The slope of the normal to the curve y = 2x^2 - 4 at P (1, -2) is : Option 1: 4 Option 2: -4 Option 3: -1/4 Option 4: 0

Answer: -1/4. Solution: dy/dx = 4x. At x = 1, slope of tangent = 4. Slope of normal = -1/(slope of tangent) = -1/4.

Calculus

Application of Derivatives

Easy

core

The slope of the normal to the curve $y = 2x^2 - 4$ at P (1, -2) is :

-4

$-\frac{1}{4}$

✅ Correct Option: 3

$dy/dx = 4x$ . At $x = 1$ , slope of tangent $= 4$ . Slope of normal $= -1/(\text{slope of tangent}) = -1/4$ .

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Match List-I with List-II

List-I	List-II
Function f(x)	Interval for increasing/decreasing of function f(x)
---	---
(A) $f(x) = x\vert x\vert$	(I) Decreases on $(0, \infty)$
(B) $f(x) = x^2 + 2x - 5$	(II) Increases on $(3, \infty)$
(C) $f(x) = x^2 - 6x + 9$	(III) Decreases on $(-\infty, -1)$
(D) $f(x) = -x^2$	(IV) Increases on $(-\infty, \infty)$

Choose the correct answer from the options given below:

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Match List-I with List-II

Consider the function f(x) = 2x³ - 21x² + 36x + 80, x∈[0, 6]. Then

List-I	List-II
(A) one of its critical points is at x =	(I) -28
(B) Its absolute maximum value is	(II) -42
(C) Its absolute minimum value is	(III) 97
(D) Its second derivative at x = 0 is	(IV) 6

Choose the correct answer from the options given below:

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The approximate volume of a cube of side a meters on increasing the side by 4% is:

The slope of the normal to the curve y=2x2−4y = 2x^2 - 4y=2x2−4 at P (1, -2) is :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The slope of the normal to the curve $y = 2x^2 - 4$ at P (1, -2) is :