f (x) = kf 200x ,then k = 17. If R denotes the set of all real numbers then the 100 + x2 function f : R R de ned f(x) = [x] (a) One-one only (a) 0.5 (b) 0.6 (c) 0.7 (d) 0.8 (b) Onto only (c) Both one-one and onto 8. e graph of the function y = f(x) is symmetrical (d) Neither one-one nor onto about the line x = 2, then (a) f(x) = –f (–x) (b) f(2 + x) = f(2 – x) (c) f(x) = f (–x) (d) f(x + 2) = f(x – 2) 18. f (x) = x + x2 is a function from R R , then f(x) is 9. Mapping f : R R which is de ned as f(x) = cosx, (a) Injective (b) Surjective x ∈ R will be (c) Bijective (d) None of these (a) Neither one-one nor onto (b) One-one 19. e period of f (x) = x – [x], if it is periodic, is (c) Onto 1 (d) One-one onto (a) f (x) is not periodic (b) 2 10. Let f : N N de ned by f(x) = x2 + x + 1, x ∈ N, (c) 1 (d) 2 then f is (a) One-one onto 20. If f(x) = ax + b and g(x) = cx + d, then f(g(x)) = g(f(x)) (b) Many one onto is equivalent to (c) One-one but not onto (a) f (a) = g (c) (b) f (b) = g (b) (d) None of these (c) f (d) = g (b) (d) f (c) = g (a) 11. Set A has 3 elements and set B has 4 elements. e 21. e domain of the function f (x) = sin−1(3 − x) is number of injection that can be de ned from A to log(| x | − 2) B is (a) 144 (b) 12 (c) 24 (d) 64 (a) [2, 4] (b) (2, 3) (3, 4] (c) [2, ∞) (d) (– ∞, –3) [2, ∞) x −m, 12. Let f : R R be a function de ned by f (x) = x −n e function f (x) = sec−1 x , where [.] denotes where m ≠ n. en x −[x] (a) f is one-one onto (b) f is one-one into 22. (c) f is many one onto (d) f is many one into the greatest integer less than or equal to x is de ned for all x belonging to 13. Which one of the following is a bijective function (a) R on the set of real numbers (b) R – {(–1, 1) (n | n ∈ Z)} (a) 2x – 5 (b) |x| (c) x2 (d) x2 + 1 (c) R+ – (0, 1) 14. Let the function f : R R be de ned by (d) R+ – {n | n ∈ N} f (x) = 2x + sin x, x ∈ R. en f is (a) One-to-one and onto 23. Domain of the function f (x) = log10 5x − x2 1/2 is (b) One-to-one but not onto 4 (c) Onto but not one-to-one (d) Neither one-to-one nor onto (a) –∞ < x < ∞ (b) 1 x 4 (c) 4 x 16 (d) –1 x 1 x 15. If f : [0, ∞) [0, ∞) and f (x) = 1 + x , then f is 24. f : [0, 1] R is a di erentiable function such that (a) One-one and onto f (0) = 0 and |f ′(x)| k |f (x)| for all x ∈ [0, 1], (b) One-one but not onto (k > 0), then which of the following is/are always (c) Onto but not one-one true ? (d) Neither one-one nor onto (a) f (x) = 0, x ∈ R (b) f (x) = 0, x ∈ [0, 1] 16. If f : R S de ned by f (x) = sin x − 3 cos x +1 is (c) f (x) ≠ 0, x ∈ [0, 1] onto, then the interval of S is (d) f (1) = k (a) [–1, 3] (b) [1, 1] (c) [0, 1] (d) [0, –1] MATHEMATICS TODAY | SEPTEMBER‘17 51
25. Domain of de nition of the function 35. e function f (x) = sin log(x + x2 +1) is 3 (a) Even function f (x) = 4 − x2 + log10 (x 3 − x) , is (b) Odd function (c) Neither even nor odd (a) (1, 2) (d) Periodic function (b) (–1, 0) (1, 2) (c) (1, 2) (2, ∞) (d) (–1, 0) (1, 2) (2, ∞) e inverse of the function ex − e−x is 26. e domain of the function 36. given by f (x) = ex + e−x +2 f (x) = x − x2 + 4 + x + 4 − x is 2 1/2 x −1 1/2 (a) [–4, ∞) (b) [–4, 4] (c) [0, 4] (d) [0, 1] (a) loge x − 1 (b) loge 3−x x − 27. e domain of the function log(x2 − 6x + 6) is (a) (–∞, ∞) (c) loge x 1/2 (d) loge x − 1 −2 (b) (−∞, 3 − 3) ∪ (3 + 3, ∞) − x + 1 2 x (c) (–∞, 1] [5, ∞) (d) [0, ∞) 37. If the function f : [1, ∞) [1, ∞) is de ned by f (x) = 2x(x – 1), then f –1 (x) is 28. e natural domain of the real valued function 1 x(x−1) de ned by f (x) = x2 −1 + x2 +1 is (a) 2 (b) 1 (1 + 1+ 4 log2 x ) (a) 1 < x < ∞ (b) –∞ < x < ∞ 2 (c) –∞ < x < –1 (d) (–∞, ∞) – (–1, 1) (c) 1 (1 − 1+ 4 log2 x ) (d) Not de ned 2 29. e domain of the function f (x) = sin−1(x − 3) is 9 − x2 38. If f (x ) = 2x −1 (x ≠ −5) , then f –1(x) is equal to (a) [1, 2) (b) [2, 3) (c) [1, 2] (d) [2, 3] x+5 30. e range of f (x) = sec π cos2 x , − ∞ < x < ∞ is (a) x+5 , x ≠ 1 (b) 5x +1 , x ≠ 2 4 2x −1 2 2−x (a) [1, 2] (b) [1, ∞) (c) 5x −1 , x ≠ 2 (d) x−5 , x ≠ 1 2−x 2x +1 2 (c) [− 2, −1]∪[1, 2] (d) (–∞, –1] [1, ∞) Multiple Correct Answer Type 31. If f (x) = a cos(bx + c) + d, then range of f (x) is (a) [d + a, d + 2a] (b) [a – d, a + d] 39. Let f(x) be a non constant polynomial satisfying (c) [d + a, a – d] (d) [d – a, d + a] the relation f (x) f (y) = f (x) + f (y) + f (xy) – 2 for all real x and y and f (0) ≠ 1, suppose f (4) = 65. 32. Range of f (x) = x2 + 34x − 71 is en x2 + 2x − 7 (a) f 1(x) is a polynomial of degree 2 (b) roots of f 1(x) = 2x + 1 are real (a) [5, 9] (b) (–∞, 5] [9, ∞) (c) xf 1(x) = 3[f (x) – 1] (c) (5, 9) (d) None of these (d) f 1 (–1) = 3 33. e function f : R R is de ned by f(x) = cos2 x + 40. If a function satis es (x – y) f (x + y) – (x + y) sin4 x for x ∈ R, then f (R) = f (x – y) = 2(x2y – y3) x, y ∈ R and f (1) = 2, then 3 3 3 3 (a) f (x) must be polynomial function (a) 4 , 1 (b) 4 , 1 (c) 4 , 1 (d) 4 , 1 (b) f (3) = 12 (c) f (0) = 0 (d) f (x) may not be di erentiable 34. If x is real, then value of the expression x2 +14x + 9 lies between x2 + 2x + 3 (a) 5 and 4 (b) 5 and –4 (c) – 5 and 4 (d) None of these 52 MATHEMATICS TODAY | SEPTEMBER‘17
41. Solutions of the equations [x] + [y] = [x] [y] is/are 46. e set A is equal to (b) [2, 5] where [.] denotes the greater integer function (a) [–5, –2] (d) [–3, –2] (a) 2 x < 3 and 2 y < 3 (c) [–5, 2] (b) 0 x < 1 and 0 y < 1 (c) 0 < x 2 and 0 < y 2 Matrix-Match Type (d) None 47. Match the following. 42. Let f (x) = a1cos (a1 + x) + a2 caonsd(ax2a+n=cxxo)s1+(a(…wn h++exr)e. Column I Column II x = 0 + an cos an = 0 If f (x) vanishes for a2 + … (A) e interval containing (p) 0, 1 kp, k Z), then the complete set of 2 x(1a)≠ a1cos ∈ + a2 cos solution of the equation a1 (b) a1sin a1 + a2 sin a2 + … + an sin an = 0 1− x2 + x = 1 are x x (c) f (x) = 0 has only two solutions 0, x1 (d) f (x) is identically zero x (B) e interval containing (q) [–1, 1]– {0} the complete set of 43. Consider the real valued function satisfying values of ‘a’, for which 2f (sin x) + f (cos x) = x, then (a + 1) x + ay – 1 = 0 (a) domain of f (x) is R is a normal to the curve (b) domain of f (x) is [–1, 1] xy =1, are (c) range of f (x) is −2π , π (C) Complete set of values (r) [0, 1] 3 6 of ‘a’ for which equation a sin2 x + |cos x| – 2a = 0 (d) range of f (x) is −2π , π has atleast one solution 3 3 belongs to the interval 44. Let f (x) = x 2 −4 x + 3 x<3 and (D) e interval containing (s) [–1, 0] x −4 x≥3 the range of the function 1 g (x) = x − 3 x<4 then f (x ) = 1 + 2 cos2 x+ 3 cos4 x (x +1)2 x≥4 + 1 + 4 cos6 x + .....∞ (a) (f − g ) 7 = −1 (b) fog(3) = 3 (t) −1, 1 2 2 (c) (fg)(2) = 1 (d) (f + g ) 7 −(f − g)(4) = 26 Integer Answer Type 2 Comprehension Type 48. Find the natural number c for which n Paragraph for Q. No. 45 and 46 f (c + r) = 16(2n − 1) where the function satis es Let f : [2, ∞) [1, ∞) de ned by f (x) = 2x4 – 4x2 and ∑ trh=e1 relation f(x + y) = f (x) . f (y) for all x, y ∈ N : π , π defined by g(x) sin x + 4 be two (natural numbers) and f (1) = 2. g 2 → A = sin x − 2 invertible function, then 49. For non negative integers m, n de ne a function as follows 45. f –1(x) is equal to n+1 if m = 0 (a) 2 + 4 − log2 x (b) 2 + 4 + log2 x f (m, n) = f (m −1,1) if m ≠ 0,n = 0 (c) 2 − 4 + log2 x (d) None of these f (m −1, f (m,n −1)) if m ≠ 0,n ≠ 0 en the value of f (1, 1) is MATHEMATICS TODAY | SEPTEMBER‘17 53
50. If function f satis es the relation 3. (c) : f(x) = cos(log x) f(x) f ′(–x) = f(–x). f ′(x) for all x and f (0) = 3, now if f (3) = 3, then the value of f (–3) is Now let y = f(x) f(4) – 1 f x + f (4x) 2 4 51. Number of real values of x, satisfying the equation [x]2 – 5[x] + 6 – sin x = 0, [.] denoting the greatest ⇒ y = cos(log x) cos(log 4) integer function is – 1 cos log x + cos (log 4x) 2 4 52. If f (x + 2) – 5f (x + 1) + 6f (x) = 0 for all, and f (x) = Aa1x + Bax2, then a1 + a2 is ⇒ y = cos(log x) cos(log 4) – 1 [cos(log x – log 4) 2 53. Let f be a function from the set of positive integers + cos(log x + log 4)] to the set of real numbers i.e., f : N R such that 1 (i) f (1) = 1 ⇒ y = cos(log x) cos(log 4) – 2 [2 cos(log x) cos (log 4)] (ii) f (1) + 2f (2) + 3f (3) + ... + nf(n) = n(n + 1) f (n); ⇒ y = 0. for n ≥ 2. Find the value of 800·f (200). 4. (b) : f(2x) = 2(2x) + |2x| = 4x + 2|x|, f (–x) = –2x + |–x| = – 2x + |x|, f (x) = 2x + |x| 54. If for all real values of u and v, 2f (u) cos v = ⇒ f (2x) + f (–x) – f (x) = 4x + 2|x| + |x| – 2x – 2x – |x| = 2|x|. f (u – v) + f (u + v) prove that for all real values of x, (i) f(x) + f(–x) = 2a cos x 5. (c) : Given f(x + ay, x – ay) = axy ...(i) (ii) f (p – x) + f(–x) = 0 Let x + ay = u and x – ay = v (iii) f (p – x) + f(x) = 2b sin x en f(x) = a cos mx + b sin nx where a and b are en x= u+v and y = u−v arbitrary constants, nd m + n. 2 2a Substituting the value of x and y in (i), we obtain ( )55. Let f(x + y + 1) = f (x) + f (y) 2 for all x, y ∈ R u2 − v2 x2 y2 and f (0) = 1. en f(x) = (x + 1)m. Find m. f(u, v) = 4 ⇒ f(x, y) = − . 4 6. (d) : f(x) = cos[p2]x + cos[–p2]x f(x) = cos(9x) + cos(–10x) = cos(9x) + cos(10x) SOLUTIONS 19x 2 x 1. (c) : Given, f(x) = log 1+ x = 2 cos cos 2 1− x 1 + 2x x2 f π = 2 cos 19π cos π ; 1+ x2 x2 2 4 4 2x 1 − +1+ 2x \\ f 1+ x2 = log 2x = log +1− 2x f π =2× −1 × 1 = −1 . 1+ x2 2 2 2 = log 1 + x 2 = 2 log 1+ x = 2f(x) 7. (a) : e f(x) = 10 + x , x ∈ (–10, 10) 1 − x 1 − x 10 − x x − 3 −3 ⇒ f(x) = log 10 + x x + 1 +1 10 − x f (x) − 3 x − 3 2. (a) : f [f(x)] = f (x) +1 == x + 1 10 + 200x − 100 + x2 x − 3 − 3x − 3 3 ⇒ f 200x = log 10 x −3+ x +1 1 100 + x2 200x = = + x 100 + x2 − x 3+ x 3+ x − 3 = log 10(10 + x) 2 = 2 log 10 + x = 2f(x) 1 − x 1− x 10(10 − x) 10 − x Now, f [f(f(x))] = f = = 3+ x 3 + x − 3 + 3x 1− x +1 f(x) = 1 200x ⇒k= 1 = 0.5. 3+ x +1−x 2 100 + x2 2 =x \\ f 54 MATHEMATICS TODAY | SEPTEMBER‘17
