Skip to content
dronacharya.ai
← JEE Main 2022 · 25 Jun · Shift 2
MathematicsSingle correct+4 / -1Question 5 of 90

JEE Main 2022 · 25 Jun · Shift 2, Mathematics question 5

lim⁡x→π2(tan⁡2x((2sin⁡2x+3sin⁡x+4)12−(sin⁡2x+6sin⁡x+2)12))\lim_{x\to\frac\pi2}\left(\tan^2x\left(\left(2\sin^2x+3\sin x+4\right)^{\frac12}-\left(\sin^2x+6\sin x+2\right)^{\frac12}\right)\right)
is equal to

Choose one option
Answer and worked solution

Official answer: A

Put s=sin⁡x→1s=\sin x\to1. The bracket is (2s2+3s+4)−(s2+6s+2)⋯+⋯=(s−1)(s−2)⋯+⋯\dfrac{(2s^2+3s+4)-(s^2+6s+2)}{\sqrt{\cdots}+\sqrt{\cdots}}=\dfrac{(s-1)(s-2)}{\sqrt{\cdots}+\sqrt{\cdots}}, and tan⁡2x=s2(1−s)(1+s)\tan^2x=\dfrac{s^2}{(1-s)(1+s)}.

Cancel (s−1)(s-1). The product is −s2(s−2)(1+s)(⋯+⋯)→12(3+3)=112\dfrac{-s^2(s-2)}{(1+s)\left(\sqrt{\cdots}+\sqrt{\cdots}\right)}\to\dfrac{1}{2(3+3)}=\dfrac1{12}.

(A)

Topics:Limits
← Previous questionNext question →