Question:
Let \(f:\mathbb{R}\to \mathbb{R}\) be defined by \(f(x)=2x+3\). If \(\alpha,\beta\) are the roots of the equation
\[
f(x^2)-2f\left(\frac{x}{2}\right)-1=0
\]
then \(\alpha^2+\beta^2=\)
Let \(f:\mathbb{R}\to \mathbb{R}\) be defined by \(f(x)=2x+3\). If \(\alpha,\beta\) are the roots of the equation
\[
f(x^2)-2f\left(\frac{x}{2}\right)-1=0
\]
then \(\alpha^2+\beta^2=\)
Show Hint
When a function is given, first evaluate each required expression separately, then substitute carefully into the equation.
Updated On: Jun 26, 2026
- \(13\)
- \(25\)
- \(5\)
- \(18\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Use the given function.
Given, \[ f(x)=2x+3 \] Therefore, \[ f(x^2)=2x^2+3 \] Also, \[ f\left(\frac{x}{2}\right)=2\left(\frac{x}{2}\right)+3 \] So, \[ f\left(\frac{x}{2}\right)=x+3 \]
Step 2: Substitute in the given equation.
The equation is \[ f(x^2)-2f\left(\frac{x}{2}\right)-1=0 \] Substituting the values, \[ (2x^2+3)-2(x+3)-1=0 \]
Step 3: Simplify the equation.
\[ 2x^2+3-2x-6-1=0 \] \[ 2x^2-2x-4=0 \] Dividing by \(2\), \[ x^2-x-2=0 \] Factorizing, \[ (x-2)(x+1)=0 \] Hence, \[ x=2 \quad \text{or} \quad x=-1 \] So, \[ \alpha=2,\quad \beta=-1 \]
Step 4: Find \(\alpha^2+\beta^2\).
\[ \alpha^2+\beta^2=(2)^2+(-1)^2 \] \[ \alpha^2+\beta^2=4+1 \] \[ \alpha^2+\beta^2=5 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{5} \]
Given, \[ f(x)=2x+3 \] Therefore, \[ f(x^2)=2x^2+3 \] Also, \[ f\left(\frac{x}{2}\right)=2\left(\frac{x}{2}\right)+3 \] So, \[ f\left(\frac{x}{2}\right)=x+3 \]
Step 2: Substitute in the given equation.
The equation is \[ f(x^2)-2f\left(\frac{x}{2}\right)-1=0 \] Substituting the values, \[ (2x^2+3)-2(x+3)-1=0 \]
Step 3: Simplify the equation.
\[ 2x^2+3-2x-6-1=0 \] \[ 2x^2-2x-4=0 \] Dividing by \(2\), \[ x^2-x-2=0 \] Factorizing, \[ (x-2)(x+1)=0 \] Hence, \[ x=2 \quad \text{or} \quad x=-1 \] So, \[ \alpha=2,\quad \beta=-1 \]
Step 4: Find \(\alpha^2+\beta^2\).
\[ \alpha^2+\beta^2=(2)^2+(-1)^2 \] \[ \alpha^2+\beta^2=4+1 \] \[ \alpha^2+\beta^2=5 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{5} \]
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