JEE Main: Absolute Maximum & Minimum on Interval — Smart Method 💡

 

❓ 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 Solution

Critical points are obtained from

$$f^{\mathrm{\prime }}(x)=0$$

We’ll first find constants $a$ and $b$, then evaluate $f(x)$ at endpoints and critical points inside the interval to get absolute max and min.

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🔹 Step 1 — Find the derivative

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

Differentiate:

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

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🔹 Step 2 — Use critical points

Given critical points:

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

So,

$$f'(-1) = 0,\quad f'(2) = 0$$

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▶ At $x = -1$

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

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▶ At $x = 2$

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

Multiply by 2:

$$24 + 8a + b = 0$$
$$8a + b = -24 \quad \text{(Equation 2)}$$

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🔹 Step 3 — Solve for $a$ and $b$

Subtract (1) from (2):

$$(8a + b) - (2a + b) = -24 - 3$$
$$6a = -27 \Rightarrow a = -\frac{9}{2}$$

Substitute into (1):

$$2\left(-\frac{9}{2}\right) + b = 3 \Rightarrow -9 + b = 3 \Rightarrow b = 12$$

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🔹 Step 4 — Write final function

$$f(x) = x^3 - \frac{9}{2}x^2 + 12\ln|x| + 1$$

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🔹 Step 5 — Check points in the interval

Interval:

$$[-2,\,-\tfrac{1}{2}]$$

Points to check:

• Endpoint: $x = -2$
  

• Critical point inside interval: $x = -1$
  

• Endpoint: $x = -\tfrac{1}{2}$

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▶ $f(-2)$

$$f(-2) = -8 - 18 + 12\ln 2 + 1 = -25 + 12\ln 2$$

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▶ $f(-1)$

$$f(-1) = -1 - \frac{9}{2} + 12\ln 1 + 1$$
$$\ln 1 = 0 \Rightarrow f(-1) = -\frac{9}{2}$$

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▶ $f(-\tfrac{1}{2})$

$$f(-\tfrac{1}{2}) = -\frac{1}{8} - \frac{9}{8} + 12\ln\!\left(\frac{1}{2}\right) + 1$$
$$= -\frac{10}{8} + 1 - 12\ln 2 = -\frac{1}{4} - 12\ln 2$$

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🔹 Step 6 — Identify absolute max & min

Numerical comparison:

• $f(-2) = -25 + 12\ln 2 \approx -16.7$
  

• $f(-1) = -4.5$
  

• $f(-\tfrac{1}{2}) \approx -8.6$
  

So:

$$M = f(-1) = -\frac{9}{2}$$
$$m = f(-2) = -25 + 12\ln 2$$

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🔹 Step 7 — Compute $|M + m|$

$$M + m = -\frac{9}{2} - 25 + 12\ln 2 = -\frac{59}{2} + 12\ln 2$$

Taking modulus:

$$|M + m| = \left|\frac{59}{2} - 12\ln 2\right|$$

Since $\frac{59}{2} > 12\ln 2$:

$$|M + m| = \boxed{\frac{59}{2} - 12\ln 2}$$

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✅ Final Answer

$$\boxed{{\displaystyle \frac{59}{2}-12\ln 2\mathrm{<br>}\mathrm{<br>}\mathrm{<br>}}}$$

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