Question 1

Passage: Let S and T be the sets where f(x)= x^3 3 - 5x^2 2 +6x+7 decreases and increases respectively. Question: What is T equal to ?

Accepted Answer

Given f(x) = x^3 3 - 5x^2 2 + 6x + 7 Differentiating with respect to x, we get: f'(x) = x^2 - 5x + 6 f'(x) = (x - 2)(x - 3) For the function to be increasing, f'(x) 0 (x - 2)(x - 3) 0 This inequality holds true for x 2 or x 3. Therefore, the set T where the function increases is \ x 2\ \ x 3\ . Answer: \ x 2\ \ x 3\

Question 2

What is the length of the longest interval in which the function f(x)=2 ^2 x-1 is decreasing ?

Accepted Answer

Given f(x) = 2 ^2 x - 1 Using the double angle formula, we can write f(x) = 2x To find the intervals where the function is decreasing, we find its derivative and set it to be less than or equal to zero: f'(x) = -2 2x For f(x) to be decreasing, f'(x) 0 -2 2x 0 2x 0 The sine function is positive in the first and second quadrants, so: 2n 2x 2n + Dividing by 2, we get: n x n + 2 The length of the longest interval in which the function is decreasing is the difference between the upper and lower bound

Question 3

Passage: Let S and T be the sets where f(x)= x^3 3 - 5x^2 2 +6x+7 decreases and increases respectively. Question: What is S equal to ?

Accepted Answer

Given f(x) = x^3 3 - 5x^2 2 + 6x + 7 Differentiating with respect to x, we get: f'(x) = x^2 - 5x + 6 f'(x) = (x - 2)(x - 3) For the function to be decreasing, f'(x) 0. (x - 2)(x - 3) 0 x [2, 3] Therefore, the set S where f(x) decreases is [2, 3]. Answer: [2,3]

Question 4

For the following three (03) items: Let y = f(x) = x ^ -1 x 1 - x^2 + 1 - x^2 . What is the slope of the tangent to the curve y = f(x) at x = 0.5?

Accepted Answer

Given y = f(x) = x ^ -1 x 1 - x^2 + 1 - x^2 We can rewrite the function as: y = x ^ -1 x 1 - x^2 + 1 2 (1 - x^2) Differentiating with respect to x using the product rule and chain rule: dy dx = d dx ( x 1 - x^2 ) ^ -1 x + x 1 - x^2 d dx ( ^ -1 x) + 1 2 1 1 - x^2 (-2x) dy dx = [ 1 1 - x^2 - x ( -x 1 - x^2 ) 1 - x^2 ] ^ -1 x + x 1 - x^2 1 1 - x^2 - x 1 - x^2 dy dx = [ 1 - x^2 + x^2 (1 - x^2)^ 3/2 ] ^ -1 x + x 1 - x^2 - x 1 - x^2 dy dx = ^ -1 x (1 - x^2)^ 3/2 To find the slope of the tangent at x =

Question 5

Consider the following statements regarding the function f(x) = 1 x - 5 : Statement-I: f(x) is decreasing on the intervals x 5. Statement-II: f'(x) > 0 for all x 5. Which one of the following is corre

Accepted Answer

Given the function f(x) = 1 x - 5 . Differentiating with respect to x, we get: f'(x) = d dx (x - 5)^ -1 = -1(x - 5)^ -2 = - 1 (x - 5)^2 For all x 5, (x - 5)^2 > 0, which implies f'(x) Since f'(x) Therefore, Statement-I is correct. Statement-II claims that f'(x) > 0 for all x 5, which contradicts our finding that f'(x) Hence, Statement-I is correct but Statement-II is not correct. Answer: Statement-I is correct but Statement-II is not correct

Question 6

Consider the following for the two (02) items that follow: Let ABC be a triangle right-angled at B and AB + AC = 3 units. What is the maximum area of the triangle?

Accepted Answer

Let AB = c, BC = a, and AC = b. Given that the triangle is right-angled at B, we have a^2 + c^2 = b^2. We are given AB + AC = 3, which means c + b = 3 b = 3 - c. Substituting b into the Pythagorean theorem: a^2 + c^2 = (3 - c)^2 a^2 + c^2 = 9 - 6c + c^2 a^2 = 9 - 6c a = 9 - 6c The area of the triangle is = 1 2 ac = 1 2 c 9 - 6c . To maximize the area, we can maximize its square, S = ^2: S = 1 4 c^2(9 - 6c) = 9 4 c^2 - 3 2 c^3 Differentiating S with respect to c and equating to zero: dS dc = 9 2

Question 7

Consider the following for the two (02) items that follow: Let ABC be a triangle right-angled at B and AB + AC = 3 units. What is A equal to if the area of the triangle is maximum?

