Question

Assertion :Let $f:X\to Y$ be a function defined by $f\left(x\right)=2\sin (x+\frac{\pi }{4})-\sqrt{2}\cos x+c$

Statement I: For set $X,x\in [0,\frac{\pi }{2}]\cup [\pi ,\frac{3\pi }{2}],f\left(x\right)$ is one-one function

Reason: Statement II: $f^{\prime }\left(x\right)\ge 0,x\in [0,\frac{\pi }{2}]$

• 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 D Statement I is false, Statement II is true

Statement I is false as
$f\left(x\right)=\sqrt{2}\sin x+\sqrt{2}\cos x-\sqrt{2}\cos x+c$
$=\sqrt{2}\sin x+c$
$\Rightarrow f\left(0\right)=f\left(\pi \right)=c$
Hence, many-one function.

Statement II is true as
$f\left(x\right)=\sqrt{2}\sin x+c\Rightarrow f^{\prime }\left(x\right)=\sqrt{2}\cos x$

Hence
$f^{\prime }\left(x\right)\ge 0\Rightarrow x\in [0,\frac{\pi }{2}]$