

Class 12 Math Ch-5 अवकलन MCQs Exam 2027 Details: नीचे दिए गए सभी Questions Bihar Board परीक्षा 2027 के लिए “Very Very Important Multiple Choice Questions (MCQs) Objective” (अत्यंत महत्वपूर्ण प्रश्न) हैं। इन सभी Class 12th के (Mathematics/गणित) = गणित भाग-1 (Hindi Medium) Book Chapter-5 अवकलन का Questions का Solve का वीडियो Youtube और Website पर Upload किया है।

$\frac{d}{dx} (\frac{x^4}{4})$ (BSEB 2022 A)
(A) $4x^3$
(B) $\frac{x^3}{4}$
(C) $x^3$
(D) $16x^3$
यदि $y = x^{20}$, तो $\frac{d^2y}{dx^2} =$ (BSEB Previous Year)
(A) $x^{18}$
(B) $20x^{19}$
(C) $380x^{18}$
(D) $x^{10}$
यदि $y = x^5$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $5x$
(B) $6x$
(C) $5x^4$
(D) $5x^2$
$\frac{d}{dx} (x^4) =$ (BSEB Previous Year)
(A) $4x^3$
(B) $12x^2$
(C) $24x$
(D) $24$
यदि $y = x^3$, तो $\frac{d^2y}{dx^2} =$ (BSEB 2026)
(A) $3x^2$
(B) $6x$
(C) $6$
(D) $0$
$\frac{d}{dx} [ \frac{1}{3x-2} ] =$ (BSEB Previous Year)
(A) $\frac{-1}{(3x-2)^2}$
(B) $\frac{-3}{(3x-2)^2}$
(C) $\frac{3}{(3x-2)^2}$
(D) $\frac{3}{3x-2}$
$\frac{d}{dx} (\sqrt{x}) =$ (BSEB Previous Year)
(A) $2\sqrt{x}$
(B) $\frac{1}{2\sqrt{x}}$
(C) $\frac{\sqrt{x}}{2}$
(D) $\frac{1}{\sqrt{x}}$
यदि $y = x^5$, तो $\frac{d^2y}{dx^2} =$ (2026)
(A) $5x^4$
(B) $20x^3$
(C) $20x^4$
(D) $x^5$
$\frac{d}{dx}(\sqrt{x}) =$ (2026)
(A) $\frac{1}{2\sqrt{x}}$
(B) $\frac{1}{\sqrt{x}}$
(C) $2\sqrt{x}$
(D) $\frac{2}{3}x^{3/2}$
यदि $y = x^3$, तो $\frac{d^2y}{dx^2} =$ (2026)
(A) $3x^2$
(B) $6x$
(C) 6
(D) 0
$\frac{d^3}{dx^3}(x^4) =$ (2026)
(A) $4x^3$
(B) $12x^2$
(C) $24x$
(D) 24
$\frac{d}{dx}(\frac{1}{x+1}) =$ (2026)
(A) $log(x+1)$
(B) $-\frac{1}{(x+1)^2}$
(C) $\frac{1}{(x+1)^2}$
(D) $-\frac{1}{x+1}$
$\frac{d}{dx} [e^{x^2}] =$ (BSEB 2025 A)
(A) $e^{x^2}$
(B) $e^{2x}$
(C) $2xe^{x^2}$
(D) $2xe^{2x}$
$\frac{d}{dx} (5^x) =$ (BSEB Previous Year)
(A) $5^x$
(B) $x \cdot 5^{x-1}$
(C) $\frac{5^x}{\log 5}$
(D) $5^x \log_e 5$
$\frac{d}{dx} [\log x] =$ (BSEB 2019 A)
(A) $\frac{1}{x}$
(B) $-\frac{1}{x^2}$
(C) $1$
(D) $\frac{1}{x^2}$
$\frac{d}{dx} (\log 3^x) =$ (BSEB 2021 A)
(A) $\frac{1}{3^x}$
(B) $\log 3$
(C) $x \log 3$
(D) $1$
$\frac{d}{dx} (a^x) =$ (BSEB 2022, 2024 A)
(A) $a^x$
(B) $\frac{a^x}{\log a}$
(C) $a^x \log a$
(D) $x \log a$
$\frac{d}{dx} (\log \sqrt{x}) =$ (BSEB Previous Year)
(A) $\frac{1}{2\sqrt{x}}$
(B) $\frac{1}{\sqrt{x}}$
(C) $\frac{1}{2x}$
(D) $\frac{\sqrt{x}}{2}$
$\frac{d}{dx} (e^{-3x}) =$ (BSEB Previous Year)
(A) $\frac{e^{-3x}}{3}$
(B) $\frac{e^{-3x}}{-3}$
(C) $3e^{-3x}$
(D) $-3e^{-3x}$
$\frac{d}{dx} (\log 5x) =$ (BSEB 2025 A)
(A) $5x$
(B) $\frac{1}{x}$
(C) $\frac{5}{x}$
(D) $\log 5 + \frac{1}{x}$
यदि $y = \log x^x$, तो $\frac{dy}{dx} =$ (BSEB 2010)
(A) 1
(B) $\log x$
(C) $1 + \log x$
(D) $\log (ex)$
$\frac{d}{dx} (11^x) =$ (BSEB Previous Year)
(A) $x11^{x-1}$
(B) $11^x \log x$
(C) $11^x \log 11$
(D) $\frac{11^x}{\log 11}$
$\frac{d}{dx} [e^{\cos x}] =$ (BSEB 2020)
(A) $(\sin x)e^{\cos x}$
(B) $-(\sin x)e^{\cos x}$
(C) $(\cos x)e^{\cos x}$
(D) $-(\cos x)e^{\cos x}$
$\frac{d}{dx} (\log_{10} x) =$ (BSEB Previous Year)
(A) $\frac{1}{10x}$
(B) $\frac{1}{x}$
(C) $10x$