8. (b) : f(x) = f(–x) ⇒ f(0 + x) = f(0 – x) is symmetrical −2 ≤ (sin x − 3 cos x) ≤ 2 about x = 0. \\ f(2 + x) = f(2 – x) is symmetrical about x = 2. −2 + 1 ≤ (sin x − 3 cos x + 1) ≤ 2 + 1 ⇒⇒Af9(.xg2a)i(ncxa=o1)tshc≠: oxLexs1ev2tx=a, 2xlsu,c1oeo,soisxot2fxfi(s2∈fx-n⇒1imR)o,t=axtgo1hfen(e=xeon2-f2)ofxn(nxple1i.e)s=xin2cobsetxw1,een –1 to 1 −1 ≤ (sin x − 3 cos x + 1) ≤ 3 i.e., range = [–1, 3] ⇒ f(R) = {f(x) : – 1 f(x) 1} \\ For f to be onto S = [–1, 3]. So other numbers of co-domain is not f-image. 17. (d) : Let f(x1) = f(x2) ⇒ [x1] = [x2] ≠ x1 = x2 f (R) ∈ R, so it is also not onto. So this mapping is {For example, if x1= 1 4, x2 = 1 5, then [1 4] = [1 5] =1} neither one-one nor onto. \\ f is not one-one. Also, f is not onto as its range I (set of integers) is a proper subset of its co-domain R. 10. (c) : Let x, y ∈ N such that f(x) = f(y) 18. (d) : We have f(x) = x + x2 = x + |x| en f(x) = f(y) ⇒ x2 + x + 1 = y2 + y + 1 Clearly f is not one-one as f(–1) = f(–2) = 0 but – 1 ≠ 2. Also f is not onto as f(x) ≥ 0 x ∈ R, ⇒ (x – y)(x + y + 1) = 0 Also range of f = [0, ∞) R. ⇒ x = y or x = –(y + 1) N (Rejected) 19. (c) : Let f(x) be periodic with period T. \\ f is one-one. Now, Let f(x) = 1 ⇒ x2 + x + 1 = 1 en, f(x + T) = f(x) for all x ∈ R ⇒ x(x + 1) = 0 ⇒ x + T – [x + T] = x – [x], for all x ∈ R ⇒ x = 0 or x = –1 (Not possible) ⇒ x + T – x = [x + T] – [x] \\ f is not onto. ⇒ [x + T] – [x] = T for all x ∈ R ⇒ T = 1, 2, 3, 4,... 11. (c) : e total number of injective functions from e smallest value of T satisfying f(x + T) = f(x) for a set A containing 3 elements to a set B containing 4 all x ∈ R is 1. elements is equal to the total number of arrangements Hence f(x) = x – [x] has period 1. of 4 by taking 3 at a time i.e., 4P3 = 24 . 20. (c) : We have f(x) = ax + b, g(x) = cx + d 12. (b) : For any x, y ∈ R, we have and f(g(x)) = g(f(x)) ⇒ f(cx + d) = g(ax + b) f(x) = f(y) ⇒ x −m = y −m ⇒x=y ⇒ a(cx + d) + b = c(ax + b) + d x −n y −n ⇒ ad + b = cb + d ⇒ f(d) = g(b). \\ f is one-one. Let a ∈ R such that f(x) = a ⇒ x − m = α 21. (b) : f(x) = sin−1(3 − x) x −n log(| x | −2) x= m − nα ⇒ 1−α Let g(x) = sin–1(3 – x) ⇒ –1 3 – x 1 Domain of g(x) is [2, 4] Clearly x R for a = 1. So, f is not onto. and let h(x) = log(|x| – 2) ⇒ |x| – 2 > 0 ⇒ |x| > 2 ⇒ x < –2 or x > 2 ⇒ (–∞, –2) 13. (a) : |x|, x2 and x2 + 1 is not one-one. But 2x – 5 We know that (2, ∞) is one-one as f(x) = f(y) ⇒ 2x – 5 = 2y – 5 ⇒ x = y Also, f(x) = 2x – 5 is onto. \\ f(x) = 2x – 5 is bijective. 14. (a) : f ′(x) = 2 + cos x > 0. So, f(x) is strictly (f/g)(x) = f (x) ∀x ∈D1 ∩ D2 − {x ∈R : g (x) = 0} monotonic increasing so, f(x) is one-to-one and onto. g (x) 15. (b) : f ′(x) = 1 > 0, x ∈ [0 ∞) and \\ Domain of f(x) = (2, 4] – {3} = (2, 3) (3, 4]. range ∈[0, 1) (1+ x)2 22. (b) : The function sec–1 x is defined for all ⇒ Function is one-one but not onto. x ∈ R – (–1, 1) and the function 1 is de ned for x −[x] 16. (a) : − 1+ (− 3)2 ≤ (sin x − 3 cos x) ≤ 1+ (− 3)2 all x ∈ R – Z. So the given function is de ned for all x ∈ R – {(–1, 1) (n | n ∈ Z)}. MATHEMATICS TODAY | SEPTEMBER‘17 55
5x x2 1/2 29. (b) : To de ne f(x), 9 – x2 > 0 ⇒ –3 < x < 3 ...(i) log10 − –1 (x – 3) 1 ⇒ 2 x 4 ...(ii) 23. (b) : We have f(x) = 4 ...(i) From (i) and (ii), 2 x < 3 i.e., [2, 3). From (i), clearly f(x) is de ned for those values of x 30. (a) : f(x) = sec π cos2 x 4 for which log10 5x − x2 0 4 ≥ We know that, 0 cos2 x 1 Also, at cos x = 0, f(x) = 1 and x2 x2 at cos x = 1, f(x) = 2 \\ 1 x 2 ⇒ x ∈ [1, 2 ]. 5x − ≥ 100 5x − ⇒ 4 ⇒ 4 ≥1 31. (d) : f(x) = a cos(bx + c) + d ...(i) For minimum cos(bx + c) = –1 from (i), f(x) = –a + d = (d – a) ⇒ x2 – 5x + 4 0 ⇒ (x – 1)(x – 4) 0 For maximum cos(bx + c) = 1 Hence domain of the function is [1, 4]. from (i), f(x) = a + d = (d + a) \\ Range of f(x) = [d – a, d + a]. 24. (b) : (f ′(x))2 – k2(f(x))2 0 ⇒ (f ′(x) – kf(x)) (f ′(x) + kf(x)) 0 ⇒ (f(x)e–kx)′ (f(x)ekx)′ 0 32. (b) : Let x2 + 34x − 71 ⇒ Exactly one of the functions x2 + 2x −7 = y Boz⇒gg⇒e12ru((rt0xog))1fgfa(((==2xt0x()))xxf0()≥==x=a=)n000e0d–f⇒(⇒kxgx)2xxfeb(dk∈x∈oxe)tcih[s[≥r00en,fa,0uo1s1nin]]nc,-gtdgio1e⇒(cn0r)egga12=s(aix0nn)gda.ngd02 2e35 ⇒ x2(1 – y) + 2(17 – y)x + (7y – 71) = 0 For real value of x, discriminant, D ≥ 0 have a value ⇒ y2 – 14y + 45 ≥ 0 ⇒ y ≥ 9, y 5. g1 increasing ⇒ f(x) 0 33. (c) : y = f(x) = cos2x + sin4x ⇒ y = f(x) = cos2x + sin2x(1 – cos2x) ⇒ y = cos2x + sin2x – sin2x cos2x 1 3 + log10(x3 – x). ⇒ y=1– sin2x cos2x ⇒ y = 1 – 4 sin22x 4 − x2 25. (d) : f(x) = \\ 3 f(x) 1, (∵ 0 sin22x 1) 4 So, 4 – x2 ≠ 0 ⇒ x ≠ ± 4 ⇒ f(R) ∈ [3/4, 1]. and x3 – x > 0 ⇒ x(x2 – 1) > 0 x2 + 14x + 9 –+– + 34. (c) : x2 + 2x + 3 = y –1 0 1 ⇒ x2(y – 1) + 2x(y – 7) + (3y – 9) = 0 \\ D = (–1, 0) (1, ∞) – { 4 } i.e., Since x is real, \\ 4(y – 7)2 – 4(3y – 9)(y – 1) ≥ 0 D = (–1, 0) (1, 2) (2, ∞). ⇒ 4(y2 + 49 – 14y) –4(3y2 + 9 – 12y) ≥ 0 26. (d) : f(x) = x − x2 + 4 + x + 4 − x ⇒ 4y2 + 196 – 56y – 12y2 – 36 + 48y ≥ 0 Clearly f(x) is de ned, if 4 + x ≥ 0 ⇒ x ≥ – 4 ⇒ 8y2 + 8y – 160 0 ⇒ y2 + y – 20 0 4–x≥0⇒x 4 ⇒ (y + 5)(y – 4) 0 \\ y lies between – 5 and 4. x(1 – x) ≥ 0 ⇒ x ≥ 0 and x 1 \\ Domain of f = (–∞, 4] [–4, ∞) [0, 1] = [0, 1]. 35. (b) : f (x) = sin log (x + 1+ x2 ) 27. (c) : log(x2 – 6x + 6) ≥ 0 i.e., x2 – 6x + 6 ≥ 1 x2 – 6x + 5 ≥ 0 ⇒ f(–x) = sin log(−x + 1+ x2 ) (x – 1)(x – 5) ≥ 0 i.e., x ∈ (–∞, 1] [5, ∞) 28. (d) : f(x) = x2 −1 + x2 +1 ⇒ f(x) = y1 + y2 ⇒ 1+ x2 − x)( 1+ x2 + x) f(–x) = sin log ( ( 1+ x2 + x) D\\x ∈om(D–a∞ionm, o∞af)iny–1o(=f–1f(,x1x))2=a−n(1d–∞D⇒,o∞mx)2a–i–n(1o–f1≥,y021)i⇒s. rexa2l ≥1 ⇒ f(–x) = sin log 1 number, (x + 1+ x2 ) 56 MATHEMATICS TODAY | SEPTEMBER‘17
⇒ f(–x) = sin log(x + 1 + x2 )−1 40. (a, b, c) : (x – y)f(x + y) – (x + y)f(x – y) = 2y((x – y)(x + y)) Let x – y = u; x + y = v ⇒ f(–x) = sin − log(x + 1+ x2 ) 2uv(v − u) uf(v) – vf(u) = 2 ⇒ f(–x) = − sin log(x + 1+ x2 ) ⇒ f(–x) = –f(x) f (v) − f (u) = v – u \\ f(x) is odd function. vu ex − e−x e2x −1 + 2 f (v) − v = f (u) − u = constant ex + e−x e2x +1 v u 36. (b) : y = +2 ⇒ y = ⇒ e2x = 1− y = y −1 ⇒ x = 1 loge y −1 Let f (x) – x = l y −3 3− y 2 3 − y x –1(y) y −1 1/2 –1(x) x −1 1/2 ⇒ f(x) = (lx + x2) 3 − y 3−x f(1) = 2 ⇒ f = loge ⇒ f = loge l + 1 = 2 ⇒ l = 1 f(x) = x2 + x 37. (b) : Given f(x) = 2x(x – 1) ⇒ x(x – 1) = log2 f(x) 41. (a, b) : Let a = [x] + [y] = [x] [y] en from the given equation, we have a + b = a b ⇒ x2 – x – log2 f(x) = 0 ⇒ x = 1± 1+ 4 log2 f (x) 2 ⇒ ab – a – b = 0 ⇒ ab – a – b + 1 = 1 ⇒ (a – 1)(b – 1) = 1. Only 1+ 1+ 4 log2 f (x) lies in the domain x= 2 is is possible if (i) a – 1 = 1, b – 1 = 1 or (ii) a – 1 = – 1, b – 1 = – 1 \\ f –1(x) = 1 [1 + 1+ 4 log2 x ]. Now, for (i), a = 2 and b = 2 2 And for (ii) a = 0 and b = 0 38. (b) : Let f(x) = y ⇒ x = f –1(y) 2x −1 us [x] = 2 and [y] = 2 or [x] = 0, [y] = 0 Now, y = x+5 ,(x ≠ −5) But [x] = 2 ⇒ 2 x < 3 and [y] = 2 ⇒ 2 y < 3. Again [x] = 0 ⇒ 0 x < 1 and [y] = 0 ⇒ 0 y < 1 xy + 5y = 2x – 1 ⇒ 5y + 1 = 2x – xy us, the solution sets are 0 x < 1 and 0 y < 1 ; 2 x < 3 and 2 y < 3. ⇒ x(2 – y) = 5y + 1 ⇒ x = 5y +1 f⇒a⇒4\\(n2xd.1)(afaaa–=(111,x(0)bsscaiio,=nn1⇒sds0aaa)i(n11:1af++1a+(cx01oaaa)+s222=asscaiio01nn2s+s⇒aaiana2222a++ac+1..o2c......so++.+.+sa..aa.2a.ann+1+nss.aii+c.nn.no.+asaas2iaanncnnon==caas=onn(00asc∵)0s2oaisxn+n1)a.x.c≠.n1o.+sn==px001) 2− y 5y +1 ⇒ f −1( y) = 2− y \\ f −1(x) = 5x +1 , x ≠ 2 2−x 39. (a, b, c, d) : Given f(x) f(y) = f(x) + f(y) + f(xy) – 2 x, y ∈ R. Put x = y = 1 then f(1)2 – 3f(1) + 2 = 0 43. (b, d) : Given 2f(sin x) + f(cos x) = x ⇒ f(1) = 1 or f(1) = 2 If f(1) = 1, then f(x) = 1 x ∈ R. A contradiction ∵ degree of f(x) is positive. Replace x by π – x, 2 \\ f(1) ≠ 1. hence, f(1) = 2. ⇒ 2f(cos x) + f(sin x) = π –x Replace 'y' with 1 then 2 x π π ⇒ f(sin x) = x – 6 ⇒ f(x) = sin–1 x – 6 f(x) f 1 = f(x) + f 1 (∵ f(1) = 2) \\ Domain of f(x) is [–1, 1] and range of f –1(x) is x x −2π −π π π π 3 π \\ f(x) must be in the form xn + 1 or –xn + 1. 2 − 6 , 2 − 6 i.e., , 3 ∵ f(4) = 65, f(x) = x3 + 1 ⇒ f 1(x) = 3x2. MATHEMATICS TODAY | SEPTEMBER‘17 57
44. (a, b, c, d) : f 7 = –0.5, g 7 = 0.5, g(3) = 0, Putting x = 1, f(1) = ek ⇒ 2 = ek. Hence from (i), f(x) = 2x. 2 2 is can also be obtained by putting y = 1 so that f(0) = 3; f(2) = –1, g(2) = –1; f(4) = 0; g(4) = 26 F(x + 1) = f(x) f(1) = 2f(x) ⇒ f(x) = 2f(x – 1). 45. (b) : ∵ –1(x) = x Putting successively x – 1, x – 2, x – 3, ...., 2 for x in the 2(f –1(x))4 – 4(f –1(x))2 = x above and multiplying them, we get f(x) = 2x. ⇒ (f –1(x))4 – 4(f –1(x))2 – log2x = 0 \\ (f –1(x))2 = 2 + 4 + log2 x n \\ Range of f –1(x) is [2, ∞). Now, ∑ f (c + r) = f(c + 1) + f(c + 2) +...+ f(c + n) \\ f –1(x) = 2 + 4 + log2 x r =1 \\ f –1(x) > 0 = 2c + 1 + 2c + 2 +...+ 2c + n = 2c 2 + 2c 22 +...+ 2c 2n = 2c 2{1 + 2 + 22 +... to n terms} 46. (a) : g(x) = sin x + 4 = 2c 2 1(2n −1) = 2c 2(2n – 1) sin x − 2 2 −1 ⇒ 16(2n – 1) = 2c + 1(2n – 1) −6 cos x π π ⇒ 2c + 1 = 16 = 24 ⇒ c + 1 = 4 \\ c = 3. ⇒ g′(x) = (sin x − 2)2 ≥0 ∵x∈ 2 , ⇒ g(x) is increasing function, hence one-one function. 49. (3) : f(1, 1) = f(0, f(1, 0)) = f(0, f(0, 1)) = f(0, 2) = 3 \\ Range is g π , g (π) lie [–5, –2]. 50. (3) : f(x) × f ′(–x) = f(–x) × f ′(x) 2 ⇒ f ′(x) × f(–x) – f(x) × f ′(–x) = 0 47. (A) (q), (B) (q, s, t), (C) (p, q, r, t), (D) (q, r) ⇒ d [ f(x)f(–x)] = 0 dx ⇒ f(x) f(–x) = k (A) 1− x2 + | x |= 1− xx +x = 1 Given, ( f(0))2 = k = 9 ⇒k=9 x x x ⇒ f(–3) = 3 1− x2 en f(3) f(–3) = 9 x 0 x2 −1 0, ≠0 ⇒ ⋅ x ≥ ⇒ ≤ x 51. (1) : [x] = 5 ± 25 + 4 sin x − 24 = 5± 1+ 4 sin x 2⋅1 2 ⇒ x ∈ [–1, 1] – {0} (B) Slope > 0 ⇒ −(a + 1) > 0 ⇒ a ∈(−1, 0) –1 sin x 1; –4 4 sin x 4; –3 1 + 4 sin 5 a (C) a – a cos2 x + |cos x| – 2a = 0 0 1+ 4 sin x 5 ⇒ [x] is an integer sin x = 0 ⇒ a = | cos x | = 1 1 ⇒ [x] = 3 ⇒ x = p 1+ cos2 x cos | cos x | + | x | 52. (5) : Let f(x) = emx be a solution. en f(x + 2) = em(x + 2) = emx e2m, | cos x | ∈[0, 1] ⇒ | cos x | + | 1 x | ∈[2, ∞) ⇒ a ∈ 0, 1 cos 2 f(x + 1) = emx em erefore, f(x + 2) – 5f(x + 1) + 6f(x) = 0 (D) If x ≠ np S = 1 + 2 cos2x + 3 cos4 x + ........... (1) ⇒ emx e2m – 5emx em + 6emx = 0 cos2 xS = cos2 x + cos4 x + ........... (2), ⇒ emx[e2m – 5em + 6] = 0 Subtracting eq. (1) and eq. (2), we get ⇒ e2m – 5em + 6 = 0; emx ≠ 0 sin2 xS = 1 + cos2 x + cos4 x + ........... Let em = u. en from the above equation, u2 – 5u + 6 = 0 ⇒ (u – 2)(u – 3) = 0 sin2 xS = 1 x ⇒ S = 1 x ⇒ f (x) = sin4 x u = 2 or u = 3. 1 − cos2 sin4 ⇒ em = 2 or em = 3 ⇒ emx = 2x or emx = 3x For x = np, f (x) is not di ned, Range of f (x) = (0, 1] Hence, f(x) = A 2x + B 3x. 48. (3) : From the given equation ...(i) 53. (2) : Given, f(1) + 2f(2) + 3f(3) +...+ nf(n) ...(i) f(x + y) = f(x) f(y), we have f(x) = ekx = n(n + 1)f(n) Where k is a constant. Replacing n by (n + 1), we get 58 MATHEMATICS TODAY | SEPTEMBER‘17
