JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

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Application +4 | -1

The maximum and minimum value of

f(x)

=a b \sin x+b \sqrt{1-a^2} \cos x+c

lie in the interval ( assuming

|a|<1, b>0)

View answer and explanation

Correct option: C

[c-b, b+c]

Option C ([c-b, b+c]) is correct because the maximum and minimum values of the function lie in the interval [c-b, b+c], which is found by rewriting the function as a sine function and using its amplitude.

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\tan \alpha=\frac{p}{q}

, where

\alpha=6 \beta

and

\alpha

being an acute angle, then

\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta]

is equal to The value of

\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}

is If the

\triangle A B C

is acute angled at

C

, then The value of the expression

{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta

equals

\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}=

\sin ^2(A / 2)+\sin ^2(B / 2)-\sin ^2(C / 2)=