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Quadratic Equation and Inequalities question

2023 · 13 Apr · Shift 2 · Q28
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  5. /2023 · 13 Apr · Shift 2 · Q28

Quadratic Equation and Inequalities question

2023 · 13 Apr · Shift 2 · Q28

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
Let α,β\alpha, \betaα,β be the roots of the equation x2−2x+2=0x^{2}-\sqrt{2} x+2=0x2−2​x+2=0. Then α14+β14\alpha^{14}+\beta^{14}α14+β14 is equal to
  1. A
    −64-64−64
  2. B
    −642-64 \sqrt{2}−642​
  3. C
    −1282-128 \sqrt{2}−1282​
  4. D
    −128-128−128
View written solutionFree

Correct answer: D

  1. Given quadratic and basic relations

The roots α,β\alpha,\betaα,β of x2−2x+2=0x^2-\sqrt{2}x+2=0x2−2​x+2=0 satisfy α+β=2,αβ=2.\alpha+\beta=\sqrt{2}, \qquad \alpha\beta=2.α+β=2​,αβ=2.

We need to find α14+β14.\alpha^{14}+\beta^{14}.α14+β14.

  1. Set up a recurrence

Let Sn=αn+βn.S_n=\alpha^n+\beta^n.Sn​=αn+βn.

Since each root satisfies r2=2 r−2,r^2=\sqrt{2}\,r-2,r2=2​r−2, for r=α,βr=\alpha,\betar=α,β, multiplying by rn−2r^{n-2}rn−2 gives rn=2 rn−1−2rn−2.r^n=\sqrt{2}\,r^{n-1}-2r^{n-2}.rn=2​rn−1−2rn−2.

Adding for r=αr=\alphar=α and r=βr=\betar=β: Sn=2 Sn−1−2Sn−2.S_n=\sqrt{2}\,S_{n-1}-2S_{n-2}.Sn​=2​Sn−1​−2Sn−2​.

Initial values: S0=α0+β0=2,S_0=\alpha^0+\beta^0=2,S0​=α0+β0=2, S1=α+β=2.S_1=\alpha+\beta=\sqrt{2}.S1​=α+β=2​.

  1. Compute successively up to S14S_{14}S14​

S2=2S1−2S0=(2)(2)−2(2)=2−4=−2.S_2=\sqrt{2}S_1-2S_0=(\sqrt{2})(\sqrt{2})-2(2)=2-4=-2.S2​=2​S1​−2S0​=(2​)(2​)−2(2)=2−4=−2.

S3=2S2−2S1=2(−2)−22=−42.S_3=\sqrt{2}S_2-2S_1=\sqrt{2}(-2)-2\sqrt{2}=-4\sqrt{2}.S3​=2​S2​−2S1​=2​(−2)−22​=−42​.

S4=2S3−2S2=2(−42)−2(−2)=−8+4=−4.S_4=\sqrt{2}S_3-2S_2=\sqrt{2}(-4\sqrt{2})-2(-2)=-8+4=-4.S4​=2​S3​−2S2​=2​(−42​)−2(−2)=−8+4=−4.

S5=2S4−2S3=2(−4)−2(−42)=−42+82=42.S_5=\sqrt{2}S_4-2S_3=\sqrt{2}(-4)-2(-4\sqrt{2})=-4\sqrt{2}+8\sqrt{2}=4\sqrt{2}.S5​=2​S4​−2S3​=2​(−4)−2(−42​)=−42​+82​=42​.

S6=2S5−2S4=2(42)−2(−4)=8+8=16.S_6=\sqrt{2}S_5-2S_4=\sqrt{2}(4\sqrt{2})-2(-4)=8+8=16.S6​=2​S5​−2S4​=2​(42​)−2(−4)=8+8=16.

S7=2S6−2S5=162−82=82.S_7=\sqrt{2}S_6-2S_5=16\sqrt{2}-8\sqrt{2}=8\sqrt{2}.S7​=2​S6​−2S5​=162​−82​=82​.

S8=2S7−2S6=2(82)−32=16−32=−16.S_8=\sqrt{2}S_7-2S_6=\sqrt{2}(8\sqrt{2})-32=16-32=-16.S8​=2​S7​−2S6​=2​(82​)−32=16−32=−16.

S9=2S8−2S7=−162−162=−322.S_9=\sqrt{2}S_8-2S_7=-16\sqrt{2}-16\sqrt{2}=-32\sqrt{2}.S9​=2​S8​−2S7​=−162​−162​=−322​.

S10=2S9−2S8=2(−322)−2(−16)=−64+32=−32.S_{10}=\sqrt{2}S_9-2S_8=\sqrt{2}(-32\sqrt{2})-2(-16)=-64+32=-32.S10​=2​S9​−2S8​=2​(−322​)−2(−16)=−64+32=−32.

S11=2S10−2S9=−322+642=322.S_{11}=\sqrt{2}S_{10}-2S_9=-32\sqrt{2}+64\sqrt{2}=32\sqrt{2}.S11​=2​S10​−2S9​=−322​+642​=322​.

S12=2S11−2S10=2(322)−2(−32)=64+64=128.S_{12}=\sqrt{2}S_{11}-2S_{10}=\sqrt{2}(32\sqrt{2})-2(-32)=64+64=128.S12​=2​S11​−2S10​=2​(322​)−2(−32)=64+64=128.

S13=2S12−2S11=1282−642=642.S_{13}=\sqrt{2}S_{12}-2S_{11}=128\sqrt{2}-64\sqrt{2}=64\sqrt{2}.S13​=2​S12​−2S11​=1282​−642​=642​.

S14=2S13−2S12=2(642)−2(128)=128−256=−128.S_{14}=\sqrt{2}S_{13}-2S_{12}=\sqrt{2}(64\sqrt{2})-2(128)=128-256=-128.S14​=2​S13​−2S12​=2​(642​)−2(128)=128−256=−128.

Thus, α14+β14=S14=−128.\alpha^{14}+\beta^{14}=S_{14}=-128.α14+β14=S14​=−128.

  1. Check with options

The value is −128,-128,−128, which matches Option D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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