(D) $\frac{1}{x} \log_{10} e$
यदि $y = a^x$, तो $\frac{d^2y}{dx^2} =$ (BSEB Previous Year)
(A) $a^x \log a$
(B) $a^x (\log a)^2$
(C) $(a^x)^2 \log a$
(D) कोई नहीं
$\frac{d^2}{dx^2} (e^{5x}) =$ (BSEB Previous Year)
(A) $e^{5x}$
(B) $10e^{5x}$
(C) $5e^{5x}$
(D) $25e^{5x}$
$\frac{d}{dx} (e^{x^3}) =$ (BSEB 2021 A)
(A) $e^{x^3}$
(B) $3x^2 e^x$
(C) $3x^2 e^{x^2}$
(D) $3x^2 e^{x^3}$
$\frac{d}{dx} (\log x^n) =$ (BSEB Previous Year)
(A) $\frac{1}{x^n}$
(B) $n$
(C) $\frac{1}{x}$
(D) $\frac{n}{x}$
$\frac{d}{dx} (e^x) =$ (BSEB Previous Year)
(A) $e^x$
(B) $e^{-x}$
(C) $e^{2x}$
(D) $e^{3x}$
$\frac{d}{dx} (2e^{2x}) =$ (BSEB Previous Year)
(A) $2e^{2x}$
(B) $e^{2x}$
(C) $4e^{2x}$
(D) $2e^x$
$\frac{d}{dx} (e^{13x}) =$ (BSEB Previous Year)
(A) $e^{13x}$
(B) $\frac{1}{13} e^{13x}$
(C) $13e^{13x}$
(D) $-13e^{13x}$
$\frac{d}{dx} (\log |x|) =$ (BSEB Previous Year)
(A) $\frac{1}{|x|}$
(B) $\frac{1}{x}$
(C) $\frac{-1}{x}$
(D) अपरिभाषित
$\frac{d}{dx} (e^{3x}) =$ (BSEB Previous Year)
(A) $e^{3x}$
(B) $\frac{e^{3x}}{3}$
(C) $3e^{3x}$
(D) $3e^{2x}$
$\frac{d}{dx} [e^{x^3}] =$ (BSEB 2024)
(A) $3x^2 e^{x^3}$
(B) $e^{x^3}$
(C) $3x^2 e^x$
(D) $x^3 e^{x^3-1}$
$\frac{d}{dx} [2^x] =$ (BSEB 2022)
(A) $x \cdot 2^{x-1}$
(B) $\frac{2^x}{\log 2}$
(C) $2^x \log 2$
(D) $2^x$
$\frac{d}{dx} (\log \sqrt{x}) =$ (BSEB 2021)
(A) $\frac{1}{2\sqrt{x}}$
(B) $\frac{1}{\sqrt{x}}$
(C) $\frac{1}{2x}$
(D) $\frac{\sqrt{x}}{2}$
$\frac{d}{dx}(e^{3x}) =$ (2026)
(A) $e^{3x}$
(B) $3e^{3x}$
(C) $\frac{1}{3}e^{3x}$
(D) $3x e^{3x-1}$
$\frac{d}{dx} (2^x) =$ (2026)
(A) $x \cdot 2^{x-1}$
(B) $2^x \log 2$
(C) $\frac{2^x}{\log 2}$
(D) $2^x$
$\frac{d}{dx} (\log_a x) =$ (2026)
(A) $\frac{1}{x \log_e a}$
(B) $\frac{1}{x}$
(C) $\frac{\log_e a}{x}$
(D) $\frac{x}{\log_e a}$
$\frac{d}{dx}(e^{x-a}) =$ (2026)
(A) $e^{x-a}$
(B) $e^x$
(C) $e^{-x}$
(D) $(x-a)e^{x-a-1}$
$\frac{d}{dx}(e^{ax}) =$ (2026)
(A) $e^{ax}$
(B) $ae^{ax}$
(C) $a^2 e^{ax}$
(D) $a$
$\frac{d}{dx}(2^x) =$ (2026)
(A) $2^x$
(B) $x \cdot 2^{x-1}$
(C) $\frac{2^x}{log 2}$
(D) $2^x log 2$
$\frac{d}{dx} [2\sin^2 \theta + 2\cos^2 \theta]$ (BSEB Previous Year)
(A) $4\sin \theta$
(B) $4\cos \theta$
(C) 0
(D) 4
$\frac{d}{dx} [\tan ax]$ (BSEB 2021)
(A) $a\tan ax$
(B) $a\sec^2 ax$
(C) $a\sec x$
(D) $a\cot ax$
$\frac{d}{dx} (\sin^2 x)$ (BSEB 2019 C, 2022 A)
(A) $\sin 2x$
(B) $\cos 2x$
(C) $\tan 2x$
(D) $\cot 2x$
यदि $y = \tan^2 x$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $\sec^2 x$
(B) $\sec^4 x$
(C) $2 \tan x \sec x$
(D) $2 \tan x \sec^2 x$
$\frac{d}{dx} (\sqrt{\cot x})$ (BSEB Previous Year)
(A) $\frac{1}{2\sqrt{\cot x}}$
(B) $\sqrt{\csc^2 x}$
(C) $\frac{-\csc^2 x}{2\sqrt{\cot x}}$
(D) $\frac{\csc^2 x}{2\sqrt{\cot x}}$
$\frac{d}{dx} (\tan \frac{x}{3})$ (BSEB 2022 A)
(A) $\sec^2 \frac{x}{3}$
(B) $\frac{1}{3} \sec^2 \frac{x}{3}$
(C) $3 \sec^2 \frac{x}{3}$
(D) $3 \cot \frac{x}{3}$
$\frac{d}{dx} [\csc x]$ (BSEB 2020)
(A) $\csc x \cot x$
(B) $-\csc x \cot x$
(C) $\csc^2 x$
(D) $-\csc^2 x$
$\frac{d}{dx} (\sin \frac{4x}{5})$ (BSEB Previous Year)
(A) $\frac{4}{5} \cos \frac{4x}{5}$
(B) $-\frac{4}{5} \cos \frac{4x}{5}$
