Calculus

Application of Derivatives

Medium

applied

The equation of the tangent line to the curve $y = x^2 - 2x + 5$ which is parallel to the line $4x - y + 1 = 0$ is

$4x - y - 1 = 0$

$4x - y - 4 = 0$

$x + 4y - 4 = 0$

$x + 4y - 1 = 0$

✅ Correct Option: 2

The line $4x - y + 1 = 0$ can be rewritten as:

$y = 4x + 1$

The slope of this line $= 4$ .

Since the tangent line must be parallel, the required tangent also has slope $= 4$ .

The curve is:

$y = x^2 - 2x + 5$

$\frac{dy}{dx} = 2x - 2$

Setting the derivative equal to $4$ :

$2x - 2 = 4$

$2x = 6$

$x = 3$

Substituting $x = 3$ back into the curve:

$y = (3)^2 - 2(3) + 5$

$y = 9 - 6 + 5$

$y = 8$

The point of tangency is $(3,\ 8)$ .

Using point-slope form with slope $= 4$ and point $(3, 8)$ :

$y - 8 = 4(x - 3)$

$y - 8 = 4x - 12$

$y = 4x - 4$

$4x - y - 4 = 0$

Therefore, the equation of the tangent line is $4x - y - 4 = 0$ .

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The function $f(x) = \sin x + \cos x$ ; $0 \leq x \leq 2\pi$ is

(A) Increasing in $\left(0, \frac{\pi}{4}\right)$

(B) Decreasing in $(0, \pi)$

(D) Decreasing in $\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$

(E) Neither increasing Nor decreasing in $\left(0, \frac{\pi}{4}\right)$

Choose the correct answer from the options given below:

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Match List-I with List-II

List-I	List-II
(A) Maximum value of $f(x) = \sin^2 x - \cos^2 x$ , $\forall x \in (\pi, 2\pi)$ is	(I) 0
(B) Minimum value of $f(x) = \sin x \cos x$	(II) 1
(C) Point of Minima of $f(x) = x^x$ $(x > 0)$	(III) $-\frac{1}{2}$
(D) Maximum value of $f(x) = -x^{2026}$	(IV) $\frac{1}{e}$

Choose the correct answer from the options given below:

The equation of the tangent line to the curve y=x2−2x+5y = x^2 - 2x + 5y=x2−2x+5 which is parallel to the line 4x−y+1=04x - y + 1 = 04x−y+1=0 is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The equation of the tangent line to the curve $y = x^2 - 2x + 5$ which is parallel to the line $4x - y + 1 = 0$ is