Q54 / 160 · WBJEE 2009 · Mathematics

If $\mathrm{P}=\frac{1}{2} \sin ^2 \theta+\frac{1}{3} \cos ^2 \theta$ then

  • Correct answer
    A
    $\frac{1}{3} \leq \mathrm{P} \leq \frac{1}{2}$
  • B
    $\mathrm{P} \geq \frac{1}{2}$
  • C
    $\quad 2 \leq \mathrm{P} \leq 3$
  • D
    $-\frac{\sqrt{13}}{6} \leq \mathrm{P} \leq \frac{\sqrt{13}}{6}$

Solution

Hints: $P=\frac{1}{2} \sin ^2 \theta+\frac{1}{3} \cos ^2 \theta=\frac{1}{2} \sin ^2 \theta+\frac{1}{3}\left(1-\sin ^2 \theta\right)=\frac{1}{3}+\frac{1}{6} \sin ^2 \theta$ $$ \begin{aligned} & 0 \leq \sin ^2 \theta \leq 1 \Rightarrow \frac{1}{3} \leq \frac{1}{3}+\frac{1}{6} \sin ^2 \theta \leq \frac{1}{3}+\frac{1}{6} \\ & \Rightarrow \frac{1}{3} \leq \mathrm{P} \leq \frac{1}{2} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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