Question

Assertion :The equation $7\cos x+5\sin x=2k+1$ has a solution for $8$ integral values of $k.$ Reason: $-\sqrt{2}\le \sin x+\cos x\le \sqrt{2}$

• 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 false but Reason is true

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Given equation $7\cos x+5\sin x=2k+1$
Let $r\cos \alpha =7,r\sin \alpha =5$
Then $r=\sqrt{74},\tan \alpha =\frac{5}{7},n\in I$

$7\cos x+5\sin x=2k+1$
$\Rightarrow r\cos (x-\alpha )=2k+1$
$\Rightarrow \sqrt{74}\cos (x-\alpha )=2k+1$
Since, $-1\le \cos (x-\alpha )\le 1$
$-\sqrt{74}\le 2k+1\le \sqrt{74}$
or $-8\le 2k+1\le 8or-9\le 2k\le 7$
$\frac{-9}{2}\le k\le \frac{7}{2}$
$\therefore k=-4,-3,-2,-1,0,1,2,3$
$\therefore$ Assertion is true as number of values of $k$ are 8
Now, $\sin x+\cos x=\sqrt{2}\cos (x-\frac{\pi }{4})$
$-1\le \cos (x-\frac{\pi }{4})\le 1$
$-\sqrt{2}\le \sqrt{2}\cos (x-\frac{\pi }{4})\le \sqrt{2}$
$\therefore -\sqrt{2}\le \sin x+\cos x\le \sqrt{2}$
$\therefore$ Reason (R) is true.
But reason is not the correct explanation for assertion