Question

Let $\{\begin{array}{cc}5-cx;& x\le -1\\ b-2x;& -1<x<0\\ 2;& x=0\\ ln\left[\right(\cos x+a\sin bx{)}^{\frac{1}{x}}];& 0<x<\frac{\pi }{2}\end{array}$ be a continuous function in its domain and if $co{s}^{-1}\left(cos\right(a+c+3b\left)\right)=m\pi +n$ (where m and n are integers), then the value of (m - n) is equal to

• A. -5
• B. 8
• C. 13
• D. 23

Answer 1

The correct option is B 8

at $x=-1$ for continuous $LHS=RHS$

$\frac{x\le -1}{f\left(x\right)=5}-c(-1)=5+c$

$f\left(x\right)=b-2(-1)=b+2$

$5+c=b+2$

$x=0$

$f\left(x\right)=6-0$

$f\left(x\right)=2$

$b=2$

$5+c=4\Rightarrow c=-1$

$f\left(x\right)=In(\cos x+a\sin bx{)}^{1/x}$

${lim}_{x\to 0}\frac{1}{x}In(\cos x+a\sin bx)$

${lim}_{x\to 0}\frac{\frac{-\sin x+2a\cos 2x}{(\cos x+a\sin bx)}}{2}=\frac{2a}{1}=2a$

$2a=2\Rightarrow a=1$

${\cos }^{-1}(\cos 1-1+6)=m\pi +n$

${\cos }^{-1}\cos 6=m\pi +n$

$={\cos }^{-1}\cos (2\pi -6)$

$m=2,n=-6$

$m-n=2+6=8$