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← JEE Main 2022 · 29 Jul · Shift 2
MathematicsSingle correct+4 / -1Question 11 of 90

JEE Main 2022 · 29 Jul · Shift 2, Mathematics question 11

Let m1m_1, m2m_2 be the slopes of two adjacent sides of a square of side a such that a2+11a+3(m12+m22)=220a^2+11a+3(m_1^2+m_2^2)=220. If one vertex of the square is (10(cos⁡α−sin⁡α), 10(sin⁡α+cos⁡α))(10(\cos\alpha-\sin\alpha),\ 10(\sin\alpha+\cos\alpha)), where α∈(0,π2)\alpha\in\left(0,\dfrac\pi2\right) and the equation of one diagonal is (cos⁡α−sin⁡α)x+(sin⁡α+cos⁡α)y=10(\cos\alpha-\sin\alpha)x+(\sin\alpha+\cos\alpha)y=10, then 72(sin⁡4α+cos⁡4α)+a2−3a+1372(\sin^4\alpha+\cos^4\alpha)+a^2-3a+13 is equal to :

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Answer and worked solution

Official answer: B

Side. The vertex gives 10[(c−s)2+(s+c)2]=2010[(c-s)^2+(s+c)^2]=20, so it is 102\frac{10}{\sqrt2} from the diagonal: half a diagonal. Hence a=10a=10.

Slopes. 100+110+3(m12+m22)=220100+110+3(m_1^2+m_2^2)=220 gives tan⁡2α+cot⁡2α=103\tan^2\alpha+\cot^2\alpha=\frac{10}3 (the sides make 45° with the diagonal), so tan⁡2α=3\tan^2\alpha=3 or 13\frac13 and sin⁡22α=34\sin^22\alpha=\frac34.

Value 72(1−38)+100−30+13=45+83=12872\left(1-\frac38\right)+100-30+13=45+83=128.

(B)

Topics:Angle between linesDistance of a point from a lineTrigonometric ratios and identities
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