Question 1

Let f: R R be a differentiable function such that f ( x+y 3 ) = f(x) + f(y) 3 for all x, y R , and f'(0) = 3. Then the minimum value of the function g(x) = 3 + e^x f(x), is:

Accepted Answer

Given f ( x+y 3 ) = f(x) + f(y) 3 Substituting x = 0 and y = 0, we get: f(0) = 2f(0) 3 f(0) = 0 Differentiating the given equation partially with respect to x, treating y as a constant: f' ( x+y 3 ) 1 3 = f'(x) 3 f' ( x+y 3 ) = f'(x) Substituting x = 0, we get: f' ( y 3 ) = f'(0) = 3 Since this is true for all y R , f'(x) = 3 for all x R . Integrating both sides with respect to x: f(x) = 3x + C Using f(0) = 0, we get C = 0. Thus, f(x) = 3x. Now, the function g(x) is given by: g(x) = 3 + e^x f(x)

Question 2

_ 0 x (16 ( x 2 ) ^3 ( x 2 ) ) is equal to:

Accepted Answer

Let y = x 2 . Since 0 x , we have 0 y 2 . The given expression becomes f(y) = 16 y ^3 y. Differentiating with respect to y: f'(y) = 16( y ^3 y + y 3 ^2 y(- y)) f'(y) = 16 ^2 y( ^2 y - 3 ^2 y) For maximum value, f'(y) = 0: ^2 y - 3 ^2 y = 0 ^2 y = 1 3 Since y [0, 2 ], y = 1 3 y = 6 . Substituting y = 6 in f(y): f ( 6 ) = 16 ( 6 ) ^3 ( 6 ) f ( 6 ) = 16 1 2 ( 3 2 )^3 f ( 6 ) = 8 3 3 8 = 3 3 The maximum value is 3 3 . Answer: 3 3

Question 3

Let f(x) be a polynomial of degree 5, and have extrema at x = 1 and x = -1. If _ x 0 ( f(x) x^3 ) = -5, then f(2) - f(-2) is equal to:

Accepted Answer

Let the polynomial of degree 5 be f(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + k. Given _ x 0 ( f(x) x^3 ) = -5, the terms of degree less than 3 must be zero, and the coefficient of x^3 must be -5. Thus, k = 0, e = 0, d = 0, and c = -5. The polynomial becomes f(x) = ax^5 + bx^4 - 5x^3. Differentiating with respect to x, we get: f'(x) = 5ax^4 + 4bx^3 - 15x^2 Since f(x) has extrema at x = 1 and x = -1, we have f'(1) = 0 and f'(-1) = 0. f'(1) = 5a + 4b - 15 = 0 f'(-1) = 5a - 4b - 15 = 0 Adding both equa

Question 4

The number of critical points of the function f(x) = cases | x x |, & x 0 \ 1, & x = 0 cases in the interval (-2 , 2 ) is equal to :

Accepted Answer

The critical points of a function are the points in its domain where the derivative is zero or does not exist. First, consider x = 0. The function is continuous at x = 0 since _ x 0 | x x | = 1 = f(0). For x (- , ), x x > 0, so f(x) = x x . The derivative at x = 0 is given by f'(0) = _ x 0 x x - 1 x = _ x 0 x - x x^2 = 0. Since f'(0) = 0, x = 0 is a critical point. Next, consider the points where f(x) = 0, which occurs at x = and x = - in the interval (-2 , 2 ). At x = , the inner function g(x)

Question 5

Let f be a differentiable function satisfying f(x)=1-2 x+ _ 0 ^ x e ^ (x-t) f(t) dt , x R and let g (x)= _ 0 ^ x (f( t )+2)^ 15 ( t -4)^ 6 ( t +12)^ 17 dt , x R . If p and q are respectively the point

Accepted Answer

Given f(x) = 1 - 2x + _0^x e^ (x-t) f(t) \, dt. Differentiating and simplifying: e^ -x f'(x) - e^ -x f(x) = -2e^ -x + (1-2x)e^ -x (-1) + e^ -x f(x) This reduces to: f'(x) - 2f(x) = 2x - 3 Solving the linear ODE dy dx - 2y = 2x - 3: y e^ -2x = e^ -2x (2x-3)\,dx On solving, we get f(x) = 1 - x. Now, g(x) = _0^x (3-t)^ 15 (t-4)^6(t+12)^ 17 \, dt g'(x) = (3-x)^ 15 (x-4)^6(x+12)^ 17 = -(x-3)^ 15 (x-4)^6(x+12)^ 17 Sign analysis of g'(x): g'(x) changes from + - at x = 3 → local maxima (q = 3) g'(x) cha

Question 6

Consider the following three statements for the function f:(0, ) R defined by f(x)= | _ e x |-|x-1| : (I) f is differentiable at all x>0. (II) f is increasing in (0,1). (III) f is decreasing in (1, ).

