If y = [2/(1+x^(b-a)+x^(c-a)) + 2/(1+x^(c-b)+x^(a-b)) + 2/(1+x^(a-c)+x^(b-c))]^x, then dy/dx is equal to : Option 1: 0 Option 2: 2^x_e 2 Option 3: _e(x^a+x^b+x^c) Option 4: 1

Answer: 2^x_e 2. Solution: Put u = x^a, v = x^b, w = x^c. The three terms become 2u/(u+v+w), 2v/(u+v+w) and 2w/(u+v+w), which sum to 2. So y = 2^x and dy/dx = 2^x_e 2.

Calculus

Continuity & Differentiability

Hard

core

If $y = \left[\dfrac{2}{1+x^{b-a}+x^{c-a}} + \dfrac{2}{1+x^{c-b}+x^{a-b}} + \dfrac{2}{1+x^{a-c}+x^{b-c}}\right]^x$ , then $\dfrac{dy}{dx}$ is equal to :

$2^x\log_e 2$

$\log_e(x^a+x^b+x^c)$

✅ Correct Option: 2

Put $u = x^a$ , $v = x^b$ , $w = x^c$ . The three terms become $\frac{2u}{u+v+w}$ , $\frac{2v}{u+v+w}$ and $\frac{2w}{u+v+w}$ , which sum to $2$ . So $y = 2^x$ and $\frac{dy}{dx} = 2^x\log_e 2$ .

2025: 13 May Shift 1

If $f(x) = x^3\log_e x$ , Then $f''(e^2)$ is equal to

2025: 13 May Shift 1

Match List-I with List-II

List-I	List-II
(Function)	(Derivative with respect to 'x')
(A) $f(x) = x^x$	(I) $ax^{a-1}$
(B) $f(x) = a^x$	(II) 0
(C) $f(x) = a^a$	(III) $a^x log_e a$
(D) $f(x) = x^a$	(IV) $x^x(1 + log_e x)$

Choose the correct answer from the options given below:

2025: 29 May Shift 2

If $x = \frac{a}{1 + t}$ and $y = \frac{a}{(1 + t)^2}$ where $a > 0$ , then $\frac{d^2y}{dx^2}$ at $t = 1$ is

2026: 21 May Shift 2

If $y = ax^{n+1} + bx^{-n}$ , then $x^2 \dfrac{d^2y}{dx^2}$ is equal to

2024: 16 May Shift 1

The second order derivative of which of the following functions is $5^{\mathrm{x}}$ ?

2025: 3 June Shift 2

If $x^m y^n = (x + y)^{m+n}$ , then $\frac{d^2y}{dx^2}$ is equal to:

If y=[21+xb−a+xc−a+21+xc−b+xa−b+21+xa−c+xb−c]xy = \left[\dfrac{2}{1+x^{b-a}+x^{c-a}} + \dfrac{2}{1+x^{c-b}+x^{a-b}} + \dfrac{2}{1+x^{a-c}+x^{b-c}}\right]^xy=[1+xb−a+xc−a2​+1+xc−b+xa−b2​+1+xa−c+xb−c2​]x, then dydx\dfrac{dy}{dx}dxdy​ is equal to :

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

If $y = \left[\dfrac{2}{1+x^{b-a}+x^{c-a}} + \dfrac{2}{1+x^{c-b}+x^{a-b}} + \dfrac{2}{1+x^{a-c}+x^{b-c}}\right]^x$ , then $\dfrac{dy}{dx}$ is equal to :