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The ratio of the greatest value of $2 - \cos x + \sin^{2} x,$ is to its least value, is

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\frac{7}{4}

\frac{11}{4}

\frac{13}{4}

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detailed solution

Correct option is C

Let $y = 2 - \cos x + \sin^{2} x .$ Then,

$\begin{array}{l} y = 2 - \cos x + 1 - \cos^{2} x \\ \Rightarrow y = 3 - (\cos^{2} x + \cos x) \Rightarrow y = \frac{13}{4} - {(\cos x + \frac{1}{2})}^{2} \end{array}$

Now,

$- 1 \leq \cos x \leq 1$ for all $x$

$\Rightarrow - \frac{1}{2} \leq (\cos x + \frac{1}{2}) \leq \frac{3}{2}$ for all $x$

$\Rightarrow 0 \leq {(\cos x + \frac{1}{2})}^{2} \leq \frac{9}{4}$ for all $x$

$\Rightarrow - \frac{9}{4} \leq - {(\cos x + \frac{1}{2})}^{2} \leq 0$ for all $x$

$\Rightarrow \frac{13}{4} - \frac{9}{4} \leq \frac{13}{4} - {(\cos x + \frac{1}{2})}^{2} \leq \frac{13}{4}$ for all $x$

$\Rightarrow 1 \leq y \leq \frac{13}{4}$

$∴ y_{max} = \frac{13}{4}$ and $y_{min} = 1$

Hence, required ratio is $\frac{13}{4}$

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The ratio of the greatest value of 2−cos⁡x+sin2⁡x, is to its least value, is