f(1) + 2f(2) + 3f(3) +...+ nf(n) + (n + 1)f(n + 1) Also, on putting u = π and v = π – x in (i), we get = (n + 1)(n + 2)f(n + 1) ...(ii) 2 2 Subtracting (i) from (ii), we get (n + 1)f(n + 1) = (n + 1)(n + 2)f(n + 1) – n(n + 1) f(n) f(p – x) + f(x) = 2f π cos π x = 2b sin x ...(iii) 2 2 − ⇒ f(n + 1) = (n + 2)f(n + 1) – nf(n) π ⇒ nf(n) = (n + 2)f(n + 1) – f(n + 1) 2 = (n + 2 – 1)f(n + 1) = (n + 1)f(n + 1). where b = f us, we have 2f(2) = 3f(3) = ... = nf(n). Now, adding (ii) and (iii), we get 2f(x) – f(–x) + f(–x) = 2a cos x + 2b sin x Hence, from (1), f(1) + (n – 1)nf(n) = n(n + 1)f(n) ⇒ 2f(x) = 2a cos x + 2b sin x (from (iii)) ⇒ f(n){n(n + 1) – n(n – 1)} = f(1) = 1 ⇒ (n2 + n – n2 + n)f(n) = 1 Hence f(x) = a cos x + b sin x. ⇒ 2nf(n) = 1 ( )55. (2) : Given, f(x + y + 1) = f (x) + f (y) 2 ⇒ f(n) = 1 \\ f(200) = 2 1 = 1 ( )Putting y = 0, we get f(x + 1) = f (x) +1 2 ; 2n × 200 400 54. (2) : Given f(u + v) + f(u – v) = 2f(u) cos v ...(i) Putting x = 0, we have f(1) = (1 + 1)2 = 22; Putting u = 0 and v = x in (i), we get f(x) + f(–x) = 2f(0) cos x ...(ii) Putting x = 1, f(2) = (2 + 1)2 = 32; = 2a cos x where a = f(0) ( )Putting x = 2, f(3) = f (2) + 1 2 = (3 + 1)2 = 42 and Now, putting u = π – x and v = π in (i), we get 2 2 so on. Proceeding in this way, we get f(x) = (x + 1)2. f(p – x) + f(–x) = 0 ∵ cos π = 0. 2 6 MATHEMATICS TODAY | SEPTEMBER‘17 59
CLASS XII Series 5 CBSE Integrals | Application of Integrals IMPORTANT FORMULAE INTEGRALS ∫ dx = x + C, where 'C ' is an arbitrary constant ∫ dx = sin−1 x + C or –cos– 1 x + C, where |x| < 1 xn + 1 1− x2 ∫ xndx n+1 –1 = + C,where n ≠ ∫ dx = tan– 1 + or – cot– 1 x+C 1+ x2 ∫ exdx = ex + C x C ∫ axdx = ax + C, where a > 0 ∫x 1 dx = sec– 1 x + C or – cosec– 1 x + C log e a x2 ∫1 −1 x dx = loge |x|+C, where x ≠ 0 ∫ dx = 1 tan −1 x +C x2 + a2 a a ∫ sinx dx = − cos x + C ∫ dx = 1 log x−a +C ∫ cosx dx = sin x + C x2 − a2 2a x+a ∫ sec2x dx = tan x + C ∫ cosec2x dx = −cot x + C ∫ dx = 1 log a+x +C a2 − x2 2a a−x ∫ sec x tan x dx = sec x + C ∫ cosec x cot x dx = −cosec x + C ∫ dx = log x + x2 + a2 + C ∫ tan x dx = log|sec x| + C or –log |cos x| + C x2 + a2 ∫ dx = log x + x2 − a2 + C x2 − a2 ∫ cot x dx = log|sin x| + C or –log cosec x + C ∫ dx = sin−1 x + C ,| x|< a a2 − x2 a ∫ sec x dx = log|sec x + tan x| + C ∫ a2 + x2 dx = x a2 + x2 + a2 log|x + a2 + x2 |+C 2 2 x π = log tan 2 + 4 +C ∫ x2 a2 dx x x2 a2 a2 log|x x2 − a2 |+C 2 2 ∫ cosecx dx = log|cosec x – cot x| + C − = − − + = log tan x +C ∫ a2 − x2 dx = x a2 − x2 + a2 sin−1 x + C 2 2 2 a 60 MATHEMATICS TODAY | SEPTEMBER ‘17
By parts (ILATE) Properties of De nite Integrals bb ∫ I ⋅ II dx = I ⋅∫ II dx − ∫ d (I)⋅ ∫ II dx dx ∫ ∫f (x)dx = f (t)dt dx aa ba a ∫ ∫ ∫f (x)dx = − f (x)dx . In particular, f (x) dx = 0 (Here I and II functions are choosen on the basis of ab a ILATE) bc b IN T E G R A T IO N B Y P A R T IA L F R A C T IO N S ∫ ∫ ∫f (x)dx = f (x)dx + f (x)dx Rational form Partial form aa c px + q (x A a) + (x B b) bb (x − a)(x − b) − − ∫ ∫f (x)dx = f (a + b − x)dx px2 + qx + r A + B + C (x − a)(x − b)(x − c) − − (x − aa aa ∫ ∫f (x)dx = f (a − x )dx 00 (x a) (x b) c) 2a a a ∫ ∫ ∫f (x)dx = f (x )dx + f (2a − x )dx px + q A B 00 0 (x − a)2 − − a)2 a (x a) + (x ∫ ∫2a 2 f (x )dx, if f (2a − x) px2 + qx + r = − f (x) f (x) (x − a)2(x − b) 0 = f (x)dx = px2 + qx + r 0 (x − a)3(x − b) A + B + C if f (2a − x) (x − a) − a)2 (x − b) 0, (x a ∫ ∫a 2 f (x ), if f (−x) = f (x) [even function] −a A B C D f (x)dx = 0 − a) − a)2 − a)3 (x −b) (x + (x + (x + 0, if f (−x) = − f (x) [odd function] px2 + qx + r A + Bx + C ∫ ex[ f (x) + f ′(x)]dx = ex f (x) + C (x − a)(x2 + bx + c) − x2 + bx + (x a) c ( f (x))n+1 n+1 where x2 + bx + c cannot be ∫( f (x))n f ′(x)dx = + C factorised further ∫ f ′(x) dx = log( f (x)) + C f (x) APPLICATION OF INTEGRALS e area of a region bounded by y2 = 4ax and e area of a region bounded by one arc of sinax or x2 = 4by is 16ab sq. units. cosax and x-axis is 2 sq. units. 3 a The area of a region bounded by y2 = 4ax and Area of an ellipse x2 + y2 = 1 is pab sq. units. 8a2 a2 b2 y = mx is 3m3 sq. units. e area of a region bounded by y = ax2 + bx + c and e area of a region bounded by y2 = 4ax and its 3 (b2 − 4ac)2 latus rectum is 8a2 sq. units. x-axis is 6a2 sq. units. 3 MATHEMATICS TODAY | SEPTEMBER ‘17 61
WORK IT OUT π/2 π/4 VERY SHORT ANSWER TYPE (ii) ∫ f (sin2x) sin x dx = 2 ∫ f (cos 2x)cos x dx ∫1. Evaluate e3log x (x4 ) dx 00 ∫2. Find −11| x |dx 18. Find the smaller of the two areas in which the circles x2 + y2 = 4 is divided by the parabola y2 = 3(2x – 1). π/2 19. Evaluate the following de nite integrals : −π/2 Evaluate log px2 + qx + r a dx ∫3. px2 − qx + r dx ∫(i) 0 x + a2 − x2 4. Solve ∫ 3 dx ∫(ii) 1 cot−1(1 − x + x2 )dx 3x − 2 0 ∫5. Evaluate (2x + 2−x )2 dx 20. Find the area enclosed between the curves y = sinx SHORT ANSWER TYPE and y = cos x that lies between the lines x = 0 and 6. Find the area bounded by the curves y = x and y = x3. x= π . 2 (ax + bx )2 Evaluate axbx SOLUTIONS ∫7. dx ∫ ∫1. e3log x (x4 )dx = x3 ⋅ x4dx ( elog x = x) Find the value of 2008 f ′(x) + f ′(−x) ∫8. −2008 (2008)x +1 dx x8 = ∫ x7dx 8 dx = +C 1− 2x − x2 9. Evaluate ∫ We have, −11| x | dx = 0 1 −1 0 ∫ ∫ ∫2. −x dx + xdx 10. Find the area of the region bounded by the curve = − x2 0 + x2 1 = − 0 − 21 + 1 − 0 = 1 x2 = y, and the line y = 4. 2 −1 2 0 2 LONG ANSWER TYPE - I px2 px2 11. Find ∫ (x2 2x − 3 dx 3. Let f (x) = log + qx + r −1)(2x + 3) − qx + r p(−x)2 Find the value of e2 log x ⇒ f (−x) = log p(−x)2 + q(−x) + r ∫12. e−1 x dx − q(−x) + r 13. Evaluate ∫ 2 sin x + 3 cos x dx = log px2 − qx + r 3 sin x + 4 cos x px2 + qx + r + qx + r −1 ∫14. Evaluate 1 exdx as the limit of a sum. = log px2 = − log px2 + qx + r = − f (x) −1 px2 − qx + r px2 15. Using integration, nd the area of the region − qx + r bounded by the triangle whose vertices are (–1, 2), (1, 5) and (3, 4). ⇒ f (x) is an odd function. LONG ANSWER TYPE - II ∫\\ π/2 log px2 + qx + r dx = 0 −π/2 px2 − qx + r ∫16. Evaluate x2 + 1(log(x2 + 1) − 2 log x) dx, x > 0 4. We have, ∫ ∫3 dx = 3 (3x − 2)−1/2 dx x4 3x − 2 17. Show that 3 (3x − 2)1/2 +C =2 3x − 2 + C. π/2 π/2 (i) ∫ f (sin2x) sin x dx = ∫ f (sin2x) cos x dx = ⋅ 1 2 ⋅ 3 00 62 MATHEMATICS TODAY | SEPTEMBER ‘17
∫ ∫5. We have, (2x + 2−x )2 dx = (22x + 2−2x + 2)dx 10. Given curves are x2 = y and y = 4 Y 22x 2−2x \\ Required area A y=4 = (log 2) × 2 + (log 2) (− 2) + 2⋅x + C = 2 area OABO x2 = y B 1 (22x − 2−2x ) + 2x + C = 2∫04 xdy = 2 log 2 = 2∫04 ydy 6. e given curves are y = x ...(i) and y = x3 ...(ii) OX Solving (i) and (ii), we get Y = 2 ⋅ 2 y3/ 2 4 = 4 (8 − 0) = 32 sq. units x3 = x ⇒ x(x2 – 1) = 0 O 3 0 3 3 y = x3 y=x ⇒ x = 0, 1, –1 P (1, 1) When x = 0, y = 0; X 11. Let 2x − 3 when x = 1, y = 1 and (x −1) (x + 1) (2x + 3) A B C = x −1 + x +1 + 2x + 3 when x = –1, y = –1. Q (–1, 1) ⇒ 2x – 3 = A (x + 1) (2x +3) + B(x – 1) (2x + 3) So, O(0, 0), P(1, 1) and Q(–1, –1) are points of + C (x –1) (x + 1) intersection of (i) and (ii). ⇒ 2x – 3 = A (2x2 + 5x + 3) + B (2x2 + x – 3) + C (x2 –1) 1 x2 x4 1 Required area = x 3 )dx 2 2 4 0 Equating the coe cients of x2, x and constant terms, ∫\\ 2 (x − = − we get 1 14 0 1 1 2A + 2B + C = 0, 5A + B = 2 and 3A – 3B – C = –3 2 − (0 − 0) 4 2 = 2 − =2× = sq. units Solving these equations, we get We have, (ax + bx )2 a2x + b2x + 2axbx A = − 1 , B = 5 and C = − 24 axbx axbx 10 2 5 ∫ ∫7. dx = dx 1 5 24 10 2 5 x x 2x − 3 − +1 − +3 − 1)(2x 2x x x a b ∴ (x2 + 3) = x −1 + x + + 2 dx b a ∫= a + b = + + 2x + C, a ≠b 2x − 3 1 dx b a log a log b (x2 −1)(2x + 3) 10 x −1 ∫ dx = − ∫ b a 2008 f ′(−x) + f ′(x) dx ∫8 . I= −2008 (2008)− +1 + 5 ∫ dx − 24 ∫ dx x 2 x+ 5 2x +3 1 2008 f ′(−x) + f ′(x) ∫= −2008 1+ (2008)x × (2008)x dx 1 5 24 log 2x +3 = − 10 log x −1 + 2 log x +1 − 5 2 +C 2I = 2008 ′(−x) + ′(x)dx ∫\\ −2008 f f = − 1 log x −1 + 5 log x +1 − 12 log 2x + 3 +C 10 2 5 ∫= 2 2008 f ′(x) + f ′(−x)dx 0 e value of e2 log x e−1 x I = [ f (x) − f (−x)]02008 = f (2008) − f (−2008) ∫12. dx hence 9. I=∫ dx = ∫ dx log x = 0 when x = 1 and negative in 1 , 1 and (x2 2 − (x2 + 2x +1) positive in (1, e2] e 1 − + 2x) =∫ dx =∫ dx ...(i) MPP-5 CLASS XII ANSWER KEY 2 − (1+ x)2 ( 2)2 − (1+ x)2 Let z = 1 + x, then dz = dx 1. (a) 2. (b) 3. (c) 4. (c) 5. (b) From (i), 6. (b) 7. (b,c,d) 8. (a,b,c) 9. (a,d) 10. (a) 1 11. (a,c,d) 12. (a,b,d,) 13. (a,c) 14. (b) 15. (c) ∫I = dz = sin−1 z +c = sin−1 +x +c 16. (b) 17. (3) 18. (1) 19. (3) 20. (1) ( 2 )2 − z2 2 2 MATHEMATICS TODAY | SEPTEMBER‘17 63
∫ ∫∴I 1 log x dx e2 log x dx = lim eh ⋅ enh −−11 = lim eh e2 −1 x x h→0 e h eh h→0 e eh −1 = + h 1 1 ∫ ∫e e2 log x dx = 1 ⋅ e0 ⋅ e2 − 1 Q lim eh −1 = 1 = e − 1 x e 1 h→0 h e = 1 − log x dx + 1/e x 1 − 1 1 1 e2 1 4 5 15. Let the vertices of the given triangle be A (–1, 2), 2 x)2 1 2 )2 1 2 (−1)2 2 2 B (1, 5) and C (3, 4). = (log + (log x = + −0= e equations of the sides AB, BC and CA respectively e are: 3x + 7 −x + 11 2 2 13. Let 2 sin x + 3 cos x = l (3 sin x + 4 cos x) + y = ...(i); y = ...(ii) m (3 cos x – 4 sin x). and y = x + 5 ...(iii) Equating the coe cients of sinx and cosx on both sides, 2 we get 2 = 3l – 4m and 3 = 4l + 3m. 1 18 25 On solving, we get l = 25 and m = \\ ∫ 2 sin x + 3 cos x dx 3 sin x + 4 cos x = ∫ l (3sin x + 4 cos x) + m (3cos x − 4 sin x) dx 3sin x + 4 cos x = l ∫ 1 dx + m∫ 3cos x − 4 sin x dx \\ Required area = Area of region bounded by trap. 3sin x + 4 cos x (put 3 sin x + 4 cos x = t ⇒ (3 cos x – 4 sin x) dx = dt) ADEB + Area of region bounded by trap. BEFC – Area of region bounded by trap. ADFC = lx + m∫ dt = lx + m log |t |+ C t 1 3x + 7 3 −x + 11 3 x + 5 −1 2 1 2 2 = 18 1 log | 3 sin 4 cos | ∫ ∫ ∫= dx + dx − dx 25 25 x + x + x + C −1 ∫ ∑14. b f (x) dx = lim n f (a + rh) ...(i) = 1 3x2 + 7x 1 + 1 x2 + 11x 3 − 1 x2 + 5x 3 a 2 2 2 − 2 2 2 h→0 h −1 1 −1 r =1 Where, nh = b – a, and n ∞ = 1 3 + 7 − 3 − 7 + 1 − 9 + 33 − − 1 +11 Here, f (x) = ex, a = –1, b = 1 2 2 2 2 2 2 \\ f (a + rh) = ea + rh = ea erh 1 9 15 1 2 2 2 Also, nh = b – a =1 – (–1) = 2 − + − − 5 ∫ ∑From (i), −11 e x dx = lim n ea ⋅ erh = 1 (14 + 22 − 4) − 1 (4 + 20) 2 2 h→0 h 1 1 2 2 n r =1 h n erh (32) (24) 16 − 12 = 4 sq. units. ∑ ∑=lim h e −1 ⋅ erh = lim e −1 ⋅ r =1 = − = h→0 r =1 h→0 x2 + 1(log(x2 + 1) 2 log x) x4 = lim 1 h(eh + e2h + e3h + ... + enh ) ∫16. − dx e h→0 ⋅ = lim h eh {1 − (eh )n } = lim eh h(1 − enh ) x 1+ 1 (log(x2 + 1) − log x2 ) e 1− eh h→0 e 1 − eh x2 x4 h→0 ∫= dx 64 MATHEMATICS TODAY | SEPTEMBER ‘17
