JEE Main 2016 — Mathematics Calculus

f(x)= {\begin{matrix}-x & x<1 \\ a+{\mathrm{cos}}^{-1}⁡(x+b) & 1 \leq x \leq 2\end{matrix}

$f(x)$f(x) is continuous at $x=1$x=1 as $f(x)$f(x) is differentiable at $x=1$x=1.

$$\Rightarrow \underset{ x\rightarrow 1 }{ \mathrm{lim} } f( x )=\underset{ x\rightarrow {1}^{+} }{ \mathrm{lim} } ( a+{ \mathrm{cos} }^{ -1 } ( x+b ) )=f( 1 )$$⇒x→1lim​f(x)=x→1+lim​(a+cos−1(x+b))=f(1)

$\Rightarrow$⇒ $-1=a+{\mathrm{cos}}^{-1}⁡(1+b)$−1=a+cos−1⁡(1+b)

${\mathrm{cos}}^{-1}⁡(1+b)=-1-a$cos−1⁡(1+b)=−1−a ........$(i)$(i)

$f(x)$f(x) is differentiable at $x=1$x=1.

$\Rightarrow$⇒ $\mathrm{LHD}=\mathrm{RHD}$LHD=RHD

$\Rightarrow$⇒ $-1=\frac{-1}{\sqrt{1-{(1+b)}^{2}}}$−1=1−(1+b)2​−1​

$\Rightarrow$⇒ $1-{(1+b)}^{2}=1$1−(1+b)2=1

$\Rightarrow$⇒ $b= -1$b=−1 ......$(\mathrm{ii})$(ii)

From $(i)$(i) $\Rightarrow$⇒ ${\mathrm{cos}}^{-1}⁡(0)= -1-a$cos−1⁡(0)=−1−a

$\therefore$∴ $-1-a=\frac{\pi }{2}$−1−a=2π​

$$a= -1-\frac{\pi }{2}$$a=−1−2π​

$a=\frac{-\pi -2}{2}$a=2−π−2​ ......$(\mathrm{iii})$(iii)

$\therefore$∴ $\frac{a}{b}=\frac{\pi +2}{2}$ba​=2π+2​