Question

The value of $a=-\lambda -\frac{\pi }{\mu }$, such that $x^2+ax+si{n}^{-1}(x^2-4x+5)+co{s}^{-1}(x^2-4x+5)=0$ has atleast one solution. Then $\lambda +\mu$ is equal to

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

The correct option is C

6

Since $si{n}^{-1}y$ is defined for $-1\le y\le 1$
$-1\le x^2-4x+5\le 1$
$\Rightarrow x^2-4x+4\le 0$ and $x^2-4x+6\ge 0$
$\Rightarrow (x-2{)}^2\le 0$ and $(x-2{)}^2+2\ge 0$
$\Rightarrow x=2$ is the only solution
On putting x = 2 in the original equation we get
$4+2a+\frac{\pi }{2}=0\Rightarrow 2a=-4-\frac{\pi }{2}$
$\Rightarrow a=-2-\frac{\pi }{4}\Rightarrow \lambda =2$ and $\mu =4$
Hence $\lambda +\mu =2+4=6$