Theory of EquationshardFree

Theory of Equations — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
Let λ\lambda be defined as:
λ=∣log⁡5100−∣log⁡2516+∣log⁡152−∣log⁡0.250∣∣∣∣\lambda = \left| \log_{\sqrt{5}} 100 - \left| \log_{25} 16 + \left| \log_{\frac{1}{5}} 2 - \left| \log_{0.2} 50 \right| \right| \right| \right|
 For some constant μ∈R\mu \in \mathbb{R}, let the equation
x3−3λx2+(λ3+1)x−μ=0x^3 - 3\lambda x^2 + (\lambda^3 + 1)x - \mu = 0
 have three distinct real roots α,β,γ\alpha, \beta, \gamma. If α,β,γ\alpha, \beta, \gamma are in arithmetic progression, then the value of α4+β4+γ4\alpha^4 + \beta^4 + \gamma^4 is:
A192192
B210210correct
C224224
D256256
Solution
Step 1: Resolve the innermost absolute value Convert all logarithms to base 55 using log⁡akb=1klog⁡ab\log_{a^k} b = \frac{1}{k} \log_a b.
∣log⁡0.250∣=∣log⁡5−150∣=∣−log⁡550∣=log⁡550|\log_{0.2} 50| = |\log_{5^{-1}} 50| = |-\log_5 50| = \log_5 50
 Step 2: Resolve the second absolute value Since log⁡1/52=−log⁡52\log_{1/5} 2 = -\log_5 2, substituting the result of Step 1 gives:
∣−log⁡52−log⁡550∣=∣−log⁡5(2×50)∣=log⁡5100\left| -\log_5 2 - \log_5 50 \right| = \left| -\log_5(2 \times 50) \right| = \log_5 100
 Step 3: Resolve the third absolute value Since log⁡2516=log⁡5242=22log⁡54=log⁡54\log_{25} 16 = \log_{5^2} 4^2 = \frac{2}{2} \log_5 4 = \log_5 4, substituting the result of Step 2 gives:
∣log⁡54+log⁡5100∣=∣log⁡5(4×100)∣=log⁡5400\left| \log_5 4 + \log_5 100 \right| = \left| \log_5(4 \times 100) \right| = \log_5 400
 Step 4: Resolve the outermost absolute value to obtain λ\lambda Since log⁡5100=log⁡51/2100=2log⁡5100=log⁡510000\log_{\sqrt{5}} 100 = \log_{5^{1/2}} 100 = 2 \log_5 100 = \log_5 10000, substituting the result of Step 3 gives:
λ=∣log⁡510000−log⁡5400∣=∣log⁡5(10000400)∣=log⁡525=2\lambda = \left| \log_5 10000 - \log_5 400 \right| = \left| \log_5 \left( \frac{10000}{400} \right) \right| = \log_5 25 = 2
 Step 5: Form the cubic equation Substituting λ=2\lambda = 2 into the given cubic yields:
x3−6x2+9x−μ=0x^3 - 6x^2 + 9x - \mu = 0
 Step 6: Use the arithmetic progression condition Let the roots in arithmetic progression be (a−d),a,(a+d)(a - d), a, (a + d). By Vieta's formulas, the sum of the roots is:
(a−d)+a+(a+d)=6  ⟹  3a=6  ⟹  a=2(a - d) + a + (a + d) = 6 \implies 3a = 6 \implies a = 2
 Thus one root is x=2x = 2. Step 7: Determine μ\mu Substituting x=2x = 2 into the equation:
23−6(22)+9(2)−μ=0  ⟹  8−24+18=μ  ⟹  μ=22^3 - 6(2^2) + 9(2) - \mu = 0 \implies 8 - 24 + 18 = \mu \implies \mu = 2
 The full cubic equation is x3−6x2+9x−2=0x^3 - 6x^2 + 9x - 2 = 0. Step 8: Find the remaining roots Since x=2x = 2 is a root, we factor:
(x−2)(x2−4x+1)=0(x - 2)(x^2 - 4x + 1) = 0
 The remaining roots come from x2−4x+1=0x^2 - 4x + 1 = 0:
x=4±16−42=2±3x = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}
 Therefore the three roots are α=2−3\alpha = 2 - \sqrt{3}, β=2\beta = 2, and γ=2+3\gamma = 2 + \sqrt{3}. Step 9: Evaluate the symmetric sum We compute S=α4+β4+γ4S = \alpha^4 + \beta^4 + \gamma^4. First, β4=24=16\beta^4 = 2^4 = 16. Next, using the identity (x−y)4+(x+y)4=2(x4+6x2y2+y4)(x - y)^4 + (x + y)^4 = 2(x^4 + 6x^2y^2 + y^4):
α4+γ4=2(24+6(22)(3)2+(3)4)=2(16+72+9)=2(97)=194\alpha^4 + \gamma^4 = 2 \left( 2^4 + 6(2^2)(\sqrt{3})^2 + (\sqrt{3})^4 \right) = 2 \left( 16 + 72 + 9 \right) = 2(97) = 194
 Summing all components:
S=194+16=210S = 194 + 16 = 210
 Answer: (2)
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