∫= 1 + 1 1/2 log 1 + 1 ⋅ 1 dx (Q f (cos 2t) cos t is an even function and f(cos 2t) sin t x2 x2 x3 is an odd function) Put 1 + 1 =t ⇒− 2 dx = dt ⇒ 1 dx − 1 dt π/4 x2 x3 x3 2 = 2 ∫ f (cos 2x) cos x dx ∫ ∫= t1/2 log t − 1 dt = − 1 log t ⋅t1/2dt 0 ...(i) 2 2 18. e given circle is x2 + y2 = 4 Its centre is (0, 0) and radius = 2. e given parabola is y2 = 3(2x – 1) ∫= 1 log t 3/2 1 t 3/2 (Integrate by parts) ...(ii) 2 3/2 t 3/2 − t ⋅ − ⋅ dt ⇒ y2 = 6 x − 1 ...(iii) 2 1 3/2 log 1 t1/2dt 1 3/2 log 1 t 3/2 ∫= − 3 t t + 3 = − 3 t t + 3 ⋅ 3/2 + C Solving (i) and (ii), we get 1 3/2 (2 3 log ) x2 + 3(2x – 1) = 4 ⇒ x2 + 6x – 7 = 0 9 = t − t + C ⇒ (x + 7) (x – 1) = 0 ⇒ x = –7, 1 = 1 1 1 3/2 2 3 log 1 1 but x ≥ 1 , from (iii) we get, x = 1 9 x2 x2 2 + − + + C When x = 1, y2 = 3 ⇒ y = ± 3 π/2 2x) sin x dx π/2 sin 2 π Y 0 2 f (sin ∫ ∫ ∫n 17. (i) = f − x π/2 0 P(1, 3) 2x) sin x dx =f sin 2 π − x sin π − x dx 2 2 0 ( )O A M (2B, 0) X π/2 π/2 1 , 0 2 = ∫ f (sin (π − 2x)) cos x dx = ∫ f (sin 2x) cos x dx 00 Q(1, − 3) (ii) Put x = π − t ⇒ t = π − x ⇒ dx = − dt 4 4 When x = 0, \\ P (1, 3), and Q (1, − 3) are points of intersection t= π −0 = π and when x = π , t = π − π = − π of (i) and (ii) 4 4 2 4 2 4 So, required area = 2 . area of the region PAB π/2 −π/4 sin 2 π t 1 2 4 − x2 dx π/4 4 f (sin 2x)sin = 2∫ 3(2x −1) dx + 2∫ 0 ∫ ∫∴ − x dx = f π (−dt ) 1/2 1 2 4 sin − t = 2⋅ 3 (2x23−⋅12)3/2 1 2 x 4 − x2 4 sin−1 x 2 2 ∫−π/4 π π + ⋅ + 2 1 4 4 1/2 = − f (cos 2t) (sin cos t − cos sin t) dt π/4 2 3 sin−1(1)) sin−1 1 3 2 ∫π/4 1 cos t − 1 sin t dt = (1 − 0) + (0 + 4 − 3 + 4 2 2 = f (cos 2t) 2 3 π π = 3 4 2 3 4 6 −π/4 1 π/4 1 π/4 + ⋅ − + ⋅ 2 −π/4 2 −π/4 f (cos 2t) cos f (cos 2t) sin ∫ ∫= t dt − t dt 2 2 3 4π 3 3 3 3 π/4 = 3 −1 + 2π − π = − + ∫= 1 ⋅ 2 f (cos 2t) cos t dt − 1 ⋅0 = 1 (4π − 3)sq. units 2 2 3 0 MATHEMATICS TODAY | SEPTEMBER‘17 65
∫1 dx 20. Given, curves are y = sin x and y = cos x; a2 − x2 19. (i) Let I = Y 0x + y = cos x y = sin x Put x = a sin t ⇒ dx = a cos t dt When x = 0, t = 0 and when x = a, t = π 2 a π/2 dx 0 a cos t I= a2 − x2 sin t + a cos 0x+ ∫ ∫\\ = dt O p p X 4 2 a t ∫= π/2 cos t dt ...(i) e curves cross each other at sin x = cos x i.e., t + cos t 0 sin π aa at tan x = 1 i.e., at x = 4 . en by using ∫ f (x) dx =∫ f (a − x) dx, we get 00 π/2 π/2 cos π − t π/2 sin t \\ Required area = ∫ | cos x − sin x |dx 2 0 cos t + sin 0 ∫ ∫I= 0 sin dt = dt π/4 π/2 π t π − t t 2 − + cos 2 ...(ii) ∫ | cos x − sin x |dx + ∫ | cos x − sin x |dx On adding (i) and (ii), we get 0 π/4 π/4 π/2 π/2 cost + sint π/2 = [t]0π/2 π −0 π 0 sint + cost 2 2 ∫ (cos x − sin x) dx + ∫ (sin x − cos x) dx ∫ ∫2I= dt = 1dt = = 0 π/4 0 = sin x + cos x0π/4 + − cos x − sin xππ//24 ⇒ π I = 4 1 1 (0 1) (0 1) 1 1 2 2 2 (ii) We have, cot–1 (1 – + x2) = tan−1 1 = + − + − + − + 2 x 1 −x+ x2 – x) x + (1− x) ( )=2 2 2 == tan−1 1− x(1− x) = tan–1 x + tan–1 (1 2 −1−1 + 2 = 2−2= 2 2 −1 sq. units. 1 11 ∫ ∫ ∫⇒ cot−1(1− x + x2 )dx = tan−1 x dx + tan−1(1− x)dx JAMMU & KASHMIR at 10 1 0 0 • Kabli Book Stall - Anantnag Mob: 8803043296, 9906460029 = ∫ tan−1 x dx +∫ tan−1(1 − (1 − x)) dx • New Valley Book Depot - Anantnag Mob: 9469183540 00 • Sharjha Book Depot - Anantnag Mob: 9419040456 11 • Harnam Dass & Bro’S - Jammu Ph: 0191-2542175, 2574428; Mob: 9419664141 • Sahitya Sangam - Jammu = ∫ tan−1 x dx +∫ tan−1 x dx 00 Ph: 0191-2579593, 2562191, 2579593; Mob: 9419190177 1 (integrate by parts) • Shiela Book Centre - Jammu Ph: 0191-2574912; Mob: 9419146803 = 2∫ tan−1 x ⋅1dx • Kapoor Sons - Srinagar Ph: 0194-2456458; Mob: 9419425757, 9419069199 0 • Kashmir Book Depot - Srinagar (tan−1 x)10 1 1 Ph: 2450262, 474440, 2475973; Mob: 9906726231, 9419761773 ∫= 2 x ⋅ − 0 1 +x 2 ⋅ x dx • Shah Book Centre - Srinagar Mob: 9906763627, 9419062444 tan−1(1) 1 2x π x2 ) |10 • Abdullah News Agency - Srinagar ∫= 2(1⋅ − 0) − 0 +x dx = 2 ⋅ 4 − | log(1+ Ph: 0194-2472621, 2435057; Mob: 9419074859 • Highway Book Shop - Srinagar Mob: 9858304786 • Paradise Book House - Srinagar Mob: 9419067856 1 2 = π − (log 2 − log 1) = π − (log 2 − 0) = π − log 2 2 2 2 66 MATHEMATICS TODAY | SEPTEMBER ‘17
Class XII This specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness. Application of Derivatives Total Marks : 80 Time Taken : 60 Min. Only One Option Correct Type 6. By LMVT, which of the following is true for x > 1? 1. e coordinates of the point P(x, y) lying in the rst (a) 1 + x ln x < x < 1 + ln x quadrant on the ellipse x2/8 + y2/18 = 1 so that the (b) 1 + ln x < x < 1 + x ln x area of the triangle formed by the tangent at P and (c) x < 1 + x ln x < 1 + ln x the coordinate axes is the smallest, are given by (d) 1 + ln x < 1 + x ln x < x (a) (2, 3) (b) ( 8 , 0) One or More Than One Option(s) Correct Type (c) ( 18 , 0) (d) none of these 7. For the function f(x) = x cos 1 , x ≥ 1, which of the 2. If f(x) = x3/2(3x – 10), x > 0, then f(x) is increasing in following is (are) correct? x (a) (–∞, –1) (1, ∞) (b) [2, ∞) (a) for at least one x in the interval [1, ∞), (c) (–∞, –1) [1, ∞) (d) (–∞, 0] (2, ∞) f(x + 2) – f(x) < 2 (b) lim f ′(x) = 1 3. e set of all values of a for which the function x→∞ f(x) = a+4 − 1 x5 – 3x + log 5 decreases for all 1−a (c) for all x in the interval [1, ∞], f(x + 2) – f(x) > 2 real x is (d) f ′(x) is strictly decreasing in the interval [1, ∞) 8. Let F : R R be a thrice di erentiable function. Suppose that F(1) = 0, F(3) = – 4 and F′(x) < 0 for 5− 27 all x ∈ (1/2, 3). Let f(x) = xF(x) for all x ∈ R. en (a) −3, 2 ∪ (2, ∞) which of the following statements is (are) correct? −4, 3 − 21 (a) f ′(1) < 0 (b) f(2) < 0 2 (1, ∞) (c) f ′(x) ≠ 0 for all x ∈ (1, 3) (b) ∪ (d) f ′(x) = 0 for some x ∈ (1, 3) (c) (–∞, ∞) 9. If f(x) = f(x) + f(2a – x) and f ′′(x) > 0, a > 0, (d) [1, ∞). 0 x 2a, then (a) f(x) increases in (a, 2a) 4. If the function f(x) = cos|x| – 2ax + b increases (b) f(x) increases in (0, a) along the entire number scale, the range of values (c) f(x) decreases in (a, 2a) of a is given by (d) f(x) decreases in (0, a) (a) a b (b) a = b/2 (c) a –1/2 (d) a ≥ 3/2 x3 + x2 −10x −1 ≤ x < 0 5. e image of the interval [–1, 3] under the maping 10. Let f(x) = sin x 0 ≤ x < π/2 f(x) = 4x3 – 12x is π/2 ≤ x ≤ π (a) [–2, 0] (b) [–8, 72] 1+ cos x (c) [–8, 0] (d) none of these then f(x) has MATHEMATICS TODAY | SEPTEMBER‘17 67
(a) local maxima at x = p/2 15. If L is the total light transmitted, then critical points (b) local minima at x = p/2 of L are (c) absolute minima at x = 0 3p (d) absolute maxima at x = p/2 (a) y= p (b) y= 12 + 3π / 2 6 + 3π / 2 11. Consider the function f : R R given by (c) y = 3p (d) y= 2p 12 + 5π/2 11+ 5π/2 f(x) = x2 − ax + 1 , 0 < a < 2. x2 + ax + 1 Matrix Match Type ∫ex f ′(t) dt , which of the following is (are) 16. Match the following. 1+ t2 If g(x) = Column I Column II not true? 0 e function f(x) = 2x3 – 9x2 (a) g′(x) is positive on (–∞, 0) and negative P. – 24x + 7 decreases on 1. (3, ∞) on (0, ∞) f(x) = 1+ 3x increases on (b) g′(x) is negative on (–∞, 0) and positive Q. 4 + 5x2 2. (–1, 4) on (0, ∞) (c) g′(x) changes sign on both (–∞, 0) and (0, ∞) 3 (d) g′(x) does not change sign on (–∞, ∞) R. f(x) = (x2 – 2x) log x – 2 x2 3. (–∞, 2] + 4x increases on 12. For x > 1, y = loge x satis es the inequality (a) x – 1> y (b) x2 –1>y 4. (e, ∞) (c) y > x – 1 (d) x −1 < y P QR x (a) 2 31 13. Suppose f ′(x) exists for each x and h(x) = (b) 2 3 4 f(x) – (f(x))2 + (f(x))3 for every real number x. en (c) 1 2 3 (a) h is increasing whenever f is increasing (d) 4 3 1 (b) h is increasing whenever f is decreasing (c) h is decreasing whenever f is decreasing Integer Answer Type (d) nothing can be said in general. 17. If the point on y = x tan α − ax2 α (a > 0), Comprehension Type 2u2 cos2 A window of xed perimeter (including the base of where the tangent is parallel to y = x has an ordinate the arch) is in the form of a rectangle surmounted by u2/4a, then 4 sin2 a is equal to a semi-circle. e semi-circular portion is tted with coloured glass, while the rectangular portion is tted 18. If the greatest and least values of the functions with clear glass. e clear glass transmits three times 1 1 as much light per square metre as the coloured glass. f(x) = arc tan x − 2 ln x on 3 , 3 are G and Suppose that y is the length and x is the breadth of the L respectively, then [G + L] = (where [·] is greater integer) rectangular portion and p the perimeter. 19. If f(x) = |x – 1| + |x – 3| + |5 – x|, x ∈ R is symmetrical about the line x = l, then l = 14. The ratio of the sides y : x of the rectangle so that the window transmit the maximum light is 20. If A > 0, B > 0 and A + B = p/3 then the maximum (a) 3 : 2 (b) 6 : 6 + p (c) 6 + p : 6 (d) 1 : 2 value of 3 tan A tan B is Keys are published in this issue. Search now! Check your score! If your score is > 90% EXCELLENT WORK ! You are well prepared to take the challenge of final exam. No. of questions attempted …… 90-75% GOOD WORK ! You can score good in the final exam. 74-60% No. of questions correct …… < 60% SATISFACTORY ! You need to score more next time. Marks scored in percentage …… NOT SATISFACTORY! Revise thoroughly and strengthen your concepts. 68 MATHEMATICS TODAY | SEPTEMBER ‘17
