JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(f: \mathbf{R}-\{0\} \rightarrow(-\infty, 1)\) be a polynomial of degree 2, satisfying \(f(x) f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)\). If \(f(K)=-2 K\), then the sum of squares of all possible values of \(K\) is :
- A \(7\)
- B \(6\)
- C \(1\)
- D \(9\)
Answer & Solution
Correct Answer
(B) \(6\)
Step-by-step Solution
Detailed explanation
as \(f(x)\) is a polynomial of degree two let it be \(f(x)=a x^2+b x+c \quad(a \neq 0)\) on satisfying given conditions we get \(C=1 \& a= \pm 1\) hence \(f(x)=1 \pm x^2\) also range \(\in(-\infty, 1]\) hence \(f(x)=1-x^2\) now \(f(k)=-2 k\)…
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