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Quadratic Equation and Inequalities question

2020 · 3 Sep · Shift 1 · Q23
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  5. /2020 · 3 Sep · Shift 1 · Q23

Quadratic Equation and Inequalities question

2020 · 3 Sep · Shift 1 · Q23

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
If α\alphaα and β\betaβ are the roots of the equation x2 + px + 2 = 0 and 1α{1 \over \alpha }α1​ and 1β{1 \over \beta }β1​ are the roots of the equation 2x2 + 2qx + 1 = 0, then (α−1α)(β−1β)(α+1β)(β+1α)\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right)(α−α1​)(β−β1​)(α+β1​)(β+α1​) is equal to :
  1. A
    94(9−q2){9 \over 4}\left( {9 - {q^2}} \right)49​(9−q2)
  2. B
    94(9+q2){9 \over 4}\left( {9 + {q^2}} \right)49​(9+q2)
  3. C
    94(9−p2){9 \over 4}\left( {9 - {p^2}} \right)49​(9−p2)
  4. D
    94(9+p2){9 \over 4}\left( {9 + {p^2}} \right)49​(9+p2)
View written solutionFree

Correct answer: C

  1. Use Vieta’s formulas for the first quadratic

Given that α,β\alpha,\betaα,β are roots of x2+px+2=0,x^2+px+2=0,x2+px+2=0, we have α+β=−p,αβ=2.\alpha+\beta=-p,\qquad \alpha\beta=2.α+β=−p,αβ=2.

  1. Use the condition on reciprocals

Now 1α,1β\dfrac1\alpha,\dfrac1\betaα1​,β1​ are roots of 2x2+2qx+1=0.2x^2+2qx+1=0.2x2+2qx+1=0. For this quadratic, sum and product of roots are 1α+1β=−2q2=−q,\frac1\alpha+\frac1\beta=-\frac{2q}{2}=-q,α1​+β1​=−22q​=−q, 1α⋅1β=12.\frac1\alpha\cdot\frac1\beta=\frac12.α1​⋅β1​=21​.

But 1α+1β=α+βαβ=−p2.\frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}=\frac{-p}{2}.α1​+β1​=αβα+β​=2−p​. Hence −q=−p2⇒p=2q.-q=-\frac p2 \quad\Rightarrow\quad p=2q.−q=−2p​⇒p=2q.

Also, 1αβ=12,\frac1{\alpha\beta}=\frac12,αβ1​=21​, which agrees with αβ=2\alpha\beta=2αβ=2.

  1. Simplify the required expression

We need to find E=(α−1α)(β−1β)(α+1β)(β+1α).E=\left(\alpha-\frac1\alpha\right)\left(\beta-\frac1\beta\right)\left(\alpha+\frac1\beta\right)\left(\beta+\frac1\alpha\right).E=(α−α1​)(β−β1​)(α+β1​)(β+α1​).

Since αβ=2\alpha\beta=2αβ=2, we get 1α=β2,1β=α2.\frac1\alpha=\frac\beta2,\qquad \frac1\beta=\frac\alpha2.α1​=2β​,β1​=2α​.

Substitute these into each factor: α−1α=α−β2=2α−β2,\alpha-\frac1\alpha=\alpha-\frac\beta2=\frac{2\alpha-\beta}{2},α−α1​=α−2β​=22α−β​, β−1β=β−α2=2β−α2,\beta-\frac1\beta=\beta-\frac\alpha2=\frac{2\beta-\alpha}{2},β−β1​=β−2α​=22β−α​, α+1β=α+α2=3α2,\alpha+\frac1\beta=\alpha+\frac\alpha2=\frac{3\alpha}{2},α+β1​=α+2α​=23α​, β+1α=β+β2=3β2.\beta+\frac1\alpha=\beta+\frac\beta2=\frac{3\beta}{2}.β+α1​=β+2β​=23β​.

Therefore, E=2α−β2⋅2β−α2⋅3α2⋅3β2.E=\frac{2\alpha-\beta}{2}\cdot \frac{2\beta-\alpha}{2}\cdot \frac{3\alpha}{2}\cdot \frac{3\beta}{2}.E=22α−β​⋅22β−α​⋅23α​⋅23β​.

Since αβ=2\alpha\beta=2αβ=2,

=\frac{9\cdot 2}{16}(2\alpha-\beta)(2\beta-\alpha) =\frac98(2\alpha-\beta)(2\beta-\alpha).$$ Now expand: $$ (2\alpha-\beta)(2\beta-\alpha)=4\alpha\beta-2\alpha^2-2\beta^2+\alpha\beta =5\alpha\beta-2(\alpha^2+\beta^2). $$ Using $$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=p^2-4,$$ because $\alpha+\beta=-p$ and $\alpha\beta=2$. So, $$ (2\alpha-\beta)(2\beta-\alpha)=5(2)-2(p^2-4)=10-2p^2+8=18-2p^2=2(9-p^2). $$ Hence, $$E=\frac98\cdot 2(9-p^2)=\frac94(9-p^2).$$ 4. **Match with the options** This is exactly $$\boxed{\frac94(9-p^2)}.$$ So the correct option is **C**. 5. **Comparison with stored answer** Stored correct answer: **C**. Our derived answer also gives **C**, so they agree.
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