Question

Let \(f: R \rightarrow R\) be defined by \(f(x) = 2x+3\). If \(\alpha, \beta\) are the roots of the equation \(f(x^2) - 2f(\frac{x}{2}) - 1 = 0\) then \(\alpha^2 + \beta^2 = \)

• A. 13
• B. 25
• C. 5
• D. 18

Hint:

Substitute the function definitions into the given equation to form a polynomial equation.
For a quadratic equation \(ax^2+bx+c=0\) with roots \(\alpha, \beta\), sum of roots \(\alpha+\beta = -b/a\) and product of roots \(\alpha\beta = c/a\).
Use the identity \(\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\).

Answer 1

The Correct Option is C

Given \(f(x) = 2x+3\). The equation is \(f(x^2) - 2f(\frac{x}{2}) - 1 = 0\). Substitute \(f(x^2)\) and \(f(\frac{x}{2})\): \(f(x^2) = 2(x^2)+3 = 2x^2+3\). \(f(\frac{x}{2}) = 2(\frac{x}{2})+3 = x+3\). So the equation becomes: \((2x^2+3) - 2(x+3) - 1 = 0\) \(2x^2+3 - 2x - 6 - 1 = 0\) \(2x^2 - 2x - 4 = 0\) Divide by 2: \(x^2 - x - 2 = 0\). This is a quadratic equation. Let its roots be \(\alpha\) and \(\beta\). From Vieta's formulas: Sum of roots: \(\alpha + \beta = -(\text{coefficient of } x) / (\text{coefficient of } x^2) = -(-1)/1 = 1\). Product of roots: \(\alpha\beta = (\text{constant term}) / (\text{coefficient of } x^2) = -2/1 = -2\). We need to find \(\alpha^2 + \beta^2\). We know that \((\alpha+\beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta\). So, \(\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\). Substitute the values: \(\alpha^2 + \beta^2 = (1)^2 - 2(-2) = 1 - (-4) = 1 + 4 = 5\). \[ \boxed{5} \]