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Question: If equation $x^4 - 4x^3 + \lambda x^2 + \mu x + 1 = 0$ has four positive real roots, then-...

If equation x4−4x3+λx2+μx+1=0x^4 - 4x^3 + \lambda x^2 + \mu x + 1 = 0 has four positive real roots, then-

A

∣λ∣+∣μ∣=2|\lambda| + |\mu| = 2.

B

λ−μ=10\lambda - \mu = 10

C

∣λ∣+∣μ∣=10|\lambda| + |\mu| = 10

D

∣λ∣−∣μ∣=2|\lambda| - |\mu| = 2

Answer

Options (B) and (C) are correct.

Explanation

Solution

Let the given equation be x4−4x3+λx2+μx+1=0x^4 - 4x^3 + \lambda x^2 + \mu x + 1 = 0. Let the four positive real roots be r1,r2,r3,r4r_1, r_2, r_3, r_4.

According to Vieta's formulas:

  1. Sum of the roots: r1+r2+r3+r4=−(−4)/1=4r_1 + r_2 + r_3 + r_4 = -(-4)/1 = 4
  2. Product of the roots: r1r2r3r4=1/1=1r_1r_2r_3r_4 = 1/1 = 1

Since the roots are positive real numbers, we can apply the AM-GM inequality: Arithmetic Mean (AM) ≥\ge Geometric Mean (GM) r1+r2+r3+r44≥r1r2r3r44\frac{r_1 + r_2 + r_3 + r_4}{4} \ge \sqrt[4]{r_1r_2r_3r_4}

Substitute the values from Vieta's formulas: 44≥14\frac{4}{4} \ge \sqrt[4]{1} 1≥11 \ge 1

The equality in AM-GM holds if and only if all the numbers are equal. Therefore, r1=r2=r3=r4r_1 = r_2 = r_3 = r_4. Since their sum is 4, we have 4r1=44r_1 = 4, which implies r1=1r_1 = 1. So, all four roots are r1=r2=r3=r4=1r_1 = r_2 = r_3 = r_4 = 1.

Now, let's find the values of λ\lambda and μ\mu using these roots: 3. Sum of the roots taken two at a time: λ=r1r2+r1r3+r1r4+r2r3+r2r4+r3r4\lambda = r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 + r_2r_4 + r_3r_4 Since all roots are 1, each term is 1×1=11 \times 1 = 1. There are (42)=6\binom{4}{2} = 6 such terms. λ=1+1+1+1+1+1=6\lambda = 1 + 1 + 1 + 1 + 1 + 1 = 6.

  1. Sum of the roots taken three at a time: −μ=r1r2r3+r1r2r4+r1r3r4+r2r3r4-\mu = r_1r_2r_3 + r_1r_2r_4 + r_1r_3r_4 + r_2r_3r_4 Since all roots are 1, each term is 1×1×1=11 \times 1 \times 1 = 1. There are (43)=4\binom{4}{3} = 4 such terms. −μ=1+1+1+1=4-\mu = 1 + 1 + 1 + 1 = 4. Therefore, μ=−4\mu = -4.

Now, let's check the given options with λ=6\lambda = 6 and μ=−4\mu = -4: (A) ∣λ∣+∣μ∣=∣6∣+∣−4∣=6+4=10|\lambda| + |\mu| = |6| + |-4| = 6 + 4 = 10. (This option is correct) (B) λ−μ=6−(−4)=6+4=10\lambda - \mu = 6 - (-4) = 6 + 4 = 10. (This option is correct) (C) ∣λ∣+∣μ∣=∣6∣+∣−4∣=6+4=10|\lambda| + |\mu| = |6| + |-4| = 6 + 4 = 10. (This option is identical to (A) and is correct) (D) ∣λ∣−∣μ∣=∣6∣−∣−4∣=6−4=2|\lambda| - |\mu| = |6| - |-4| = 6 - 4 = 2. (This option is incorrect)

Both options (B) and (C) (which is identical to A) are correct.