Skip to main content
Back to DIFFERENTIATION

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium
1 of 1
Analyzing +4 | -1
f:(0,∞)→(0,∞)f:(0, \infty) \rightarrow(0, \infty) is a twice differentiable function and satisfies the functional relation f(x+1)=xf(x)f(x+1)=x f(x) for all positive xx. If g(x)=log⁡(f(x))g(x)=\log (f(x)), then for any positive integer nn
g′′(n+12)−g′′(12)=g^{\prime \prime}\left(n+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=
View answer and explanation
Correct option: A
−4(1+19+125+⋯+1(2n−1)2)-4\left(1+\frac{1}{9}+\frac{1}{25}+\cdots+\frac{1}{(2 n-1)^2}\right)
Option A is correct because by differentiating the given functional relation twice and substituting x=1,2,...,nx = 1, 2, ..., n, the expression simplifies to −4(1+19+125+⋯+1(2n−1)2)-4\left(1+\frac{1}{9}+\frac{1}{25}+\cdots+\frac{1}{(2 n-1)^2}\right).

More Questions from DIFFERENTIATION

Practice All Chapter MCQs →
If f(x)=∣x∣∣tan⁡x∣f(x)=|x|^{|\tan x|}, then f′(−π6)f^{\prime}\left(\frac{-\pi}{6}\right) is equal to Let xx and yy be two real numbers such that x>0x>0 and xy=1x y=1. The minimum value of x+yx+y is Let f(x)=[x]sin⁡(πx)f(x)=[x] \sin (\pi x) where [x][x] denotes the integer part of xx. Then at x=k,kx=k, k being an integer, the left derivative of f(x)f(x) is If y=sec⁡−1[cosecx]+cosec−1[sec⁡x]+sin⁡−1[cos⁡x]+cos⁡−1[sin⁡x],y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x], then dydx\frac{{dy}}{{dx}} is equal to ddxsin⁡−1(3x−4x3)\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right) is equal to If z=y+f(v),z = y + f\left( v \right), where v=(xy),v = \left( {\frac{x}{y}} \right), then v∂z∂x+∂z∂yv\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}} is
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →