ƒ(b) ( a ) If a continuous function y = (x) is increasing in the closed interval [a, b], then (a) is the least value and (b) is the greatest value of (x) in [a, b] (figure-1) ƒ(a) Oa b figure-1 ƒ(a) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 ( b ) If a continuous function y = (x) is decreasing in [a, b], then (b) is the least ƒ(b) and (a) is the greatest value of (x) in [a, b]. (figure-2) Oa b figure-2 y ( c ) If a continuous function y = ƒ(x) is increasing/decreasing in the (a,b), x then no greatest and least value exist. 0 y 0x Illustration 6 : Show that f(x) = sin 1 x – nx is decreasing in x 1, 3 . Also find its range. 1 x2 3 54 E

JEE-Mathematics f(x) = sin 1 x – nx = tan–1x – nx 1 1 1 x2 x Solution : 1 x2 f '(x) = = 1 x2 x x 1 x2 f '(x) < 0 x 1 3 , 3 f(x) is decreasing f(x) max f 1 1 n3 & f(x) min f 3 1 n3 3 62 32 Range of f(x) = 1 n 3 , 1 n3 Ans. 3 2 6 2 Do yourself - 3 : ( i ) Let f(x) = x 1 . Find the greatest and least value of f(x) for x (0, 4). x 4. SPECIAL POINTS : ( a ) Critical points : The points of domain for which f'(x) is equal to zero or doesn't exist are called critical points. ( b ) Stationary points: The stationary points are the points of domain where f'(x) = 0 . Note : Every stationary point is a critical point but vice-versa is not true. Illustration 7 : Find the critical point(s) & stationary point(s) of the function f(x) = (x – 2)2/3(2x + 1) Solution : f(x) = (x – 2)2/3(2x + 1) 2 21 f'(x) = (x – 2)2/3 . 2 + (2x + 1) (x – 2)–1/3 = 2(x – 2)2/3 + (2x + 1) 3 3 21 / 3 x = 2( x 2 ) 2 (2 x 1) 1 = 2 5x 5 3 3 x 2 1 / 3 x 21 / 3 f'(x) does not exist at x = 2 and f'(x) = 0 at x = 1 x = 1, 2 are critical points and x = 1 is stationary point. Do yourself - 4 : ex ( i ) Find the critical points and stationary point of the function ƒ (x) = x NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 5 . PROVING INEQUALITIES USING MONOTONICITY : Comparison of two functions f(x) and g(x) can be done by analysing their monotonic behaviour. Illustration 8 : For x 0, prove that sin x < x < tan x Solution : 2 Let f(x) = x – sinx f'(x) = 1 – cosx f'(x) > 0 for x 0, 2 f(x) is M.I. f(x) > f(0) g'(x) = 1 – sec2x x – sinx > 0 x > sinx Similarly consider another function g(x) = x – tanx g'(x) < 0 for x 0, g(x) is M.D. 2 Hence g(x) < g(0) x – tanx < 0 x < tan x sinx < x < tan x Hence proved E 55

JEE-Mathematics Illustration 9 : For x (0, 1) prove that x – x3 tan 1 x < x – x3 & hence or otherwise find lim tan 1 x 3 6 x 0 x Solution : x3 1 x4 Let f(x) = x – – tan–1x f'(x) = 1 – x2 – 1 x2 = – 1 x2 3 f'(x) < 0 for x (0, 1) f(x) is M.D. f(x) < f(0) x– x3 – tan–1x < 0 x3 ......... (i) x – < tan–1x 3 3 x3 Similarly g(x) = x – – tan–1x 6 x2 1 x2 (1 x2 ) g'(x) = 1 – 2 1 x2 = 2(1 x2 ) g'(x) > 0 for x (0, 1) g(x) is M.I. x3 x3 x – –tan–1x > 0 x – > tan–1x ......... (ii) 66 Hence Proved x3 x3 from (i) and (ii), we get x – < tan–1x < x – 36 x2 tan 1 x x2 tan1 x 1 Now, 1 1 lim 3x 6 x0 x lim tan 1 x but 0 x 0 x Do yourself - 5 : (i) Find the larger of n(1 + x) & tan 1 x , for x > 0 . 1x x3 x5 (ii) Show that : sin x < x – , for x > 0. 6 120 (iii) For x > 1, y = nx satisfies (A) x – 1 > y (B) x2 – 1 > y (C) y > x – 1 (D) x 1 y x 6 . SIGNIFICANCE OF THE SIGN OF IInd ORDER DERIVATIVE : NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 The sign of the 2nd order derivative determines the concavity of the curve. If f''(x) > 0 x (a, b) then graph of f(x) is concave upward in (a, b). Similarly if f''(x) < 0 x (a, b) then graph of f(x) is concave downward in (a, b). concave concave upwards downwards ab ab 7 . PROVING INEQUALITIES USING GR APHS & CONCAVITY : Generally these inequalities involve comparison between values of two function at some particular points. Note : y (rx1+(1–r)x2, r f(x1)+(1–r)f(x2)) y= f(x) (i) If function f(x) is concave upward, then f rx1 (1 r )x2 rf(x1 ) (1 r) f(x2 ) distinct (x2,f(x2)) x & x domain of f(x) and r > 0. (x1,f(x1)) (rx1 + (1–r)x2, f(rx1+(1–r)x2)) 1 2 x2 x x1 (fig. 1) 56 E

JEE-Mathematics (ii) If function f(x) is concave downward, then y (rx1 + (1–r)x2, f(rx1+(1–r)x2)) f rx1 (1 r )x2 rf(x1 ) (1 r) f(x2 ) distinct (x2,f(x2)) (x1,f(x1)) (rx1+(1–r)x2, r f(x1)+(1–r)f(x2)) x& x domain of f(x) and r > 0. 1 2 Note : Equality hold when x and x coincide. x1 x2 x 12 y = ex Illustration 10 : Prove that for any two numbers x1 & x2, 3e x1 e x2 3(fxi1g.x22 ) Solution : 4 (x2 , e x2 ) e 4 . Q Assume f(x) = ex and let x & x 3 12 be two points on the curve y = ex. R 3 x1 +x2 , 3 e x1 +e x2 Let R be another point which divides 44 line segment PQ in ratio 1 : 3. 3e x1 e x2 (x1 , e x1 ) 1 S y coordinate of point R is 4 and P 3 x1 x2 3 x1 +x2 y coordinate of point S is e 4 . Since e4 f(x) = ex is always concave up, hence point R x1 3 x1 +x2 4 will always be above point S. 3e x1 e x2 3 x1 x2 x2 4 e 4 (C,sinC) x Illustration 11: In any ABC prove that sinA + sinB + sinC < 33 . sin x Solution : 2 A B C sin A sin B sin C are 3 , 3 Co-ordinates of centroid G from figure we have y , 3 PM > GM 32 sin A B C sin A sin B sin C P 3 3 (B,sinB) G (A, sinA) sin sin A sin B sin C 3 3 OM 33 sinA + sinB + sinC < A+B3+C=3 2 equality holds when triangle is equilateral. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 Do yourself - 6 : ( i ) In any triangle ABC, prove that cos A cos B cos C 3 . 2 8. ROLLE'S THEOREM : Let f be a function that satisfies the following three hypotheses : ( a ) f is continuous on the closed interval [a, b]. ( b ) f is differentiable on the open interval (a, b) ( c ) f(a) = f(b) Then there exist at least one number c in (a, b) such that f'(c) = 0. y yyy AB AB A B AB O a c1 c2 b x Oa cbx O a c1 c2 b x Oa c bx (a) (b) (c) (d) E 57

JEE-Mathematics Note : If f is differentiable function then between any two consecutive roots of f(x) = 0, there is atleast one root of the equation f'(x) = 0. (d) Geometrical Interpretation : Geometrically, the Rolle's theorem says that somewhere between A and B the curve has at least one tangent parallel to x-axis. Illustration 12 : Verify Rolle's theorem for the function f(x) = x3 – 3x2 + 2x in the interval [0, 2]. Solution : Here we observe that (a) f(x) is polynomial and since polynomial are always continuous, as well as differentiable. Hence f(x) is continuous in the [0,2] and differentiable in the (0, 2). & (b) f(0) = 0, f(2) = 23 – 3. (2)2 + 2(2) = 0 f(0) = f(2) Thus, all the condition of Rolle's theorem are satisfied. So, there must exists some c (0, 2) such that f'(c) = 0 1 f'(c) = 3c2 – 6c + 2 = 0 c = 1 ± 3 1 where both c = 1 ± (0, 2) thus Rolle's theorem is verified. 3 Illustration 13 : Show that between any two roots of e–x – cosx = 0 there exists at least one root of sin x – e–x = 0. Solution : If x = a and x = b are two distinct roots of e–x – cosx = 0 then e–a – cosa = 0 and e–b – cosb = 0 ......... (1) and let f(x) = e–x – cosx We observe that (i) e–x and cos x are continuous as well as differentiable in [a, b] then f(x) is also continuous in [a, b] & differentiable in (a,b). (ii) f(a) ea cos a 0 {from (1)} and f(b) eb cos b 0 i.e. f(a) = f(b) = 0 Thus f satisfies all the three conditions of Rolle's theorem in [a, b]. Hence there is at least one value of x in (a, b), say c such that f'(c) = 0. Now f'(c) = 0 –e–c + sinc = 0 sinc – e–c = 0 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 c is a root of the equation sin x – e–x = 0. Hence between any two roots of the equation e–x – cosx = 0 there is at least one root of the equation sinx – e–x = 0. Do yourself - 7 : (i) Verify Rolle's theorem for y = 1 – x4/3 on the interval [–1,1] (ii) Show that between any two roots of tanx = 1 there exists atleast one root of tanx = –1. 9 . LAGR ANGE'S MEAN VALUE THEOREM (LMVT) : Tangent parallel to chord Let f be a function that satisfies the following hypotheses: y ( i ) f is continuous in [a, b] ( i i ) f is differentiable in (a, b). Slope f'(c) B (b,f(b)) Then there is a number c in (a, b) such that f '(c) f(b) f(a ) ba (a,f(a)) f(b) –f(a) A Slope b–a 58 0a x y= f(x) cb E

JEE-Mathematics (a) Geometrical Interpretation : Geometrically, the Mean Value Theorem says that somewhere between A and B the curve has at least one tangent parallel to chord AB. (b) Physical Interpretations : If we think of the number (f(b) – f(a))/(b – a) as the average change in f over [a, b] and f'(c) as an instantaneous change, then the Mean Value Theorem says that at some interior point the instantaneous change must equal the average change over the entire interval. Illustration 14 : Find c of the Lagrange's mean value theorem for the function f(x) = 3x2 + 5x + 7 in the interval [1, 3]. Solution : Given f(x) = 3x2 + 5x + 7 ...... (i) f(1) = 3 + 5 + 7 = 15 and f(3) = 27 + 15 + 7 = 49 Again f'(x) = 6x + 5 Here a = 1, b = 3 Now from Lagrange's mean value theorem f(b) f(a) f(3) f(1) 49 15 f'(c) = 6c + 5 = = = 17 or c = 2. ba 3 1 2 Illustration 15 : If f(x) is continuous and differentiable over [–2, 5] and –4 f'(x) 3 for all x in (–2, 5), then the greatest possible value of f(5) – f(–2) is - (A) 7 (B) 9 (C) 15 (D) 21 Apply LMVT Solution : f(5) f(2) f'(x) = 5 (2) for some x in (–2, 5) f(5) f(2) Now, –4 7 3 –28 f(5) – f(–2) 21 Greatest possible value of f(5) – f(–2) is 21. Illustration 16 : If functions f(x) and g(x) are continuous in [a, b] and differentiable in (a, b), show that there will f(a) f(b) f(a) f '(c) (b a) be at least one point c, a < c < b such that g(a) g(b) g(a) g '(c) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 Solution : Let F(x) = f(a) f(x) = f(a) g(x) – g(a) f(x) ......(i) g(a) g(x) F '(x) = f(a) g'(x) – g(a) f'(x) ......(ii) Since f(x) and g(x) are continuous in [a, b] and differentiable in (a, b), therefore, from (i) and (ii) it follows that F(x) is continuous in [a, b] and differentiable in (a, b). Also from (i), F(a) = f(a)g(a) – g(a)f(a) = 0 And F(b) = f(a) g(b) – g(a) f(b) Now by mean value theorem for F(x) in [a, b], there will be at least one point c, a < c < b F(b) F(a) such that F '(c) = ba f(a) g'(c) –g(a) f'(c) = f(a)g(b) g(a)f(b) 0 ba or f(a)g(b) – g(a)f(b) = (b – a){f(a)g'(c) – g(a)f'(c)} or f(a) f(b) =(b – a) f(a) f '(c) g(a) g(b) g(a ) g '(c) E 59

