Question

Let $f\left(x\right)=x^2+\lambda x+\mu \cos x,\lambda$ is $+ve$ integer $\mu$ is a real number. The number of ordered pairs $(\lambda ,\mu )$ for which $f\left(x\right)=0$ and $f\left(f\right(x\left)\right)=0$ have same set of real roots.

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

$f\left(x\right)=x^2+\lambda x+\mu cosx;\lambda$ the integer

$\mu$ real number.

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

$x^2+\lambda x+\mu cosx=0...\left(i\right)$

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

$\left(f\right(x){)}^2+\lambda f(x)+\mu cos(f\left(x\right))=0...(ii)$

For $f\left(x\right)=0$ & $f\left(f\right(x\left)\right)=0$ have same root

then use $f\left(x\right)=0$ in eq (ii)

$\left(f\right(x){)}^2+\lambda f(x)+\mu cos(f\left(x\right))=0$

$0+0+\mu cos0=0$

$\mu \times 1=0$

$\mu =0$

From eq (i)

$x^2+\lambda x=0$

$x(x+\lambda )=0;x\ne 0$ that's why

$x=-\lambda$ for fixed value of $\mu$

The Number of ordered pair of $(\lambda ,\mu )=1$