QJEEM23Ma76ac: JEE Main 2023 Mathematics - Class 11 JEEM23Ma76ac Solution: The set of all values of for which the equation has a real…

Question

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The set of all values of $\lambda$ for which the equation ${\cos ^2}2x - 2{\sin ^4}x - 2{\cos ^2}x = \lambda$ has a real solution $x$ , is :

Accepted Answer

The given equation is

\cos ^2 2x - 2 \sin ^4 x - 2 \cos ^2 x = \lambda

Using the trigonometric identities

\cos^2x = 1 - \sin^2x

and

\cos^22x = 1 - 2\sin^2x

, we can rewrite the equation in terms of

\cos^2x

:

\lambda = (2 \cos ^2 x - 1)^2 - 2(1 - \cos ^2 x)^2 - 2 \cos ^2 x

Simplify this equation :

\lambda = 4 \cos ^4 x - 4 \cos ^2 x + 1 - 2(1 - 2 \cos ^2 x + \cos ^4 x) - 2 \cos ^2 x

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

We can factor out a 2 and rewrite this as :

\lambda = 2(\cos ^4 x - \cos ^2 x - \frac{1}{2})

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

So

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

(maximum value) and

\lambda_{\min }=2\left[0-\frac{3}{4}\right]=-\frac{3}{2}(

minimum value

)

So range of the value of

x

is

\left[\frac{-3}{2},-1\right]

.

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