JEE-Mathematics Do yourself - 8 : ( i ) If ƒ (x) = x2 in [a, b], then show that there exist atleast one c in (a, b) such that a, c, b are in A.P. ( i i ) Using LMVT, prove that x n(1 x) x for x > 0 1x 10 . SPECIAL NOTE : Use of Monotonicity in identifying the number of roots of the equation in a given interval. Suppose a and b are two real numbers such that, ( a ) Let f(x) is differentiable & either MI or MD for 0 < x < b. & ( b ) f(a) and f(b) have opposite signs. Then there is one & only one root of the equation f(x) = 0 in (a, b). Miscellaneous Illustrations : Illustration 17 : If g(x) = f(x) + f(1 – x) and f''(x) < 0; 0 x 1, show that g(x) increasing in x [0, 1/2] and decreasing in x [1/2, 1] Solution : f''(x) < 0 f'(x) is decreasing function. Now, g(x) = f(x) + f(1 – x) g'(x) = f'(x) – f'(1 – x) ......... (i) Case I : If x (1 – x) x 1/2 f'(x) f'(1 – x) f'(x) – f'(1 – x) 0 g'(x) 0 g(x) decreases in x 1 , 1 2 Case II : If x (1 – x) x 1/2 f'(x) f'(1 – x) f'(x) – f'(1 – x) 0 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 g'(x) 0 g(x) increases in x [0, 1/2] Illustration 18 : Prove that if 2a 2 < 15a, all roots of x5 – a0x4 + 3ax3 + bx2 + cx + d = 0 can't be real. It is 0 given that a0, a, b, c, d R. Solution : Let f(x) = x5 – a0x4 + 3ax3 + bx2 + cx + d f'(x) = 5x4 –4a0x3 + 9ax2 + 2bx + c f\"(x) = 20x3 – 12a0x2 + 18ax + 2b f\"'(x) = 60x2 – 24a0x + 18a = 6(10x2 – 4a0x + 3a) Now, discriminant = 1 6 a 2 – 4. 10. 3a = 8( 2 a 2 – 15a) < 0 0 0 as 2a02 – 15a < 0 is given. 60 E

JEE-Mathematics Hence the roots of f\"'(x) = 0 can not be real. f''(x) have one real root and f'(x) = 0 have at most two real roots so f(x) = 0 have at most three real roots. Therefore all the roots of f(x) = 0 will not be real. Illustration 19 : Show by using mean value theorem that < tan–1 – tan–1 where > > 0. 1 2 < 1 2 Solution : Take f(x) = tan–1x 1 f'(x) = 1 x2 . By mean value theorem for f(x) in [, ] f() f() = f'(c) = 1 where < c < ........ (i) 1 c2 Now, c > 11 1 c2 < 1 2 11 as c < 1 c2 > 1 2 1 f() f() 1 from (i), 1 2 < < 1 2 or 1 2 f() f() 1 2 < tan–1 – tan–1 < Hence, 1 2 1 2 Illustration 20 : Compare which of the two is greater (100)1/100 or (101)1/101. Solution : Assume f(x) = x1/x and let us examine monotonic nature of f(x) , (x > 0) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 f'(x) = x1/x . 1 nx x2 e1/e f'(x) > 0 x (0, e) (100)1/100 (101)1/101 and f'(x) < 0 x (e, ) 1 e 100 101 Hence f(x) is M.D. for x e and since 100 < 101 f(100) > f(101) (100)1/100 > (101)1/101 Illustration 21 : For x [0, 1], let the second derivative f\"(x) of a function f(x) exist and satisfy |f\"(x)| 1. If f(0) = f(1) then show that |f'(x)| < 1 for all x in [0, 1]. E 61

JEE-Mathematics Solution : Since f\"(x) exists for all x in [0, 1] f(x) and f'(x) are differentiable as well as continuous for all x in [0, 1] Now f(x) is continuous in [0, 1] and differentiable in (0, 1) and f(0) = f(1) By Rolle's theorem there is at least one c such that f'(c) = 0, where 0 < c < 1. Case I : Let x = c then f'(x) = f'(c) = 0 |f'(x)| = |0| = 0 < 1 Case II : Let x > c. By Largrange's mean value theorem for f'(x) in [c, x] f '(x) f '(c) f \"() for at least one , c < < x x c or f'(x) = (x – c) f ''() ( f'(c) = 0) or |f'(x)| = |x – c||f''()| But x [0, 1], c (0, 1) |x – c| < 1 – 0 or |x – c| < 1 and given |f''(x)| 1 x [0, 1] |f''()| 1 |f'(x)| < 1.1. ( |f'(x)| = |x – c||f''()| or |f'(x)|< 1 x [0, 1] Case III: Let x < c then f '(x) f '(c) f \"() x c |–f'(x)| = |c – x||f''()| |f'(x)|< 1.1 or |f''(x)| < 1 Combining all cases, we get |f'(x)| < 1 x [0, 1] ANSWERS FOR DO YOURSELF NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 1 : (i) –1 2 : (i) Increasing (ii) B 3 : (i) Not defined 4 : ( i ) x = 1 is a critical point as well as stationary point (Note x = 0 is not in the domain of ƒ (x)) 5 : (i) n(1+x) (iii) A,B,D 7 : ( i ) Rolles theorem is valid 62 E

EXERCISE - 01 JEE-Mathematics CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . Function f x x2 x 22 is - (A) increasing in 0,1 2, (B) decreasing in 0,1 2, (C) decreasing function (D) increasing function 2 . The function f x tan x x (A) always increases (B) always decreases (C) never decreases (D) sometimes increases and sometimes decreases 3 . The function f, defined by f(x ) (x 2)e x is - (A) decreasing for all x (B) decreasing in (, 1) and increasing in (1, ) (C) increasing for all x (D) decreasing in (1, ) and increasing in (, 1) 4 . Function f x x3 6 x2 9 2k x 1 is increasing function if 3 3 3 3 (A) k (B) k (C) k (D) k 2 2 2 2 5 . The function f(x) cos x 2px is monotonically decreasing for (A) p 1 (B) p 1 (C) p < 2 (D) p > 2 2 2 6 . The value of 'a' for which the function f(x) = sinx – cosx – ax + b decreases for all real values of x, is - (A) a 2 (B) a 2 (C) a 2 (D) a 2 7 . Let f(x) be a quadratic expression which is positive for all real values of x. If g(x) = f(x) + f'(x) +f''(x), then for any real x - (A) g(x) < 0 (B) g(x) > 0 (C) g(x) = 0 (D) g(x) 0 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 f(x) = 1 1 dx then f is 8 . 2 1 x 2 1 - x2 (A) increasing in (0, ) and decreasing in ( –, 0) (B) increasing in (–, 0) and decreasing in (0, ) (C) increasing in (– ) (D) decreasing in ( –, ) 4x for 0 < x < 4 9. Let f(x) = 2 x for x 4 4 for 4 x 6 E 16 3x Which of the following properties does f have on the interval (0, 6) ? I. n f(x) exists ; II. f is continuous III. f is monotonic (A) I only (B) II only (C) III only (D) none 63

JEE-Mathematics 1 0 . The length of largest continuous interval in which function f (x) = 4x – tan2x is monotonic, is - (A) /2 (B) /4 (C) /8 (D) /16 1 1 . The largest set of real values of x for which ln (1 + x) x is (A) (–1, ) (B) (1, 0) (0, ) (C) [0, ) (D) (0, ) 1 2 . The true set of real values of x for which the function, f(x) x n x x 1 is positive is (A) (1, ) (B) (1 / e,) (C) [e,) (D) (0,1) (1,) 13. Number of solution of the equation 3tan x + x3 = 2 in 0, is 4 (A) 0 (B) 1 (C) 2 (D) 3 1 4 . Rolle's theorem in the indicated intervals will not be valid for which of the following function- sin x x0 LM(B) g(x) 1 cos x x0 x0 ; x [ 2, 2 ] Lx ; x [ 1, 1 ] MMN0 x M(A) f(x) x0 NMM1 1 cos x x0 MMMLN x sinHFG 1xJKI x0 MNL 1 1 QPO ; x [ 2 , 2] 0 ; 2 L x2 (D) k(x) x , MM(C) h(x) x0 x0 MNM 1 2 x3 2x2 5x 6 if x 1 , then 15. Consider the function for x = [–2, 3], f(x) x 1 6 if x 1 (A) f is discontinuous at x = 1 Rolle's theorem is not applicable in [–2, 3] (B) f(–2) f(3) Rolle's theorem is not applicable in [–2, 3] (C) f is not derivable in (–2, 3) Rolle's theorem is not applicable (D) Rolle's theorem is applicable as f satisfies all the conditions and c of Rolle's theorem is 1/2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 1 6 . If the function f(x) = 2x2 + 3x + 5 satisfies LMVT at x = 2 on the closed interval [1, a] then the value of 'a' is equal to - (A) 3 (B) 4 (C) 6 (D) 1 1 7 . Consider the function f(x) = 8x2 – 7x + 5 on the interval [–6, 6]. The value of c that satisfies the conclusion of the mean value theorem, is - (A) –7/8 (B) –4 (C) 7/8 (D) 0 1 8 . Let f be a function which is continuous and differentiable for all real x. If f(2) = – 4 and f '(x) 6 for all x [2, 4] then (A) f(4) < 8 (B) f(4) 8 (C) f(4) 12 (D) none of these 1 9 . If f(x) = x 3+ 7x – 1 then f(x) has a zero between x = 0 and x = 1. The theorem which best describes this, is - (A) Rolle's theorem (B) mean value theorem (C) maximum-minimum value theorem (D) intermediate value theorem 64 E

JEE-Mathematics 2 0 . Consider f(x) =|1 – x|, 1 x 2 and g(x) = f(x) + b sin 2 x, 1 x 2 then which of the following is correct ? 3 (A) Rolles theorem is applicable to both f, g and b = 2 1 (B) LMVT is not applicable to f and Rolles theorem if applicable to g with b = 2 (C) LMVT is applicable to f and Rolles theorem is applicable to g with b = 1 (D) Rolles theorem is not applicable to both f, g for any real b. SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 2 1 . Let h(x) = f(x) – (f (x))2 + (f(x))3 for every real number x. Then. [JEE 98] (A) h is increasing whenever f is increasing (B) h is increasing whenever f is decreasing (C) h is decreasing whenever f is decreasing (D) nothing can be said in general 2 2 . Let f(x) = ex (x 1)(x 2)dx . Then 'f' increases in the interval - (A) (–, –2) (B) (–2, –1) (C) (1, 2) (D) (2, ) 3x2 12x 1 , 1 x 2 2 3 . If f(x) = 37 x then - , 2x3 (A) f(x) is increasing on (–1, 2) (B) f(x) is continuous on [–1, 3] (C) f'(2) does not exist (D) f(x) has the maximum value at x = 2 2 4 . Let f(x) = 8x3 – 6x2 – 2x + 1, then - (A) f(x) = 0 has no root in (0, 1) (B) f(x) = 0 has at least one root in (0, 1) (C) f'(c) vanishes for some c (0,1) (D) none 2 5 . Let f and g be two functions defined on an interval I such that f(x) 0 and g(x) 0 for all x I and f is strictly decreasing on I while g is strictly increasing on I then - (A) the product function fg is strictly increasing on I (B) the product function fg is strictly decreasing on I (C) fog(x) is monotonically increasing on I (D) fog(x) is monotonically decreasing on I NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 CHECK YOUR GRASP ANSWER KEY EXERCISE-1 Que. 1 2 3 4 5 6 7 8 9 10 Ans. A A D A B D B C B B Que. 11 12 13 14 15 16 17 18 19 20 Ans. A D B D D A D B D C Que. 21 22 23 24 25 Ans. A,C A,B,D A,B,C,D B,C A,D E 65

