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Application of Derivatives Part 2

Section: Mathematics

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Question : 85 of 100

Marks: +1, -0

The maximum value of

f(x)=\begin{vmatrix} \sin^{2} x & 1+\cos^{2} x & \cos 2 x \\ 1+\sin^{2} x & \cos^{2} x & \cos 2 x \\ \sin^{2} x & \cos^{2} x & \sin 2 x \end{vmatrix}, x \in R

[16 Mar 2021 Shift 2]

$\sqrt{7}$
$\; \frac{3}{4}$
$\sqrt{5}$
5

Solution:

Given,

f(x)=\begin{vmatrix} \sin^{2} x & 1+\cos^{2} x & \cos 2 x \\ 1+\sin^{2} x & \cos^{2} x & \cos 2 x \\ \sin^{2} x & \cos^{2} x & \sin 2 x \end{vmatrix}

On applying

C_{1} \rightarrow C_{1}+C_{2}

, we get

f(x)=\begin{vmatrix} \sin^{2} x+1+\cos^{2} x & 1+\cos^{2} x & \cos 2 x \\ 1+\sin^{2} x+\cos^{2} x & \cos^{2} x & \cos 2 x \\ \sin^{2} x+\cos^{2} x & \cos^{2} x & \sin 2 x \end{vmatrix}

f(x)=\begin{vmatrix} 2 & 1+\cos^{2} x & \cos 2 x \\ 2 & \cos^{2} x & \cos 2 x \\ 1 & \cos^{2} x & \sin 2 x \end{vmatrix}

On applying

R_{1} \rightarrow R_{1}-R_{2}

f(x)=\begin{vmatrix} 0 & 1 & 0 \\ 2 & \cos^{2} x & \cos 2 x \\ 1 & \cos^{2} x & \sin 2 x \end{vmatrix}

f(x)=-1(2 \sin 2 x-\cos 2 x)

As, we know that, if

f(\theta)=A \sin \theta + B \cos \theta

Then,

-\sqrt{A^{2}+B^{2}} \leq f(\theta) \leq \sqrt{A^{2}+B^{2}}

Here, we have,

f(x)=\cos 2 x-2 \sin 2 x

-\sqrt{2^{2}+1^{2}} \leq f(x) \leq \sqrt{2^{2}+1^{2}}

-\sqrt{5} \leq f(x) \leq \sqrt{5}

So, maximum value of

f(x)

is

\sqrt{5}

.

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