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Objective Type Questions:

42. Total number of parallel tangents of f1(x) = x2 – x + 1 and f2(x) = x3 – x2 –2x + 1 is equal to

     (A) 2                                           (B) 3

     (C) 4                                           (D) None of these

SOLUTION: 

(D) f1 (x1) = 2x1 – 1, f2(x2) = 3x22 – 2x2 – 2 .

Let tangents drawn to the curves y = f1(x) and y = f2(x) at (x1, f(x1)) and (x2, f(x2)) are parallel, then 2x1 – 1 = 3x22 – 2x2 – 2, which is possible for infinite number of order pair (x1, x2).

43. The function 2tan3x – 3tan2x + 12tanx + 3, x Î  is

            (A) Increasing

            (B) Decreasing

            (C) Increasing in (0, p/4) and decreasing in (p/4, p/2)

            (D) None of these

SOLUTION: 

(A)       Let f(x) = 2 tan3x – 3tan2x + 12 tanx + 3

            f'(x) = (6tan2x – 6tanx + 12) sec2x

                   = 6sec2x (tan2x – tanx + 2) > 0

Hence f(x) is always increasing.

44. Let f (x) = (4–x2)2/3 , then f has a

            (A) a local maxima at x = 0                   (B) a local maxima at x = 2

            (C) a local maxima at x = –2                 (D) none of these

SOLUTION: 

(A)       f(x) = (4 – x2)2/3

            f'(x) = –4x / 3(4–x2)1/3 = 4x / 3(x2–4)1/3

            at x = –2         local minima

                x = 0            local maxima

                x = 2            local minima

45. Let f (x) = x3 – 6x2 + 9x + 18, then f (x) is strictly decreasing in

            (A) (–, 1]                                             (B) [3, )

            (C) (–, 1] U [3, )                                 (D) [1, 3]               

SOLUTION: 

(D)       f(x) = x3 – 6x2 + 9x + 18

            f'(x) = 3x2 – 12x + 9

                   = 3(x2 – 4x + 3)

                   = 3(x – 1)(x – 3) £ 0

                x  [1, 3]

46. The absolute minimum value of x4 – x2 – 2x+ 5

            (A) is equal to 5                                    (B) is equal to 3

            (C) is equal to 7                                    (D) does not exist

SOLUTION: 

(B)       f(x) = x4 – x2 – 2x + 5

            f'(x) = 4x3 – 2x – 2

                   = (x – 1) (4x2 + 4x + 2)

Clearly at x = 1, we will set the minimum value which is 3.

47. Equation of the tangent to the curve y = e–|x| at the point where it cuts the line x=1

            (A) is ey + x =2                   (B) is x + y = e

            (C) is ex + y = 1                  (D) does not exist

SOLUTION: 

(A) y = e–|x| cut the line x = 1 at (1, 1/e)

                        (dy/dx)(11,1/e) = –1/e

            Tangent y = –1/e  = –1/e (x – 1)

            =>        ey + x = 2

48. Rolle’s theorem holds for the function x3 + bx2 + cx, 1 < x < 2 at the point 4/3, the value of b and c are;

            (A) b = 8, c = - 5                       (B) b = -5, c = 8   

            (C) b = 5, c = -8                        (D) b = -5, c = -8.

SOLUTION: 

(D)       f'(4/3) = 0

            =>        16 + 8b + 3c = 0

            Also f(1) = f(2)

            =>        3b + c + 7 = 0

            Hence b = – 5, c = –8

49. The number of value of k for which the equation x3 – 3x + k = 0 has two different roots lying in the interval (0, 1) are

            (A) 3                                          (B) 2

            (C) infinitely many                         (D) no value of k satisfies the requirement.

SOLUTION: 

(C)       Let f(x) = x3 – 3x + k

            f'(x) = 3(x – 1) (x + 1)

                873_equation.JPG   

            only are root may lie between –1 and 1

50. From mean value theorem: f(b) – f(a) = (b –a) f' (x1); a < x1 < b if f(x) = 1/x , then x1 =

            (A) √ab                          (B) a+b/2

            (C) 2ab/q+b                    (D) b–a/b+a

SOLUTION: 

(A)       Since f(x) = 1/x

            f'(x1) = f(b) – f(a) / b – a

899_equation.JPG  

13.1    OBJECTIVE LEVEL – 2

51. The minimum value of ax + by, where xy = r2, is (r, ab >0)

            (A) 2r √ab                                (B) 2ab√r

            (C)       –2r √ab                        (D) None of these

SOLUTION: 

(A)       Let f(x) = ax + br2 / x 

            f ’(x) = a – br2 / x = 0

            x = √b / √a r 

            f (√b / √a r) = a√br / √a + √br2 / √br √a = 2r √ab 

52. If a, b, c, d are four positive real numbers such that abcd =1, then minimum value of (1+ a) (1 + b) (1 + c) (1 + d) is

            (A) 8                                    (B) 12

            (C) 16                                   (D) 20

SOLUTION: 

(C)       Applying AM ³GM

            769_equation.JPG

Multiplying all focus (1 + a) (1 + b)(1 + c)(1 + d) ³ 16

53. A lamp of negligible height is placed on the ground ‘l1’ metre away from a wall. A man ‘l2’ metre tall is walking at a speed of l1/10 m/sec. from the lamp to the nearest point on the wall. When he is mid-way between the lamp and the wall, the rate of change in the length of this shadow on the wall is

            (A) –5l2 / 2 m/sec.                   (B) –2l2 / 5 m/sec.

            (C) –l2 / 2 m/sec.                     (D) –l2 / 5 m/sec.

SOLUTION: 

(B)       Clearly h / l1 = l2 / l1 – x

            => hl1 – hx = l1l2        (Since h is decreasing put a – ve sign)

439_equation.JPG

296_mid-way between the lamp.JPG

            at mid point x1 = l1/2, h = 2l2

            => dh/dt = 2l2/5 

            f ’(x) = 2ex + ae-x + 2a + 1> 0 clear a > 0

54. Let f (x) = 1936_equation.JPG. If f (x) has a local minima at x = 1, then

            (A) a > 5                          (B) a > 5

            (C)       a > 0                    (D) none of these

SOLUTION: 

            (A) f (x) = 1936_equation.JPG

            Local minimum value of f (x) at x = 1, will be 5

            i.e. 1 –x + a ³ 5 at x = 1 => a ³ 5

55. Global minimum value of f (x) = x8 + x6 –x4 –2x3 –x2 –2x + 9 is

            (A) 0                                 (B) 1

            (C) 5                                 (D) 9

SOLUTION: 

(C) We will get the minimum value of f(x) at x = 1 which is 5

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