JEE-Mathematics BRAIN TEASERS EXERCISE - 02 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1 . A differentiable function f(x) is strictly increasing x R , Then - (A) f '(x) 0 x R . (B) f '(x) 0 x R , provided it vanishes at finite number of points. (C) f '(x) 0 x R provided it vanishes at discrete points though the number of these discrete points may not be finite. (D) f '(x) 0 x R provided it vanishes at discrete points and the number of these discrete points must be infinite. 2. The function 2x 1 (x 2) - y x 2 (A) is its own inverse (B) decreases for all values of x (C) has a graph entirely above x-axis (D) is bound for all x 3 . If f(0) = f(1) = f(2) = 0 & function f(x) is twice differentiable in (0, 2) and continuous in [0, 2]. Then which of the following is/are definitely true - (A) f ''(c) 0 ; c (0 , 2) (B) f '(c) 0 ; for atleast two c (0, 2) (C) f '(c) 0 ; for exactly one c (0, 2) (D) f '' (c) 0 ; for atleast one c (0, 2) 4. Consider the function f(x) x sin for x 0 then the number of points in (0, 1) where the derivative f'(x) x vanishes, is - 0 for x 0 (A) 0 (B) 1 (C) 2 (D) infinite 5 . Let (x) = (f(x))3 – 3(f(x))2 + 4f(x) + 5x + 3sinx + 4 cosx x R, then - (A) is increasing whenever f is increasing (B) is increasing whenever f is decreasing (C) is decreasing whenever f is decreasing (D) is decreasing if f'(x) = –11 6 . If f(x ) x and g(x ) x , where 0 < x 1, then in this interval - NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 sin x tan x (A) both f(x) and g(x) are increasing functions (B) both f(x) and g(x) are decreasing functions (C) f(x) is an increasing function (D) g(x) is an increasing function 7 . f(x) > x ; x R. Then the equation f (f(x)) – x = 0 has– (A) Atleast one real root (B) More than one real root (C) no real root if f(x) is a polynomial & one real root if f(x) is not a polynomial (D) no real root at all 11 8. A function y f(x) is given by x & y for all t > 0 then f is : 1 t2 t(1 t2 ) (A) increasing in (0,3/2) & decreasing in (3/2, ) (B) increasing in (0,1) (C) increasing in (0,) (D) decreasing in (0,1) 66 E

JEE-Mathematics 9 . Suppose that f is differentiable for all x such that f'(x) 2 for all x. If f(1) = 2 and f(4) = 8 then f(2) has the value equal to - (A) 3 (B) 4 (C) 6 (D) 8 10. The function f(x) n( x) is - n(e x) (A) increasing on (0, ) (B) decreasing on (0, ) (C) increasing on (0, e), decreasing on (/e, ) (D) decreasing on (0, e), increasing on (/e, ) 1 1 . Number of solution(s) satisfying the equation, 3x2 2x3 log2 (x2 1) log2 x is - (A) 1 (B) 2 (C) 3 (D) none 1 2 . Let g (x) = 2f (x/2) + f (1 – x) and f''(x) < 0 in 0 x 1 then g (x) - (D) increases in (2/3,1] (A) decreases in [0,2/3) (B) decreases in (2/3,1] (C) increases in [0,2/3) 1 3 . If f(x ) a a| x| sgn x ; g(x) a a|x| sgn x for a > 0, a 1 and x R , where { } & [ ] denote the fractional part and integral part functions respectively, then which of the following statements holds good for the function h (x), where ( n a)h(x) ( n f(x) n g(x)) - (A) ‘h’ is even and increasing for a > 1 (B) ‘h’ is odd and decreasing for a < 1 (C) ‘h’ is even and decreasing for a < 1 (D) ‘h’ is odd and increasing for a > 1 1 4 . Number of roots of the equation x2 .e2|x| 1 is - (A) 2 (B) 4 (C) 6 (D) infinite 1 15. Equation 3x sin x = 0 has - (x 1)3 (A) no real root (B) two real and distinct roots (D) exactly one root between –1 and 1 (C) exactly one negative root p4 x5 3x n 16. The values of p for which the function f(x) 1 5 decreases for all real x is 1p NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 (A) (, ) (B) 3 21 (1, ) (C) 5 27 (2, ) (D) [1, ) 4, 3, 2 2 BRAIN TEASERS ANSWER KEY EXERCISE-2 Que. 1 2 3 4 5 6 7 8 9 10 Ans. C A,B B,D D A,D C D B B B Que. 11 12 13 14 15 16 Ans. A B,C D B B,C,D B E 67

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 OR MORE statement(s) in Column-II. 1 . Column-I Column-II (A) The equation x log x = 3 – x has at least (p) [0, 1] one root in (B) If 27a + 9b + 3c + d = 0, then the equation (q) [1, 3] 4ax3 + 3bx2 + 2cx + d = 0 has at least one root in 1 (r) [0, 3] (C) If c = 3 & f(x) = x + then interval in which (s) [–1, 1] x LMVT is applicable for f(x) is 1 (D) If c = & f(x) = 2x – x2, then interval in which 2 LMVT is applicable for f(x) is ASSERTION & REASON These questions contains, Statement I (assertion) and Statement II (reason). (A) Statement-I is true, Statement-II is truez ; 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 : The quadratic equation 10x2–28x + 17 = 0 has atleast one root in [1,2]. Because : Statement-II : f(x) = e10x(x – 1)(x – 2) satisfies all the conditions for Rolle's theorem in [1,2] (A) A (B) B (C) C (D) D 2 . Let f : R R, f(x) = x3 + x2 + 3x + sin x. Statement-I : f(x) is one-one. Because : Statement-II : f(x) is decreasing function. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 (A) A (B) B (C) C (D) D 3 . Statement-I : The greatest of the numbers 1, 21/2, 31/3, 41/4, 51/5, 61/6, 71/7 is 31/3. Because : Statement-II : x1/x is increasing for 0 < x < e and decreasing for x > e. (A) A (B) B (C) C (D) D COMPREHENSION BASED QUESTIONS Comprehension # 1 f : (0, ) , be defined as, f(x) = arc tan(nx) 2 2 On the basis of above information, answer the following questions : 1 . The above function can be classified as (A) injective but not surjective (B) surjective but not injective (C) neither injective nor surjective (D) both injective as well as surjective 68 E

2 . The graph of y = f(x) is best represented as - y JEE-Mathematics /2 y y y /2 /2 (A) 0 1 x (B) 0 1 x (C) 0 1 x (D) 0 1 x /2 /2 3 . If x , x and x are the points at which g(x) = [f(x)] is discontinuous where [.] denotes greatest integer function, 12 3 then x + x + x is - 123 (A) equal to 2 (B) equal to 3 (C) greater than 3 (D) greater than 2 but less than 3 Comprehension # 2 Consider the polynomial function f(x) = 3x4 – 4x3 – 12x2 + 5. This function has monotonicity as given below : in (–, a) decreasing 1 in (a , a ) increasing 12 in (a , a ) decreasing 23 in (a , ) increasing 3 A rectangle ABCD is formed such that (AB) = portion of the tangent to the curve y = f(x) at x = a , intercepted between the lines x = a 11 & x = a. 3 (BC) = portion of the line x = a intercepted between the curve & x-axis. 3 On the basis of above information, answer the following questions : 1 . Triplet (a , a , a ) is given by - 123 (A) (–1, 0, 2) (B) (0, –1, 2) (C) (2, –1, 0) (D) (2, 0, –1) 2 . Area of rectangle ABCD - (A) 51 (B) 57 (C) 87 (D) 81 3 . The equation f(x) = 0 has - (A) 2 real, 2 imaginary roots (B) 2 complex, 2 irrational roots (C) 4 real & distinct roots (D) 2 real coincident roots & 2 irrational roots NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE-3 Match the Column 1 . (A) (q,r), (B) (r), (C) (q), (D) (p) Assertion & Reason 1. A 2. C 3. A Comprehension Based Questions Comprehension # 1 : 1. D 2. C 3. C Comprehension # 2 : 1. A 2. D 3. D E 69

JEE-Mathematics CONCEPTUAL SUBJECTIVE EXERCISE EXERCISE - 04 [A] 1 . Find the intervals of monotonicity for the following functions & represent your solution set on the number line. (a) f(x ) 2.e x2 4 x (b) f(x) = ex/x (c) f(x) = x2e–x 2 . Let f(x) = 1 – x – x3. Find all real values of x satisfying the inequality, 1 – f(x) – f3(x) > f(1 – 5x) 3 . Show that the function y = arctanx – x decreases everywhere. 4 . Find the intervals of monotonicity of the function : y = n(x 1 x2 ) 10 5 . Find the intervals of monotonicity of the function : y = 4 x3 9 x2 6 x 6 . x2 1 n t 1 is increasing and decreasing. Find the value of x > 1 for which the function F(x) x t 32 dt 7 . Find the range of values of 'a' for which the function f(x) = x3 + (2a + 3)x2 + 3(2a + 1)x + 5 is monotonic in R. Hence find the set of values of 'a' for which f(x) is invertible. a 2 1 3 8. If f(x) = x3 + (a –1)x2 + 2x + 1 is monotonic increasing for every x R, then find the range of values of ‘a’. 9 . Find the set of all values of the parameter ‘a’ for which the function f(x) = sin 2x – 8(a + 1) sin x + (4a2 + 8a – 14)x increases for all x R and has no critical points for all x R. 1 0 . Find the greatest & the least values of the following functions in the given interval if they exist. (a) f(x) = 12x4/3 – 6x1/3, x [–1, 1] (b) y = x5 – 5x4 + 5x3 + 1 in [–1, 2] (c) y sin 2x x , (d) y 2 tan x tan2 x 0, 2 2 2 1 1 . Prove the following inequalitities : (b) 2xnx > 4(x –1) –2 n x for x > 1. (a) x2 – 1 > 2x nx for x > 1 (c) tan2x + 6 n secx + 2cos x + 4 > 6 sec x for x 3 , 2 2 1 2 . Let f, g be differentiable on R and suppose that f(0) = g(0) and f'(x) g'(x) for all x 0. Show that f(x) g(x) for NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 all x 0. tan x2 x2 0 x1 x2 13. Using monotonicity prove that tan x1 x1 for 2 1 e2/3 1 2 / 3 1 4 . Identify which is greater : e1 / 3 or 1 / 3 1 5 . Verify Rolles theorem for f(x) = (x – a)m(x – b)n on [a, b]; m, n being positive integer. 1 6 . Check the validity of Rolle's theorem for the function y = x3 + 4x2 – 7x – 10 in the interval [–1, 2]. 1 7 . Let f(x) = 4x3 – 3x2 – 2x + 1, use Rolle's theorem to prove that there exist c, 0 < c < 1 such that f(c) = 0. 18. If the equation a xn + a xn–1 + ........+ ax = 0 has a positive root , prove that the equation 0 1 n-1 na xn–1 + (n – 1)a xn–2 + ........... + a = 0 also has a positive root smaller that . 0 1 n–1 1 9 . f(x) and g(x) are differentiable functions for 0 x 2 such that f(0) = 5, g(0) = 0, f(2) = 8, g(2) = 1. Show that there exists a number c satisfying 0 < c < 2 and f'(c) = 3 g'(c). 70 E

JEE-Mathematics 2 0 . If f, , are continuous in [a,b] and derivable in ]a,b[ then show that there is a value of c lying between a & b such f(a ) f(b) f '(c) that, (a) (b) '(c) 0 (a ) (b) '(c) 2 x2 2 1 . The function y x4 takes on equal values at the end-points of the interval [–1, 1]. Make sure that the derivative of this function does not vanish at any point of the interval [–1, 1], and explain this deviation from Rolle's theorem. 22. A (0, 1), B , 1 are two points on the graph given by y = 2 sinx + cos2x. Prove that there exists a point P on 2 the curve between A & B such that tangent at P is parallel to AB. Find the co-ordinates of P. 23. With the aid of Lagrange's formula prove the inequalities a b n a a b , for the condition 0 < b a. a b b 2 4 . Let f : [a, b] R be continuous on [a, b] and differentiable on (a, b). If f(a) < f(b), then show than f'(c) > 0 for some c (a, b). 2 5 . Show that the function ƒ (x) = xn + px + q cannot have more than two real roots if n is even and more than three if n is odd. Investigate the behaviour of the following functions and construct their graph : (Q. 26 to 34) 2 6 . y = x3 – 3x + 2 2 7 . y = x4 – 10x2 + 9 2 8 . y = (x –1)2(x – 2)3 2 9 . y = (x + 3)/(x – 1) 3 0 . y = x + sinx 3 1 . y = 2.e x2 4 x 3 2 . y = ex/x 3 3 . y = x2e–x 34. y x2 3x 4 x2 3x 4 3 5 . Investigate the behaviour of the function y = (x3 + 4)/(x + 1)3 and construct its graph. How many solutions does the equation (x3 + 4)/(x + 1)3 = c possess ? CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 1 . (a) I in (2, ) & D in (–, 2) (b) I in (1, ) & D in (–, 0)(0, 1) (c) I in (0, 2) & D in (–, 0)(2, ) 2 . (–2, 0) (2, ) 4 . Increases monotonically. 5. decreases in (–, 0) 0, 1 (1, ) increases in 1 , 1 6 . I in (3, ) and D in (1,3) 2 2 3 8 . a (–, –3] [1, ) 9 . a < – (2 5 ) or a > 5 7. 0 a 2 1 0 . (a) Maximum at x = –1 and f(–1) = 18; Minimum at x = 1/8 and f(1/8) = –9/4 (b) 2 & –10 (d) The greatest value is equal to 1, no least value. (c) and – 22 1 2 / 3 15. c = mb na which lies between a & b 22. , 3 1 4 . 1 / 3 6 2 m n E 71

JEE-Mathematics y y 9 y 4 27. –5 5 28. 0 1A2 x –3 –1 0 1 3 x 26. 2 HG KJ–8 A 7, – 108 –2 –1 0 1 2 x 5 3125 y –16 y 3 y 29. 1 3 x 30. x 31. –3 –1 0 1 2/e4 –1 –3 2x y y 7 32. e x 33. y 2 1 1 4/e2 3 4 . x 1/7 2x –2 y 35. One root for c , 4 4, ; 9 two roots for c 4 ,1, 4 ; (–2,4)A (2,4/9) 9 –4 B x –34 0 2 three roots for c 4 , 1 (1, 4) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 9 72 E

JEE-Mathematics EXERCISE - 04 [B] BRAIN STORMING SUBJECTIVE EXERCISE dg 1 . Let a + b = 4, where a < 2 and let g(x) be a differentiable function. If >0 for all x, prove that dx ab g(x)dx g(x)dx increases as (b – a) increases. 00 2 . Find all the values of the parameter 'a' for which the function; f(x) = 8ax – a sin 6x – 7x – sin 5x increases & has no critical points for all x R. 3 . Prove that if f is differentiable on [a, b] and if f(a) = f(b) = 0 then for any real there is an x (a, b) such that f(x) + f'(x) = 0. 4 . Let f be continuous on [a, b] and differentiable on (a, b). If f(a) = a and f(b) = b, show that there exist distinct c , 1 c , in (a, b) such that f'(c ) + f'(c ) = 2. 2 12 5. Prove the inequality ex > (1 + x) using LMVT for all x R and use it to determine which of the two numbers e 0 and e is greater. 6. For x 0, identify which is greater (2sinx + tanx) or (3x). Hence find lim 2 sin 3x where [.] denote 2 x tan x , x 0 the greatest integer function. 7 . Suppose that on the interval [–2, 4] the function f is differentiable, f(–2) = 1 and|f'(x)| 5. Find the bounding function of f on [–2, 4], using LMVT. 8. Using LMVT prove that : (a) tan x > x in 0, , (b) sin x < x for x > 0. 2 b 9 . If ax2 + c for all positive x where a > 0 and b > 0, then show that 27ab2 4c3. x 1 0 . Show that 1 xn x x2 1 1 x2 for all x R. 11. Let f(x) xeax , x 0 , where a is positive constant. Find the interval in which f'(x) is increasing. x ax2 x3, x 0 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 1 2 . If b > a, find the minimum value of |(x – a)3|+ |(x – b)3|, x R. 1 3 . f(x) is a differentiable function and g(x) is a double differentiable function such that |f(x)| 1 and f'(x) = g(x). If f2(0) + g2(0) = 9. Prove that there exists some c (–3, 3) such that g(c). g''(c) < 0 BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 2 . (6, ) 5 . e 6 . 2sinx + tanx, limit = 0 7 . y = –5x – 9 and y = 5x + 11 11. 2 , a 1 2 . (b – a)3/4 a 3 E 73

JEE-Mathematics JEE-[MAIN] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [A] 1 . f(x) and g(x) are two differentiable function in [0, 2] such that f \"(x) – g\"(x) = 0, f '(1) = 2, g'(1) = 4, f(2) = 3, g(2) = 9, then f(x) – g(x) at x = 3/2 is- [AIEEE-2002] (1) 0 (2) 2 (3) 10 (4) –5 2 . A function y = f(x) has a second order derivative f \"(x) = 6(x – 1). If its graph passes through the point (2,1) and at that point the tanget to the graph is y = 3x – 5, then the function, is- [AIEEE-2004] (1) (x – 1)2 (2) (x – 1)3 (3) (x + 1)3 (4) (x + 1)2 3 . A function is matched below against an interval where it is supposed to be increasing. which of the following pairs is incorrectly matched ? [AIEEE-2005] interval function (1) (–, ) x3 – 3x2 + 3x + 3 (2) [2, ) 2x3 – 3x2 – 12x + 6 (3) –, 1 3x2 – 2x + 1 3 (4) (–, –4) x3 + 6x2 + 6 4 . The function f(x) = tan–1(sinx + cosx) is an increasing function in- [AIEEE-2007] (4) (–/2, /2) (1) (/4, /2) (2) (–/2, /4) (3) (0, /2) 1 5 . Let f : R R be a continuous function defined by f(x) = ex 2ex 1 Statement–1 : f(c) = , for some c R. 3 1 [AIEEE-2010] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 Statement–2 : 0 < f(x) 2 2 , for all x R. (1) Statement–1 is true, Statement–2 is true ; Statement–2 is a correct explanation for Statement–1. (2) Statement–1 is true, Statement–2 is true ; Statement–2 is not a correct explanation for statement–1. (3) Statement–1 is true, Statement–2 is false. (4) Statement–1 is false, Statement–2 is true. PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] 1. 4 2. 2 3. 3 4. 2 5. 1 E 74

EXERCISE - 05 [B] JEE-Mathematics JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS 1. (a) For all x (0, 1 ) : 2. (A) ex < 1 + x (B) loge (1+x ) < x (C) sin x > x (D) loge x > x 3. 4. (b) Consider the following statements S and R : [JEE 2000 (Screening) 1+1M out of 35] 5. 6. S : Both sin x & cos x are decreasing functions in the interval ( / 2, ) . 7. R : If a differentiable function decreases in an interval (a, b), then its derivative also decreases in (a, b) 8. Which of the following is true ? 9. (A) both S and R are wrong (B) both S and R are correct, but R is not the correct explanation for S E (C) S is correct and R is the correct explanation for S (D) S is correct and R is wrong. Let f(x) = ex (x 1)(x 2) dx then f decreases in the interval - [JEE 2000, Screening, 1M out of 35] (A) (–, 2) (B) (–2, –1) (C) (1, 2) (D) (2, ) [JEE 2001 (Screening ) 1M out of 35] If f(x) = xex(1– x), then f(x) is - (A) increasing on 1 ,1 (B) decreasing on R (C) increasing on R (D) decreasing on 1 ,1 2 2 Let – 1 p 1. Show that the equations 4x3 – 3x – p = 0 has a unique root in the interval 1 ,1 and 2 identify it. [JEE 2001 (Mains), 5M] The length of a longest interval in which the function 3sinx – 4 sin3x is increasing, is - [JEE 2002 (Screening ), 3] 3 (D) (A) (B) (C) 3 2 2 (a) Using the relation 2(1 – cosx) < x2, x 0 or otherwise, prove that sin(tanx) x x 0, . 4 ( b ) Let f : [0, 4] R be a differentiable function. [JEE 2003 (Mains), 4+4M out of 60] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 (i) Show that there exist a, b [0, 4], (f(4))2 – (f(0))2 = 8 f'(a) f(b) 4 (ii) Show that there exist with 0 < < < 2 such that f(t)dt 2(f(2 ) f(2 )) 0 xnx, x 0 . Rolle's theorem is applicable to f for x [0, 1], if = [JEE 2004, Screening] Let f(x)= 0, x 0 (A) –2 (B) –1 (C) 0 (D) 1/2 If p (x) = 51 x101 – 2323 x100 – 45x + 1035, using Rolle’s theorem prove that atleast one root of p(x) = 0 lies between. 1 . [JEE 2004 (Mains), 2M out of 60] 45 100 , 46 Prove that sin x 2x 3x (x 1) x 0, (Justify the inequality, If any used.) 2 [JEE 2004 (Mains), 4M out of 60] 75

JEE-Mathematics (C) has a local minima [JEE 2004, Screening] (D) has bounded area 1 0 . If f(x) = x3 + bx2 + cx + d, 0 < b2 < c, then f(x) (A) is strictly increasing (B) has a local maxima 1 1 . If f(x) is twice differentiable function f(1) = 1, f(2) = 4, f(3) = 9 [JEE 2005, Screening] (A) f''(x) = 2, for atleast one in x (1, 3) (C) f''(x) = f'(x) = 5, for some x (2, 3) (C) f''(x) = 3, (2, 3) (D) f''(x) = 2, for some x (1, 2) 1 2 . If f(x) is a twice differentiable function such that f(a) = 0, f(b) = 2, f(c) = –1, f(d) = 2 , f(e) = 0 where a < b < c < d < e then the minimum number of zeros of g(x) = (f'(x))2 + f''(x) f(x) in the interval [a, e] is. [JEE 2006, 6M out of 184] 1 3 . Let f(x) = 2 + cos x for all real x. Statement-1 : For each real t, there exists a point c in [t, t + ] such that f'(c) = 0 because Statement-2 : f(t) = f(t + 2) for each real t. [JEE 2007, 3M] (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. Comprehension (Q. 14 to 16) If a continuous function f defined on the real line R, assumes positive and negative values in R then the equation f(x) = 0 has a root in R. For example, if it is known that a continuous function f on R is positive at some point and its minimum value is negative then the equation f(x) = 0 has a root in R. Consider f(x) = kex – x for all real x where k is a real constant. 1 4 . The line y = x meets y = kex for k 0 at :- [2007, 4M] (A) no point (B) one point (C) two point (D) more than two points 1 5 . The positive value of k for which kex – x = 0 has only one root is :- [2007, 4M] 1 (B) 1 (C) e (D) log 2 (A) e e 1 6 . For k > 0, the set of all values of k for which kex – x = 0 has two distinct roots is :- [2007, 4M] (A) 0, 1 (B) 1 ,1 (C) 1 , (D) (0, 1) e e e 17. Let the function g : (–, ) , be given by g(u) = 2tan–1(eu) – NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 2 2 . Then, g is - 2 (A) even and is strictly increasing in (0, ) [JEE 2008, 3M, –1M] (B) odd and is strictly decreasing in (– ) (C) odd and is strictly increasing in (– ) (D) neither even nor odd, but is strictly increasing in (– ) 1 1 8 . Let ƒ(x) be a non-constant twice differentiable function defined on (–, ) such that ƒ(x) = ƒ(1 – x) and ƒ ' 4 =0. Then, [JEE 2008, 4M] (A) ƒ \"(x) vanishes at least twice on [0, 1] 1 (B) ƒ ' 2 = 0 (C) 1/2 ƒ x 1 sin x dx 0 1/2 1 1 / 2 2 (D) ƒ(t)esin tdt = ƒ(1 t)esin tdt 0 1/2 E 76

