Question

If the equation $si{n}^{-1}(x^2+x+1)+co{s}^{-1}(\lambda x+1)=\frac{\pi }{2}$ has exactly two solutions, then integral value of $\lambda$ is equal to

• A. 1
• B. -1
• C. 2

Answer 1

According to equation
$x^2+x+1=dx+1x^2+x(1-\lambda )=0x=0orx=\lambda -1$
Now,
$-1\le x^2+x+1\le +1$
$-1\le x^2+x+1$ $\to$ always true
$x^2+x+1\le 1x^2+x+1\le 1x^2+x\le 0x(x+1)\le 0xϵ[-10]-1\le x\le 0-1\le \lambda -1<00\le \lambda <1$
Hence integral value of $\lambda =0$