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Quadratic Equation and Inequalities question

2023 · 12 Apr · Shift 1 · Q30
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  5. /2023 · 12 Apr · Shift 1 · Q30

Quadratic Equation and Inequalities question

2023 · 12 Apr · Shift 1 · Q30

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
Let α,β\alpha, \betaα,β be the roots of the quadratic equation x2+6x+3=0x^{2}+\sqrt{6} x+3=0x2+6​x+3=0. Then α23+β23+α14+β14α15+β15+α10+β10\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}α15+β15+α10+β10α23+β23+α14+β14​ is equal to :
  1. A
    72
  2. B
    9
  3. C
    729
  4. D
    81
View written solutionFree

Correct answer: D

  1. Find the roots structurally

Given x2+6x+3=0x^2+\sqrt{6}x+3=0x2+6​x+3=0 with roots α,β\alpha,\betaα,β.

For this quadratic, α+β=−6,αβ=3.\alpha+\beta=-\sqrt{6},\qquad \alpha\beta=3.α+β=−6​,αβ=3.

Now check the discriminant: D=(6)2−4⋅1⋅3=6−12=−6.D=(\sqrt{6})^2-4\cdot 1\cdot 3=6-12=-6.D=(6​)2−4⋅1⋅3=6−12=−6. So the roots are complex conjugates.

A more useful observation is: x2+6x+3=0  ⟺  x2+6x+62=0.x^2+\sqrt{6}x+3=0 \iff x^2+\sqrt{6}x+\frac{6}{2}=0.x2+6​x+3=0⟺x2+6​x+26​=0. This suggests writing roots in polar form.

Using quadratic formula, x=−6±i62=62(−1±i).x=\frac{-\sqrt{6}\pm i\sqrt{6}}{2}=\frac{\sqrt{6}}{2}(-1\pm i).x=2−6​±i6​​=26​​(−1±i). Now 62(−1±i)=3 −1±i2.\frac{\sqrt{6}}{2}(-1\pm i)=\sqrt{3}\,\frac{-1\pm i}{\sqrt{2}}.26​​(−1±i)=3​2​−1±i​. Since −1+i2=cos⁡3π4+isin⁡3π4,\frac{-1+i}{\sqrt{2}}=\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4},2​−1+i​=cos43π​+isin43π​, we get

\qquad \beta=\sqrt{3}\,\text{cis}\left(-\frac{3\pi}{4}\right).$$ Thus for any integer $n$, $$\alpha^n+\beta^n=(\sqrt{3})^n\left(\text{cis}\frac{3n\pi}{4}+\text{cis}\left(-\frac{3n\pi}{4}\right)\right) =2(\sqrt{3})^n\cos\frac{3n\pi}{4}.$$ Let $$S_n=\alpha^n+\beta^n=2(\sqrt{3})^n\cos\frac{3n\pi}{4}.$$ --- 2. **Compute the required terms** We need $$\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}} =\frac{S_{23}+S_{14}}{S_{15}+S_{10}}.$$ Now evaluate each cosine: ### (i) $S_{23}$ $$S_{23}=2(\sqrt{3})^{23}\cos\left(\frac{69\pi}{4}\right).$$ Since $$\frac{69\pi}{4}=16\pi+\frac{5\pi}{4},$$ so $$\cos\left(\frac{69\pi}{4}\right)=\cos\frac{5\pi}{4}=-\frac{1}{\sqrt{2}}.$$ Hence $$S_{23}=2(\sqrt{3})^{23}\left(-\frac{1}{\sqrt{2}}\right) =-\sqrt{2}(\sqrt{3})^{23}.$$ ### (ii) $S_{14}$ $$S_{14}=2(\sqrt{3})^{14}\cos\left(\frac{42\pi}{4}\right) =2(\sqrt{3})^{14}\cos\left(\frac{21\pi}{2}\right).$$ Now $$\cos\left(\frac{21\pi}{2}\right)=0,$$ so $$S_{14}=0.$$ ### (iii) $S_{15}$ $$S_{15}=2(\sqrt{3})^{15}\cos\left(\frac{45\pi}{4}\right).$$ Since $$\frac{45\pi}{4}=10\pi+\frac{5\pi}{4},$$ so $$\cos\left(\frac{45\pi}{4}\right)=\cos\frac{5\pi}{4}=-\frac{1}{\sqrt{2}}.$$ Thus $$S_{15}=2(\sqrt{3})^{15}\left(-\frac{1}{\sqrt{2}}\right) =-\sqrt{2}(\sqrt{3})^{15}.$$ ### (iv) $S_{10}$ $$S_{10}=2(\sqrt{3})^{10}\cos\left(\frac{30\pi}{4}\right) =2(\sqrt{3})^{10}\cos\left(\frac{15\pi}{2}\right).$$ Again, $$\cos\left(\frac{15\pi}{2}\right)=0,$$ so $$S_{10}=0.$$ --- 3. **Substitute into the expression** Therefore, $$\frac{S_{23}+S_{14}}{S_{15}+S_{10}} =\frac{-\sqrt{2}(\sqrt{3})^{23}+0}{-\sqrt{2}(\sqrt{3})^{15}+0} =(\sqrt{3})^{23-15}=(\sqrt{3})^8=3^4=81.$$ --- 4. **Check options** The value is $$81.$$ So the correct option is **D**.
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