Question

Assertion :If $x$ and $y$ are real numbers such that $x^2+y^2=27$, then the maximum possible value of $x-y$ is $3\sqrt{6}$. Reason: $-1\le \cos (\theta +\pi /4)\le 1$.

• 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. Assertion is incorrect but Reason is correct

Answer 1

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

$x^2+y^2=27$
$x^2+y^2=(3\sqrt{3}{)}^2$
$\Rightarrow x=3\sqrt{3}\cos \theta ,y=3\sqrt{3}\sin \theta$
$x-y=3\sqrt{3}(\cos \theta -\sin \theta )$
$\Rightarrow x-y=3\sqrt{6}\cos (\theta +\frac{\pi }{4})$
Now, $-1\le \cos (\theta +\frac{\pi }{4})\le 1$
$\Rightarrow -3\sqrt{6}\le 3\sqrt{6}\cos (\theta +\frac{\pi }{6})\le 3\sqrt{6}$
Hence, maximum possible value is $3\sqrt{6}$
Hence, assertion and reason both are correct . Reason is correct explanation for assertion