Skip to main content
Back to DIFFERENTIATION

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy
1 of 1
Analyzing +4 | -1
If f(x)=cos⁡−1(4x3−3x),x∈[−1,1]f(x)=\cos ^{-1}\left(4 x^3-3 x\right), x \in[-1,1] then
View answer and explanation
Correct option: D
f′(x)={31−x2,      if∣x∣<12−31−x2,        if−1<x<−12,12<x<1f'(x) = \left\{ {\begin{array}{llllllllllllllllllll} {\frac{3}{{\sqrt {1 - {x^2}} }},\mathrm{ }\,\,\,\,\,\,\mathrm{if }|x| < \frac{1}{2}}\\ {\frac{{ - 3}}{{\sqrt {1 - {x^2}} }},\mathrm{ }\,\,\,\,\,\,\,\,\mathrm{if } - 1 < x < \frac{{ - 1}}{2},\frac{1}{2} < x < 1} \end{array}} \right.
Option D is correct because the derivative of arccos(4x3-3x) is piecewise, being 3/sqrt(1-x2) for |x|<1/2 and -3/sqrt(1-x2) for 1/2<|x|<1 due to the properties of arccos.

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 →