JEE-Mathematics 1 [JEE 2009, 4M, –1M] 1 9 . For the function f(x) = x cos x , x 1 , (A) for at least one x in the interval [1, ) , f(x+2) – f(x) < 2 (B) lim f'(x) = 1 x (C) for all x in the interval [1, ) , f(x+2) –f (x) > 2 (D) f'(x) is strictly decreasing in the interval [1, ) x 2 0 . Let f be a real-valued function defined on the interval (0, ) by f(x) nx 1 sin tdt . Then which of 0 the following statement(s) is (are) true? [JEE 10, 3M] (A) f''(x) exists for all x (0, ) (B) f'(x) exists for all x (0, ) and f' is continuous on (0, ), but not differentiable on (0, ) (C) there exists > 1 such that |f'(x)| < |f(x)| for all x (, ) (D) there exists > 0 such that |f(x)|+|f'(x)|< for all x (, ) 21. Let ƒ : (0,1) R be defined by ƒ(x ) bx , 1 bx where b is a constant such that 0 < b < 1. Then (A) ƒ is not invertible on (0,1) (B) ƒ ƒ–1 on (0,1) and ƒ '(b) 1 ƒ '(0) (C) ƒ = ƒ–1 on (0,1) and ƒ '(b) 1 (D) ƒ–1 is differentiable on (0,1) ƒ '(0) 2 2 . The number of distinct real roots of x4 – 4x3 + 12x2 + x – 1 = 0 is [JEE 2011, 4M] [JEE 2011, 4M] Paragraph for Question 23 and 24 g(x) x 2 t 1 nt ƒ( t ) dt Let ƒ(x) = (1 – x)2sin2x + x2 for all x IR, and let 1 for all x (1,). t 1 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 2 3 . Consider the statements : P : There exists some x IR such that ƒ(x) + 2x = 2(1 + x2) Q : There exists some x IR such that 2ƒ(x) + 1 = 2x(1 + x) Then [JEE 2012, 3M, –1M] (A) both P and Q are true (B) P is true and Q is false (C) P is false and Q is true (D) both P and Q are false 2 4 . Which of the following is true ? [JEE 2012, 3M, –1M] (A) g is increasing on (1,) (B) g is decreasing on (1,) (C) g is increasing on (1,2) and decreasing on (2,) (D) g is decreasing on (1,2) and increasing on (2,) E 77

JEE-Mathematics x [JEE 2012, 4M] 2 5 . If ƒ(x) et2 t 2 t 3 dt for all x (0,), then - 0 (A) ƒ has a local maximum at x = 2 (B) ƒ is decreasing on (2,3) (C) there exists some c (0,) such that ƒ\"(c) = 0 (D) ƒ has a local minimum at x = 3 [JEE 2013, 2M] 2 6 . The number of points in (– ), for which x2 – xsinx – cosx = 0, is (A) 6 (B) 4 (C) 2 (D) 0 2 7 . Let ƒ(x) = x sin x, x > 0. Then for all natural numbers n, ƒ '(x) vanishes at - [JEE 2013, 4M, –1M] (A) a unique point in the interval n, n 1 2 (B) a unique point in the interval n 1 ,n 1 2 (C) a unique point in the interval (n, n + 1) (D) two points in the interval (n, n + 1) PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#07\\Eng\\03-Monotonocity.p65 1. (a) B (b) D 2. C 3. A 4. cos 1 cos 1 p 5. A 7. D 10. A 1 1 . A 1 2 . 6 13. B 14. B 15. A 16. A 3 19. B,C,D 2 0 . B,C 17. C 1 8 . A,B,C,D 2 1 . A 2 2 . 2 2 3 . C 2 4 . B 2 5 . A,B,C,D 26. C 27. B,C 78 E

JEE-Mathematics PARABOLA 1. CONIC SECTIONS : A conic section, or conic is the locus of a point which moves in a plane so that its distance from a fixed point is in a constant ratio to its perpendicular distance from a fixed straight line. ( a ) The fixed point is called the focus. ( b ) The fixed straight line is called the directrix. ( c ) The constant ratio is called the eccentricity denoted by e. ( d ) The line passing through the focus & perpendicular to the directrix is called the axis. ( e ) A point of intersection of a conic with its axis is called a vertex. 2 . GENER AL EQUATION OF A CONIC : FOCAL DIRECTRIX PROPERTY : The general equation of a conic with focus (p, q) & directrix lx + my + n = 0 is : (l2 + m2) [(x – p)2 + (y – q)2] = e2 (lx + my + n)2 ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 3 . DISTINGUISHING BETWEEN THE CONIC : The nature of the conic section depends upon the position of the focus S w.r.t. the directrix & also upon the value of the eccentricity e. Two different cases arise. Case (i) When the focus lies on the directrix : In this case D abc + 2fgh – af2 – bg2 – ch2 = 0 & the general equation of a conic represents a pair of straight lines and if : e > 1 the lines will be real & distinct intersecting at S. e = 1 the lines will coincident. e < 1 the lines will be imaginary. Case (ii) When the focus does not lie on the directrix : The conic represents: a parabola an ellipse a hyperbola a rectangular hyperbola e = 1 ; D 0 0 < e < 1 ; D 0 D 0 ; e > 1 ; e > 1 ; D 0 h2 = ab h2 ab h2 > ab h2 > ab ; a + b = 0 4. PAR A BOLA : E A parabola is the locus of a point which moves in a plane, such that its distance from a fixed point (focus) is NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 equal to its perpendicular distance from a fixed straight line (directrix). Standard equation of a parabola is y2 = 4ax. For this parabola : (i) Vertex is (0, 0) (ii) Focus is (a, 0) (iii) Axis is y = 0 (iv) Directrix is x + a = 0 ( a ) Focal distance : The distance of a point on the parabola from the focus is called the focal distance of the point. ( b ) Focal chord : A chord of the parabola, which passes through the focus is called a focal chord. ( c ) Double ordinate : A chord of the parabola perpendicular to the axis of the symmetry is called a double ordinate. ( d ) Latus rectum : A double ordinate passing through the focus or a focal chord perpendicular to the axis of parabola is called the latus rectum. For y2 = 4ax. 1

JEE-Mathematics Length of the latus rectum = 4a. Length of the semi latus rectum = 2a. Ends of the latus rectum are L(a, 2a ) & L'(a, – 2a ) Note that : (i) Perpendicular distance from focus on directrix = half the latus rectum. (ii) Vertex is middle point of the focus & the point of intersection of directrix & axis. (iii) Two parabolas are said to be equal if they have the same latus rectum. 5 . PAR A METRIC REPRESENTATION : The simplest & the best form of representing the co-ordinates of a point on the parabola is (at2, 2at) . The equation x = at2 & y = 2at together represents the parabola y2 = 4ax , t being the parameter. 6 . TYPE OF PAR A BOL A : Four standard forms of the parabola are y2 = 4ax ; y2 = – 4ax ; x2 = 4ay ; x2 = –4ay Z Y Y Y M L P(x,y) Y Z (0,a) S y=a (a,0) X (–a,0) 0 X Z' 0 Z X' T A S N X' S X' 0 X Z' y=–a Z X' X\\ S L' Y' P' (0,–a) x2 = 4ay x=–a Y' Y' Z' Y' Z' y2 = –4ax x2 = –4ay y2 = 4ax Parabola Ver tex Focus Axis Directrix Length of Ends of Parametric Focal y=0 Latus rectum Latus rectum e q u a ti o n l e ng t h y=0 y2 = 4ax (0,0) (a,0) x=0 x=–a 4a (a, ±2a) (at2,2at) x+a y2 =–4ax (0,0) (–a,0) x=0 (–a, ±2a) x2 = +4ay (0,0) (0,a) y=k x=a 4a (±2a, a) (–at2,2at) x–a x2 = –4ay (0,0) (0,–a) x=p (±2a, –a) (y–k)2 = 4a(x–h) (h,k) (h+a,k) y=–a 4a (h+a, k±2a) (2at,at2) y+a (x–p)2 = 4b(y–q) (p,q) (p, b+q) (p±2a,q+a) y=a 4a (2at, –at2) y–a x+a– h =0 4a (h+at2,k+2at) x–h+a y+b–q=0 4b (p+2at,q+at2) y–q+b Illustration 1 : Find the vertex, axis, directrix, focus, latus rectum and the tangent at vertex for the parabola NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 9y2 – 16x – 12y – 57 = 0. Solution : The given equation can be rewritten as y 22 16 x 61 which is of the form Y2 = 4AX. 3 9 16 Hence the vertex is 61 , 2 16 3 22 The axis is y – = 0 y= 3 3 61 4 613 The directrix is X + A = 0 x + 16 9 0 x 144 The focus is X = A and Y = 0 61 4 and 2 x y 0 16 9 3 focus = 485 , 2 144 3 2E

16 JEE-Mathematics Length of the latus rectum = 4A = Ans. 9 61 The tangent at the vertex is X = 0 x . 16 Illustration 2 : The length of latus rectum of a parabola, whose focus is (2, 3) and directrix is the line x – 4y + 3 = 0 is - 7 14 7 14 (A) 17 (B) 21 (C) 21 (D) 17 Solution : The length of latus rectum = 2 × perp. from focus to the directrix 2 2 4(3) 3 14 (1)2 (4 )2 17 Ans. (D) Illustration 3 : Find the equation of the parabola whose focus is (–6, –6) and vertex (–2, 2). Solution : Let S(–6, –6) be the focus and A(–2, 2) is vertex of the parabola. On SA take a point K(x , y ) such 11 that SA = AK. Draw KM perpendicular on SK. Then KM is the directrix of the parabola. Since A bisects SK, 6 x1 , 6 y1 2, 2 2 2 –6 + x = –4 and –6 + y = 4 or (x , y ) = (2, 10) P M 1 1 11 (x,y) AK Hence the equation of the directrix KM is S (–2, 2) (x1, y1) (–6, –6) y – 10 = m(x – 2) ......... (i) Also gradient of SK = 10 (6) = 16 = 2; m 1 2 (6) 8 2 y – 10 = 1 x 2 (from (i)) 2 x + 2y – 22 = 0 is the directrix Next, let PM be a perpendicular on the directrix KM from any point P(x, y) on the parabola. From SP = PM, the equation of the parabola is x 2y 22 x 6 2 y 62 12 22 or 5(x2 + y2 + 12x + 12y + 72) = (x + 2y – 22)2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 or 4x2 + y2 – 4xy + 104x + 148y – 124 = 0 or (2x – y)2 + 104x + 148y – 124 = 0. Ans. Illustration 4 : The extreme points of the latus rectum of a parabola are (7, 5) and (7, 3). Find the equation of the parabola. Solution : Focus of the parabola is the mid-point of the latus rectum. S is (7, 4). Also axis of the parabola is perpendicular to the latus rectum and passes through the focus. Its equation is y – 4 = 0 x 7 y 4 5 3 Length of the latus rectum = (5 – 3) = 2 Hence the vertex of the parabola is at a distance 2/4 = 0.5 from the focus. We have two parabolas, one concave rightwards and the other concave leftwards. The vertex of the first parabola is (6.5, 4) and its equation is (y – 4)2 = 2(x – 6.5) and it meets the x-axis at (14.5, 0). The equation of the second parabola is (y – 4)2 = –2(x – 7.5). It meets the x-axis at (–0.5, 0). Ans. E3

JEE-Mathematics Do yourself - 1 : ( i ) Name the conic represented by the equation ax by 1 , where a, b R, a, b, > 0. ( i i ) Find the vertex, axis, focus, directrix, latus rectum of the parabola 4y2 + 12x – 20y + 67 = 0. (i i i ) Find the equation of the parabola whose focus is (1, –1) and whose vertex is (2, 1). Also find its axis and latus rectum. ( i v ) Find the equation of the parabola whose latus rectum is 4 units, axis is the line 3x + 4y = 4 and the tangent at the vertex is the line 4x – 3y + 7 = 0. 7 . POSITION OF A POINT REL ATIVE TO A PAR A BOL A : The point (x , y ) lies outside, on or inside the parabola y2 = 4ax according as the expression y 2 – 4ax 11 11 is positive, zero or negative. Illustration 5 : Find the value of for which the point ( – 1, ) lies inside the parabola y2 = 4x. Solution : Point ( – 1, ) lies inside the parabola y2 = 4x y2 –4ax < 0 11 2 – 4( – 1) < 0 2 – 4 + 4 < 0 ( – 2)2 < 0 Ans. 8 . CHORD JOINING TWO POINTS : The equation of a chord of the parabola y2 = 4ax joining its two points P(t ) and Q(t ) is 12 y(t + t ) = 2x + 2at t 12 12 Note : (i) If PQ is focal chord then t t = –1. 12 (ii) Extremities of focal chord can be taken as (at2, 2at) & a , 2 a t2 t Illustration 6 : Through the vertex O of a parabola y2 = 4x chords OP and OQ are drawn at right angles to one Solution : another. Show that for all position of P, PQ cuts the axis of the parabola at a fixed point. The given parabola is y2 = 4x ...... (i) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 Let P t12 , 2 t1 ,Q t 2 , 2t2 2 Slope of OP = 2t1 2 and slope of OQ = 2 t12 t1 t2 Since OP OQ, 4 1 or t t = –4 ...... (ii) t1 t2 12 The equation of PQ is y(t + t ) = 2 (x + t t ) 12 12 4 [from (ii)] y t1 t1 = 2(x – 4) 4 L + L = 0 2(x – 4) – y t1 t1 = 0 12 variable line PQ passes through a fixed point which is point of intersection of L = 0 & L = 0 12 i.e. (4, 0) Ans. 4E

