Question

Assertion: In a $△ABC,$ is ${\cos }^2\frac{A}{2}+{\cos }^2\frac{B}{2}+{\cos }^2\frac{C}{2}=y(x^2+\frac{1}{x^2}),$ then the maximum value of $y$ is $\frac{9}{8}$

Reason: In a $△ABC,\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}\le \frac{1}{8}$

• 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. None of these

Answer 1

Answer: (D)

None of these

${\cos }^2\frac{A}{2}+{\cos }^2\frac{B}{2}+{\cos }^2\frac{C}{2}=\frac{1+\cos A}{2}+\frac{1+\cos B}{2}+\frac{1+\cos C}{2}$

$=\frac{3}{2}+\frac{1}{2}(\cos A+\cos B+\cos C)=\frac{3}{2}+\frac{1}{2}[1+4\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}]$

$=2+2\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}$ ...(1)

Also $0<\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}\le \frac{1}{8}$

$\therefore {\cos }^2\frac{A}{2}+{\cos }^2\frac{B}{2}+{\cos }^2\frac{C}{2}\le 2+2.\frac{1}{8}=\frac{9}{8}$

$\Rightarrow y(x^2+\frac{1}{x^2})\le \frac{9}{8}\Rightarrow 0<y\le \left(\frac{9/8}{x^2+\frac{1}{x^2}}\right)\le \frac{9}{16}$

$[\because x^2+\frac{1}{x^2}\ge 2\Rightarrow 0<\frac{1}{(x^2+\frac{1}{x^2})}\le \frac{1}{2}]$

$\therefore$ Assertion is incorrect, and reason is correct.