NIMCET 2023 — Mathematics PYQ
NIMCET | Mathematics | 2023Largest value of cos2θ−6sinθcosθ+3sin2θ+2
Choose the correct answer:
- A.
4
- B.
0
- C.
4+10
(Correct Answer) - D.
4−10
4+10
Explanation
\begin{align*}
&\cos^2 \theta - 6\sin \theta \cos \theta + 3\sin^2 \theta + 2 \\
&= 2\sin^2 \theta - 6\sin \theta \cos \theta + 3 \\
&= 3 - 3\sin 2\theta + \left[1 - \cos 2\theta \right] \\
&= 4 - \left[3\sin 2\theta + \cos 2\theta \right] \\
&-\sqrt{10} \leq 3\sin 2\theta + \cos 2\theta \leq \sqrt{10}
\end{align*}
For maximum value of given expression, 3sin2θ+cos2θ should be minimum.
Hence, maximum value is 4+10.
Explanation
\begin{align*}
&\cos^2 \theta - 6\sin \theta \cos \theta + 3\sin^2 \theta + 2 \\
&= 2\sin^2 \theta - 6\sin \theta \cos \theta + 3 \\
&= 3 - 3\sin 2\theta + \left[1 - \cos 2\theta \right] \\
&= 4 - \left[3\sin 2\theta + \cos 2\theta \right] \\
&-\sqrt{10} \leq 3\sin 2\theta + \cos 2\theta \leq \sqrt{10}
\end{align*}
For maximum value of given expression, 3sin2θ+cos2θ should be minimum.
Hence, maximum value is 4+10.