(C) $\frac{5}{4} \cos \frac{4x}{5}$
(D) $-\frac{5}{4} \cos \frac{4x}{5}$
$\frac{d}{dx} (\sin 2x)$ (BSEB Previous Year)
(A) $\cos 2x$
(B) $\frac{\cos 2x}{2}$
(C) $2 \sin 2x$
(D) $2 \cos 2x$
$\frac{d}{dx} (\sin \sqrt{x})$ (BSEB 2021 A)
(A) $\cos \sqrt{x}$
(B) $\frac{\cos \sqrt{x}}{\sqrt{x}}$
(C) $\frac{1}{\sqrt{x}} \cos \sqrt{x}$
(D) $\frac{1}{2\sqrt{x}} \cos \sqrt{x}$
$\frac{d}{dx} (\tan x)$ (BSEB 2019 A)
(A) $\sec^2 x$
(B) $\sec x$
(C) $\cot x$
(D) $-\sec^2 x$
$\frac{d}{dx} [\cos(\pi x + \sin \pi)]$ (BSEB 2025 A)
(A) $-\sin(\pi x + \sin \pi)$
(B) $-\pi \sin(\pi x)$
(C) $-\sin \pi x$
(D) $\sin x$
यदि $y = \sin x^2$, तो $\frac{dy}{dx} =$ (BSEB 2017 C)
(A) $2x \sin x^2$
(B) $x \sin x$
(C) $x \cos x^2$
(D) $2x \cos x^2$
$\frac{d^2}{dx^2} (\sin 2x) =$ (BSEB Previous Year)
(A) $4 \sin 2x$
(B) $4 \cos^2 2x$
(C) $-4 \sin 2x$
(D) $2 \sin 4x$
$\frac{d}{dx} (\tan x^2) =$ (BSEB Previous Year)
(A) $\sec x^2$
(B) $2x \sec^2 x^2$
(C) $2x^2 \sec^2 x^2$
(D) $\frac{\sec x^2}{2x}$
$\frac{d}{dx} (2\cos \frac{3x}{4}) =$ (BSEB Previous Year)
(A) $-2\sin \frac{3x}{4}$
(B) $-\frac{3}{8} \sin \frac{3x}{4}$
(C) $-\frac{3}{2} \sin \frac{3x}{4}$
(D) $-\frac{3}{4} \sin \frac{3x}{4}$
$\frac{d}{dx} (\sin x + \sin^2 x) =$ (BSEB Previous Year)
(A) $\cos x + \sin 2x$
(B) $\cos x + \cos 2x$
(C) $\cos x + \sin 2x$
(D) $\cos x – \sin 2x$
$\frac{d}{dx} (\tan kx) =$ (BSEB Previous Year)
(A) $\sec^2 kx$
(B) $k \sec^2 x$
(C) $\frac{\sec^2 kx}{k}$
(D) $k \sec^2 kx$
$\frac{d}{dx} (\sin x)$ (BSEB 2022 A)
(A) $\cos x$
(B) $-\sin x$
(C) $-\cos x$
(D) $\tan x$
$\frac{d}{dx} [ \frac{1}{4} \sec 4x ] =$ (BSEB Previous Year)
(A) $\sec 4x \tan 4x$
(B) $\sec^2 4x$
(C) $\tan^2 4x$
(D) $\frac{1}{16} \sec 4x \tan 4x$
$\frac{d}{dx} (\sec x) =$ (BSEB Previous Year)
(A) $\sec^2 x$
(B) $\tan^2 x$
(C) $\sec x \tan x$
(D) $0$
$\frac{d}{dx} (\cos 2x) =$ (BSEB Previous Year)
(A) $-\sin 2x$
(B) $2\sin 2x$
(C) $-2\sin 2x$
(D) $-\frac{1}{2} \sin 2x$
यदि $y = \sin^2 x$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $2\sin x$
(B) $\cos^2 x$
(C) $2\sin x \cos x$
(D) $\sin x \cos x$
$\frac{d}{dx} (\cot x) =$ (BSEB Previous Year)
(A) $\tan x$
(B) $\csc^2 x$
(C) $-\csc^2 x$
(D) $\csc x \cot x$
$\frac{d}{dx} (\cos x) =$ (BSEB Previous Year)
(A) $\cos x$
(B) $\sin x$
(C) $-\cos x$
(D) $-\sin x$
$\frac{d}{dx} (\sin \frac{x}{5}) =$ (BSEB Previous Year)
(A) $\cos \frac{x}{5}$
(B) $\frac{1}{5} \cos \frac{x}{5}$
(C) $5 \cos \frac{x}{5}$
(D) $-\frac{1}{5} \sin \frac{x}{5}$
$\frac{d}{d\theta} (\cos^3 \theta) =$ (BSEB Previous Year)
(A) $-3\sin^2 \theta$
(B) $3\sin^2 \theta \cos \theta$
(C) $-3\cos^2 \theta \sin \theta$
(D) $3\cos^2 \theta \sin \theta$
$\frac{d}{dx} (\sqrt{\tan x}) =$ (BSEB Previous Year)
(A) $2\sqrt{\tan x}$
(B) $\frac{\sec^2 x}{2\sqrt{\tan x}}$
(C) $2\tan x \sec x$
(D) $\frac{\sec x}{2\sqrt{\tan x}}$
$\frac{d}{dx} [\sin 4x] =$ (BSEB 2020)
(A) $4 \sin 4x$
(B) $4 \cos 4x$
(C) $4x \sin 4x$
(D) $4x \cos 4x$
$\frac{d}{dx} [\sqrt{\sin x}] =$ (BSEB 2021)
(A) $\frac{\cos x}{2\sqrt{\sin x}}$
(B) $\frac{1}{2\sqrt{\sin x}}$
(C) $\frac{\sin x}{2\sqrt{\cos x}}$
(D) $\cos \sqrt{x}$
