CAT 2024 Slot 2 · QA · Medium · Algebra
The roots α,β\alpha, \betaα,β of the equation 3x2+λx−1=03x^2 + \lambda x - 1 = 03x2+λx−1=0, satisfy 1α2+1β2=15\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15α21+β21=15. The value of (α3+β3)2(\alpha^3 + \beta^3)^2(α3+β3)2, is
9
16
4
1
C
For 3x2+λx−1=03x^2 + \lambda x - 1 = 03x2+λx−1=0: α+β=−λ3,αβ=−13\alpha + \beta = -\frac{\lambda}{3}, \quad \alpha \beta = -\frac{1}{3}α+β=−3λ,αβ=−31
Given: 1α2+1β2=α2+β2(αβ)2=(α+β)2−2αβ(αβ)2=15\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha \beta)^2} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} = 15α21+β21=(αβ)2α2+β2=(αβ)2(α+β)2−2αβ=15 (α+β)2−2(−13)=15(−13)2(\alpha+\beta)^2 - 2\left(-\frac{1}{3}\right) = 15 \left(-\frac{1}{3}\right)^2(α+β)2−2(−31)=15(−31)2 (α+β)2+23=159=53(\alpha+\beta)^2 + \frac{2}{3} = \frac{15}{9} = \frac{5}{3}(α+β)2+32=915=35 (α+β)2=1 ⟹ α+β=±1(\alpha+\beta)^2 = 1 \implies \alpha + \beta = \pm 1(α+β)2=1⟹α+β=±1
Now, for α3+β3\alpha^3 + \beta^3α3+β3: α3+β3=(α+β)((α+β)2−3αβ)=(±1)(1−3(−13))=±2\alpha^3 + \beta^3 = (\alpha+\beta)((\alpha+\beta)^2 - 3\alpha\beta) = (\pm 1)\left(1 - 3\left(-\frac{1}{3}\right)\right) = \pm 2α3+β3=(α+β)((α+β)2−3αβ)=(±1)(1−3(−31))=±2
Thus, (α3+β3)2=(±2)2=4(\alpha^3 + \beta^3)^2 = (\pm 2)^2 = 4(α3+β3)2=(±2)2=4.
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