Applied Mathematics

Calculus

Easy

applied

Let $f : R \to R$ be defined such that $f(x) = 16x^2 - 16x + 12$

(A) Maximum value of $f(x)$ is 8

(B) Minimum value of $f(x)$ is 8

(C) Minimum value of $f(x)$ is 16

(D) No maximum value of $f(x)$

Choose the correct answer from the options given below :

(A), (C) Only

(B), (D) Only

(A), (B), (C) Only

(A), (B) Only

✅ Correct Option: 2

$f'(x) = 32x - 16 = 0 \Rightarrow x = \frac{1}{2}$ . $f(1/2) = 4 - 8 + 12 = 8$ . Since the parabola opens upward, this is a minimum and there is no maximum. Hence (B) and (D) only.

2023: 11 Aug Shift 1

The order and degree Of the differential equation $\frac{d^2 y}{dx^2} + 2 e^{-x} = 0$ , respectively are

2026: 14 May Shift 2

Match the LIST-I with LIST-II

	LIST-I (Differential equation)		LIST-II (Degree)
A.	$\left(\dfrac{dy}{dx}\right)^2 + \dfrac{1}{\dfrac{dy}{dx}} = 5$	I.	$2$
B.	$\dfrac{d^3y}{dx^3} = \left\{3 + \left(\dfrac{dy}{dx}\right)^2\right\}^{3/2}$	II.	$4$
C.	$\dfrac{d^2y}{dx^2} = \left(\dfrac{dy}{dx}\right)^{5/4}$	III.	$1$
D.	$y\dfrac{d^2x}{dy^2} = y^2 + 1$	IV.	$3$

Choose the correct answer from the options given below:

2023: 7 June Shift 1

If the cost function and the profit function for a company is given by $C = 10 - 0.3x^2$ and $P = 0.3x^2 + 2x - 10$ respectively , where X represent units of output, then the revenue function is given by :

2023: 11 Aug Shift 1

If $x = 2at, y = at^2$ , where 'a' is a constant, then $\frac{d^2 y}{dx^2}$ at $x = 2$ is :

2023: 15 June Shift 2

A product costs the manufacturer ₹ 20 per unit. The demand function is given by $p(x) = 1000 - 20x$ , then the quantity for maximum profit is :

2023: 15 June Shift 2

If $x = 3at^2$ , $y = 3at^4$ then $\frac{dy}{dx}$ is :

Solution

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