JEE Main 2016 (Online) 9th AprilQuestion 1
The number of $x\in[0,2\pi]$ for which $\left|\sqrt{2\sin^{4}x+18\cos^{2}x}-\sqrt{2\cos^{4}x+18\sin^{2}x}\right|=1$ is :
Trigonometric Ratio & Identities
Solution
Concept. Put $w=\cos2x$. The two quantities under the roots have a simple sum and difference in $w$, which pins down $w^2$. Formulas used. (1) $\sin^2x=\frac{1-w}{2}$, $\cos^2x=\frac{1+w}{2}$ where $w=\cos2x$(2) $(p-q)^2(p+q)^2=\left(p^2-q^2\right)^2$Step 1 - Name the roots. Let $p^2=2\sin^4x+18\cos^2x$ and $q^2=2\cos^4x+18\sin^2x$. By (1), $$p^2-q^2=2(\sin^2x-\cos^2x)-18(\sin^2x-\cos^2x)=16w,$$ $$p^2+q^2=2\left(\sin^4x+\cos^4x\right)+18=(1+w^2)+18=w^2+19.$$ Step 2 - Use $|p-q|=1$. Then $p^2+q^2-2pq=1$, so $2pq=w^2+18$ and $$(p+q)^2=p^2+q^2+2pq=2w^2+37.$$ Step 3 - Find $w^2$. By (2), $1\cdot(2w^2+37)=256w^2$, so $$w^2=\frac{37}{254},\qquad\cos2x=\pm\sqrt{\tfrac{37}{254}}\approx\pm0.38.$$ Step 4 - Count. For $x\in[0,2\pi]$, $2x$ runs over $[0,4\pi]$, two full turns. Each of the two values of $\cos2x$ (strictly between $-1$ and $1$) is taken twice per turn, so $2\cdot2\cdot2=8$ solutions. Answer: (D)