If y = (_e x)/x, then d^2y/dx^2 = Option 1: (2_e x - 3)/x^3 Option 2: (2_e x + 3)/x^3 Option 3: (3 - 2_e x)/x^3 Option 4: (3_e x - 3)/x^3

Answer: (2_e x - 3)/x^3. Solution: Using quotient rule: dy/dx = ((1/x)x - x)/x^2 = (1 - x)/x^2 d^2y/dx^2 = ((-1/x)x^2 - (1- x)(2x))/x^4 = (-x - 2x + 2x x)/x^4 = (2 x - 3)/x^3.

Calculus

Continuity & Differentiability

Medium

If $y = \frac{\log_e x}{x}$ , then $\frac{d^2y}{dx^2} =$

$\frac{2\log_e x - 3}{x^3}$

$\frac{2\log_e x + 3}{x^3}$

$\frac{3 - 2\log_e x}{x^3}$

$\frac{3\log_e x - 3}{x^3}$

✅ Correct Option: 1

Using quotient rule:

$\frac{dy}{dx} = \frac{(1/x)x - \log x}{x^2} = \frac{1 - \log x}{x^2}$

$\frac{d^2y}{dx^2} = \frac{(-1/x)x^2 - (1-\log x)(2x)}{x^4} = \frac{-x - 2x + 2x\log x}{x^4} = \frac{2\log x - 3}{x^3}$ .

2022: 16 July Shift 2

Match List I with List II

List I	List II
A. If $f(x) = 2x$ and $g(x) = \frac{x^2}{2} + 1$ , then $\frac{g(x)}{f(x)}$ is	I. discontinuous at exactly three points.
B. The function $f(x) = \frac{4-x^2}{4x-x^3}$ is	II. continuous everywhere
C. The function $f(x) = \lvert x \rvert + \lvert x-1 \rvert$ is	III. discontinuous at $x = 0$ .
D. The function $f(x) = \lvert \sin x \rvert$ is	IV. continuous at $x = 0$ and $x = 1$

Choose the correct answer from the options given below:

2026: 14 May Shift 2

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2026: 22 May Shift 1

Let a function $f(x)=\alpha+(\beta^2+5\beta+6)|x|+\gamma|x|^4$ , where $\alpha, \beta$ and $\gamma$ are real number. Then the function $f(x)$ is differentiable at $x=0$ if

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If $y = \cot^{-1}\left(\frac{1-x}{1+x}\right)$ then $\frac{dy}{dx}$ equals

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2026: 12 May Shift 2

If $y=7^x\log(5x^3)$ , then $\dfrac{dy}{dx}$ is equal to: (consider $\log_e x = \log x$ )

If y=log⁡exxy = \frac{\log_e x}{x}y=xloge​x​, then d2ydx2=\frac{d^2y}{dx^2} =dx2d2y​=

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