यदि $y = \sin(x^3)$, तो $\frac{dy}{dx} =$ (BSEB 2015)
(A) $x^3 \cos(x^3)$
(B) $3x^2 \sin(x^3)$
(C) $3x^2 \cos(x^3)$
(D) $\cos(x^3)$
$\frac{d}{dx} [3 \sin x – 4 \sin^3 x] =$ (BSEB Previous Year)
(A) $3 \cos 3x$
(B) $\cos 3x$
(C) $3 \sin 3x$
(D) $-3 \cos 3x$
$\frac{d}{dx} [\sec x] =$ (BSEB 2020)
(A) $\sec x \cot x$
(B) $\sec x \tan x$
(C) $\tan x$
(D) $\sec^2 x$
$\frac{d}{dx} [\tan x] =$ (BSEB 2019 A)
(A) $\sec^2 x$
(B) $\sec x$
(C) $\cot x$
(D) $-\sec^2 x$
$\frac{d}{dx} (\sin^2 x) =$ (BSEB 2019 C)
(A) $\sin 2x$
(B) $\cos 2x$
(C) $\tan 2x$
(D) $\cot 2x$
$\frac{d}{dx} (\tan \frac{x}{3}) =$ (BSEB 2022 A)
(A) $\sec^2 \frac{x}{3}$
(B) $\frac{1}{3} \sec^2 \frac{x}{3}$
(C) $3 \sec^2 \frac{x}{3}$
(D) $3 \cot \frac{x}{3}$
$\frac{d}{dx} (\sin \sqrt{x}) =$ (BSEB 2021 A)
(A) $\cos \sqrt{x}$
(B) $\frac{\cos \sqrt{x}}{\sqrt{x}}$
(C) $\frac{1}{\sqrt{x}} \cos \sqrt{x}$
(D) $\frac{1}{2\sqrt{x}} \cos \sqrt{x}$
$\frac{d}{dx}(\cos x^3) =$ (2026)
(A) $-3x^2 \sin x^3$
(B) $\sin x^3$
(C) $3x^2 \sin x^3$
(D) $3x^2$
$\frac{d}{dx}(\tan 5x) =$ (2026)
(A) $5 \sec^2 5x$
(B) $\sec^2 5x$
(C) $5 \sec 5x$
(D) $-5 \sec^2 5x$
$\frac{d}{dx}(\tan x^2) =$ (2026)
(A) $2x \sec^2 x^2$
(B) $\sec^2 x^2$
(C) $2x \tan x^2$
(D) $2 \sec^2 x^2$
$\frac{d}{dx}(\sqrt{\sin x}) =$ (2026)
(A) $\frac{\cos x}{2\sqrt{\sin x}}$
(B) $\frac{\sin x}{2\sqrt{\cos x}}$
(C) $\frac{\cos x}{\sin x}$
(D) $\frac{1}{2\sqrt{sin x}}$
$\frac{d}{dx}(\sin 2x + \cos 2x) =$ (2026)
(A) $2 \cos 2x – 2 \sin 2x$
(B) $2 \cos 2x + 2 \sin 2x$
(C) $\cos 2x – \sin 2x$
(D) 0
$\frac{d^2}{dx^2}(\sin 2x) =$ (2026)
(A) $4 \sin 2x$
(B) $-4 \sin 2x$
(C) $2 \cos 2x$
(D) $-2 \cos 2x$
$\frac{d}{dx}(\sin 2x) =$ (2026)
(A) $\cos 2x$
(B) $2 \cos 2x$
(C) $-2 \cos 2x$
(D) $\frac{cos 2x}{2}$
$\frac{d}{dx}(\tan kx) =$ (2026)
(A) $k \sec^2 kx$
(B) $\sec^2 kx$
(C) $\frac{sec^2 kx}{k}$
(D) $\tan k$
यदि $y = \log \cos x^2$, तो $x = \sqrt{\pi}$ पर $\frac{d}{dx}$ का मान है: (BSEB 2018 A)
(A) $1$
(B) $\frac{\pi}{4}$
(C) $0$
(D) $\sqrt{\pi}$
यदि $y = \sin(\log x)$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $\frac{1}{x} \cos(\log x)$
(B) $\frac{1}{x} \sin(\log x)$
(C) $0$
(D) $1$
$\frac{d}{dx} [\log (\tan x)] =$ (BSEB Previous Year)
(A) $\sec^2 x$
(B) $\frac{1}{\sin x \cos x}$
(C) $2\csc 2x$
(D) (B) और (C) दोनों
$\frac{d}{dx} (e^x + \cos 5x) =$ (BSEB Previous Year)
(A) $e^x + \cos 5x$
(B) $e^x + 5\sin 5x$
(C) $e^x – 5\sin 5x$
(D) $e^x – 5\cos 5x$
$\frac{d}{dx} [\log(\sec x + \tan x)] =$ (BSEB 2018 A, 2024 A)
(A) $\sec x + \tan x$
(B) $\sec x$
(C) $\tan x$
(D) $\sec x + \tan x$
$\frac{d}{dx} (\sqrt{x^2 + ax + 1}) =$ (BSEB Previous Year)
(A) $\frac{x + a}{2\sqrt{x^2 + ax + 1}}$
(B) $\frac{2x + a}{2\sqrt{x^2 + ax + 1}}$
(C) $\frac{2x + a}{\sqrt{x^2 + ax + 1}}$
(D) $\frac{1}{2\sqrt{x^2 + ax + 1}}$
$\frac{d}{dx} (\frac{1+e^x}{\sin x}) =$ (BSEB 2021 A)
(A) $-\frac{1+e^x}{\sin^2 x}$
(B) $\csc x + e^x$
(C) $-\csc x \cot x + e^x$
(D) $\csc x \cot x + e^x$
$\frac{d}{dx} (x^3 + e^x) =$ (BSEB Previous Year)
(A) $3x^2 + e^x$
(B) $3x^2 + 3e^x$
(C) $3x^2 + e^x$
(D) $3x^2 + e^x$
$\frac{d}{dx} (\log \cos x) =$ (BSEB Previous Year)
