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Let d $$ \in $$ R, and $$A = \left[ {\begin{matrix} { - 2} & {4 + d} & {\left( {\sin \theta } \right) - 2} \\ 1 & {\left( {\sin \theta } \right) + 2} & d \\ 5 & {\left( {2\sin \theta } \right) - d} & {\left( { - \sin \theta } \right) + 2 + 2d} \\ \end{matrix} } \right],$$ $$\theta \in \left[ {0,2\pi } \right]$$ If the minimum value of det(A) is 8, then a value of d is -

JEE · Math · previous-year question

  1. A.$$-$$ 7
  2. B.$$2\left( {\sqrt 2 + 2} \right)$$
  3. C.$$-$$ 5correct
  4. D.$$2\left( {\sqrt 2 + 1} \right)$$

Answer

C. $$-$$ 5

Explanation

$$\det A = \left| {\begin{matrix} { - 2} & {4 + d} & {\sin \theta - 2} \\ 1 & {\sin \theta + 2} & d \\ 5 & {2\sin \theta - d} & { - \sin \theta + 2 + 2d} \\ \end{matrix} } \right|$$ (R1 $$ \to $$ R1 + R3 $$-$$ 2R2) $$ = \left| {\begin{matrix} 1 & 0 & 0 \\ 1 & {\sin \theta + 2} & d \\ 5 & {2\sin \theta - d} & {2 + 2d - \sin \theta } \\ \end{matrix} } \right|$$ = (2 + sin $$\theta $$) ( 2 + 2d $$-$$ sin$$\theta $$) $$-$$ d(2sin$$\theta $$ $$-$$ d) = 4 + 4d $$-$$ 2sin$$\theta $$ + 2sin$$\theta $$ + 2dsin$$\theta $$ $$-$$ sin2$$\theta $$ $$-$$ 2dsin$$\theta $$ + d2 d2 + 4d + 4 $$-$$ sin2$$\theta $$ = (d + 2)2 $$-$$ sin2$$\theta $$ For a given d, minimum value of det(A) = (d + 2)2 $$-$$ 1 = 8 $$ \Rightarrow $$ d = 1 or $$-$$ 5

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