JEE-Mathematics 9 . LINE & A PARABOLA : ( a ) The line y = mx + c meets the parabola y2 = 4ax in two points real, coincident or imaginary according a as a cm condition of tangency is, c = . m Note : Line y = mx + c will be tangent to parabola x2 = 4ay if c = – am2. ( b ) Length of the chord intercepted by the parabola y2 = 4ax on the line y = mx + c is : 4 a(1 m 2 )(a mc) . m 2 Note : Length of the focal chord making an angle with the x - axis is 4a cosec2 . Illustration 7 : If the line y = 3x + intersect the parabola y2 = 4x at two distinct points then set of values of is - Solution : (A) (3, ) (B) (–, 1/3) (C) (1/3, 3) (D) none of these Putting value of y from the line in the parabola - (3x + )2 = 4x 9x2 + (6 – 4)x + 2 = 0 line cuts the parabola at two distinct points D>0 4(3 – 2)2 – 4.92 > 0 92 – 12 + 4 – 92 > 0 < 1/3 Hence, (–, 1/3) Ans.(B) Do yourself - 2 : ( i ) Find the value of 'a' for which the point (a2 – 1, a) lies inside the parabola y2 = 8x. ( i i ) The focal distance of a point on the parabola (x –1)2 = 16(y – 4) is 8. Find the co-ordinates. ( i i i ) Show that the focal chord of parabola y2 = 4ax makes an angle with x-axis is of length 4a cosec2. ( i v ) Find the condition that the straight line ax + by + c = 0 touches the parabola y2 = 4kx. ( v ) Find the length of the chord of the parabola y2 = 8x, whose equation is x + y = 1. 1 0 . LENGTH OF SUBTANGENT & SUBNORMAL : P(at2, 2at) PT and PG are the tangent and normal respectively at the point P to T NG the parabola y2 = 4ax. Then NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 TN = length of subtangent = twice the abscissa of the point P (Subtangent is always bisected by the vertex) NG = length of subnormal which is constant for all points on the parabola & equal to its semilatus rectum (2a). 1 1 . TANGENT TO THE PAR ABOLA y2 = 4ax : (a) Point form : Equation of tangent to the given parabola at its point (x1, y1) is yy1 = 2a (x + x1) (b) Slope form : Equation of tangent to the given parabola whose slope is 'm', is y = mx + a , m 0 m Point of contact is a , 2a m 2 m E5

JEE-Mathematics (c) Parametric form : Equation of tangent to the given parabola at its point P(t), is ty = x + at2 Note : Point of intersection of the tangents at the point t & t is [ at t , a(t + t )]. 12 12 1 2 Illustration 8 : A tangent to the parabola y2 = 8x makes an angle of 45° with the straight line y = 3x + 5. Find its Solution : equation and its point of contact. Let the slope of the tangent be m 3m tan45° = 1 3m 1 3m (3 m) 1 m = –2 or 2 As we know that equation of tangent of slope m to the parabola y2 = 4ax is y = mx + a and point m of contact is a , 2a m2 m for m = –2, equation of tangent is y = –2x – 1 and point of contact is 1 , 2 2 11 Ans. for m = , equation of tangent is y = x + 4 and point of contact is (8, 8) 22 Illustration 9 : Find the equation of the tangents to the parabola y2 = 9x which go through the point (4, 10). Solution : Equation of tangent to parabola y2 = 9x is 9 y = mx + 4m Since it passes through (4, 10) 9 16m2 – 40 m + 9 = 0 10 = 4m + 4m m = 1, 9 44 equation of tangent's are y= x & 9 Ans. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 9 y = x 1 4 4 Illustration 10 : Find the locus of the point P from which tangents are drawn to the parabola y2 = 4ax having Solution : slopes m and m such that - 12 (i) m 2 m 2 (constant) (ii) – = (constant) 1 2 120 where 1 and 2 are the inclinations of the tangents from positive x-axis. Equation of tangent to y2 = 4ax is y = mx + a/m Let it passes through P(h, k) m2h – mk + a = 0 (i) m 2 m 2 1 2 (m + m )2 – 2m m = 1 2 12 k2 a 2. h2 h locus of P(h, k) is y2 – 2ax = x2 6E

JEE-Mathematics (ii) 1 – 2 = 0 tan( – ) = tan 12 0 m1 m2 tan 0 1 m1m2 (m + m )2 – 4m m = tan20(1 + m m )2 1 2 12 12 k2 4a tan2 0 1 a 2 h2 h h k2 – 4ah = (h + a)2 tan20 Ans. locus of P(h, k) is y2 – 4ax = (x + a)2tan2 0 Do yourself - 3 : (i ) Find the equation of the tangent to the parabola y2 = 12x, which passes through the point (2, 5). Find also the co-ordinates of their points of contact. ( i i ) Find the equation of the tangents to the parabola y2 = 16x, which are parallel and perpendicular respectively to the line 2x – y + 5 = 0. Find also the co-ordinates of their points of contact. ( i i i ) Prove that the locus of the point of intersection of tangents to the parabola y2 = 4ax which meet at an angle is (x + a)2 tan2 = y2 – 4ax. 1 2 . NORMAL TO THE PAR ABOLA y2 = 4ax : ( a ) Point form : Equation of normal to the given parabola at its point (x , y ) is 11 y – y = – y 1 (x – x ) 1 2a 1 ( b ) Slope form : Equation of normal to the given parabola whose slope is 'm', is y = mx – 2am – am3 foot of the normal is (am2, – 2am) ( c ) Parametric form : Equation of normal to the given parabola at its point P(t), is y + tx = 2at + at3 Note : (i) Point of intersection of normals at t & t is (a(t 2 + t 2 + t t + 2), – at t (t + t )). 12 12 12 12 1 2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 (ii) If the normal to the parabola y2 = 4ax at the point t , meets the parabola again at the point t , 12 2 then t = – t1 . 2 t1 (iii) If the normals to the parabola y2 = 4ax at the points t & t intersect again on the parabola at the point 12 ‘t ’ then t t = 2 ; t = – (t + t ) and the line joining t & t passes through a fixed point (–2a, 0). 3 12 3 12 12 (iv) If normal drawn to a parabola passes through a point P(h,k) then k = mh – 2 am – am3, i.e. am3 + m (2a – h) + k = 0. 2a h k This gives m1 + m2 + m3 = 0 ; m1m2+m2m3+m3m1 = a ; m1m2m3 = a where m , m , & m are the slopes of the three concurrent normals : 12 3 Algebraic sum of slopes of the three concurrent normals is zero. Algebraic sum of ordinates of the three co-normal points on the parabola is zero. Centroid of the formed by three co-normal points lies on the axis of parabola (x-axis). E7

JEE-Mathematics Illustration 11 : Prove that the normal chord to a parabola y2 = 4ax at the point whose ordinate is equal to abscissa subtends a right angle at the focus. Let the normal at P at12, 2at1 Q 2 , Solution : meet the curve at a t 2 2at2 PQ is a normal chord. 2 ..........(i) y P (at21 , 2at1) and t2 t1 t1 By given condition 2at1 at12 t = 2 from equation (i), t = –3 x' x AS 12 then P(4a, 4a) and Q(9a, –6a) but focus S(a, 0) y' Q (at22 , 2at2) Slope of SP = 4a 0 = 4a 4 4a a 3a 3 6a 0 6a 3 and Slope of SQ = = 9a a 8a 4 43 1 Slope of SP × Slope of SQ 34 PSQ = /2 i.e. PQ subtends a right angle at the focus S. Illustration 12 : If two normals drawn from any point to the parabola y2 = 4ax make angle and with the axis Solution : such that tan . tan = 2, then find the locus of this point. Let the point is (h, k). The equation of any normal to the parabola y2 = 4ax is y = mx – 2am – am3 passes through (h, k) ....(i) k = mh – 2am – am3 am3 + m(2a – h) + k = 0 m , m , m are roots of the equation, then m . m . m = – k 123 123 a k but m m = 2, m = – 1 2 3 2a m is root of (i) a k 3 k (2a h) k 0 k2 = 4ah 3 2a 2a Thus locus is y2 = 4ax. Ans. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 Illustration 13 : Three normals are drawn from the point (14, 7) to the curve y2 – 16x – 8y = 0. Find the coordinates of the feet of the normals. Solution : The given parabola is y2 – 16x – 8y = 0 ......... (i) Let the co-ordinates of the feet of the normal from (14, 7) be P( ). Now the equation of the tangent at P( ) to parabola (i) is y – 8(x + ) – 4(y + ) = 0 or ( – 4)y = 8x + 8a + 4 ......... (ii) 8 Its slope = 4 4 E Equation of the normal to parabola (i) at ( ) is y – = 8 (x – ) It passes through (14, 7) 8

JEE-Mathematics 7–= 4 (14 – ) 6 .......... (iii) ......... (iv) 8 4 Also ( ) lies on parabola (i) i.e. 2 – 16 – 8 = 0 Putting the value of from (iii) in (iv), we get 2 – 96 8 0 4 2( – 4) – 96 – 8( – 4) = 0 (2 – 4 – 96 – 8 + 32) = 0 (2 – 12 – 64) = 0 ( – 16)( + 4) = 0 = 0, 16, – 4 from (iii), = 0 when = 0; = 8, when = 16 ; = 3 when = –4 Hence the feet of the normals are (0, 0), (8, 16) and (3, –4) Ans. Do yourself - 4 : ( i ) If three distinct and real normals can be drawn to y2 = 8x from the point (a, 0), then - (A) a > 2 (B) a (2, 4) (C) a > 4 (D) none of these ( i i ) Find the number of distinct normal that can be drawn from (–2, 1) to the parabola y2 – 4x – 2y – 3 = 0. ( i i i ) If 2x + y + k = 0 is a normal to the parabola y2 = –16x, then find the value of k. ( i v ) Three normals are drawn from the point (7, 14) to the parabola x2 – 8x – 16y = 0. Find the co-ordinates of the feet of the normals. 1 3 . AN IMPORTANT CONCEPT : If a family of straight lines can be represented by an equation 2P + Q + R = 0 where is a parameter and P, Q, R are linear functions of x and y then the family of lines will be tangent to the curve Q2 = 4PR. Illustration 14 : If the equation m2(x + 1) + m(y – 2) + 1 = 0 represents a family of lines, where 'm' is parameter then find the equation of the curve to which these lines will always be tangents. Solution : m2(x + 1) + m(y – 2) + 1 = 0 The equation of the curve to which above lines will always be tangents can be obtained by equating its discriminant to zero. (y – 2)2 – 4(x + 1) = 0 y2 – 4y + 4 – 4x – 4 = 0 y2 = 4(x + y) Ans. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 1 4 . PAIR OF TANGENTS : The equation of the pair of tangents which can be drawn from any point P(x1, y1) outside the parabola to the parabola y2 = 4ax is given by : SS = T2 where : 1 S y2 – 4ax ; S y2 – 4ax ; T yy – 2a (x + x ). 1 1 1 1 1 15 . DIRECTOR CIRCLE : Locus of the point of intersection of the perpendicular tangents to the parabola y2 = 4ax is called the director circle. It’s equation is x + a = 0 which is parabola’s own directrix. Illustration 15 : The angle between the tangents drawn from a point (–a, 2a) to y2 = 4ax is - (A) /4 (B) /2 (C) /3 (D) /6 Solution : The given point (–a, 2a) lies on the directrix x = –a of the parabola y2 = 4ax. Thus, the tangents are at right angle. Ans. (B) Illustration 16 : The circle drawn with variable chord x + ay – 5 = 0 (a being a parameter) of the parabola y2 = 20x as diameter will always touch the line - (A) x + 5 = 0 (B) y + 5 = 0 (C) x + y + 5 = 0 (D) x – y + 5 = 0 E9

