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Explain solution RD Sharma class 12 chapter Determinants exercise multiple choise question 20 maths

Answers (1)


Correct option (b)


Simply first solve the determinant.


Here the distinct real roots of

        \left|\begin{array}{ccc} \cos e c x & \sec x & \sec x \\ \sec x & cos\, ecx & \sec x \\ \sec x & \sec x & \cos e c x \end{array}\right|=0

lies in the interval

        \frac{-\pi }{4}\leq x\leq \frac{\pi }{4}

We have to find the number of roots of the given determinant.


        Here\; \; \left|\begin{array}{ccc} \cos e c x & \sec x & \sec x \\ \sec x & cos\, ecx & \sec x \\ \sec x & \sec x & \cos e c x \end{array}\right|=0

Applying C1→C1-C3; C2→C2-C3

Taking common cos sec x-sec sec x  from column 1 and 2, we get

        \Rightarrow(cos\, \, ecx-\sec x)^{2}\left|\begin{array}{ccc} 1 & 0 & \sec x \\ 0 & 1 & \sec x \\ -1 & -1 & \cos e c x \end{array}\right|=0

Expanding along R1 , we get

          \Rightarrow(cos\, \, ecx-\sec x)^{2}[(cos\, ecx+sec\, x)+sec\, x(0+1)]=0

        \Rightarrow(cos\, \, ecx-\sec x)^{2}(cos\, ecx+2sec\, x)=0

        \Rightarrow(cos\, \, ecx-\sec x)^{2}=0\; \; or\; \; (cos\, ecx+2sec\, x)=0

        \Rightarrow cos\, \, ecx-\sec x =0\; \; or\; \; cos\, ecx=-2sec\, x

        \Rightarrow cos\, \,cos\, x-sin sin\, x =0\; \; or\; \; cos \, cos\, x=-2sin\, sin\, x

        \Rightarrow sin\, x =cos\, x\; \; or\; \;\frac{cos\, cos\, x}{sin\, sin\, x}=-2

        \Rightarrow \frac{cos\, cos\, x}{sin\, sin\, x}=1\; \;or\, \, cot\, cot\, x=-2

        \Rightarrow cot\, cot\, x=1\; \;or\, \, cot\, cot\, x=-2

        \Rightarrow tan\, tan\, x=1\; \;or\, \, tan\, tan\, x=-\frac{1}{2}

Hence, there are two solutions.

Posted by

Gurleen Kaur

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