Maths Musing was started in January 2003 issue of Mathematics Today. The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material. During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new pattern by changing the style of problems. Some of the Maths Musing problems have been adapted in JEE benefitting thousand of our readers. It is heartening that we receive solutions of Maths Musing problems from all over India. Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them. We do hope that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced. JEE MAIN if balls are di erent but boxes are identical? (a) 25 (b) 15 (c) 10 (d) 35 1. If x2 – y2 – 84y = 2012, x, y ∈ N, then x – 3y = (a) 2 (b) 3 (c) 6 (d) 7 COMPREHENSION Let g(x) (x −1)n Let f be a function satisfying integers, l≠og0c,onsm>(x0−, 2. = 1) ;0 <x< 2, m and n are f (x) = a a = ga (x)(a > 0) m and let p be the le hand + ax 1995 derivative of |x – 1| at x = 1. If lim g(x) = p, then ∑7. f 19r96 (b) n = x1→, m1+ = –1 Let f(x) = g9(x), then the value of r=1 (a) n = 1, m = 1 is (where [.] denotes the greatest integer function) (c) n = 2, m = 2 (d) n > 2, m = n (a) 995 (b) 996 (c) 997 (d) 998 3. Given, x = a cost cos 2t and 2n If the value of r =0 r 1 987, then the value of n is 2n +1 1+ y = a sint cos 2t (a > 0), then (1+ (dy / dx)2 )3/2 at ∑8. f = a + d2 y / dx2 π is given by (a) 493 (b) 494 (c) 987 (d) 988 6 2 2a (a) a (b) a 2 (c) 3a (d) 3 INTEGER TYPE 3 sin3 x /2 9. e remainder when 22003 is divided by 17 is ∫4. x dx is equal to MATRIX MATCH 2 cos cos3 x + cos2 x + cos x 10. Match the following. (a) cos−1 sec x + tan x +1 + c List-I List-II 1/x 1. e (b) tan−1 sin x + cos x +1 + c P. lim tan x = 2. 1 x (c) sin−1 tan x + sec x +1 + c x→0 (d) tan−1 cos x + sec x +1 + c Q. e number of points at which f (x) = |x|−1 is not di erentiable, is giveenlimbyitxin2 g+pyo2i+nt2sgoxf +thce+cola(xxi2a+l syy2st+em2fyo+f circles 5. k) = 0 tan x 4t dt cot x 4 dt 1/e 1 + t2 1/e t(1+ t2 ) subtend a right angle at the origin, if ∫ ∫R. + = 3. 2 (a) − c − k =2 (b) c + k = −2 S. The degree of the differential 4. 3 g2 f2 g2 f2 equation satisfied by all circles of radius r is (c) c − k =2 (d) c + k =2 5. 4 g2 f2 g2 f2 P QR S JEE ADVANCED (a) 3 25 4 (b) 1 24 3 6. Five balls are to be placed in three boxes. Each can (c) 3 24 5 hold all the ve balls. In how many di erent ways (d) 2 45 3 can we place the balls so that no box remains empty, See Solution Set of Maths Musing 176 on page no 85 MATHEMATICS TODAY | SEPTEMBER ‘17 69
TWO DIMENSIONAL GEOMETRY Time : 1 hr. Marks : 60 MULTIPLE CORRECT CHOICE TYPE (a) cos−1 1 (b) 2 cos−1 1 e e This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which (c) sin−1 1 (d) None of these ONE or MORE may be correct. [Correct ans. 3 marks & wrong e ans., no negative mark] 1. If e1 and e2 are the eccentricities of the conic sections 6. e circle x2 + y2 + 4x – 6y + 3 = 0 is one of the 16x2 + 9y2 = 144 and 9x2 – 16y2 = 144,then circles of a coaxial system of circles having as radical axis the line 2x – 4y + 1 = 0. en the equation (a) e12 + e22 = 3 (b) e12 + e22 > 3 of the circle of the system which touches the line (d) e12 − e22 < 0 (c) e12 + e22 < 3 x + 3y – 2 = 0 is (a) x2 + y2 + 2x – 2y + 2 = 0 2. e equation(s) to the tangent(s) to the conic (b) x2 + y2 + 2x + 6y = 0 x2 + 4xy + 3y2 – 5x – 6y + 3 = 0, which are parallel to (c) x2 + y2 – 2x + 6y = 0 x + 4y = 0 are (d) x2 + y2 + 2x – 6y = 0 (a) x + 4y – 1 = 0 (b) x + 4y – 3 = 0 (c) x + 4y – 5 = 0 (d) x + 4y – 8 = 0 7. If a circle of constant radius 3k passes through the origin and meets the axes at ‘A’ and ‘B’, the locus of 3. Consider the parabola y2 = 4ax and x2 = 4by. e the centroid of DOAB is straight line b1/3 y + a1/3x + a2/3 b2/3 = 0 (a) x2 + y2 = k2 (b) x2 + y2 = 2k2 (a) touches y2 = 4ax (c) x2 + y2 = 3k2 (d) None of these (b) touches x2 = 4by (c) intersects both parabolas in real points 8. P2 x2 6 + P2 y2 5 = 1 will represent the (d) touches rst and intersects other −P − 6P − + 4. e co-ordinates of a point on the parabola y2 = 8x ellipse if P lies in the interval (a) (−∞, −2) (b) (1, ∞) whose distance from the circle x2 + (y + 6)2 = 1 is (c) (3, ∞) (d) (5, ∞) minimum is (a) (2, 4) (b) (2, –4) (c) (18, –12) (d) 8, 8 9. If the eccentric angles of the extremities of a focal x2 y2 5. e angle between the asymptotes of the hyperbola chord of an ellipse a2 + b2 = 1 are a and b , then x2 − y2 = 1 is (a) e = cos α + cos β (b) e = sinα + sinβ a2 b2 cos(α + β) sin(α + β) By : Vidyalankar Institute, Pearl Centre, Senapati Bapat Marg, Dadar (W), Mumbai - 28. Tel.: (022) 24306367 70 MATHEMATICS TODAY | SEPTEMBER ‘17
(c) cos α − β = e cos α + β on OB and AB. If the ratio of Area (∆AMN) = 3 , 2 2 then nd the value of AN . Area (∆OAB) 8 (d) tan α tan β = e −1 BN 2 2 e +1 19. A(0, 0), B(4, 2) and C(6, 0) are the vertices of a 10. If b and c are the lengths of the segments of any triangle ABC and BD is its altitude. e line through focal chord of a parabola y2 = 4ax, then the length D parallel to the side AB intersects the side BC at of the semi-latus rectum is a point E. Find the product of areas of DABC and DBDE. (a) b+c (b) bc 2 b+c 20. Find the number of integral values of l if (l , 2) (c) 2bc (d) bc is an interior point of DABC formed by, x + y = 4, b+c 3x – 7y = 8, 4x – y = 31. ONE INTEGER VALUE CORRECT TYPE ANSWER KEY This section contains 10 questions. Each question, when 1. (c, d) 2. (c, d) 3. (a, b) 4. (b) worked out will result in one integer from 0 to 9 (both 8. (a, d) inclusive). [Correct ans. 3 marks & wrong ans., no negative 5. (b) 6. (a, c) 7. (d) 12. (3) mark] 16. (2) 9. (b, c, d) 10. (c) 11. (5) 20. (1) 11. e locus of the centre of the circle for which one end of diameter is (3, 3) while the other end lies on 13. (3) 14. (6) 15. (7) the line x + y = 4 is x + y = k, then k equals_____ 17. (6) 18. (3) 19. (8) 12. e greatest distance of the point (10, 7) from the circle x2 + y2– 4x –2y – 20 = 0 is 5a , then a For detailed solution to the Sample Paper, equals____ visit our website www. vidyalankar.org 13. Angle between the tangents drawn from (1, 4) to HARYANA at the parabola y2 = 4x is π , where m equals_____ • Bharat Book Depot - Ambala Ph: 0171-2643568; Mob: 9896316290 m • Gulati Book Palace - F aridabad Ph: 0129-2295960, 4007870; Mob: 9871450002 • Book Care Centre - Gurgaon 14. If the straight line y = 2x + c is a tangent to the Ph: 0124-6521670, 4251670; Mob: 9873621670, 9873401670 ellipse x2 + y2 = 1, then |c| equals_____ 8 4 • Brilliant E ducational Aid - Gurgaon x2 y2 15. e foci of the ellipse 16 + b2 = 1 and the Ph: 0124-406725600; Mob: 9811092425, 9873092425 of b2 hyperbola x2 − y2 = 1 coincide, then value • Mangla And Company Book Seller - Gurgaon equals____1_44 81 25 Ph: 0172-2725216, 4642255; Mob: 9814006655 16. If the angle between the two lines represented by 2x2+ 5xy +3y2+ 7y + 4 = 0 is tan–1(m), then the • Modern Book Shop - Hissar Mob: 9416346209 value of 10 m must be_____ • New Gulati Book Depot - Jind 17. From a point, common tangents are drawn to the Ph: 01681-252586; Mob: 9416947565, 9416166838 circle x2 + y2 = 8 and parabola y2 = 16x. If the area of the quadrilateral formed by the common tangents, • Books Ally Incorporation - Karnal Mob: 9812440050 the chord of contact of the circle and the chord of • Balbir And Brothers - Kurukshetra Mob: 9996400044 contact of the parabola is 10k, then nd k. • Hanuman Book Depot - Narnaul Mob: 9416178859, 9813194705 • Holkaran Dass Hemraj - Narnaul 18. e line x + y = a meets the x-axis at A and y-axis at B. A DAMN is inscribed in the DOAB, O being the Ph: 01282-252008; Mob: 9355552008, 9416268011 origin,with right angle at N; M and N lie respectively • Kamal Book Depot - Narnaul Ph: 252305; Mob: 9416382533, 9050632762 • Anamika Trading - Panchkula Ph: 2579143, 5011377; Mob: 9316002471 • Y adav Book Depot - Rewari Ph: 1281245052; Mob: 9896126652, 9813074160 • Manish Traders - Rohtak Ph: 01262-268715; Mob: 9812556687 • Sunrise - Rohtak Mob: 8199944420, 8950350733 • Tinku Book Depot - Rohtak Ph: 01262-211532; Mob: 9812375783 • F air Deal Book Centre - Sirsa Mob: 9896054814, 9255102978 • Sachdeva News Agency - Y amuna Nagar Mob: 9896377026 MATHEMATICS TODAY | SEPTEMBER‘17 71
Series-3 Time: 1 hr 15 min. The entire syllabus of Mathematics of JEE MAIN is being divided into eight units, on each unit there will be a Mock Test Paper (MTP) which will be published in the subsequent issues. The syllabus for module break-up is given below: Unit Topic Syllabus In Details No. 3 Permutations Fundamental principle of counting, permutation as an arrangement and combination as and selection, meaning of P(n, r) and C(n, r), simple applications. Combinations General solution and Properties of Triangle. Trigonometry Circles: Standard form of equation of a circle, general form of the equation of a circle, its Co-ordinate radius and centre, equations of a circle when the end points of a diameter are given, points Geometry-2D of intersection of a line and a circle with the centre at the origin and condition for a line to be tangent to a circle, equation of the tangent. 1. e number of positive terms in the sequence of getting exactly 4 correct answers by a candidate answering all the questions is 195 n+3P3 , is (a) 46 – 32 (b) 135 xn = 4 ⋅ nPn − n+1Pn+1 n ∈N 3 (c) 55 (d) 120. (a) 2 (b) (c) 4 (d) none of these. 6. Number of ways in which 3 boys and 3 girls(all are of di erent heights) can be arranged in a line so that boys 2. e number of numbers greater than 50,000 that as well as girls among themselves are in decreasing can be formed by using the digits 3, 5, 6, 6, 7 is (a) 36 (b) 48 order to their height (from le to right) is (a) 720 (b) 72 (c) 54 (d) none of these. (c) 10 (d) 20. 3. e rank of the word 'FLOWER' is 7. e number of permutations of the letters of the (a) 165 (b) 155 (c) 145 (d) none of these. word HINDUSTAN such that neither the pattern 'HIN' nor 'DUS' nor 'TAN' appears, are 4. A person write letters to his 4 friends and addressed (a) 166674 (b) 169194 the corresponding envelopes. e number of ways at least two of them are in the wrong envelopes is (c) 166680 (d) 181434. (a) 23 (b) 19 (c) 17 (d) 14. 8. e number of ways in which four letters of the word MATHEMATICS can be arranged is given by 5. A test consists of 6 multiple choice questions each 8! having 4 alternative answers of which only one is (a) 4!4! (b) 2454 correct. Also only one of the alternatives must be marked by each candidate. e number of ways (c) 2464 (d) 2474. By : Sankar Ghosh, S.G.M.C, Mob : 09831244397. 72 MATHEMATICS TODAY | SEPTEMBER ‘17
