Calculus

Application of Derivatives

Hard

applied

If the critical points (extreme values) of the function $f(x)$ given by $f(x) = -\dfrac{3}{4}x^4 - ax^3 - \dfrac{bx^2}{2} + 15$ are -3 and -5, then the value of $f''(-1)$ is :

-3

-6

$\dfrac{3}{4}$

✅ Correct Option: 3

$f'(x)=-3x^3-3ax^2-bx=-x(3x^2+3ax+b)$ .

Roots $-3$ and $-5$ satisfy $3x^2+3ax+b=0$ : sum $-a=-8$ gives $a=8$ ; product $\dfrac{b}{3}=15$ gives $b=45$ .

$f''(x)=-9x^2-48x-45$ , so $f''(-1)=-9+48-45=-6$ .

2022: 16 July Shift 2

The equation of the normal to the curve $y = x - \frac{1}{x}$ at $(1,0)$ is:

2026: 21 May Shift 1

Given a function $f(x) = ax + \frac{b}{x}$ ; $a > 0, b > 0$ , then

Match List-I with List-II

List-I	List-II
(A) $f(x)$ has local maxima at $x$ is equal to	(I) $\sqrt{\frac{b}{a}}$
(B) $f(x)$ has local minima at $x$ is equal to	(II) $2\sqrt{ab}$
(C) Local Maxima of $f(x)$	(III) $-\sqrt{\frac{b}{a}}$
(D) Local Minima of $f(x)$	(IV) $-2\sqrt{ab}$

Choose the correct answer from the options given below:

2025: 16 May Shift 1

For x ∈ ℝ - {0}, the function f(x) = $\frac{3}{x}$ + 7 is decreasing when

2025: 3 June Shift 2

Which of the following functions has a local minima at $x = 0$ ?

(A) $f(x) = x^3$

(B) $f(x) = |x|$

(D) $f(x) = x^{-2}$

Choose the correct answer from the options given below:

2022: 25 May Shift 1

The equation of the tangent, to the curve $y = x^2 - 2x - 3$ which is perpendicular to the line $x + 2y + 3 = 0$ , is

2025: 15 May Shift 1

The function $f(x) = x^2 - 4x + 6$ is

(A) Strictly decreasing on $(-\infty, 2) \cup (2, \infty)$

(B) Strictly increasing on $(2, \infty)$

(D) Strictly decreasing on $(-\infty, 2)$

Choose the correct answer from the options given below:

If the critical points (extreme values) of the function f(x)f(x)f(x) given by f(x)=−34x4−ax3−bx22+15f(x) = -\dfrac{3}{4}x^4 - ax^3 - \dfrac{bx^2}{2} + 15f(x)=−43​x4−ax3−2bx2​+15 are -3 and -5, then the value of f′′(−1)f''(-1)f′′(−1) is :