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

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

(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

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, ∞)

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

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

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

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

Solution:

f'(x) = 3(x – 1) (x + 1)

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

Solution:

(A) Since f(x) = 1/x

f'(x₁) = f(b) – f(a) / b – a

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

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

Solution:

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.