Set, Relations and FunctionsPYQ Sep 24Question 1973 of 217
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If f(x)=x2+x−1\displaystyle f(x) = x^2+x-1 and 4f(x)=f(2x)\displaystyle 4f(x) = f(2x), then find the value of 'x'.

Options

A2/3\displaystyle 2/3
B3/2\displaystyle 3/2
C3/4\displaystyle 3/4
D4/3\displaystyle 4/3
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Correct Answer

✅ Option b — 3/2\displaystyle 3/2

All Options:

  • A2/3\displaystyle 2/3
  • B3/2\displaystyle 3/2
  • C3/4\displaystyle 3/4
  • D4/3\displaystyle 4/3

Detailed Solution & Explanation

Given:
f(x)=x2+x−1f(x) = x^2 + x - 1
We are given the equation:
4f(x)=f(2x)4f(x) = f(2x)

**Step 1: Compute 4f(x)\displaystyle 4f(x)**
4f(x)=4(x2+x−1)=4x2+4x−44f(x) = 4(x^2 + x - 1) = 4x^2 + 4x - 4

**Step 2: Compute f(2x)\displaystyle f(2x)**
f(2x)=(2x)2+(2x)−1=4x2+2x−1f(2x) = (2x)^2 + (2x) - 1 = 4x^2 + 2x - 1

**Step 3: Equate the two expressions and solve for x\displaystyle x**
4x2+4x−4=4x2+2x−14x^2 + 4x - 4 = 4x^2 + 2x - 1
Subtract 4x2\displaystyle 4x^2 from both sides:
4x−4=2x−14x - 4 = 2x - 1
Subtract 2x\displaystyle 2x from both sides:
2x−4=−12x - 4 = -1
Add 4\displaystyle 4 to both sides:
2x=32x = 3
x=32x = \frac{3}{2}

Hence, **Option B** is the correct answer.

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