Accepted Answer

For 0 For x > 1: f(x) = x - x + 1, f'(x) = 1 x - 1 = 1-x x At x = 1: LHD = 1-1 1 = 0, RHD = 1-1 1 = 0. So f is differentiable at x = 1. (I) TRUE: f is differentiable for all x > 0. (II) FALSE: f is decreasing in (0, 1). (III) TRUE: f is decreasing in (1, ). Only (I) and (III) are TRUE.

Question 7

Let (2 , ) be the largest interval in which the function f(t)= |t+1| t^ 2 , t 2, is \_\_\_\_

Accepted Answer

For t For -1 f(t) = t+1 t^2 , so f'(t) = -(t+2) t^3 > 0 (increasing). For t f(t) = -t-1 t^2 , so f'(t) = t+2 t^3 . For -2 f'(t) For t f'(t) > 0 (increasing). Largest decreasing interval is (-2, -1) = (2 , ), so = -1. For g(x) = 2 (x-2) - x^2 + 4x + 1: g'(x) = 2 x-2 - 2x + 4 = 0 gives (x-2)^2 = 1, so x = 3. Since g''(x) = - 2 (x-2)^2 - 2 g(3) = 2 (1) - 9 + 12 + 1 = 4

Question 8

Let f(x)=x^ 2025 -x^ 2000 , x [0,1] and the minimum value of the function f(x) in the interval [0,1] be (80)^ 80 (n)^ -81 . Then n is equal to

Accepted Answer

f(x) = x^ 2025 - x^ 2000 , f'(x) = x^ 1999 (2025x^ 25 - 2000) = 0. Critical point: x^ 25 = 2000 2025 = 80 81 , i.e., x = ( 80 81 )^ 1/25 . f(0) = f(1) = 0. At the critical point: f = x^ 2000 (x^ 25 -1) = ( 80 81 )^ 80 ( 80 81 -1 ) = - 80^ 80 81^ 81 . This equals (80)^ 80 (n)^ -81 , so n^ -81 = -81^ -81 n = -81.

Question 9

Let f: R R be a twice differentiable function such that f''(x) > 0 for all x R and f'(a-1) = 0, where a is a real number. Let g(x) = f ( ^ 2 x - 2 x + a ), \; 0 Consider the following two statements:

Accepted Answer

Given f''(x) > 0 for all x R , which means f'(x) is a strictly increasing function. We are given f'(a-1) = 0. Since f'(x) is strictly increasing, f'(x) 0 for x > a-1. The function g(x) is defined as g(x) = f( ^2 x - 2 x + a) for x (0, /2). Let u(x) = ^2 x - 2 x + a = ( x - 1)^2 + a - 1. Differentiating g(x) with respect to x: g'(x) = f'(u(x)) u'(x) = f'(( x - 1)^2 + a - 1) (2 x ^2 x - 2 ^2 x) g'(x) = f'(( x - 1)^2 + a - 1) 2 ^2 x ( x - 1). Case 1: x (0, /4). In this interval, 0 Also, ( x - 1)^2

Question 10

Let f: R R be a twice differentiable function such that the quadratic equation f(x) m ^ 2 -2 f^ (x) m +f^ (x)=0 in m, has two equal roots for every x R . If f(0)=1, f^ (0)=2, and ( , ) is the largest

Accepted Answer

For equal roots in f(x)m^2 - 2f'(x)m + f''(x) = 0, discriminant = 0: [f'(x)]^2 = f(x)f''(x) This gives f''(x) f'(x) = f'(x) f(x) Integrating: f'(x) = kf(x), so f(x) = Ae^ kx Using f(0) = 1 and f'(0) = 2: A = 1, k = 2 f(x) = e^ 2x For g(x) = f( x - x): g'(x) = 2e^ 2( x - x) ( 1 x - 1 ) g'(x) > 0 when 1-x x > 0, i.e., 0 ( , ) = (0, 1) + = 1

Application of Derivatives — Previous Year Questions & Solutions