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Please solve RD Sharma class 12 chapter Determinants exercise multiple choise question 21 maths textbook solution

Answers (1)

Answer:

Correct option (d)

Hint:

First solve the given determinant then find |A| for \Theta =0 and for

        \theta =\frac{\pi }{2},\frac{3\pi }{2},.......

Given:

Given that,

        Let\; \; A=\begin{vmatrix} 1 &sin\theta &1 \\ -sin\theta &1 &sin\theta \\ -1 &-sin\theta &1 \end{vmatrix}

W\! here\; \; 0\leq \theta \leq 2\pi

We have to find the |A|

Solution:

Here\; \; A=\begin{vmatrix} 1 &sin\theta &1 \\ -sin\theta &1 &sin\theta \\ -1 &-sin\theta &1 \end{vmatrix}

Expanding along R1, we get

        \left | A \right |=1+sin^{2}\theta -sin\theta (-sin\theta +sin\theta )+sin^{2}\theta +1

        \Rightarrow \left | A \right |=2+2sin^{2}

        \Rightarrow \left | A \right |=2(1+\theta )

Given\; that\; \; 0\leq \theta \leq 2\pi

For \theta=0

        \Rightarrow \left | A \right |=2

F\! or\; \theta =\frac{\pi }{2},\frac{3\pi }{2},.......

        \Rightarrow \left | A \right |=2(1+1)

        \Rightarrow \left | A \right |=4

Hence \left | A \right |\varepsilon [2,4]

Posted by

Gurleen Kaur

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