9. e number of ways of distributing of 50 identical BC internally in the ratio 1:3, then sin ∠BAD things among 8 persons in such a way that three of equals sin ∠CAD them get 8 things each, two of them get 7 things each, and remaining three of them get 4 things 1 (b) 1 (c) 1 2 each, is equal to (a) 3 3 6 (d) 3 (a) (50!)(8!) 16. e ratio of the sides of a triangle is 19:16:5, then (8 !)3 (3!)2 (7 !)2 (4 !)3 (2 !) cot A : cot B : cot C equals (50!)(8!) 2 2 2 (b) (8 !)3 (7 !)3 (4 !)3 (a) 1:15:4 (b) 15:1:4 (c) (50!) (d) 8! ⋅1 (c) 4:1:15 (d) 1:4:15. (8 !)3 (7 !)2 (4 !)3 (3!)2 ⋅2 !3! 17. If the sides a, b, c of a triangle ABC are the roots of the equation x3 – 15x2 + 47x – 82 = 0 then the value 10. Consider seven digit number x1, x2, ...., x7 where x1, x2 , ..., x7 =/ 0 having the property that x4 is the of cos A + cos B + cosC equals greatest digit and digits towards the le and right abc 225 131 169 of x4 are in decreasing order. en total number of (a) 164 (b) 164 (c) 131 (d) 82 such numbers in which all digits are distinct is 82 (a) 9C7 · 6C3 (b) 9C6 · 5C3 18. If two sides of a triangle are 2 3 − 2 and 2 3 + 2 (c) 10C7 · 6C3 (d) none of these. and their included angle is 60°, then the other angles are 11. e most general solution of the equation (a) 75°, 45° (b) 105°, 15° logcosqtanq +logsinqcotq = 0, is (c) 60°, 60° (d) 90°, 30° (a) nπ + π (b) nπ − π 19. e maximum value of a cos A + b cos B + c cosC is 4 4 a+b+c (c) 2nπ − π (d) 2nπ + π (a) sin 2A (b) 4 sin A sin B sin C 4 4 2 2 2 x 12. e general solution of 2 − cos x = 2 tan 2 is (c) sin C (d) none of these. 2 (a) (2n + 1) π (b) (4n + 1) π 2 2 20. Let a, b, c are the sides of a triangle and (c) 2np (d) (4n + 1)p. (sinA + sinB + sinC)(sinA + sinC – sinB) = sinA 13. e general solution of sinC where sinA = ak, sinB = bk, sinC = ck then the sinx + cosx = min{1,a2 − 4a + 6} is range of is (a) [0, 1] (b) [–4, 4] a∈R (c) (0, 4) (d) [0, 4] (a) nπ + (−1)n π (b) 2nπ + (−1)n π 21. A line passing through the point (11, –2) and 2 4 4 touching the circle x2 + y2 = 25 are (a) 4x + 3y = 38, 7x – 24y = 125 (c) nπ + (−1)n+1 π (d) nπ + (−1)n π − π (b) 3x + 4y = 25, 7x – 24y = 125 4 4 4 (c) 3x – 4y = 41, 7x + 24y = 125 (d) 7x – 24y = 125, 4x – 3y = 38 14. e equation cos4q + sin4q + l = 0 has real solution for q, if 3 1 (a) 4 < λ < 1 (b) −1 < λ < − 2 22. Two circles each of radius 5, have a common (c) 0 < l < 1 d) l < –1 tangent at (1, 1) whose equation is 4x + 3y – 7 = 0 e centres are π π (a) (–4, 4), (6, 2) (b) (–3, 4), (5, –2) 15. In a triangle ABC, if ∠B = 3 , ∠C = 4 and D divides (c) (5, 4), (–3, –2) (d) (4, 2), (–2, 0) MATHEMATICS TODAY | SEPTEMBER‘17 73
23. e equation of the circles touching the line 30. If (–4, 3) and (12, –1) are the ends of diameter of a x + 2y = 0 and passing through the points circle which makes an intercept of 2l on the y-axis, of intersection of the circle x2 + y2 = 4 and then l = x2 + y2 – 2x – 4y + 4 = 0 is (a) 13 (b) 4 13 (c) 3 13 (d) 2 13 (a) x2 + y2 + x + 2y = 0 (b) x2 + y2 – x + 2y = 0 SOLUTIONS (c) x2 + y2 + x – 2y = 0 (d) x2 + y2 – x – 2y = 0. 1. (c): Given that xn = 195 − n+3P3 , n ∈N 4 ⋅nPn n+1Pn+1 24. A circle which passes through the point (1, 1) and cuts orthogonally the two circles x2 + y2 – 8x – 2y = 195 − (n + 2)(n +1) = 195 − (n + 3)(n + 2) + 16 = 0 and x2 + y2 –4x – 4y + 1 = 0 . If its centre 4 ⋅n! (n +1)! 4 ⋅n! n! (a, b), then a + b = 1 1 Q⇒= 1x9n45ni−s2p4+on22s4i0nt−ni!v2−e0∴1n7−112<7410=−1447n⋅1n2−!−442n⋅0n2n!−>200n 2 2 (a) 0 (b) –1 (c) (d) − 25. e x + y + 1 = 0 meets the circle x2 + y2 + 3x + y – 6 = 0 at the points A and B. If C is a point on the circle, then the locus of the orthocentre of the triangle ABC is e above inequality is true for n = 1, 2, 3 and 4. (a) x2 + y2 – x + y – 8 = 0 (b) x2 + y2 – x + y = 0 2. (b) : Here, we have 5 digits among which 6 is (c) x2 + y2 + x – y – 8 = 0 repeated twice and number greater than 50,000 are (d) x2 + x – y = 0 consists of 5 digits. soofmpeernmuumtabteiorsnam= o25n!! g= t1h22e0se=6600 26. e circle C1 touches the line x – y = 0 at the origin \\ Number numbers and the circle C2 touches the line x+y = 0 at the But there are origin. Let C be the circle x2 + y2 – 4x + 2y – 6 = 0. which are started with 3 and those are less than 50,000, so they are to be rejected. e common chord of C and C1 pass through A and 4! common chords of C and C2 pass through B. en Such number of numbers = 2! = 12 AB = \\ e required number of numbers = 60 – 12 = 48 (a) 5 (b) 5 (c) 2 5 (d) 3 5 3. (b) : Words before FLOWER are 1 × 5! + 1 × 4! + 1 × 3! + 2 × 2! = 154 27. e triangle PQR is inscribed in the circle \\ Rank of the word FLOWER is 154 + 1 = 155. x2 + y2 = 25 if Q and R have co-ordinates (3, 4) and (–4, 3), then ∠QPR is equal to 4th.in(gas)go: eLsettoDwrrdoenngoptelasctehse. number of ways in which r If r goes to wrong place out (a) π (b) π (c) π d) π of n then (n – r) goes to correct places. 2 3 4 6 \\ Dn = nCn – rDr 28. In an equilateral triangle, 3coins of radii 1 are kept so that they touch each other and al∴so Dthne=sindCens−orfDr 1 1 1 1 1 the triangle. Area of the triangle is wWhheerreeDDrr == r!! 2 − 3! + 4! − 5! + ... + (−1)r r ! (a) 4 + 2 3 (b) 6 + 4 3 ∑ ∑∴ r Dn = r nCn−r Di (r < n) (c) 12 + 7 43 i=1 i=1 (d) 3 + 7 3 4 Now for our problem 29. Consider a family of circles which are passing (Dr )r = 4 = 4 ! 1 − 1 + 1 = 12 − 4 +1= 9 = D4 through the point (–1, 1) and are tangent to x-axis. 2 3! 4! If (h, k) is the centre of circle, then kk<≥1212 1 1 (Dr )r =3 3! 1 1 3−1= 2 (a) (b) − 2 ≤k ≤ 2 = 2 − 3! = = D3 (c) (d) 0 k< < 1 (Dr )r =2 = 2! 1 =1= D2 2 2! 74 MATHEMATICS TODAY | SEPTEMBER ‘17
Now no. of ways at least two goes to the wrong places, erefore the required no. is n = 4, r > 2 4! 4! 3C2 × 2!2! + 3C1 × 7C2 × 2! + 8C4 × 4! = 2454. ∑4 nCn−r Dr = 4C2D2 + 4C1D3 + 4C0D4 9. (d) : Number of ways of dividing 8 persons in three i=2 groups rst having 3 persons, second having 2 persons 8! = 6D2 + 4D3 + D4 = 6 + 8 + 9 = 23 and third having 3 persons = 3!2!3! So, 23 are the number of ways in which at least two things goes to wrong place. Since all the 50 things are identical 5. (b) : e number of ways by which 4 questions with \\ Required number = 8! ⋅1 correct answers can be chosen in 6C4. Since only one of (3!)2 ⋅2!3! the four alternatives is correct, the wrong answers can be 10. (a) : Number of selections of 7 digits out of digits given in 3 di erent ways for each of the two remaining 1, 2, 3, ..., 9 = 9C7 questions attempted unsuccessfully by the candidate. Out of these 7 selected digits excluding the greatest digit = 6, the 6 digits can be divided in two groups e desired number of ways = 6C4 · 32 = 135 6. (d) : Since order of boys and girls are to be each having 3 digits = 3!× 6! 2! = 6C3 × 1 maintained in any of the di erent arrangements, the 3!× 2! 6! required number = 3!3! But the three digits on one side can go on the other side 7. (b) : Total number of paerermaulwtaatyiosntso=ge92t!!h;ern=o. of \\ Required number = 9C7 ⋅ 6C3 1 .2! = 9C7 ⋅ 6C3 7!; 11. (a) : e given equation is 2! permutations where 'HIN' no. of permutations where 'DUS' are always together logcosqtanq + logsinqcotq = 0 t=og72e!!thaenrd= no. of permutations where 'TAN' are always ⇒ logcos θ tan θ − logsinθ tan θ = 0 7!; lo(θbgll=oo)cgogn:sctπθaon+ssieθθnπ4gθ=,iv=nleloon1∈gg⇒eZtsqai.nnucoaθθtsiθon⇒=issin2lθlo−o⇒ggccosoistnasxnθθ=θ=2=1tatannx2π4 Now the number of permutations where 'HIN' and ⇒ 'DUS' are always together = 5! ⇒ Number of permutations where 'HIN' and 'TAN' are ∴ always together = 5! 12. Number of permutations where 'TAN' and 'DUS' are always together = 5! Number of permutations where 'HIN', 'DUS' and 'TAN' 1− tan2 x are always together = 3! ⇒ 2 1 − tan x = cos x ⇒ 2 1 − tan x = 1+ tan2 2 2 2 x Required number of permutations 2 9 ! 7! −or112++−tta1an1n+2+2xtx2taann2=2xx20 = 2 ! − 7 !+ 7 !+ 2! + 3 × 5!− 3! = 169194 ⇒ 1− tan x 2 8. (b) : e word 'MATHEMATICS' consists of 11 Either 1− tan 2 letters in which there are 2M's, 2A's and 2 T's x e following cases are possible. 2 Case I: Four letters having a pair of similar letters. =0 = 0 is can be done in 3C2 ways and no. of permutation 4! of such four letters is 2!2! ⇒ tan x =1 or 2 tan2 x − tan x +1= 0 2 2 2 Case II: one similar pair and 2 di erent letters. is can be done in 3C1 × 7C2 ways and then no. of possible ⇒ tan x = 1 = tan π ⇒ x = nπ + π 2 4 2 4 4! permutations = 2! ⇒ x = 2nπ + π ⇒ x = (4n + 1) π 2 2 Case III: All di erent letters are taken. is can be and 2 tan2 x − tan x +1= 0 done in 8C4 ways and then no. of permutation = 4! 2 2 MATHEMATICS TODAY | SEPTEMBER‘17 75
