Find fogx Using Functional Equation Trick

 

❓ Question

If

$$g(x)=3x^2+2x-3$$
$$f(0)=-3$$

and

$$4g(f(x))=3x^2-32x+72$$

then

$$f(g(2))$$

is equal to:

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Composite functions
👉 Comparing polynomial expressions
👉 Functional equations.

Main idea:

Use:

$$g(t)=3t^2+2t-3$$

inside:

$$g(f(x))$$

and compare coefficients.

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🔷 Step 1 — Write Given Relation 💯

Given:

$$g(x)=3x^2+2x-3$$

So:

$$g(f(x)) = 3f(x)^2+2f(x)-3$$

Also:

$$4g(f(x)) = 3x^2-32x+72$$

Thus:

$$4(3f(x)^2+2f(x)-3) = 3x^2-32x+72$$

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🔷 Step 2 — Simplify Equation

Expand:

$$12f(x)^2+8f(x)-12 = 3x^2-32x+72$$

Bring all terms together:

$$12f(x)^2+8f(x) = 3x^2-32x+84$$

Divide by:

$$4$$
$$3f(x)^2+2f(x) = \frac34x^2-8x+21$$

Add and subtract 3:

$$3f(x)^2+2f(x)-3 = \frac34x^2-8x+18$$

Notice:

$$3a^2+2a-3=g(a)$$

So:

$$g(f(x)) = \frac34x^2-8x+18$$

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🔷 Step 3 — Identify Perfect Square Pattern

Observe:

$$\frac34x^2-8x+18 = 3\left(\frac x2-\frac83\right)^2 + 2\left(\frac x2-\frac83\right)-3$$

Thus:

$$f(x)=\frac x2-\frac83$$

Now use:

$$f(0)=-3$$

Substitute:

$$f(0)=c=-3$$

Hence actual linear form becomes:

$$f(x)=\frac x2-3$$

Check:

$$g(f(x)) = 3\left(\frac x2-3\right)^2 + 2\left(\frac x2-3\right)-3$$
$$= \frac34x^2-8x+18$$

Correct.

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🔷 Step 4 — Find $g(2)$

Given:

$$g(x)=3x^2+2x-3$$

So:

$$g(2)=3(4)+4-3$$
$$=12+4-3$$
$$=13$$

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🔷 Step 5 — Find $f(g(2))$

$$f(13) = \frac{13}{2}-3$$
$$= \frac{13-6}{2}$$
$$= \frac72$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ $g(f(x))$ expand na karna

❌ Composite function ko direct substitution without comparison kar dena

❌ $f(0)=-3$ condition ignore kar dena

Remember:

$$\boxed{ g(f(x))=3f(x)^2+2f(x)-3 }$$

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✅ Final Answer

$$\boxed{ \frac72 }$$

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📚 Related Topics

• Range of Function Using Discriminant Method
• Polynomial with Given Extrema Find Value at x Equal 2
• Critical Points and Absolute Extrema in Log Function