JEE-Mathematics Solution : Clearly x + ay – 5 = 0 will always pass through the focus of y2 = 20x i.e. (5, 0). Thus the drawn circle will always touch the directrix of the parabola i.e. the line x + 5 = 0. Ans.(A) Do yourself - 5 : ( i ) If the equation 2x + y – 2 + 2 + 7 = 0 represents a family of lines, where '' is parameter, then find the equation of the curve to which these lines will always be tangents. ( i i ) Find the angle between the tangents drawn from the origin to the parabola, y2 = 4a(x – a). 1 6 . CHORD OF CONTACT : Equation of the chord of contact of tangents drawn from a point P(x , y ) is yy = 2a(x + x ) 11 1 1 Note : The area of the triangle formed by the tangents from the point (x , y ) & the chord of contact is 11 y2 3 /2 i.e. S 1 3 / 2 , also note that the chord of contact exists only if the point P is not inside. 1 4ax1 2a 2a Illustration 17 : If the line x – y – 1 = 0 intersect the parabola y2 = 8x at P & Q, then find the point of intersection of tangents at P & Q. Solution : Let (h, k) be point of intersection of tangents then chord of contact is yk = 4(x + h) 4x – yk + 4h = 0 ....... (i) But given line is x–y–1=0 ....... (ii) Comparing (i) and (ii) 4 k 4h h = – 1, k = 4 1 1 1 point (–1, 4) Ans. Illustration 18 : Find the locus of point whose chord of contact w.r.t. to the parabola y2 = 4bx is the tangent of the parabola y2 = 4ax. Solution : Equation of tangent to y2 = 4ax is y = mx + a ......... (i) m Let it is chord of contact for parabola y2 = 4bx w.r.t. the point P(h, k) Equation of chord of contact is yk = 2b(x + h) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 2b 2bh ......... (ii) y = x kk From (i) & (ii) 2b a 2bh 4b2h m= , a km k k2 locus of P is y2 = 4b2 Ans. x. a 1 7 . CHORD WITH A GIVEN MIDDLE POINT : Equation of the chord of the parabola y2 = 4ax whose middle point is (x , y ) is y – y 2a x x1 . 11 1 = y1 This reduced to T = S, where T yy – 2a (x + x) & S y2 – 4ax . 1 1 1 1 1 1 10 E

JEE-Mathematics Illustration 19 : Find the locus of middle point of the chord of the parabola y2 = 4ax which pass through a given (p, q). Solution : Let P(h, k) be the mid point of chord of the parabola y2 = 4ax, so equation of chord is yk – 2a(x + h) = k2 – 4ah. Since it passes through (p, q) qk – 2a(p + h) = k2 – 4ah Required locus is y2 – 2ax – qy + 2ap = 0. Illustration 20 : Find the locus of the middle point of a chord of a parabola y2 = 4ax which subtends a right angle at the vertex. Solution : The equation of the chord of the parabola whose middle point is ( ) is y – 2a(x + ) = 2 – 4a y – 2ax = 2 – 2a y 2ax 1 or 2 2a ......... (i) Now, the equation of the pair of the lines OP and OQ joining the origin O i.e. the vertex to the points of intersection P and Q of the chord with the parabola y2 = 4ax is obtained by making the equation homogeneous by means of (i). Thus the equation of lines OP and OQ is y2 4ax y 2ax = 2 2a y2(2 – 2a) – 4axy + 8a2x2 = 0 If the lines OP and OQ are at right angles, then the coefficient of x2 + the coefficient of y2 = 0 Therefore, 2 – 2a + 8a2 = 0 2 = 2a( – 4a) Hence the locus of ( ) is y2 = 2a(x – 4a) Do yourself - 6 : ( i ) Find the equation of the chord of contacts of tangents drawn from a point (2, 1) to the parabola x2 = 2y. ( i i ) Find the co-ordinates of the middle point of the chord of the parabola y2 = 16x, the equation of which is 2x – 3y + 8 = 0 (ii i) Find the locus of the mid-point of the chords of the parabola y2 = 4ax such that tangent at the extremities of the chords are perpendicular. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 18 . DIAMETER : The locus of the middle points of a system of parallel chords of a Parabola is called a DIAMETER. Equation to the diameter of a parabola is y = 2a/m, where m = slope of parallel chords. 1 9 . IMPORTANT HIGHLIGHTS : Y (h,k) P ( a ) If the tangent & normal at any point ‘P’ of the parabola intersect the axis at T & G then ST = SG = SP where ‘S’ is the focus. In K S other words the tangent and the normal at a point P on the (a,0) parabola are the bisectors of the angle between the focal radius X' ) G X SP & the perpendicular from P on the directrix. From this we conclude that all rays emanating from S will become parallel to T the axis of the parabola after reflection. x= –a Y' ( b ) The portion of a tangent to a parabola cut off between the directrix P M & the curve subtends a right angle at the focus. Q S E 11

JEE-Mathematics ( c ) The tangents at the extremities of a focal chord intersect at right angles on the directrix, and a circle on any focal chord as diameter touches the directrix. Also a circle on any focal radii of a point P (at2, 2at) as diameter touches the tangent at the vertex and intercepts a chord of length a 1 t2 on a normal at the point P. ( d ) Any tangent to a parabola & the perpendicular on it from the focus meet on the tangent at the vertex. ( e ) Semi latus rectum of the parabola y2 = 4ax, is the harmonic mean between segments of any focal chord of the parabola is ; 2a = 2bc 1 1 1 i.e. . b c bc a P ( f ) If the tangents at P and Q meet in T, then : T (i) TP and TQ subtend equal angles at the focus S. Q S(a,0) (ii) ST2 = SP . SQ & (iii) The triangles SPT and STQ are similar. L ( g ) Tangents and Normals at the extremities of the latus rectum of a parabola (a,2a) y2 = 4ax constitute a square, their points of intersection being (–a, 0) N G & (3a, 0). (–a,0) S (3a,0) Note : (a,–2a) L' (i) The two tangents at the extremities of focal chord meet on the foot of the directrix. (ii) Figure LNL'G is square of side 2 2a ( h ) The circle circumscribing the triangle formed by any three tangents to a parabola passes through the focus. Do yourself - 7 : ( i ) The parabola y2 = 4x and x2 = 4y divide the square region bounded by the line x = 4, y = 4 and the co-ordinates axes. If S , S , S are respectively the areas of these parts numbered from top to bottom; 123 then find S : S : S . 123 ( i i ) Let P be the point (1, 0) and Q a point on the parabola y2 = 8x, then find the locus of the mid point of PQ. Miscellaneous Illustrations : NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 Illustration 21 : The common tangent of the parabola y2 = 8ax and the circle x2 + y2 = 2a2 is - (A) y = x + a (B) x + y + a = 0 (C) x + y + 2a =0 (D) y = x + 2a Solution : 2a Any tangent to parabola is y = mx + m Solving with the circle x2 + (mx + 2a )2 = 2a2 x2 (1 + m2) + 4ax + 4a2 – 2a2 = 0 m m2 B2 – 4AC = 0 gives m = ± 1 Tangent y = ± x ± 2a Ans. (C,D) Illustration 22 : If the tangent to the parabola y2 = 4ax meets the axis in T and tangent at the vertex A in Y and the rectangle TAYG is completed, show that the locus of G is y2 + ax = 0. Solution : Let P(at2, 2at) be any point on the parabola y2 = 4ax. Then tangent at P(at2, 2at) is ty = x + at2 Since tangent meet the axis of parabola in T and tangent at the vertex in Y. E 12

Co-ordinates of T and Y are (–at2, 0) and G JEE-Mathematics (0, at) respectively. x' y Let co-ordinates of G be (x , y ). T P 11 Y (at2, 2at) Since TAYG is rectangle. x Mid-points of diagonals TY and GA is same A x1 0 at2 0 x1 at2 .......... (i) 22 y' and y1 0 0 at y1 at .......... (ii) 22 y1 2 a Eliminating t from (i) and (ii) then we get x = – a 1 or y 2 ax1 or y12 ax1 0 1 The locus of G(x , y ) is y2 + ax = 0 11 Illustration 23 : If P(–3, 2) is one end of the focal chord PQ of the parabola y2 + 4x + 4y = 0, then the slope of the normal at Q is - (A) 1/2 (B) 2 (C) 1/2 (D) 2 Solution : The equation of the tangent at (–3, 2) to the parabola y2 + 4x + 4y = 0 is 2y + 2(x – 3) + 2(y + 2) = 0 or 2x + 4y – 2 = 0 x + 2y – 1 = 0 Since the tangent at one end of the focal chord is parallel to the normal at the other end, the slope 1 Ans.(A) of the normal at the other end of the focal chord is – 2 . Illustration 24 : Prove that the two parabolas y2 = 4ax and y2 = 4c(x – b) cannot have common normal, other than the axis unless b/(a – c) > 2. Solution : Given parabolas y2 = 4ax and y2 = 4c(x – b) have common normals. Then equation of normals in terms of slopes are y = mx – 2am – am3 and y = m(x – b) – 2cm – cm3 respectively then normals must be identical, compare the co-efficients 1 2am am3 mb 2cm cm3 m[(c – a)m2 + (b + 2c – 2a)] = 0, m 0 ( other than axis) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 and m2 = 2a 2c b , m 2 a c b ca ca or m 2 c b a –2– b 0 ca or – 2 + b 0 b 2 ac ac Illustration 25 : If r , r be the length of the perpendicular chords of the parabola y2 = 4ax drawn through the 12 4/3 16a2 vertex, then show that r1 r2 r2 / 3 r2/3 . 1 2 Solution : Since chord are perpendicular, therefore if one makes an angle then the other will make an angle (90° – ) with x-axis E 13

JEE-Mathematics Let AP = r and AQ = r y 12 P(r1cos, r1sin ) If PAX = then QAX = 90° – r1 Co-ordinates of P and Q are (r cos, r sin) x' x 1 1 A r2 90° – and (r sin, – r cos) respectively. 2 2 Q Since P and Q lies on y2 = 4ax y' (r2sin, –r2cos) r12 sin2 4ar1 cos and r22 cos2 4ar2 sin r= 4a cos and r = 4a sin 1 2 sin2 cos2 4a cos 4a sin 4 / 3 16a2 4/3 sin2 cos2 4/3 r1 r2 . sin cos ....... (i) and 4a cos 2/3 4a sin 2 / 3 16a2. r2/3 r22 / 3 16a2 sin2 cos2 1 =16a2 . 4 a 2 / 3 cos 2 / 3 sin 2 / 3 = 16a2 .4a 2 / 3 cos2 sin2 cos 4/3 sin 4 / 3 sin 4 /3 cos 4 / 3 16a2 . 4a 2 / 3 16a2 4/3 = sin cos 4 / 3 cos cos = r1r2 4 / 3 {from (i)} Illustration 26 : The area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points. Solution : Let the three points on the parabola be (a t12 , 2a t1 ), ( a t 2 , 2a t 2 ) and ( a t 2 , 2 a t 3 ) 2 3 The area of the triangle formed by these points 1 = 1 [ at12 (2at – 2at ) + at 2 (2at – 2at ) + at22 (2a – 2at )] 2 3 2 3 1 1 2 2 = – a2(t – t )(t – t )(t – t ). 2 33 11 2 The points of intersection of the tangents at these points are (at t , a(t + t )), (at t , a(t + t )) and (at t , a(t + t )) 23 2 3 31 3 1 12 1 2 The area of the triangle formed by these three points NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 1 2 = 2 at2 t3 (at3 at2 ) at3 t1 (at1 at3 ) at1t2 (at2 at1 ) = 1 a2 (t2 t3 )( t 3 t1 )(t1 t2 ) 2 Hence 1 = 22 Illustration 27 : Prove that the orthocentre of any triangle formed by three tangents to a parabola lies on the Solution : directrix. Let the equations of the three tangents be t1 y x at12 ................(i) t2y x at 2 ................(ii) 2 and t3 y x at23 ................(iii) The point of intersection of (ii) and (iii) is found, by solving them, to be (at t , a(t + t )) E 23 2 3 14

