JEE Main 2014 · 11 Apr · Shift 1 · Mathematics Q61

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MathematicsSingle correct+4 / -1Question 61 of 90

Let $f$ be an odd function defined on the set of real numbers such that for $x\ge0$ ,
$f(x)=3\sin x+4\cos x.$
Then $f(x)$ at $x=-\dfrac{11\pi}6$ is equal to :

\dfrac32+2\sqrt3

-\dfrac32+2\sqrt3

\dfrac32-2\sqrt3

-\dfrac32-2\sqrt3

Answer and worked solution

Official answer: C

$f$ is odd, so
$f\left(-\tfrac{11\pi}6\right)=-f\left(\tfrac{11\pi}6\right)=-\left(3\left(-\tfrac12\right)+4\cdot\tfrac{\sqrt3}2\right)=\frac32-2\sqrt3.$

Topics:Types of functions Trigonometric ratios and identities

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JEE Main 2014 · 11 Apr · Shift 1, Mathematics question 61