Question

Assertion :

$f$ is a function defined on the interval $[-1,1]$ such that $f\left(\sin 2x\right)=\sin x+\cos x$

Statement I: If $x\in [-\frac{\pi }{4},\frac{\pi }{4}]$, then $f\left({\tan }^{2}x\right)=\sec x$ Reason: Statement II: $f\left(x\right)=\sqrt{1+x},\forall x\in [-1,1]$

• A. Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I
• B. Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I
• C. Statement I is true, Statement II is false
• D. Statement I is false, Statement II is true

Answer 1

The correct option is A Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I

$\left(f\right(\sin 2x){)}^2=(\sin x+\cos x{)}^2={\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x$
$=1+\sin 2x$

$f\left(sin2x\right)=\sqrt{1+sin2x}$
Put $\sin 2x=x,weget$

$\Rightarrow f\left(x\right)=\sqrt{1+x},\forall x\in [-1,1]$

Put $x={\tan }^{2}x$

$\Rightarrow$ If $x\in [-\frac{\pi }{4},\frac{\pi }{4}]$, then

$f\left({\tan }^{2}x\right)=\sqrt{1+{\tan }^{2}x}=\sec x$

$\therefore$ Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I