Question

Assertion :The range of $f\left(x\right)=\sin x-\cos x$ is $[-\sqrt{2},\sqrt{2}]$. Reason: The range of $g\left(x\right)=a\sin x+b\cos x$ is $[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}]$ and $f\left(x\right)=\sin x-\cos x$ is the particular case of $g\left(x\right)$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

The range of $a\sin x+b\cos x$ is

$-\sqrt{a^2+b^2}\le a\sin x+b\cos x\le \sqrt{a^2+b^2}$

Putting $a=1,b=-1$

$-\sqrt{2}\le \sin x-\cos x\le \sqrt{2}$

$\Rightarrow$ Assertion is true, Reason is true & Reason (R) is the correct explanation of Assertion (A).