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  • Explain solution RD Sharma class 12 chapter Determinants exercise multiple choise question 28 maths

Explain solution RD Sharma class 12 chapter Determinants exercise multiple choise question 28 maths

Answers (1)

Answer:

Correct option (a)

Hint:

First solve the given determinant then find the maximum value of sin function, which is 1.

Given:

Given that,

        \Delta =\begin{vmatrix} 1 &1 &1 \\ 1 &1+sin\, \theta &1 \\ 1+cos\, \theta &1 &1 \end{vmatrix}

We have to find the maximum value of \Delta.

Solution:

Here\: \: \Delta =\begin{vmatrix} 1 &1 &1 \\ 1 &1+sin\, \theta &1 \\ 1+cos\, \theta &1 &1 \end{vmatrix}

Applying R1 → R1 - R2

        \Rightarrow \Delta =\begin{vmatrix} 1 &1 &1 \\ 1 &1+sin\, \theta &1 \\ 1+cos\, \theta &1 &1 \end{vmatrix}

Expanding along R1

        \begin{aligned} &\Rightarrow \quad \Delta=\sin \theta[1-1-\cos \theta] \\ &\Rightarrow \quad \Delta=-\sin \theta \cos \theta \\ &\Rightarrow \quad \Delta=\frac{-2 \sin \theta \cos \theta}{2} \\ &\Rightarrow \quad \Delta=\frac{-\sin 2 \theta}{2} \end{aligned}            [\because \sin \sin 2 \theta=2 \sin \sin \theta \cos \cos \theta]

We know that the maximum value of sin function is 1.

Hence, maximum value of

        \Delta =\frac{1}{2}

Posted by

Gurleen Kaur

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