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← JEE Main 2014 · 12 Apr · Shift 1
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JEE Main 2014 · 12 Apr · Shift 1, Mathematics question 89

If f(θ)=∣1cos⁡θ1−sin⁡θ1−cos⁡θ−1sin⁡θ1∣f(\theta)=\begin{vmatrix}1&\cos\theta&1\\-\sin\theta&1&-\cos\theta\\-1&\sin\theta&1\end{vmatrix} and A and B are respectively the maximum and the minimum values of f(θ)f(\theta), then (A, B) is equal to :

Choose one option
Answer and worked solution

Official answer: C

Expanding along the first row:
f=(1+sin⁡θcos⁡θ)+cos⁡θ(sin⁡θ+cos⁡θ)+(1−sin⁡2θ)=2+sin⁡2θ+cos⁡2θ.f=(1+\sin\theta\cos\theta)+\cos\theta(\sin\theta+\cos\theta)+(1-\sin^2\theta)=2+\sin2\theta+\cos2\theta.
This ranges over 2±22\pm\sqrt2, so (A,B)=(2+2, 2−2)(A,B)=(2+\sqrt2,\,2-\sqrt2).

Topics:Determinants and propertiesTrigonometric ratios and identities
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