JEE Main 2016 · 9 Apr · Shift 1 · Mathematics Q71

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MathematicsSingle correct+4 / -1Question 71 of 90

If the function $f(x)=\begin{cases}-x, & x<1\\ a+\cos^{-1}(x+b), & 1\le x\le2\end{cases}$ is differentiable at x = 1, then $\dfrac ab$ is equal to :

\dfrac{\pi-2}2

\dfrac{-\pi-2}2

\dfrac{\pi+2}2

-1-\cos^{-1}(2)

Answer and worked solution

Official answer: C

Differentiability needs the right derivative $-\frac1{\sqrt{1-(1+b)^2}}$ to equal −1, so $b=-1$ . Continuity gives $-1=a+\cos^{-1}0$ , so $a=-1-\frac\pi2$ . Then $\frac ab=\frac{\pi+2}2$ .

Topics:Differentiability

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JEE Main 2016 · 9 Apr · Shift 1, Mathematics question 71