JEE-Mathematics The equation of the straight line through this point & perpendicular to (i) is y – a(t + t ) = –t (x – at t ) 23 1 23 i.e. y + t x = a(t + t + t t t ) ................(iv) 1 2 3 123 Similarly, the equation of the straight line through the point of intersection of (iii) and (i) & perpendicular to (ii) is y +t x = a(t + t + t t t ) ................(v) 2 3 1 123 and the equation of the straight line through the point of intersection of (i) and (ii) & perpendicular to (iii) is y + t x = a(t + t + t t t ) ................(vi) 1 1 2 123 The point which is common to the straight lines (iv), (v) and (vi) i.e. the orthocentre of the triangle, is easily seen to be the point whose coordinates are x = –a, y = a(t + t + t + t t t ) 1 2 3 123 and this point lies on the directrix. ANSWERS FOR DO YOURSELF 1:NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 (i) Parabola (ii) Vertex : 7, 5 , Axis : y = 5 , Focus : 17 , 5 , Directrix : x 11 ; LR = 3 2 2 2 4 2 4 2: (i i i ) 4x2 + y2 – 4xy + 8x + 46y – 71 = 0; Axis : 2x – y = 3; LR = 4 5 unit 3: ( i v ) (3x + 4y – 4)2 = 20(4x – 3y + 7) 4: (i) 8 8 (ii) (–7, 8), (9, 8) (iv) kb2 = ac (v) 8 3 5: , 7 7 , 6: 7: (i) x – y + 3 = 0, (3, 6); 3x – 2y + 4 = 0, 4 , 4 3 E ( i i ) 2x – y + 2 = 0, (1, 4) ; x + 2y + 16 = 0, (16, –16) (i) C (ii) 1 (iii) 48 ( i v ) (0, 0), (–4, 3) and (16, 8) ( i ) (y + 2)2 = 28(x – 1) (ii) /2 ( i ) 2x = y + 1 (ii) (14, 12) (i i i ) y2 = 2a(x – a) (i) 1 : 1 : 1 ( i i ) y2 – 4x + 2 = 0 15

JEE-Mathematics CHECK YOUR GRASP EXERCISE - 01 SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . Latus rectum of the parabola whose focus is (3, 4) and whose tangent at vertex has the equation x + y = 7 5 2 is - (A) 5 (B) 10 (C) 20 (D) 15 2 . Directrix of a parabola is x + y = 2. If it's focus is origin, then latus rectum of the parabola is equal to - (A) 2 units (B) 2 units (C) 2 2 units (D) 4 units 3 . Which one of the following equations represents parametrically, parabolic profile ? (A) x = 3 cos t ; y = 4 sint t (B) x2 – 2 = –cost ; y = 4 cos2 2 (C) x tan t ; y sec t (D) x 1 sin t ; y sin t cos t 22 4 . Let C be a circle and L a line on the same plane such that C and L do not intersect. Let P be a moving point such that the circle drawn with centre at P to touch L also touches C. Then the locus of P is - (A) a straight line parallel to L not intersecting C (B) a circle concentric with C (C) a parabola whose focus is centre of C and whose directrix is L. (D) a parabola whose focus is the centre of C and whose directrix is a straight line parallel to L. 5 . If (t2, 2t) is one end of a focal chord of the parabola y2 = 4x then the length of the focal chord will be- (A) t 1 2 (B) t 1 t 2 1 (C) t 1 t 2 1 (D) none t t t2 t t2 6 . From the focus of the parabola y2 = 8x as centre, a circle is described so that a common chord of the curves is equidistant from the vertex and focus of the parabola. The equation of the circle is - (A) (x – 2)2 + y2 = 3 (B) (x – 2)2 + y2 = 9 (C) (x + 2)2 + y2 = 9 (D) x2 + y2 – 4x = 0 7 . The point of intersection of the curves whose parametric equations are x = t2+1, y = 2t and x = 2s, y = 2/s is given by - (A) (4, 1) (B) (2, 2) (C) ( –2, 4) (D) (1, 2) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 8 . If M is the foot of the perpendicular from a point P of a parabola y2= 4ax to its directrix and SPM is an equilateral triangle, where S is the focus, then SP is equal to - (A) a (B) 2a (C) 3a (D) 4a 9 . Through the vertex ‘O’ of the parabola y2 = 4ax, variable chords OP and OQ are drawn at right angles. If the variable chord PQ intersects the axis of x at R, then distance OR : (A) varies with different positions of P and Q (B) equals the semi latus rectum of the parabola (C) equals latus rectum of the parabola (D) equals double the latus rectum of the parabola 1 0 . The triangle PQR of area ‘A’ is inscribed in the parabola y2 = 4ax such that the vertex P lies at the vertex of the parabola and the base QR is a focal chord. The modulus of the difference of the ordinates of the points Q and R is - A A 2A 4A (A) (B) (C) (D) 2a a a a 16 E

JEE-Mathematics 1 1 . Point P lies on y2 = 4ax & N is foot of perpendicular from P on its axis. A straight line is drawn parallel to the axis to bisect NP and meets the curve in Q. NQ meets the tangent at the vertex in a point T such that AT = k NP, then the value of k is : (where A is the vertex) (A) 3/2 (B) 2/3 (C) 1 (D) none 1 2 . The tangents to the parabola x = y2 + c from origin are perpendicular then c is equal to - 1 (B) 1 (C) 2 1 (A) (D) 2 4 1 3 . The locus of a point such that two tangents drawn from it to the parabola y2 = 4ax are such that the slope of one is double the other is - (A) y2 9 ax (B) y2 9 (C) y2 9ax (D) x2 4ay 2 ax 4 1 4 . T is a point on the tangent to a parabola y2 = 4ax at its point P. TL and TN are the perpendiculars on the focal radius SP and the directrix of the parabola respectively. Then - (A) SL = 2 (TN) (B) 3 (SL) = 2 (TN) (C) SL = TN (D) 2 (SL) = 3 (TN) 1 5 . The equation of the circle drawn with the focus of the parabola (x – 1 )2 – 8y = 0 as its centre and touching the parabola at its vertex is : (A) x2 + y2 – 4 y = 0 (B) x2 + y2 – 4 y + 1 = 0 (C) x2 + y2 – 2x – 4 y = 0 (D) x2 + y2 – 2x – 4 y + 1 = 0 1 6 . Length of the normal chord of the parabola, y2 = 4x, which makes an angle of with the axis of x is- 4 (A) 8 (B) 8 2 (C) 4 (D) 4 2 1 7 . Tangents are drawn from the point (–1, 2) on the parabola y2 = 4x . The length , these tangents will intercept on the line x = 2 : (A) 6 (B) 6 2 (C) 2 6 (D) none of these 1 8 . Locus of the point of intersection of the perpendiculars tangent of the curve y2 + 4y – 6x – 2 = 0 is : (A) 2x – 1 = 0 (B) 2x + 3 = 0 (C) 2y + 3 = 0 (D) 2x + 5 = 0 1 9 . Tangents are drawn from the points on the line x – y + 3 = 0 to parabola y2 = 8x. Then the variable chords NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 of contact pass through a fixed point whose coordinates are- (A) (3, 2) (B) (2, 4) (C) (3, 4) (D) (4, 1) 2 0 . The line 4x – 7y + 10 = 0 intersects the parabola, y2 = 4x at the points A & B. The co-ordinates of the point of intersection of the tangents drawn at the points A & B are : (A) 7 , 5 (B) 5 , 7 (C) 5 , 7 (D) 7 , 5 2 2 2 2 2 2 2 2 2 1 . From the point (4, 6) a pair of tangent lines are drawn to the parabola, y2 = 8x. The area of the triangle formed by these pair of tangent lines & the chord of contact of the point (4, 6) is (A) 2 (B) 4 (C) 8 (D) none 2 2 . TP & TQ are tangents to the parabola, y2=4ax at P & Q. If the chord PQ passes through the fixed point (-a, b) then the locus of T is - (A) ay = 2b (x – b) (B) bx = 2a (y – a) (C) by = 2a (x – a) (D) ax = 2b (y – b) E 17

JEE-Mathematics 2 3 . If the tangent at the point P (x ,y ) to the parabola y2 = 4ax meets the parabola y2 = 4a (x + b) at Q & R, then 11 the mid point of QR is - (A) (x + b, y + b) (B) (x – b, y – b) (C) (x , y ) (D) (x + b, y ) 11 11 11 11 2 4 . Let PSQ be the focal chord of the parabola, y2 = 8x. If the length of SP=6 then, l(SQ) is equal to(where S is the focus) - (A) 3 (B) 4 (C) 6 (D) none 2 5 . Two parabolas y2 = 4a(x – l1) and x2 = 4a(y – l2) always touch one another , the quantities l1 and l2 are both variable. Locus of their point of contact has the equation - (A) xy = a2 (B) xy = 2a2 (C) xy = 4a2 (D) none SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 2 6 . Equation x2 – 2x – 2y + 5 = 0 represents - (A) a parabola with vertex (1, 2) (B) a parabola with vertex (2, 1) (C) a parabola with directrix y 3 (D) a parabola with directrix y2 2 5 2 7 . The normals to the parabola y2 = 4ax from the point (5a, 2a) are - (A) y = –3x + 33a (B) x = –3y + 3a (C) y = x – 3a (D) y = –2x + 12a 2 8 . The equation of the lines joining the vertex of the parabola y2 = 6x to the points on it whose abscissa is 24, is - (A) 2y + x + 1 = 0 (B) 2y – x + 1 =0 (C) x + 2y = 0 (D) x – 2y = 0 2 9 . The equation of the tangent to the parabola y2 = 9x which passes through the point (4, 10) is - (A) x + 4y + 1 = 0 (B) x – 4y + 36 = 0 (C) 9x – 4y + 4 = 0 (D) 9x + 4y + 4 = 0 3 0 . Consider the equation of a parabola y2 = 4ax, (a < 0) which of the following is false - (A) tangent at the vertex is x = 0 (B) directrix of the parabola is x = 0 (C) vertex of the parabola is at the origin (D) focus of the parabola is at (–a, 0) CHECK YOUR GRASP ANSWER KEY EXERCISE-1 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 Que. 1 23 4 5 6 7 8 9 10 11 12 13 14 15 Ans. C CB DABBDCCBDACD Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. B BD C C A C C A C A,C C,D C,D B,C B,D 18 E

Search

### Read the Text Version

- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
- 40
- 41
- 42
- 43
- 44
- 45
- 46
- 47
- 48
- 49
- 50
- 51
- 52
- 53
- 54
- 55
- 56
- 57
- 58
- 59
- 60
- 61
- 62
- 63
- 64
- 65
- 66
- 67
- 68
- 69
- 70
- 71
- 72
- 73
- 74
- 75
- 76
- 77
- 78
- 79
- 80
- 81
- 82
- 83
- 84
- 85
- 86
- 87
- 88
- 89
- 90
- 91
- 92
- 93
- 94
- 95
- 96
- 97
- 98
- 99
- 100
- 101
- 102
- 103
- 104
- 105
- 106
- 107
- 108
- 109
- 110
- 111
- 112
- 113
- 114
- 115
- 116
- 117
- 118
- 119
- 120
- 121
- 122
- 123
- 124
- 125
- 126
- 127
- 128
- 129
- 130
- 131
- 132
- 133
- 134
- 135
- 136
- 137
- 138
- 139
- 140
- 141
- 142
- 143
- 144
- 145
- 146
- 147
- 148
- 149
- 150
- 151
- 152
- 153
- 154
- 155
- 156
- 157
- 158
- 159
- 160
- 161
- 162
- 163
- 164
- 165
- 166
- 167
- 168
- 169
- 170
- 171
- 172
- 173
- 174
- 175
- 176
- 177
- 178
- 179
- 180
- 181
- 182
- 183
- 184
- 185
- 186
- 187
- 188
- 189
- 190
- 191
- 192
- 193
- 194
- 195
- 196
- 197
- 198
- 199
- 200
- 201
- 202
- 203
- 204
- 205
- 206
- 207
- 208
- 209
- 210
- 211
- 212
- 213
- 214
- 215
- 216
- 217
- 218
- 219
- 220
- 221
- 222
- 223
- 224
- 225
- 226
- 227
- 228
- 229
- 230
- 231
- 232
- 233
- 234
- 235
- 236
- 237
- 238
- 239
- 240
- 241
- 242
- 243
- 244
- 245
- 246
- 247
- 248
- 249
- 250
- 251
- 252
- 253
- 254
- 255
- 256
- 257
- 258
- 259
- 260
- 261
- 262
- 263
- 264
- 265
- 266
- 267
- 268
- 269
- 270
- 271
- 272
- 273
- 274
- 275
- 276
- 277
- 278
- 279
- 280
- 281
- 282
- 283
- 284
- 285
- 286
- 287
- 288
- 289
- 290
- 291
- 292
- 293
- 294
- 295
- 296
- 297
- 298
- 299
- 300
- 301
- 302
- 303
- 304
- 305
- 306
- 307
- 308
- 309
- 310
- 311
- 312
- 313
- 314
- 315
- 316
- 317
- 318
- 319
- 320
- 321
- 322
- 323
- 324
- 325
- 326
- 327
- 328
- 329
- 330
- 331
- 332
- 333
- 334
- 335
- 336
- 337
- 338
- 339
- 340
- 341
- 342
- 343
- 344
- 345
- 346