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Quadratic Equation and Inequalities question

2018 · Shift 1 · Q31
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  5. /2018 · Shift 1 · Q31

Quadratic Equation and Inequalities question

2018 · Shift 1 · Q31

JEE AdvancedMathematicsQuadratic Equation and InequalitiesNumerical+3 / −1
Let a, b, c three non-zero real numbers such that the equation 3acos⁡x+2bsin⁡x=c,x∈[−π2,π2]\sqrt 3 a\cos x + 2b\sin x = c,x \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]3​acosx+2bsinx=c,x∈[−2π​,2π​], has two distinct real roots α\alphaα and β\betaβ with α+β=π3\alpha + \beta = {\pi \over 3}α+β=3π​. Then, the value of ba{b \over a}ab​ is ............
Numerical answer
View written solutionFree

Correct answer: 0.5

  1. Rewrite the given equation in standard form

We have 3acos⁡x+2bsin⁡x=c.\sqrt{3}a\cos x+2b\sin x=c.3​acosx+2bsinx=c.

Since a≠0a\neq 0a=0, divide throughout by aaa: 3cos⁡x+2(ba)sin⁡x=ca.\sqrt{3}\cos x+2\left(\frac ba\right)\sin x=\frac ca.3​cosx+2(ab​)sinx=ac​.

Let m=ba.m=\frac ba.m=ab​. Then the equation becomes 3cos⁡x+2msin⁡x=ca.\sqrt{3}\cos x+2m\sin x=\frac ca.3​cosx+2msinx=ac​.

  1. Use the fact that α,β\alpha,\betaα,β are roots

Since α\alphaα and β\betaβ satisfy the equation, 3cos⁡α+2msin⁡α=ca,\sqrt{3}\cos\alpha+2m\sin\alpha=\frac ca,3​cosα+2msinα=ac​, 3cos⁡β+2msin⁡β=ca.\sqrt{3}\cos\beta+2m\sin\beta=\frac ca.3​cosβ+2msinβ=ac​.

Subtracting, 3(cos⁡α−cos⁡β)+2m(sin⁡α−sin⁡β)=0.\sqrt{3}(\cos\alpha-\cos\beta)+2m(\sin\alpha-\sin\beta)=0.3​(cosα−cosβ)+2m(sinα−sinβ)=0.

Now use identities: cos⁡α−cos⁡β=−2sin⁡α+β2sin⁡α−β2,\cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2},cosα−cosβ=−2sin2α+β​sin2α−β​, sin⁡α−sin⁡β=2cos⁡α+β2sin⁡α−β2.\sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}.sinα−sinβ=2cos2α+β​sin2α−β​.

So, 3(−2sin⁡α+β2sin⁡α−β2)+2m(2cos⁡α+β2sin⁡α−β2)=0.\sqrt{3}\left(-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}\right)+2m\left(2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}\right)=0.3​(−2sin2α+β​sin2α−β​)+2m(2cos2α+β​sin2α−β​)=0.

Factor out 2sin⁡α−β22\sin\dfrac{\alpha-\beta}{2}2sin2α−β​: 2sin⁡α−β2[−3sin⁡α+β2+2mcos⁡α+β2]=0.2\sin\frac{\alpha-\beta}{2}\left[-\sqrt{3}\sin\frac{\alpha+\beta}{2}+2m\cos\frac{\alpha+\beta}{2}\right]=0.2sin2α−β​[−3​sin2α+β​+2mcos2α+β​]=0.

Since roots are distinct, α≠β\alpha\neq\betaα=β, so sin⁡α−β2≠0.\sin\frac{\alpha-\beta}{2}\neq 0.sin2α−β​=0. Hence, −3sin⁡α+β2+2mcos⁡α+β2=0.-\sqrt{3}\sin\frac{\alpha+\beta}{2}+2m\cos\frac{\alpha+\beta}{2}=0.−3​sin2α+β​+2mcos2α+β​=0.

Thus, 2mcos⁡α+β2=3sin⁡α+β2,2m\cos\frac{\alpha+\beta}{2}=\sqrt{3}\sin\frac{\alpha+\beta}{2},2mcos2α+β​=3​sin2α+β​, m=32tan⁡α+β2.m=\frac{\sqrt{3}}{2}\tan\frac{\alpha+\beta}{2}.m=23​​tan2α+β​.

  1. Use the given sum of roots

Given α+β=π3,\alpha+\beta=\frac\pi3,α+β=3π​, so α+β2=π6.\frac{\alpha+\beta}{2}=\frac\pi6.2α+β​=6π​.

Therefore, m=32tan⁡π6.m=\frac{\sqrt{3}}{2}\tan\frac\pi6.m=23​​tan6π​.

Since tan⁡π6=13,\tan\frac\pi6=\frac1{\sqrt{3}},tan6π​=3​1​, we get m=32⋅13=12.m=\frac{\sqrt{3}}{2}\cdot\frac1{\sqrt{3}}=\frac12.m=23​​⋅3​1​=21​.

Thus, ba=12.\frac ba=\frac12.ab​=21​.

  1. Final answer

The required value is 12.\boxed{\frac12}.21​​.

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