Accepted Answer

Let AB = c, BC = a, and AC = b. Given that the triangle is right-angled at B, we have b^2 = a^2 + c^2. We are also given AB + AC = 3 c + b = 3 b = 3 - c. Substituting b in the Pythagoras theorem: a^2 = (3 - c)^2 - c^2 = 9 - 6c + c^2 - c^2 = 9 - 6c The area of the triangle is = 1 2 ac = 1 2 c 9 - 6c . To maximize the area, we can maximize its square: S = ^2 = 1 4 c^2(9 - 6c) = 9 4 c^2 - 3 2 c^3 Differentiating S with respect to c and equating to zero for maximum area: dS dc = 9 2 c - 9 2 c^2 = 0

Question 8

A wire of length 20 cm is to be bent into a rectangle. Which of the following statements is/are correct? I. The rectangle of the largest area is the square. II. It is possible to form a rectangle of a

Accepted Answer

Let the length and breadth of the rectangle be x and y respectively. Given the perimeter of the rectangle is 20 cm . 2(x + y) = 20 x + y = 10 The area of the rectangle is A = xy. Substituting y = 10 - x, we get: A = x(10 - x) = 10x - x^2 To find the maximum area, we can complete the square: A = 25 - (x - 5)^2 The maximum value of A is 25 cm ^2, which occurs when x = 5 cm . When x = 5 cm , y = 10 - 5 = 5 cm . Since x = y, the rectangle of the largest area is a square. Thus, statement I is correct

Question 9

Consider the following for the two (02) items that follow: Let the function f(x) = x^2 + 9. Consider the following statements: I. f(x) is an increasing function. II. f(x) has local maximum at x = 0. W

Accepted Answer

Given f(x) = x^2 + 9. Differentiating with respect to x, we get f'(x) = 2x. For x For x > 0, f'(x) > 0, which means f(x) is strictly increasing in (0, ). Since f(x) is not increasing on its entire domain, Statement I is incorrect. To find local extrema, we set f'(x) = 0, which gives x = 0. The second derivative is f''(x) = 2. Since f''(0) = 2 > 0, f(x) has a local minimum at x = 0, not a local maximum. Thus, Statement II is also incorrect. Both statements are incorrect. Answer: Neither I nor II

Question 10

For the following two (02) items: Consider the function f(x) = 1 - [3] (x-1)^2 . The function has

Accepted Answer

Given f(x) = 1 - (x-1)^ 2/3 Differentiating with respect to x: f'(x) = - 2 3 (x-1)^ -1/3 = -2 3(x-1)^ 1/3 For x 0. Thus, f(x) is strictly increasing. For x > 1, x-1 > 0 f'(x) Since f'(x) changes sign from positive to negative as x passes through 1, f(x) has a local maximum at x = 1. Alternatively, (x-1)^ 2/3 0 for all real x. 1 - (x-1)^ 2/3 1 f(x) f(1) for all x. Hence, f(x) has a maximum at x = 1. Answer: a maximum at x = 1

Question 11

Consider the following for the two (02) items that follow: The slope of the tangent to the curve y = f(x) at (x, f(x)) is 4 for every real number x and the curve passes through the origin. What is the

Accepted Answer

Given that the slope of the tangent to the curve y = f(x) is 4 for all x. dy dx = 4 Integrating both sides with respect to x: y = 4x + C Since the curve passes through the origin (0, 0), substituting x = 0 and y = 0 gives: 0 = 4(0) + C C = 0 The equation of the curve is y = 4x. This represents a straight line. Checking the given points, for x = 1, y = 4(1) = 4. Thus, the straight line passes through the point (1, 4).

Question 12

For the following three (03) items: Consider the function f(x) = x|x|. Consider the following statements: I. The function is increasing in the interval (- , ). II. The function is differentiable at x

Accepted Answer

The given function is f(x) = x|x|. This can be redefined as: f(x) = x^2 for x 0 f(x) = -x^2 for x Differentiating with respect to x, we get: f'(x) = 2x for x > 0 f'(x) = -2x for x At x = 0, the left-hand derivative is _ h 0^- -h^2 - 0 h = 0 and the right-hand derivative is _ h 0^+ h^2 - 0 h = 0. Since the left-hand derivative equals the right-hand derivative, f(x) is differentiable at x = 0. Thus, statement II is correct. Also, f'(x) = 2|x| 0 for all x (- , ). Since f'(x) > 0 for all x 0 and f(x

Question 13

Consider the following statements: Statement-I: The function f(x) = x^3 + 128 x has a minimum value 48 at x = 4. Statement-II: As x increases through 4, f'(x) changes sign from positive to negative. W

Accepted Answer

Given f(x) = x^3 + 128 x = x^2 + 128 x Differentiating with respect to x, we get: f'(x) = 2x - 128 x^2 = 2(x^3 - 64) x^2 For critical points, f'(x) = 0 x^3 - 64 = 0 x = 4 At x = 4, f(4) = 4^3 + 128 4 = 192 4 = 48 Now, checking the sign of f'(x) around x = 4: For x For x > 4, x^3 > 64 f'(x) > 0 (positive) Since f'(x) changes sign from negative to positive as x increases through 4, x = 4 is a point of local minimum. The minimum value is 48. Thus, Statement-I is correct. Statement-II claims that f'

Application of Derivatives — Previous Year Questions & Solutions