The slope of normal to the curve y = kx^2 - 3x + 2 at x = 1/2 is 5. The value of 'k' is Option 1: 8 Option 2: 16/5 Option 3: 14/5 Option 4: -1/8

Answer: 14/5. Solution: dy/dx = 2kx - 3. At x = 1/2: slope of tangent = k - 3. Slope of normal = -1/(k - 3) = 5. So k - 3 = -1/5, giving k = 3 - 1/5 = 14/5.

Calculus

Application of Derivatives

Easy

The slope of normal to the curve $y = kx^2 - 3x + 2$ at $x = \frac{1}{2}$ is 5. The value of 'k' is

$\frac{16}{5}$

$\frac{14}{5}$

$-\frac{1}{8}$

✅ Correct Option: 3

$\frac{dy}{dx} = 2kx - 3$ .

At $x = 1/2$ : slope of tangent $= k - 3$ .

Slope of normal $= -\frac{1}{k - 3} = 5$ .

So $k - 3 = -1/5$ , giving $k = 3 - 1/5 = 14/5$ .

2025: 22 May Shift 1

Match List-I with List-II

List-I	List-II
Function	Increasing on the interval
(A) $f(x) = -x^2 - 2x + 1$	(I) $(-\infty, -1)$
(B) $f(x) = x^2 + 1$	(II) $(1, \infty)$
(C) $f(x) = x^2 - 2x + 3$	(III) $(-\infty, 0)$
(D) $f(x) = -x^2$	(IV) $(0, \infty)$

Choose the correct answer from the options given below:

2023: 15 June Shift 2

Angle between tangents to the curve $y = x^2 - 5x + 6$ at the points (2, 0) and (3, 0) is :

2026: 18 May Shift 1

If the function $f(x) = x^2-2kx+7$ is increasing on [1, 5] then $k$ belongs to

2026: 12 May Shift 2

The function $f(x) = x^3 + 2x^2 - 1$ is

(A) increasing in $\left(\dfrac{-4}{3}, 0\right)$

(B) decreasing in $\left(\dfrac{-4}{3}, 0\right)$

(D) increasing in $(0, \infty)$

Choose the correct answer from the options given below:

2022: 4 Aug Shift 1

If x is real, then minimum value of $x^2 - 8x + 17$ is :

2022: 30 Aug Shift 1

Match List-I with List-II

List-I	List-II
(a) If $x = t^2$ and $y = t^3$ , then $\frac{d^2y}{dx^2}$ at $t = 1$	(i) $-2$
(b) If $f(x) = \sqrt{x} + 1$ , then $f''(1)$	(ii) $-1$
(c) The minimum value of $f(x) = 9x^2 + 12x + 2$ is	(iii) $\frac{3}{4}$
(d) The point of inflexion of the function $f(x) = (x-2)^4 (x+1)^3$ is	(iv) $-\frac{1}{4}$

Choose the correct answer from the options given below

The slope of normal to the curve y=kx2−3x+2y = kx^2 - 3x + 2y=kx2−3x+2 at x=12x = \frac{1}{2}x=21​ is 5. The value of 'k' is

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CUET Mathematics Past Year Questions (Topic-Wise)

The slope of normal to the curve $y = kx^2 - 3x + 2$ at $x = \frac{1}{2}$ is 5. The value of 'k' is