IIT-JEE 2012 · Paper 2 · Mathematics Q60

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MathematicsMultiple correct+4Question 60 of 60

Let $f:(-1,1)\to\mathbb R$ be such that $f(\cos4\theta)=\dfrac2{2-\sec^2\theta}$ for $\theta\in\left(0,\dfrac\pi4\right)\cup\left(\dfrac\pi4,\dfrac\pi2\right)$ . Then the value(s) of $f\left(\dfrac13\right)$ is (are)

1-\sqrt{\dfrac32}

1+\sqrt{\dfrac32}

1-\sqrt{\dfrac23}

1+\sqrt{\dfrac23}

Answer and worked solution

Official answer: Dropped: full marks to everyone

This question was dropped in the official key (marks to all). Below is the working; it shows that $f$ is not well defined.

Simplify.
$\frac2{2-\sec^2\theta}=\frac{2\cos^2\theta}{2\cos^2\theta-1}=\frac{1+\cos2\theta}{\cos2\theta}=1+\sec2\theta.$

Solve $\cos4\theta=\frac13$ . $2\cos^22\theta-1=\frac13$ , so $\cos2\theta=\pm\sqrt{\frac23}$ , and both signs occur for $\theta$ in the given range ( $2\theta<\frac\pi2$ or $2\theta>\frac\pi2$ ).

Values. $1+\sec2\theta=1\pm\sqrt{\frac32}$ : options (A) and (B). Since one input $\frac13$ gives two outputs, no function $f$ satisfies the condition, which is presumably why the question was dropped.

(A), (B) (dropped)

Topics:Trigonometric ratios and identities Types of functions

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IIT-JEE 2012 · Paper 2, Mathematics question 60