Critical Points and Absolute Extrema in Log Function

 

❓ Question

Let

$$f(x) = x^3 + ax^2 + b\log_e|x| + 1,\quad x \ne 0$$

Given that

$$x = -1 \text{ and } x = 2$$

are the critical points of $f(x)$.

Let $m$ and $M$ respectively be the absolute minimum and absolute maximum values of $f(x)$ in the interval

$$[-2,-\frac{1}{2}]\mathrm{.}$$

Find:

$$\mathrm{\mid }M+m\mathrm{\mid }$$

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Critical points
👉 Maxima-minima
👉 Logarithmic differentiation.

Main idea:

Use critical point condition:

$$f'(x)=0$$

to find $a,b$, then evaluate function at interval endpoints and critical points.

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🔷 Step 1 — Differentiate the Function 💯

Given:

$$f(x)=x^3+ax^2+b\log|x|+1$$

Derivative:

$$f'(x)=3x^2+2ax+\frac{b}{x}$$

Critical points are:

$$x=-1,\ 2$$

Thus:

$$f'(-1)=0$$

and

$$f'(2)=0$$

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🔷 Step 2 — Use $x=-1$

$$3(-1)^2+2a(-1)+\frac{b}{-1}=0$$
$$3-2a-b=0$$
$$\boxed{ 2a+b=3 }$$

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🔷 Step 3 — Use $x=2$

$$3(2)^2+2a(2)+\frac{b}{2}=0$$
$$12+4a+\frac b2=0$$

Multiply by 2:

$$24+8a+b=0$$
$$\boxed{ 8a+b=-24 }$$

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🔷 Step 4 — Solve for $a,b$

Subtract equations:

$$(8a+b)-(2a+b)=-24-3$$
$$6a=-27$$
$$a=-\frac92$$

Now:

$$2\left(-\frac92\right)+b=3$$
$$-9+b=3$$
$$b=12$$

Thus:

$$\boxed{ a=-\frac92,\quad b=12 }$$

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🔷 Step 5 — Write Function

$$f(x)=x^3-\frac92x^2+12\log|x|+1$$

Interval:

$$\left[-2,-\frac12\right]$$

Critical point inside interval:

$$x=-1$$

Now check:

$$x=-2,\ -1,\ -\frac12$$

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🔷 Step 6 — Compute Values

At $x=-2$

$$f(-2) = -8-\frac92(4)+12\log2+1$$
$$=-8-18+8.4+1$$
$$=-16.6$$

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At $x=-1$

$$f(-1) = -1-\frac92+12\log1+1$$
$$=-\frac92 =-4.5$$

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At $x=-\frac12$

$$f\left(-\frac12\right) = -\frac18-\frac92\cdot\frac14+12\log\frac12+1$$

Since:

$$\log\frac12=-\log2=-0.7$$
$$= -\frac18-\frac98-8.4+1$$
$$=-0.125-1.125-8.4+1$$
$$=-8.65$$

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🔷 Step 7 — Find Absolute Min & Max

Values:

$x$	$f(x)$
$-2$	$-16.6$
$-1$	$-4.5$
$-\frac12$	$-8.65$

Thus:

$$m=-16.6$$
$$M=-4.5$$

Now:

$$|M+m| = |-4.5-16.6|$$
$$=21.1$$

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🔷 Step 8 — JEE Trap Alert 🚨

❌ $\log|x|$ derivative wrong kar dena

Remember:

$$\boxed{ \frac d{dx}\log|x|=\frac1x }$$

for $x\ne0$.

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✅ Final Answer

$$\boxed{ 21.1 }$$

(Option 1)

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