(A) $\tan x$
(B) $-\tan x$
(C) $\cot x$
(D) $-\cot x$
$\frac{d}{dx} (\sin 2x + e^x – \cos x) =$ (BSEB Previous Year)
(A) $\cos 2x + e^x – \sin x$
(B) $2\cos 2x + e^x + \sin x$
(C) $2\cos 2x + e^x – \sin x$
(D) $-2\cos 2x + e^x + \sin x$
$\frac{d}{dx} [x^2 \cdot e^x] =$ (BSEB 2023)
(A) $e^x(x^2 + 2x)$
(B) $e^x(x^2 – 2x)$
(C) $2xe^x$
(D) $x^2e^x$
यदि $y = \cos(\log x)$, तो $\frac{dy}{dx} =$ (BSEB 2016)
(A) $-\sin(\log x)$
(B) $\frac{-\sin(\log x)}{x}$
(C) $\frac{\cos(\log x)}{x}$
(D) $-\sin(\log x) \cdot \log x$
$\frac{d}{dx} [e^x \sin x] =$ (BSEB Previous Year)
(A) $e^x (\sin x + \cos x)$
(B) $e^x (\sin x – \cos x)$
(C) $e^x \cos x$
(D) $e^x \sin x$
$\frac{d}{dx} [x^2 \sin \frac{1}{x}] =$ (BSEB Previous Year)
(A) $2x \sin \frac{1}{x} – \cos \frac{1}{x}$
(B) $2x \sin \frac{1}{x} + \cos \frac{1}{x}$
(C) $x \sin \frac{1}{x} + \cos \frac{1}{x}$
(D) $2x \cos \frac{1}{x}$
$\frac{d}{dx} (\log \cos x) =$ (BSEB Previous Year)
(A) $\tan x$
(B) $-\tan x$
(C) $\cot x$
(D) $-\cot x$
$\frac{d}{dx} [ \sqrt{x^2+ax+1} ] =$ (BSEB 2023)
(A) $\frac{2x+a}{2\sqrt{x^2+ax+1}}$
(B) $\frac{x+a}{2\sqrt{x^2+ax+1}}$
(C) $\frac{2x+a}{\sqrt{x^2+ax+1}}$
(D) $\frac{1}{2\sqrt{x^2+ax+1}}$
यदि $y = \log(\sin x)$, तो $\frac{dy}{dx} =$ (2026)
(A) $\cot x$
(B) $\tan x$
(C) $\frac{1}{\sin x}$
(D) $\cos x$
यदि $y = \sin(\log x)$, तो $\frac{dy}{dx} =$ (2026)
(A) $\frac{\cos(\log x)}{x}$
(B) $\cos(\log x)$
(C) $\frac{\sin(\log x)}{x}$
(D) $-\frac{\cos(\log x)}{x}$
यदि $y = \cos( \sin x)$, तो $\frac{dy}{dx} =$ (2026)
(A) $-\sin(\sin x) \cdot \cos x$
(B) $\sin(\sin x) \cdot \cos x$
(C) $-\sin(\sin x)$
(D) $\cos(\cos x)$
$\frac{d}{dx}(\log \cos x) =$ (2026)
(A) $\tan x$
(B) $-\tan x$
(C) $\cot x$
(D) $-\cot x$
$\frac{d}{dx}(e^{\cot x}) =$ (2026)
(A) $-\csc^2 x \cdot e^{\cot x}$
(B) $\sin x \cdot e^{\cot x}$
(C) $\cos x \cdot e^{\cot x}$
(D) $e^{\cot x}$
यदि $y = \cos(\log x)$, तो $\frac{dy}{dx} =$ (2026)
(A) $-sin(log x)$
(B) $-\frac{sin(log x)}{x}$
(C) $\frac{sin(log x)}{x}$
(D) $-\frac{cos(log x)}{x}$
$\frac{d}{dx} [\sin^{-1} (3x – 4x^3)] =$ (BSEB 2021 A)
(A) $\frac{1}{\sqrt{1-x^2}}$
(B) $\frac{1}{\sqrt{1-(3x-4x^3)^2}}$
(C) $\frac{3}{\sqrt{1-x^2}}$
(D) $\frac{2}{\sqrt{1-x^2}}$
यदि $y = \tan^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}$, तो $\frac{dy}{dx} =$ (BSEB 2018 C)
(A) $\frac{1}{2}$
(B) $-\frac{1}{2}$
(C) $\frac{1}{1 + \tan^2 \frac{x}{2}}$
(D) $\frac{1}{1+x^2}$
$\frac{d}{dx} (\sec^{-1} x + \csc^{-1} x) =$ (BSEB 2024 A)
(A) $\frac{1}{\sqrt{1-x^2}}$
(B) $\frac{-1}{\sqrt{1-x^2}}$
(C) $\frac{1}{\sqrt{1-x^2}}$
(D) $0$
$\frac{d}{dx} (\cot^{-1} x) =$ (BSEB 2020 A)
(A) $\frac{1}{1+x^2}$
(B) $\frac{1}{1-x^2}$
(C) $-\frac{1}{1+x}$
(D) $-\frac{1}{1+x^2}$
यदि $y = \tan^{-1} (\frac{\sin x + \cos x}{\cos x – \sin x})$, तो $\frac{dy}{dx} =$ (BSEB 2022)
(A) $0$
(B) $1$
(C) $1/2$
(D) $x$
$\frac{d}{dx} (2\tan^{-1} x) =$ (BSEB 2021 A)
(A) $\frac{1}{1+x^2}$
(B) $\frac{2}{1+x^2}$
(C) $\frac{1}{2(1+x^2)}$
(D) $\frac{1}{2(1-x^2)}$
यदि $y = \tan^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $-\frac{1}{2}$
(B) $0$
(C) $\frac{1}{2}$
