Question

Let $f\left(x\right)=\sqrt{2-x-x^2}$ and $g\left(x\right)=cosx$. Check given statement is true or not?

I. Domain of $f\left(\right(g\left(x\right){)}^2)$ = Domain of $f\left(g\right(x\left)\right)$

Answer 1

$f\left(x\right)=\sqrt{2-x-x^2}$

$g\left(x\right)=\left(\cos x\right)$

Now, $y_1=f{\left(g\left(x\right)\right)}^2=f\left(co{s}^{2}x\right)$

$y_1=\sqrt{2-{\cos }^{2}x-{\cos }^{4}x}$

For $y_1$ to exist,

$2-{\cos }^{2}x-{\cos }^{4}x\ge 0$

$\Rightarrow (x-0)(x-\pi )\le 0$

$\Rightarrow x\in [0,\pi ]$

Now, $y_2=f\left(g\left(x\right)\right)$

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

$=\sqrt{2-\cos x-{\cos }^2}$

For existance of $y_2,$ $2-\cos x-{\cos }^{2}x\ge 0$

$\Rightarrow {\cos }^{2}x+\cos x-2\le 0$

$\Rightarrow (\cos x-1)\le 0$

$\Rightarrow x\in R$ (as $\cos x\in [-1,1]\forall x\in R$)

So, $D(y_1\ne D(y_2)$

This solution is false.