x 1± 12 − 4 ⋅ 2 ⋅1 1 ±i 7 (no solution) 17. (b) : We are given that, a, b, c are the roots of the 2 2⋅2 4 equation x3 – 15x2 + 47x – 82 = 0 ⇒ tan = = \\ a + b + c = 15, ab + bc + ca = 47, abc = 82 13. (d) : Here, a2 – 4a + 6 = (a – 2)2 + 2 > 2 Now cos A + cos B + cosC ∴ min{1,a2 − 4a + 6} = 1 abc b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2 a2 + b2 + c2 a∈R 1 sinx + 1 cosx = 1 = 2abc + 2abc + 2abc = 2abc 2 2 2 Now, sin x + cos x = 1 ⇒ ⇒ sin x + π = sin π ∴ x + π = nπ + (−1)n π = (a + b + c)2 − 2(ab + bc + ca) = (15)2 − 2(47) = 131 4 4 4 4 2abc 2(82) 164 ∴ x = nπ + (−1)n π − π 18. (b) : Let b = 2 3 + 2, c = 2 3 − 2 and A = 60° 4 4 14. (b) : We have, cos4 q + sin4 q + l = 0 B − C b − c A 4 ∴ tan 2 = b + c cot 2 = 43 cot 30° =1= tan 45° ⇒ 1− 2sin2 θcos2 θ = −λ 1 1 ⇒ B − C = 90° Also, B + C = 120° (Q A = 60°) ⇒ 1 − 2 sin2 2θ = − λ ⇒ 1− 4 (1 − cos 4θ) = −λ erefore the other two angles are B = 105°andC = 15° 3 1 ⇒ 4 + 4 cos 4θ = −λ 19. (b) : a cos A + b cos B + c cosC a+b+c Q −1 < cos 4θ < 1 ∴ − 1 < 1 cos 4θ < 1 2 sin cos A + 2sin B cos B + 2sinC cos 4 4 4 A 2(sin A + sin B + sinC) C = 3 1 3 1 3 1 ⇒ 4 − 4 < 4 + 4 cos 4θ < 4 + 4 2CQ a b c = 2R sin A sin B sin C 1 1 1 1 = = 2 2 ⇒ < − λ < ⇒ − < λ < − sin 2A + sin 2B + sin 2(sin A + sin B 15. (c): From ∆ABD we have, BD AD = + sinC) ∠BAD sin B sin = 4 sin Asin B sinC A B C 2 2 2 From ∆ACD we have, CD AD . = A B C = 4 sin sin sin sin ∠CAD sin C 2 2 2 = 2 4 cos cos cos QBD : CD = 1: 3 ∴ AD sin ∠BAD : AD sin ∠CAD = 1: 3 20. (c) : (sinA + sinB + sinC)(sinA + sinC – sinB) sin B sin C = sinAsinC (given) sin ∠BAD : sin ∠CAD =1: 3 (a + b + c) (a + c – b) = ac ⇒ π π Q a = b = c = 2R 3 4 sin A sin B sinC sin sin ⇒ (a + c)2 – b2 = ac ⇒ a2 + c2 – b2 + 2ac = ac ⇒ 2 sin ∠BAD = 1 ⇒ sin ∠BAD = 1 ⇒ a2 + c2 − b2 +1= µ 3 sin ∠CAD 3 sin ∠CAD 6 2ac 2 16. (d) : Given a : b : c = 19 : 16 : 5 ⇒ 1+ cos B = µ Q cos B = a2 + c2 − b2 \\ 2s = a + b + c = 40 2 2ac \\ cot A : cot B : cos C ⇒ cos2 B2<=4µ4 ⇒ 0< µ <1 (∵ ∠B ≠ 0 and ∠B ≠ 180°) 2 2 2 \\ 0< 4 = (s s(s − a) c) : (s s(s − b) c) : s(s − c) 21. (b) : Let the equation of the line be − b)(s − − a)(s − (s − a)(s − b) = (s – a) : (s – b) : (s – c) y + 2 = m(x – 11) or mx – y – (11m + 2) = 0 ...(i) Since the line (i) touches the circle x2 + y2 = 25 = a + b + c − a : a + b + c − b : a + b + c − c 2 2 2 − (11m + 2) ∴ =5⇒m= − 3 , 7 40 19 40 16 40 15 m2 +1 4 24 = 2 − : 2 − : 2 − = 1 : 4 : 15 erefore (i) becomes 3x + 4y = 25 or 7x – 24y = 125 . 76 MATHEMATICS TODAY | SEPTEMBER ‘17
22. (c) : e line joining the centres passing through (1, 1) 25. (c) : Let (h, k) be the orthocentre. e image of (h, k) on the line x + y + 1 = 0 is given by and perpendicular to the tangent 4x + 3y – 7 = 0 is −2(h + k +1) 3x – 4y + 1 = 0 ...(i) x−h = y−k = 1+1 1 1 Let the abscissa of the centre be a and its ordinate can e image is (–1 –k, –1 – h) be obtained from (i) It lies on the circumcircle of triangle ABC. 3α + 1 \\ (1 + k)2 + (1 + h)2 – 3(1 + k) – (1 + h) – 6 = 0 ∴ Centre C α, 4 ⇒ h2 + k2 + h − k − 8 = 0 e distance between the centre C and (1, 1) is 5 erefore the locus is x2 + y2 + x – y – 8 = 0 12 ∴ (α − 1)2 + 3α +1 − = 25 26. (c) : Here C1 = x2 + y2 + l(x – y) = 0, l = R, 4 C2 : x2 + y2 + λ(x + y) = 0, λ ∈R and C : x2 + y2 – 4x + 2y – 6 = 0 On solving the above equation we get a = 5, –3 therefore C = (5, 4), (–3, –2) e common chord of C and C1 are l(x – y) + 4x – 2y + 6 = 0 which are concurrent at 23. (d) : e equation of the circle passing through the x − y = 0, ⇒ 4x − 2x + 6 = 0 ⇒ x = −3 ∴ A = (−3, −3) intersection of the circle x2 + y2 = 4 and x2 + y2 – 2x – 4y + 4 = 0 is e common chord of C and C2 are (x2 + y2 – 2x – 4y + 4) + l(x2 + y2 – 4) = 0 l(x + y) + 4x – 2y + 6 = 0 ⇒ (1+ λ)x2 + (1+ λ)y2 − 2x − 4 y + 4(1− λ) = 0 ⇒ x = − y, − 6y + 6 = 0 ⇒ y = 1 ∴ B = (−1,1) ⇒ x2 + y2 − 2x − 4y + 4(1− λ) = 0 ...(i) ∴ AB = 42 + 22 = 2 5 1+ λ 1+ λ 1+ λ Since the line x + 2y = 0 touches the circle (i) 27. (c) : e centre of the circle x2 + y2 = 25 is (0, 0). Now slope of QO × slope of RO 1 + 4 1 2 2 2 4(1− λ) 4−0 3−0 4 3 1 + λ 1 + λ 1 + 1 + 1+ λ 3−0 −4 − 0 3 4 −1 ∴ = + − = × = × − = 12 + 22 λ λ \\ erefore angle at the centre is ∠QOR π , 5 1 = 2 5 − (4)(1− λ2 ) 2 × π ⇒ 1+ λ = (1+ λ)2 Angle at circumference ∠QPR = ∠QOR = 4 5 28. (b) : Length of the side 1+ 4λ2 1+ 4λ2 ⇒ 5 = (1+ λ)2 ⇒ 5 = (1+ λ)2 = 2 + 2 cot π = 2(1 + 3) 1+ λ (1+ λ)2 6 ⇒ 1+ 4λ2 = 5 ⇒ λ2 = 1 ⇒ λ = 1 ( ( ))Area =3 3 2 =6+4 4 2 1+ 3 Now put the value of l = 1 in equation (i) we get 29. (a) : Circle with centre (h, k) and touching x-axis is x2 + y2 – x – 2y = 0 x2 + y2 – 2hx – 2ky + h2 = 0 ....(i) 24. (d) : Let the equation of the circle be Q (−1,1) lies on (i) ∴ 2 + 2h − 2k + h2 = 0 x2 + y2 + 2gx + 2fy + c = 0. ⇒ (h2 + 2h +1) +1− 2k = 0 ⇒ 2k = (h +1)2 +1 It cuts the two given circles orthogonally 1 1 1 \\ –8g – 2 f = c + 16 and –4g – 4f = c + 1 2 2 1)2 2 ⇒ 8g + 2 f + c = −16 ...(i) ⇒ k= + (h + ⇒ k > and 4g + 4 f + c = −1 ...(ii) 30. (d) : e equation of circle having extremities of the diameter (–4, 3) and (12, –1) is e circle passes through (1, 1) (x + 4)(x – 12) + (y – 3)(y + 1) = 0 \\ 2g + 2f + c = –2 ...(iii) 7 17 ⇒ x2 + y2 − 8x − 2y − 51 = 0 Solving (i), (ii) and (iii) we get g = − 3 , f = 6 , c = −3 Now, length of the chord intercepted on y-axis is \\ e centreis 7 , − 17 = (a, b) ⇒ a = 7 , b = − 17 , a +b = − 1 2 f 2 − c = 2 (−1)2 − (−51) = 2 52 = 4 13 3 6 3 6 2 ∴ 2λ = 4 13 ⇒ λ = 2 13 MATHEMATICS TODAY | SEPTEMBER‘17 77
hallen ing PROBLEMS CON ALCULUS 1. Consider a function f : (0, ∞) → R and a real 6. Let f, g : R → R be two periodic functions with respective periods and and number a > 0 such that f (a) = 1 and lim ( f (x) − g(x)) = 0 then T1 T2 f (x)⋅ f (y) + f a ⋅ f a = 2 f (xy) x→∞ x y (a) T1 = T2 (b) T1 = 2T2 for all x, y ∈(0, ∞) and f(2) = 1 then f(3) = (c) 2T1 = T2 (d) T1 = 1 + T2 (a) 1 (b) 6 (c) 9 (d) 16 7. Let f :[1, ∞] → R be the function de ned by 2. Consider the polynomial P(x) with integral coe cients such that P(P′(x)) = P′(P(x)) and f (x) = [x] + {x} . P(4) = 4 then P(2) = x e smallest number k such (a) 2 (b) 3 (c) 4 (d) 5 that f(x) < k for all x > 1 is 3. All polynomials P of degree n having only real zeroes (a) 1 (b) 2 (c) 3 (d) 2 n2 x1, x2, x3, ..., xn such that n 1 xi = xp′(x) for 8. e real parameter m such that the graph of P(x) − the function f (x) = 3 8x3 + mx2 − nx has the ∑ horizontal asymptote y = 1 is (a) 10 (b) 12 (c) 14 (d) 16 i=1 all non-zero real numbers x is of the form P(x) = (a) 2xn + 3xn – 1 (b) 2xn + 3xn – 2 (c) 2xn (d) 2xn – 1 + 3xn – 2 + 4xn 9. f is continuous function R → R, f (0) = 1 and 4. Consider the polynomial with real coe cients f(2x) – f(x) = x x ∈ R then f(2) = P(x) = a00xhnas+ aal1lxints–1r+oo..t.s. +reaanl,aannd 0. If the equation P(x) = distinct then the (a) 1 (b) 2 (c) 3 (d) 4 equation x2P′′(x) + 3xP′(x) + P(x) = 0 has 10. Real functions f, g, h : R → R are such that for all real x, y, (x – y)f(x) + h(x) – xy + y2 < h(y) < (x – y) (a) real and distinct roots g(x) + h(x) – xy + y2 then h(x) is (b) real and equal roots (a) a linear function (c) not real roots (b) a quadratic function (d) not all real roots (c) a reciprocal function (d) an exponential function 5. Let f, g : R → R be periodic functions of periods 'a' (x) g(x) and 'b' such that lim f x = l and lim x = m , 11. Let f(x) be a function which contains x→0 x→0 element 2 in its domain and range. Suppose that l ∈R, m ∈R − {0}. en lim f ((3 + 7)na) = f(f(x)) · (1 + f(x)) = –f(x) for all numbers x in the (a) 0 n→∞ g((2 + 2)nb) domain of f, then f(2) = (c) –1 (b) 1 (a) 1 (b) 2/3 (d) does not exist (c) –1 (d) –2/3 By : Tapas Kr. Yogi, Visakhapatnam Mob : 09533632105 78 MATHEMATICS TODAY | SEPTEMBER ‘17
12. Let a0 ∈ (–1, 1) and de ned recursively, (b) lim cn does not exist 1 + an−1 . Let = 4n(1 – an) then n→∞ 2 an = , n > 0 An (c) lim dn does not exist lim n→∞ n→∞ An = (d) lim (cn + dn) does not exist (a) (cos–1a0)2 (b) (cos−1 a0)2 n→∞ 2 SOLUTIONS (c) 2(cos–1a0)2 (d) 3(cos–1a0)2 1. (a) : Putting x = 1 = y gives f(1) = 1 13. lim 1 ⋅ n 2n − 2 n = ___ Putting y = 1 gives, f (x) = f a , x >0 n k k x n→∞ ∑ a a k =1 Putting x gives, (x) ⋅ x 1 ([·] denotes the greatest integer function) y = f f = (a) log4 (b) log4 + 1 so that, gives, f 2(x) = 1 (c) log4 – 1 (d) 2log2 – 2 f 2( 2 a 14. Let f(x) be a continuous function on (a, b) and Putting x = y = t gives t)+ f t = 2 f (t) lim f (x) = +∞ , lim f (x) = −∞ and As L.H.S. is positive so R.H.S is positive. n→a+ n→b− Hence f(x) = 1 for all x. f ′(x) + f 2(x) ≥ −1 for x ∈(a, b) then minimum 2. (a) : Consider a polynomial value of (b – a) is P(x) =a1a,0ax2n, +....a, 1axnn – 1 + ... + an, a0 =/ 0 (a) p (b) p/2 (c) 2p (d) p/4 where a0, ∈ I, then 15. Let f(x, y) be a function satisfying the functional P′(x) = na0xn−1 + (n −1)a1xn−2 + ... + an−1 equation f(x, y) = f(2x + 2y, 2y – 2x) for all real number x, y. De ne g(x) by g(x) = f(2x, 0). en By comparing coe cient of xn(n – 1) in the relation P(P′(x)) = P′(P(x)) , we have a0n+1 ⋅nn = a0n ⋅n period of g(x) is (a) 3 (b) 4 (c) 6 (d) 12 i.e., a0 = 1 . 16. e number of real solutions to the equation nn−1 3 5 x = 2 + 16 (x − 3) is/are PNSouo,twtPi,n(sxgi)nb=caecxak+0tihasi1sanvailnuteegoefr ⇒ n= 1 and a0 =1 x +1 P(x) in the given relation, (a) 1 (b) 2 (c) 3 (d) 5 we have 1 + = 1, = 0 So, P(x) = x 17. Consider the function, a1 i.e. a1 f (x) = log(1 + x + x 2), x ≤b n P′(x) n2 x >b P(x) − xi x ax + c, 3. (c) : e given equation is ∑ = If the graph of f(x) is concave in R then n i=1 −(1 + 3) Integrating, ∑ log | P(x) − xi | = n2 log c | x |, c > 0 2 (b) c + ab = log(b + b2) i=1 (a) b > n cn2 |n2 2b + 1 2b + 1 or, log ∏ | P(x) − x | = log | x b2 + b + b2 + b + i=1 (c) a ≤ (d) a > 1 1 n |n2 , i.e. | P(x) − xi | = k | x k > 0 18. Consider an = sin(np/2) and the three sequences ∏ i=1 i.e. |P(P(x)| = k|x|n2 bn = nan (n > 0), cn = an (n ≥ 1) and dn = an cos nπ Hence, P(x) = axn n 2 (n > 1) then 4. (a) : De ne, Q(x) = xP(x). As an =/ 0, the polynomial Q has distinct real roots. So, (a) lim bn does not exist Q'(x) has distinct real roots as well. n→∞ MATHEMATICS TODAY | SEPTEMBER‘17 79
