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STD 11 - Trigonometrical equations — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - Trigonometrical equations4 Marks
MCQ
The set of all values of $\lambda$ for which the equation $\cos ^2 2 x-2 \sin ^4 x-2 \cos ^2 x=\lambda$
  • A
    $[-2,-1]$
  • B
    $\left[-2,-\frac{3}{2}\right]$
  • C
    $\left[-1,-\frac{1}{2}\right]$
  • ✓
    $\left[-\frac{3}{2},-1\right]$
✓

Answer

Correct option: D.
$\left[-\frac{3}{2},-1\right]$
d
$\lambda=\cos ^2 2 x-2 \sin ^4 x-2 \cos ^2 x$

$\text { convert all in to } \cos x \text {. }$

$\lambda=\left(2 \cos ^2 x-1\right)^2-2\left(1-\cos ^2 x\right)^2-2 \cos ^2 x$

$=4 \cos ^4 x-4 \cos ^2 x+1-2\left(1-2 \cos ^2 x+\cos ^4 x\right)-$

$2 \cos ^2 x$

$=2 \cos ^4 x-2 \cos ^2 x+1-2$

$=2 \cos ^4 x-2 \cos ^2 x-1$

$=2\left[\cos ^4 x-\cos ^2 x-\frac{1}{2}\right]$

$=2\left[\left(\cos ^2 x-\frac{1}{2}\right)^2-\frac{3}{4}\right]$

$\left.\lambda_{\max }=2\left[\frac{1}{4}-\frac{3}{4}\right]=2 \times\left(-\frac{2}{4}\right)=-1 \text { (max Value }\right)$

$\left.\lambda_{\min }=2\left[0-\frac{3}{4}\right]=-\frac{3}{2} \text { (MinimumValue }\right)$

$\text { So, Range }=\left[-\frac{3}{2},-1\right]$

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