Question

Assertion :Let $f\left(\theta \right)=5\cos \theta +3\cos (\theta +\frac{\pi }{3}),then-4\le f\left(0\right)\le 10$ Reason: The maximum & minimum values of $a\cos x\pm b\sin x+c$ are $c-\sqrt{a^2+b^2}$ and $c+\sqrt{a^2+b^2}$

• 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 false but Reason is true

Answer 1

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

Let $P\left(x\right)=a\cos x\pm b\sin x,leta=r\cos \alpha ,b=r\sin \alpha$

$\therefore r=\sqrt{a^2+b^2},\tan \alpha =\frac{b}{a}$

$P\left(x\right)=r(\cos \alpha \cos x\pm \sin \alpha \sin x)=r\cos (x\pm \alpha )$

$\Rightarrow -r\le p\left(x\right)\le r\forall$ values of $x$

Now $f\left(\theta \right)=5\cos \theta +3\cos (\theta +\frac{\pi }{3})+3$

$=5\cos \theta +\frac{3}{2}\cos \theta -\frac{3\sqrt{3}}{2}\sin \theta +3$

$=\frac{13}{2}\cos \theta -\frac{3\sqrt{3}}{2}\sin \theta +3$

$\therefore 3-\sqrt{{\left(\frac{13}{2}\right)}^2+{\left(\frac{3\sqrt{3}}{2}\right)}^2}\le f\left(\theta \right)\le 3+\sqrt{{\left(\frac{13}{2}\right)}^2+{\left(\frac{3\sqrt{3}}{2}\right)}^2}$

$\Rightarrow -4\le f\left(\theta \right)\le 10$