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Quadratic Equation and Inequalities question

2019 · 10 Apr · Shift 1 · Q28
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  5. /2019 · 10 Apr · Shift 1 · Q28

Quadratic Equation and Inequalities question

2019 · 10 Apr · Shift 1 · Q28

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
If α\alphaα and β\betaβ are the roots of the quadratic equation, x2 + x sin θ\thetaθ- 2 sin θ\thetaθ= 0, θ∈(0,π2)\theta \in \left( {0,{\pi \over 2}} \right)θ∈(0,2π​), then α12+β12(α−12+β−12).(α−β)24{{{\alpha ^{12}} + {\beta ^{12}}} \over {\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right).{{\left( {\alpha - \beta } \right)}^{24}}}}(α−12+β−12).(α−β)24α12+β12​ is equal to :
  1. A
    212(sin⁡θ−8)6{{{2^{12}}} \over {{{\left( {\sin \theta - 8} \right)}^6}}}(sinθ−8)6212​
  2. B
    26(sin⁡θ+4)12{{{2^6}} \over {{{\left( {\sin \theta + 4} \right)}^{12}}}}(sinθ+4)1226​
  3. C
    212(sin⁡θ+8)12{{{2^{12}}} \over {{{\left( {\sin \theta + 8} \right)}^{12}}}}(sinθ+8)12212​
  4. D
    212(sin⁡θ−4)12{{{2^{12}}} \over {{{\left( {\sin \theta - 4} \right)}^{12}}}}(sinθ−4)12212​
View written solutionFree

Correct answer: C

Let E=α12+β12(α−12+β−12)(α−β)24E=\frac{\alpha^{12}+\beta^{12}}{\left(\alpha^{-12}+\beta^{-12}\right)(\alpha-\beta)^{24}}E=(α−12+β−12)(α−β)24α12+β12​ where α,β\alpha,\betaα,β are roots of x2+xsin⁡θ−2sin⁡θ=0,θ∈(0,π2).x^2+x\sin\theta-2\sin\theta=0,\qquad \theta\in\left(0,\frac\pi2\right).x2+xsinθ−2sinθ=0,θ∈(0,2π​).

We use standard relations between roots.

1. Sum and product of roots

For x2+xsin⁡θ−2sin⁡θ=0,x^2+x\sin\theta-2\sin\theta=0,x2+xsinθ−2sinθ=0, we have α+β=−sin⁡θ,αβ=−2sin⁡θ.\alpha+\beta=-\sin\theta,\qquad \alpha\beta=-2\sin\theta.α+β=−sinθ,αβ=−2sinθ.

Also, (α−β)2=(α+β)2−4αβ.(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta.(α−β)2=(α+β)2−4αβ. So,

(α−β)2=sin⁡2θ−4(−2sin⁡θ)=sin⁡2θ+8sin⁡θ=sin⁡θ(sin⁡θ+8).(\alpha-\beta)^2=\sin^2\theta-4(-2\sin\theta)=\sin^2\theta+8\sin\theta =\sin\theta(\sin\theta+8).(α−β)2=sin2θ−4(−2sinθ)=sin2θ+8sinθ=sinθ(sinθ+8).

Hence,

(\alpha-\beta)^{24}=\left((\alpha-\beta)^2\right)^{12}=ig[\sin\theta(\sin\theta+8)\big]^{12}.

2. Simplify α−12+β−12\alpha^{-12}+\beta^{-12}α−12+β−12

We have

α−12+β−12=1α12+1β12=α12+β12(αβ)12.\alpha^{-12}+\beta^{-12}=\frac{1}{\alpha^{12}}+\frac{1}{\beta^{12}} =\frac{\alpha^{12}+\beta^{12}}{(\alpha\beta)^{12}}.α−12+β−12=α121​+β121​=(αβ)12α12+β12​.

Therefore the denominator of EEE becomes

(α−12+β−12)(α−β)24=α12+β12(αβ)12(α−β)24.\left(\alpha^{-12}+\beta^{-12}\right)(\alpha-\beta)^{24} =\frac{\alpha^{12}+\beta^{12}}{(\alpha\beta)^{12}}(\alpha-\beta)^{24}.(α−12+β−12)(α−β)24=(αβ)12α12+β12​(α−β)24.

So,

E=α12+β12α12+β12(αβ)12(α−β)24=(αβ)12(α−β)24.E=\frac{\alpha^{12}+\beta^{12}}{\frac{\alpha^{12}+\beta^{12}}{(\alpha\beta)^{12}}(\alpha-\beta)^{24}} =\frac{(\alpha\beta)^{12}}{(\alpha-\beta)^{24}}.E=(αβ)12α12+β12​(α−β)24α12+β12​=(α−β)24(αβ)12​.

3. Substitute root values

Since αβ=−2sin⁡θ,\alpha\beta=-2\sin\theta,αβ=−2sinθ, we get

(αβ)12=(−2sin⁡θ)12=212sin⁡12θ.(\alpha\beta)^{12}=(-2\sin\theta)^{12}=2^{12}\sin^{12}\theta.(αβ)12=(−2sinθ)12=212sin12θ.

And from above,

(α−β)24=[sin⁡θ(sin⁡θ+8)]12=sin⁡12θ(sin⁡θ+8)12.(\alpha-\beta)^{24}=\big[\sin\theta(\sin\theta+8)\big]^{12} =\sin^{12}\theta(\sin\theta+8)^{12}.(α−β)24=[sinθ(sinθ+8)]12=sin12θ(sinθ+8)12.

Thus,

E=212sin⁡12θsin⁡12θ(sin⁡θ+8)12=212(sin⁡θ+8)12.E=\frac{2^{12}\sin^{12}\theta}{\sin^{12}\theta(\sin\theta+8)^{12}} =\frac{2^{12}}{(\sin\theta+8)^{12}}.E=sin12θ(sinθ+8)12212sin12θ​=(sinθ+8)12212​.

4. Match with the options

This is exactly 212(sin⁡θ+8)12.\boxed{\frac{2^{12}}{(\sin\theta+8)^{12}}}.(sinθ+8)12212​​. So the correct option is:

C\boxed{\text{C}}C​

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