Consider the following statements in respect of the determinant | array cc ^2 (α)/(2) & ^2 (α)/(2) ^2 (β)/(2) & ^2 (β)/(2) array | where α, β are complementary angles 1. The value of the determinant is 1 2 ((α-β)/(2)) .

Correct answer: C. Both 1 and 2 aligned & α+β=90^ ^ & | array ll ^2(α)/(2) & ^2(α)/(2) ^2 (β)/(2) & ^2 (β)/(2) array | & = ^2 (α)/(2) ^2 (β)/(2)- ^2 (α)/(2) ^2 (β)/(2) & = & ( (α)/(2) (β)/(2)+ (α)/(2) (β)/(2))( (α)/(2) (β)/(2)- (α)/(2) (β)/(2)) & = ((α-β))/(2) × ((α+β))/(2) & = ((α-β))/(2) × (90^ ^ ) 2 = ((α-β))/(2) × 1 2 aligned Maximum value of ((α-β)/(2))is 1 . So maximum value of determinent is ( 1 2 )So both 1 and 2 are correct.

NDA 2015

Exam: NDA
Year: 2015
Subject: Mathematics
Chapter: Determinants

NDA 2015 · Mathematics · Determinants

Consider the following statements in respect of the determinant |...

Consider the following statements in respect of the determinant

\left|

\begin{array}{cc} \cos ^2 \frac{\alpha}{2} & \sin ^2 \frac{\alpha}{2} \\ \sin ^2 \frac{\beta}{2} & \cos ^2 \frac{\beta}{2} \end{array}

\right|

where \alpha, \beta

are complementary angles 1. The value of the determinant is \frac{1}{\sqrt{2}} \cos \left(\frac{\alpha-\beta}{2}\right)

2. The maximum value of the determinant is \frac{1}{\sqrt{2}}.

Which of the above statements is/ are correct?

Options

A1 only
B2 only
CBoth 1 and 2
DNeither 1 nor 2

Correct answer

C. Both 1 and 2

Step-by-step solution

\begin{aligned} & \alpha+\beta=90^{{}^{\circ}} \\ & \left|\begin{array}{ll} \cos ^2\frac{\alpha}{2} & \sin ^2\frac{\alpha}{2} \\ \sin ^2 \frac{\beta}{2} & \cos ^2 \frac{\beta}{2} \end{array}\right| \\ & =\cos ^2 \frac{\alpha}{2} \cos ^2 \frac{\beta}{2}-\sin ^2 \frac{\alpha}{2} \sin ^2 \frac{\beta}{2} \\ & = \\ & \left(\cos \frac{\alpha}{2} \cos \frac{\beta}{2}+\sin \frac{\alpha}{2} \sin \frac{\beta}{2}\right)\left(\cos \frac{\alpha}{2} \cos \frac{\beta}{2}-\sin \frac{\alpha}{2} \sin \frac{\beta}{2}\right) \\ & =\cos \frac{(\alpha-\beta)}{2} \times \cos \frac{(\alpha+\beta)}{2} \\ & =\cos \frac{(\alpha-\beta)}{2} \times \cos \frac{\left(90^{{}^{\circ}}\right)}{2}=\cos \frac{(\alpha-\beta)}{2} \times \frac{1}{\sqrt{2}} \end{aligned}

Maximum value of

\cos \left(\frac{\alpha-\beta}{2}\right)is 1 . So maximum value

of determinent is \left(\frac{1}{\sqrt{2}}\right)So both 1 and 2 are correct.

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