(D) $1$
$\frac{d}{dx} (\sin^{-1} x) =$ (BSEB Previous Year)
(A) $\frac{1}{1+x^2}$
(B) $\frac{1}{1-x^2}$
(C) $\frac{1}{\sqrt{1-x^2}}$
(D) $\sqrt{\frac{1}{1+x^2}}$
$\frac{d}{dx} (\sec^{-1} x) =$ (BSEB Previous Year)
(A) $\frac{1}{\sqrt{1-x^2}}$
(B) $\frac{1}{x\sqrt{x^2-1}}$
(C) $\frac{1}{1+x^2}$
(D) $\frac{1}{x\sqrt{x^2-1}}$
यदि $y = \tan^{-1} \sqrt{x}$, तो $\frac{dy}{dx} =$ (BSEB 2017 C)
(A) $\frac{1}{2\sqrt{x}(1-x)}$
(B) $\frac{1}{x(1+x)}$
(C) $\frac{1}{x^2(1+x)}$
(D) $\frac{1}{2\sqrt{x}(1+x)}$
$\frac{d}{dx} (\sin^{-1} \sqrt{x} + \cos^{-1} \sqrt{x}) =$ (BSEB Previous Year)
(A) $\pi/2$
(B) $0$
(C) $1$
(D) $\sqrt{x} \pi/2$
$\frac{d}{dx} (\tan^{-1} x + \cot^{-1} x) =$ (BSEB 2016 A)
(A) $\frac{2}{1+x^2}$
(B) $0$
(C) $1$
(D) $2$
$\frac{d}{dx} (\cos^{-1} x) =$ (BSEB 2016 A)
(A) $\frac{1}{2\sqrt{1-x^2}}$
(B) $\frac{1}{\sqrt{1-x^2}}$
(C) $\frac{-1}{\sqrt{1-x^2}}$
(D) $\frac{-1}{2\sqrt{1-x^2}}$
$\frac{d}{dx} [\tan^{-1} (\frac{1-x}{1+x})] =$ (BSEB 2017)
(A) $\frac{1}{1+x^2}$
(B) $\frac{-1}{1+x^2}$
(C) $\frac{1}{1-x^2}$
(D) $\frac{-1}{1-x^2}$
$\frac{d}{dx} [\cos^{-1}(\sin x)] =$ (BSEB Previous Year)
(A) $-1$
(B) $\frac{x}{\sqrt{1-x^2}}$
(C) $\frac{\sin x}{\sqrt{1-x^2}}$
(D) $\frac{\pi-x}{2}$
यदि $y = \tan^{-1} (\frac{1+x}{1-x})$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $\frac{2}{1+x^2}$
(B) $\frac{1}{1+x^2}$
(C) $\frac{1}{1-x^2}$
(D) कोई नहीं
यदि $y = \sec(\tan^{-1} x)$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $\frac{xy}{\sqrt{1+x^2}}$
(B) $\frac{-x}{\sqrt{1+x^2}}$
(C) $\frac{x}{\sqrt{1-x^2}}$
(D) कोई नहीं
$\frac{d}{dx} [\tan^{-1} (\frac{\cos x + \sin x}{\cos x – \sin x})] =$ (BSEB Previous Year)
(A) 0
(B) 1
(C) $1/2$
(D) $-1$
$\frac{d}{dx} [\sin^{-1} x] =$ (BSEB 2015, 2019)
(A) $\frac{1}{\sqrt{1-x^2}}$
(B) $\frac{-1}{\sqrt{1-x^2}}$
(C) $\frac{1}{1-x^2}$
(D) $\frac{1}{1+x^2}$
$\frac{d}{dx} [\tan^{-1} \sqrt{1+x^2} – \cot^{-1} (-\sqrt{1+x^2})] =$ (BSEB 2018 A)
(A) $\pi$
(B) $1$
(C) $0$
(D) $\frac{2x}{\sqrt{1+x^2}}$
$\frac{d}{dx} (\sin^{-1} x + \cos^{-1} x) =$ (BSEB 2021)
(A) $0$
(B) $1$
(C) $\pi/2$
(D) $\frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx} \sin^{-1}(3x – 4x^3) =$ (2026)
(A) $\frac{3}{\sqrt{1-x^2}}$
(B) $\frac{-3}{\sqrt{1-x^2}}$
(C) $\frac{1}{\sqrt{1-x^2}}$
(D) $\frac{-1}{\sqrt{1-x^2}}$
यदि $y = \tan^{-1} x$, तो $\frac{dy}{dx} =$ (2026)
(A) $\frac{1}{1+x^2}$
(B) $\frac{1}{1-x^2}$
(C) $\frac{-1}{1+x^2}$
(D) $\frac{1}{\sqrt{1-x^2}}$
$x$ के सापेक्ष $\cos^{-1} x$ का अवकलज है: (2026)
(A) $\frac{1}{\sqrt{1-x^2}}$
(B) $\frac{-1}{\sqrt{1-x^2}}$
(C) $\frac{1}{1+x^2}$
(D) $\frac{-1}{1+x^2}$
$\frac{d}{dx}(\sin^{-1} x + \cos^{-1} x) =$ (2026)
(A) $\frac{2}{\sqrt{1-x^2}}$
(B) 0
(C) $\pi/2$
(D) 1
$\frac{d}{dx}(\tan^{-1} \sqrt{x}) =$ (2026)
(A) $\frac{1}{2\sqrt{x}(1+x)}$
(B) $\frac{1}{1+x}$
(C) $\frac{\sqrt{x}}{1+x}$
(D) $\frac{2}{\sqrt{x}(1+x)}$
यदि $x^m y^n = (x+y)^{m+n}$, तो $\frac{dy}{dx} =$ (BSEB 2018 C)
(A) $\frac{x}{y}$
(B) $\frac{y}{x}$
(C) $\frac{-y}{x}$
(D) $\frac{-x}{y}$
यदि $x = a \sec \theta, y = b \tan \theta$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $\frac{b}{a} \sec \theta$
