AP EAMCET · Maths · Quadratic Equation
Let \(\phi(x)=\frac{x}{\left(x^2+1\right)(x+1)}\). If \(a, b\) and \(c\) are the roots of the equation \(x^3-3 x+\lambda=0,(\lambda \neq 0)\). Then, \(\phi(a) \phi(b) \phi(c)=\)
- A \(\lambda\)
- B \(\frac{-\lambda}{(\lambda+2)\left(\lambda^2+16\right)}\)
- C \(\frac{\lambda}{(\lambda+2)}\)
- D \(\frac{\lambda}{(\lambda+2)\left(\lambda^2+16\right)}\)
Answer & Solution
Correct Answer
(D) \(\frac{\lambda}{(\lambda+2)\left(\lambda^2+16\right)}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text {Given, } \phi(x)=\frac{x}{\left(x^2+1\right)(x+1)} \\ & \therefore \phi(a) \phi(b) \phi(c)=\frac{a b c}{(1+a)(1+b)(1+c)\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)} \\ & =\frac{a b c}{(1+a+b+c+a b+b c+c a+a b c)\left(1+a^2+b^2+c^2\right.} \\ &…
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