IIT-JEE 2011 · Paper 2 · Mathematics Q49

MathematicsMultiple correct+4Question 49 of 60

If
$f(x)=\begin{cases}-x-\dfrac\pi2, & x\le-\dfrac\pi2\\[4pt]-\cos x, & -\dfrac\pi2<x\le0\\[4pt]x-1, & 0<x\le1\\[2pt]\ln x, & x>1,\end{cases}$
then

f(x)

is continuous at

x=-\dfrac\pi2

f(x)

is not differentiable at

x=0

f(x)

is differentiable at

x=1

f(x)

is differentiable at

x=-\dfrac32

Answer and worked solution

Official answer: A, B, C, D

(A) At $-\frac\pi2$ : left value $0$ , right limit $-\cos(-\frac\pi2)=0$ . Continuous ✓
(B) At 0: left derivative $\sin0=0$ , right derivative 1. Not differentiable ✓
(C) At 1: $x-1$ and $\ln x$ both equal 0 there, with derivatives $1$ and $1/x=1$ . Differentiable ✓
(D) $-\frac32>-\frac\pi2$ , so $x=-\frac32$ lies inside the $-\cos x$ piece. Differentiable ✓

(A), (B), (C), (D)

Topics:Differentiability Continuity

IIT-JEE 2011 · Paper 2, Mathematics question 49