(B) $\frac{b}{a} \csc \theta$
(C) $\frac{b}{a} \cot \theta$
(D) $\frac{b}{a}$
यदि $y = e^{x + e^{x + e^{x + \dots \infty}}}$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $\frac{y}{y+1}$
(B) $\frac{y}{y-1}$
(C) $\frac{y}{1-y}$
(D) कोई नहीं
यदि $x = a \cos \theta, y = a \sin \theta$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $\tan \theta$
(B) $-\cot \theta$
(C) $-\tan \theta$
(D) $\sec^2 \theta$
यदि $x = \sin \theta, y = \cos \theta$, तो $\frac{dy}{dx} =$ (BSEB 2019 C)
(A) $\tan \theta$
(B) $-\tan \theta$
(C) $\cot \theta$
(D) इनमें से कोई नहीं
यदि $y + x = \sin(y+x)$, तो $\frac{dy}{dx} =$ (BSEB 2018 C)
(A) $\frac{1-\cos(y+x)}{1+\cos(y+x)}$
(B) $1$
(C) $-1$
(D) $0$
यदि $y = \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}$, तो $\frac{dy}{dx} =$ (BSEB 2024 A)
(A) $\frac{\cos x}{2y + 1}$
(B) $\frac{\cos x}{2y – 1}$
(C) $\frac{\sin x}{2y + 1}$
(D) $\frac{\sin x}{2y – 1}$
यदि $x^2y^3 = (x+y)^5$, तो $\frac{dy}{dx} =$ (BSEB 2018 A)
(A) $\frac{x}{y}$
(B) $\frac{y}{x}$
(C) $-\frac{y}{x}$
(D) $-\frac{x}{y}$
यदि $x = a \cos^2 \theta, y = b \sin^2 \theta$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $\frac{b}{a}$
(B) $-\frac{b}{a}$
(C) $\frac{b}{a} \sin 2\theta$
(D) $-\frac{b}{a} \tan^2 \theta$
यदि $x = a \cos \theta, y = b \cos \theta$, तो $\frac{dy}{dx} =$ (BSEB 2017)
(A) $b/a$
(B) $-b/a$
(C) $a/b$
(D) $-a/b$
$\frac{d}{dx} (x^x) =$ (BSEB Previous Year)
(A) $x \cdot x^{x-1}$
(B) $x^x (1 + \log x)$
(C) $x^x \log x$
(D) कोई नहीं
यदि $x = \sin t, y = \cos t$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $-\tan t$
(B) $-\cot t$
(C) $\tan t$
(D) $\cot t$
यदि $x = t + \frac{1}{t}, y = t – \frac{1}{t}$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $\frac{2t}{t^2+1}$
(B) $\frac{t^2+1}{t^2-1}$
(C) $\frac{t^2-1}{t^2+1}$
(D) $\frac{2t}{1-t^2}$
यदि $x\sqrt{1+y} + y\sqrt{1+x} = 0$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $\frac{x+1}{x}$
(B) $\frac{1}{1+x}$
(C) $\frac{-1}{(1+x)^2}$
(D) कोई नहीं
यदि $y = \sqrt{\sin x + y}$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $\frac{\cos x}{2y – 1}$
(B) $\frac{\cos x}{1 – 2y}$
(C) $\frac{\sin x}{1 – 2y}$
(D) $\frac{\sin x}{2y – 1}$
यदि $y = x \tan y$, तो $\frac{dy}{dx} =$ (BSEB Previous Year)
(A) $\frac{\tan y}{x – x^2 – y^2}$
(B) $\frac{y^2}{x – x^2 – y^2}$
(C) $\frac{\tan y}{y – x}$
(D) $\frac{\tan x}{x – y}$
यदि $y = \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}$, तो $\frac{dy}{dx} =$ (2026)
(A) $\frac{1}{2y-1}$
(B) $\frac{\cos x}{2y-1}$
(C) $\frac{\sin x}{2y-1}$
(D) $\frac{2y-1}{\cos x}$
यदि $x^n + y^n = a^n$ तो $\frac{dy}{dx} =$ (2026)
(A) $-\frac{x^{n-1}}{y^{n-1}}$
(B) $\frac{x^{n-1}}{y^{n-1}}$
(C) $-\frac{y^{n-1}}{x^{n-1}}$
(D) $nx^{n-1}$
यदि $x = a(1 – \cos \theta)$, $y = a(\theta + \sin \theta)$, तो $\frac{dy}{dx} =$ (2026)
(A) $\tan \frac{\theta}{2}$
(B) $-\tan \frac{\theta}{2}$
(C) $\cot \frac{\theta}{2}$
(D) $-\cot \frac{\theta}{2}$
यदि $y = x^x$ तो $\frac{dy}{dx} =$ (2026)
(A) $x^x(\log x + 1)$
(B) $\log x$
(C) $(\log x + 1)$
(D) $nx^{n-1}$
यदि $x = a \cos \theta, y = a \sin \theta$, तो $\frac{dy}{dx} =$ (2026)