De ne f(x) = x Q′(x). f (x) = m 2/3 m m 1/3 Again, we observe that H' has distinct real roots and x 2 8 x since H′(x) = x2P′′(x) + 3xP′(x) + P(x) has distinct real 8 + + + + 4 roots. As | x | → ∞, (x) m =1 (A.T.Q) 5. (a) : Numerator = f ((3 + 7)na) f = 12 = f (3 + 7)na + (3 − 7)na − (3 − 7)na = f [Ma − (3 − 7)na] So, m = 12 where (3 + 7)n + (3 − 7)n = integer (M) 9. (c) : f(x) – f(x/2) = x/2 f(x/2) – f(x/4) = x/4 = f (−(3 − 7)na) [as f is periodic] f x − f x = x 2n−1 2n 2n Similarly, denominator = g((2 + 2)nb) Adding and n → ∞, f (x) − f (0) = x + x + ... + ∞ = g(−(2 − 2)nb) 2 4 Hence, given limit Hence, f(x) = x + 1 = a lim f (−(3 − 7)n) × −(2 − 2)n × (3 − 7)n 10. (b) : Putting y = x – 1 in L.H.S. and R.H.S, we have b (−(3 − 7)n) g(−(2 − 2)n) (2 − 2)n f (x) ≤ g(x). n→∞ Putting y = x + 1 in L.H.S. and R.H.S., f (x) ≥ g(x) Hence, for all x, f(x) = g(x) and so inequality becomes = zero as 3− 7 <1 equality and (x – y)f(x) + h(x) – xy + y2 = h(y) 2− 2 For x = 0, h(y) = y2 – y f(0) + h(0) for all y ∈ R. Hence, h(x) is a quadratic function. 6. (a) : lim f (x0 + nT1 + T2) − g(x0 + nT1 + T2) = 0 n→∞ lim f (x0 + T2) − g(x0 + nT1) = 0 11. (d) : Let f(x0) = 2 then at x = x0, f(f(x0))·(1 + f(x0)) = –f(x0) n→∞ = g(x +quneTs1ti)on, af(cxc0or+diTn2g) to the ... (1) i.e. f(2)(1 + 2) = –2 i.e. But lim g(x0 nT1) − (x0 + nT1) 0 −2 3 n→∞ + f = Hence, f (2) = So, lim g (x0 + nT1) − f (x0) = 0 12. (b) : Let a0 = cosq, q ∈ (0, p) n→∞ i.e. fT(2x0is+pTer2i)o=d fo(fx0f()x) as well. (By (1)) So, a1 = 1+ cos θ = cos θ i.e. 2 2 7. (b) : Let [x] = a, {x} = b then x = a + b Similarly, a2, a3, ..., an = cos(q/2n) and f (x) = a + b Hence, An = 4n 1 − cos θ a+b 2n Squaring: ( f (x))2 = a + b+2 ab = 1 + ab θ2 sin(θ / 2n) 2 a+b +b θ / 2n a = θ 2n 2 1 + cos Using A.M. G.M., we have f (x) ≤ 2 θ2 θ2 1 2 2 2 8. (b) : lim = 1 (given) So, as n→∞, An = (1) = = (cos−1 a0)2 x→+∞ x→−∞ f (x) = x3(8 − n3) + mx2 Solution Sender of Maths Musing mx2)2 + nx ⋅ 3 8x3 + mx2 3 (8x3 + + n2x2 SET-176 (8 – n3) must be zero and then 1. Devjit Acharjee (West Bengal) 80 MATHEMATICS TODAY | SEPTEMBER ‘17
MATHEMATICS TODAY | SEPTEMBER‘17 81
13. (c) : e given limit = 1 2 − 2 1 ⋅ dx 16. (a) : Let f (x) = x 1 . Domain (−1, − ∞) x x x+ ∫ Notice, f ′(x) > 0 and f ′′(x) < 0 . So, f is strictly 2 2 0 2n +1 2n (x) 2 2 1 0 if x ∈ , increasing and concave in its domain. x x if f = − = ∈ 2 2 Now, f(3) = 3/2 and f '(3) = 5/16. So, the tangent line 2n + 2n + 1 x 2 , 1 to graph of f at (3, 3/2) is y = 3 + 5 (x − 3). Since, f is 2 16 So, required integral 3 5 2 2 2 2 2 2 2 16 = 3 − 4 + 5 − 6 + 7 − 8 + .... strictly concave, so f (x) < + (x − 3) 1 1 1 .... 1 Hence, the only solution is x = 3. 3 4 5 2 = 2 − + − = 2 log 2 − 1 + 17. (c) : Let g(x) = log(1 + x + x2). Domain is x ∈ R. = log4 – 1 g ′′(x) < 0 ⇒ x > 3 −1 or < −(1 + 3) 2 2 14. (a) : Given inequation becomes ′(x) (1 + 3) d (x + tan−1 f (x)) = f f 2(x) +1 ≥ 0 So, f is concave, when b < 2 dx 1+ and lim f (x) = f (b) for continuity of f gives So, x + tan–1f(x) is non-dec. x→b Hence, π +a ≤− π + b, so b−a≥ π log(1 + b + b2) = ab + c and if f is concave then 2 2 2b +1 f ′(b−) ≥ f ′(b+) ⇒ 1+ b + b2 ≥ a Notice that a = 0, b = x, f (x) = cotx gives the minimum condition. 18. (a) : e subsequence of bn corresponding to odd 15. (d) : g(x) = f(2x, 0) = f(2x + 1, –2x + 1) integers is b2n + 1 = (–1)n (2n + 1), which does not have = f(0, –2x + 3) = f(–2x + 4, –2x + 4) (as1eq/nxu)ee.dnScloiemlciimnt. iiHst tepenrncodedsubtcnotazolseforobd.oouensdneodt have a limit. e = f(–2x + 6, 0) = f(–2x + 7, 2x + 7) sequence an and = f(0, 2x + 9) = f(2x + 10, 2x + 10) = f(2x + 12, 0) = g(x + 12) And dn = an cos nπ = 1 sin(nπ) = 0, n ∈N. 2 2 Hence, g(x) is periodic with period = 12 82 MATHEMATICS TODAY | SEPTEMBER ‘17
10 BEST PROBLEMS Math Archives, as the title itself suggests, is a collection of various challenging problems related to the topics of JEE Main & Advanced Syllabus. is section is basically aimed at providing an extra insight and knowledge to the candidates preparing for JEE Main & Advanced. In every issue of MT, challenging problems are o ered with detailed solution. e readers’ & comments and suggestions regarding the problems and solutions o ered are always welcome. 1. e eccentricity of the ellipse which meets the deviations were 3 and 4 respectively. e variance of x y combined sample of size 500 is straight line 7 + 2 = 1 on the axis of x and the straight (a) 64 (b) 65.2 (c) 67.2 (d) 64.2 line x − y = 1 on the axis of y and whose axes lie along 6. e mean and variance of random variable X 3 5 the axes of coordinates is having a binomial distribution are 4 and 2 respectively, (a) 26 (b) 32 (c) 6 (d) 42 then P(X = 1) is 7 7 7 7 (a) 1 (b) 1 (c) 1 (d) 1 x−3 16 8 4 32 If the lines x −1 y +1 z −1 and 1 2. 2 = 3 = 4 7. e only statement among the following that is a tautology is: = y −k = z intersect, then k is equal to 2 1 (a) 9/2 (c) 0 (d) –1 (d) 2/9 (a) A ∧ (A ∨ B) (b) A ∨ (A ∧ B) 3. e vectors a and b are not perpendicular and (c) [A ∧ (A → B)] → B (d) B → [A ∧ (A → B)] c and d are the vectors satisfying b × c = b × d and a ⋅d = 0. en the vector d is equal to: 8. InaDPQR,if3sinP+4cosQ=6and4sinQ+3cosP=1, en the angle R is equal to (a) p/4 (b) 3p/4 (c) 5p/6 (d) p/6 (a) c − a ⋅c b (b) b − b ⋅c c 9. e equation 2cos–1x = sin–1 2x 1− x2 is valid a ⋅b a ⋅b for values of x satisfying all a ⋅c b ⋅c (c) c + a ⋅b b (d) b + a ⋅b c (a) –1 < x < 1 (b) 0 < x < 1 4. e number of seven digit integers, with sum of (c) 0 < x < 1/ 2 (d) 1/ 2 < x < 1 the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is 10. Let f(x) = (x – 4)(x – 5)(x – 6) then (a) 55 (b) 66 (c) 77 (d) 88 (a) f ′(x) = 0 has four real roots (b) three roots of f ′(x) = 0 lie in (4,5) ∪ (5,6) ∪(6,7) 5. e mean of two samples of sizes 200 and 300 (c) the equation f ′(x) has only two roots were found to be 25, 10 respectively. eir standard (d) three roots of f ′(x) lie in (3, 4) ∪ (4,5) ∪ (5,6) B y : Prof. Shyam Bhushan, D i r e tc o r , N a r a ay n a I I T A ac d e m y , aJ m hs e d p u r . Mo b . : 0 9 3 3 4 8 7 0 0 2 1 83 MATHEMATICS TODAY | SEPTEMBER‘17
SOLUTIONS 6. (d) : We are given, np = 4, npq = 2 1. (a) : x2 + y2 =1 ...(i) ⇒ q= 2 = 1 ∴ n 1 = 4⇒n=8 a2 b2 4 2 2 8 x + y =1 meets x-axis at A(7,0), line x − y = 1 meets ∴ P(X = 1) = nC1 pqn−1 = 8 pq7 = 8 1 = 1 7 2 3 5 2 32 y-axis at B(0,–5) Eq. (i) passes through A and B 7. (c) : Note that A ∧ (A ∨ B) is F when A = F 49 25 A ∨ (A ∧ B) is F when A = F, B = F, and ⇒ a2 + 0 = 1, 0 + b2 = 1 B → [A ∧ (A → B)] is F when A = F, B = T \\ We check only (c) ⇒ a2 = 49,b2 = 25;b2 = a2(1− e2 ) ⇒ 25 = 49(1− e2 ) [A ∧ (A → B)] → B ≡ [A ∧ (~ A ∨ B)] → B ≡ [(A ∧ ( A))]∨ (A ∧ B) → B ⇒ e2 = 24 ⇒ e = 26 ≡ A ∧ B → B ≡~ (A ∧ B) ∨ B ≡~[(A ∧ B) ∧ ( B)] 49 7 ≡~[A ∧ (B∧ B)] ≡~[A ∧ F] ≡~ F ≡ T Thus [A ∧ (A → B)] → B is a tautology. 2. (a) : Any point on the rst line is A(2nre1dl+ionn1e,st3hwre1ils–lei1cno,t4nerdr1sel+icnt1e)wishe(nr2 + 3, 2r2 + k, r2) 8. (d) : Squaring and adding the given relations we 2r1 + 1 = r2 + 3, 3r1 – 1 = 2r2 + k, 4r1 + 1 = r2 get 16 + 9 + 24 sin(P + Q) = 37 ⇒ 2r1 − r2 = 2, 4r1 − r2 = −1 ⇒ r1 = −3 / 2,r2 = −5 ⇒ sin(P + Q) = 1/ 2 ⇒ sin R = 1/ 2 And k = 3r1 – 2r2 – 1 = –9/2 + 10 – 1 = 9/2 3. (a) : b × c = b × d ⇒ b × (c − d) = 0 ⇒ R = π / 6 or 5π / 6 ⇒ c − d ||b ⇒ c − d = αb for some α ∈R ⇒ d = c − αb If R = 5p/6, then P < p/6 Also, a ⋅d = a ⋅c − αa ⋅b ⇒ 0 = a ⋅c − αa ⋅b ⇒ 3sin P < 3 / 2 ⇒ 3sin P + 4 cosQ < 3 / 2 + 4 < 6 ⇒ α = a ⋅c Thus, d = c − a ⋅c b So, R ≠ 5π / 6 a⋅b a⋅b 4. (c) : ere are only two possibilities for sum of the 9. (d) : If we denote cos–1x by y, then since digits equal to 10. 0 < cos−1 x < π ⇒ 0 < 2y < 2π, Case (i): 1,1,1,1,1,2,3 ...(i) Number of seven digit integers = 7! = 42 Also since − π / 2 < sin−1 2x 1− x2 < π / 2 Case (ii): 1,1,1,1,2,2,2 5! Number of seven digit integers = 7! = 35 ⇒ − π / 2 < sin−1 sin(2y) < π / 2, ...(ii) 4!3! \\ Total number of integers = 42 + 35 = 77 − π < 2y < π 2 2 n1x1 + n2x2 5. (c) : Combined mean x = n1 + n2 From (i) and (ii) we nd = 0 < 2y < π / 2 ⇒ 0 < y < π / 4 ⇒ 0 < cos−1 x < π / 4 200 × 25 + 300 × 10 = 16 500 Which holds if 1/ 2 < x < 1 Let d1 = x1 − x = 25 −16 = 9, d2 = x2 − x = 10 −16 = −6 10. (b) : Since f(4) = f(5) = f(6) = f(7) = 0, so by Rolle’s ( ) ( )Now we know that σ2 = n1 σ12 + d12 + n2 σ22 + d22 theorem applied to the intervals [4, 5], [5, 6], [6, 7] n1 + n2 there exist x1 ∈(4,5), x2 ∈(5,6), x3(6,7) such that f ′(x1) = f ′(x2) = f1(x3) = 0. Since f ′ is a polynomial of degree = 200(9 + 81) + 300(16 + 36) = 33600 = 67.2 3 so cannot have four roots. 500 500 84 MATHEMATICS TODAY | SEPTEMBER ‘17
⇒ x(x + 1) x − 1 x − 1 < 0 3 2 1 1 SOLUTION SET-176 ∴ x ∈ (−1, 0) ∪ 3 , 2 1. (b) : a + b = ab + 2 ⇒ b = a +2 7. (a) : z1 = cosq + isinq 2 (a, b) = (1, 9), (4, 16), (9, 25), (16, 36), (25, 49), (36, 64), z2 = cos2q + isin2q – (cosq + isinq) |==z2(2|c2–o=s22((qccoo–ss22cqoqsc–qo)csqo+s+qi()ss2iinn+22(qqsis–nins2iqqn)q–=)s2in–q)22cosq (49, 81), (64, 100), (81, 121) = 9 pairs. = 4 sin2 θ 2 2. 1+(3d3)++:2.3...+.e+..n.(.2t+hntn−e3r1m) ,=tnn2(nn42+ 1)2 |z2| = 2 sin θ = (n + 1)2 \\ 2 4 1 = 8. (c) : z2 = (cos2q – cosq) + i(sin2q – sinq) ∑ ∑ ∑9 9 (n + 1)2 1 10 n2 − 1 = −2 sin 3θ sin θ + i 2 cos 3θ sin θ 4 4 n=1 2 2 2 2 tn = n=1 = n=1 = 2i sin θ cos 3θ + i sin 3θ { }= 2 2 2 1 10 ⋅11 ⋅ 21 − 1 = 1 {385 − 1} = 1 × 384 = 96 4 6 4 4 arg z2 = arg 2 i sin θ + arg cos 3θ + i sin 3θ = π + 3θ 3. (a) : e area of the pentagon 2 2 2 2 2 = Area of BCDE + Area of DEAB = 4ab + a2 – ab Since, 4np < q < (4n + 2)p = a(3b + a) = 451 = 11 × 41 ⇒ a = 11, b = 10 \\ 2nπ < θ < (2n +1)π ⇒ sin θ > 0 \\ a–b=1 2 2 4. (c) : y2 = 4ax ⇒ yy′ = 2a 9. (5) : m = 22013 = 4 22011 2m – 1 = (24)22011 – 1 = 162.2011 – 1 Last digit is 6 – 1 = 5 Eliminating a, 2x y′ = y f (x) − f (0) 1 −1 x x Replacing y′ by y′ we get, 2x + yy′ = 0 10. (b) : P. f ′(0) = lim = lim x α −1 sin = 0 x→0 x→0 Solving, x2 + y2 = c, c e∈ccRentricityf=′(0)1=−xl12i→m=0 f (x) − f (0) = lim x α−1 sin 1 = 0 if a > 1 2 with x x sin − 1x x→0 1 ese are ellipses 2 x 1 ≠ 0 ⇒ f ′(x) = αxα−1 xα−2 cos x 5. (c) : lim f ′(x) = 0 if α > 2 \\ a=3 x→0 Q. y(1) = 2 ⇒ 2 = a + b + c, y(0) = 0 ⇒ c = 0 Also, y′(0) =1⇒ b=1\\ = 1 Hence, y = x2 + x ⇒ y(–1) a 0 = π/2 BM = 2, C1M = 6 − 4 = 2 = MC2 R. f(x) = sinx(1 + a), where a = ∫ costdt = 1 AM = 4 cos30° = 2 3 ∴ b = 2 3 + 2 π/2 0 1 Area = 2 bc sin A = b = 2 3+ 2 ∴ f (x) = 2 sin x, ∫ f (x)dx = 2. 0x 2 6. (b, c, d) : 1 2(x−1)/(2x2 +x−1) < 1 1/x S. e hyperbolas 2 − y2 = 1 and y2 − x2 =1 have 2 2 2 common tangents with slopes ±1. e four tangents 2(x − 1) 1 1 − 3x 0 are x + y ± 1 = 0, x – y ± 1 = 0. ey form a square of ⇒ 2x2 + x − 1 > x ⇒ x + 1) > (2x − 1)(x area 2 × 2 = 2 ⇒ (1 – 3x)x(2x – 1)(x + 1) > 0 MATHEMATICS TODAY | SEPTEMBER‘17 85
86 MATHEMATICS TODAY | SEPTEMBER ‘17
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