(A) $tan \theta$
(B) $cot \theta$
(C) $-cot \theta$
(D) $-tan \theta$
यदि $x = at^2, y = 2at$, तो $\frac{dy}{dx} =$ (2026)
(A) $t$
(B) $\frac{1}{t}$
(C) $-t$
(D) $-\frac{1}{t}$
$\frac{d}{dx} [\lim_{x \to 0} \cos 3x]$ (BSEB Previous Year)
(A) $-\sin 3x$
(B) $1$
(C) $-\sin 3x$
(D) $0$
$\frac{d}{dx} [\lim_{x \to 1} \frac{x^n – 1}{x – 1}] =$ (BSEB 2018 C)
(A) $n$
(B) $nx^{n-1}$
(C) $1$
(D) $0$
$\frac{d^2}{dx^2} [x^2 + 3x + 2] =$ (BSEB Previous Year)
(A) $4$
(B) $4x$
(C) $2x+3$
(D) $2$
$\frac{d}{dx} [\lim_{x \to a} \frac{x^n – a^n}{x – a}] =$ (BSEB 2021 A)
(A) $na^{n-1}$
(B) $1$
(C) $0$
(D) $n$
यदि $y = \log(\sin x)$, तो $\frac{dy}{dx} =$ (2026)
(A) $\cot x$
(B) $\tan x$
(C) $\frac{1}{\sin x}$
(D) $\cos x$
यदि $y = \sin(\log x)$, तो $\frac{dy}{dx} =$ (2026)
(A) $\frac{\cos(\log x)}{x}$
(B) $\cos(\log x)$
(C) $\frac{\sin(\log x)}{x}$
(D) $-\frac{\cos(\log x)}{x}$
$\frac{d}{dx}(\sin^{-1} x + \cos^{-1} x) =$ (2026)
(A) $\frac{2}{\sqrt{1-x^2}}$
(B) 0
(C) $\pi/2$
(D) 1
| Q.No. | Ans | Q.No. | Ans | Q.No. | Ans | Q.No. | Ans |
| 1 | (C) | 42 | (D) | 83 | (A) | 124 | (A) |
| 2 | (C) | 43 | (C) | 84 | (A) | 125 | (B) |
| 3 | (C) | 44 | (B) | 85 | (B) | 126 | (A) |
| 4 | (A) | 45 | (A) | 86 | (B) | 127 | (B) |
| 5 | (B) | 46 | (D) | 87 | (A) | 128 | (A) |
| 6 | (B) | 47 | (C) | 88 | (C) | 129 | (C) |
| 7 | (B) | 48 | (B) | 89 | (A) | 130 | (A) |
| 8 | (B) | 49 | (B) | 90 | (D) | 131 | (A) |
| 9 | (A) | 50 | (A) | 91 | (C) | 132 | (A) |
| 10 | (B) | 51 | (D) | 92 | (B) | 133 | (B) |
| 11 | (C) | 52 | (D) | 93 | (B) | 134 | (B) |
| 12 | (B) | 53 | (A) | 94 | (C) | 135 | (A) |
| 13 | (C) | 54 | (B) | 95 | (A) | 136 | (B) |
| 14 | (D) | 55 | (D) | 96 | (B) | 137 | (B) |
| 15 | (A) | 56 | (C) | 97 | (B) | 138 | (B) |
| 16 | (B) | 57 | (B) | 98 | (A) | 139 | (B) |
| 17 | (C) | 58 | (C) | 99 | (B) | 140 | (B) |
| 18 | (C) | 59 | (A) | 100 | (A) | 141 | (C) |
| 19 | (D) | 60 | (D) | 101 | (A) | 142 | (B) |
| 20 | (B) | 61 | (A) | 102 | (B) | 143 | (B) |
| 21 | (C) | 62 | (A) | 103 | (A) | 144 | (B) |
| 22 | (C) | 63 | (C) | 104 | (A) | 145 | (A) |
| 23 | (B) | 64 | (C) | 105 | (A) | 146 | (B) |
| 24 | (D) | 65 | (C) | 106 | (A) | 147 | (A) |
| 25 | (B) | 66 | (C) | 107 | (B) | 148 | (C) |
| 26 | (D) | 67 | (D) | 108 | (A) | 149 | (C) |
| 27 | (D) | 68 | (B) | 109 | (B) | 150 | (A) |
| 28 | (D) | 69 | (C) | 110 | (C) | 151 | (B) |
| 29 | (A) | 70 | (B) | 111 | (A) | 152 | (B) |
| 30 | (C) | 71 | (B) | 112 | (D) | 153 | (A) |
| 31 | (C) | 72 | (A) | 113 | (D) | 154 | (C) |
| 32 | (B) | 73 | (C) | 114 | (B) | 155 | (A) |
| 33 | (C) | 74 | (A) | 115 | (B) | 156 | (C) |
| 34 | (A) | 75 | (B) | 116 | (A) | 157 | (B) |
| 35 | (C) | 76 | (A) | 117 | (C) | 158 | (D) |
| 36 | (C) | 77 | (A) | 118 | (D) | 159 | (D) |
| 37 | (B) | 78 | (B) | 119 | (D) | 160 | (D) |
| 38 | (B) | 79 | (D) | 120 | (B) | 161 | (C) |
| 39 | (A) | 80 | (A) | 121 | (B) | 162 | (A) |
| 40 | (A) | 81 | (A) | 122 | (C) | 163 | (A) |
| 41 | (B) | 82 | (A) | 123 | (B) | 164 | (B) |
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