Let f(x)=(x+3)^2(x-2)^3, x \in[-4,4] . If M and m are the maximum and minimum values of f , respectively in [-4,4] , then the value of M-m is

Application of Derivatives PYQ — JEE Main 2024 · 30 Jan · Shift 2 · Q44 with Solution

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Correct answer: C

We need the maximum and minimum values of

\, x\in[-4,4].$$ Since $f$ is continuous on a closed interval, both maximum and minimum exist and occur either at critical points or at endpoints. 2. Differentiate $f(x)$. Using product rule: $$f(x)=(x+3)^2(x-2)^3$$ $$f'(x)=2(x+3)(x-2)^3+(x+3)^2\cdot 3(x-2)^2.$$ Factor common terms: $$f'(x)=(x+3)(x-2)^2\left[2(x-2)+3(x+3)\right].$$ Now simplify the bracket: $$2(x-2)+3(x+3)=2x-4+3x+9=5x+5=5(x+1).$$ Hence, $$f'(x)=5(x+3)(x-2)^2(x+1).$$ 3. Critical points are given by $$f'(x)=0 \implies x=-3,-1,2.$$ All lie in the interval $[-4,4]$. 4. Evaluate $f(x)$ at endpoints and critical points. - At $x=-4$: $$f(-4)=(-4+3)^2(-4-2)^3=( -1)^2(-6)^3=1\cdot (-216)=-216.$$ - At $x=-3$: $$f(-3)=(0)^2(-5)^3=0.$$ - At $x=-1$: $$f(-1)=(2)^2(-3)^3=4\cdot (-27)=-108.$$ - At $x=2$: $$f(2)=(5)^2(0)^3=0.$$ - At $x=4$: $$f(4)=(7)^2(2)^3=49\cdot 8=392.$$ 5. Identify maximum and minimum. From the values: $$f(-4)=-216,\, f(-3)=0,\, f(-1)=-108,\, f(2)=0,\, f(4)=392.$$ So, $$M=392,\qquad m=-216.$$ 6. Compute $M-m$: $$M-m=392-(-216)=392+216=608.$$ 7. Therefore, the correct option is $$\boxed{608}$$ which is option $\boxed{\text{C}}$. 8. Comparison with stored correct answer: Stored correct answer is C, which matches our result.

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View written solutionFree

Correct answer: C

We need the maximum and minimum values of

\, x\in[-4,4].$$ Since $f$ is continuous on a closed interval, both maximum and minimum exist and occur either at critical points or at endpoints. 2. Differentiate $f(x)$. Using product rule: $$f(x)=(x+3)^2(x-2)^3$$ $$f'(x)=2(x+3)(x-2)^3+(x+3)^2\cdot 3(x-2)^2.$$ Factor common terms: $$f'(x)=(x+3)(x-2)^2\left[2(x-2)+3(x+3)\right].$$ Now simplify the bracket: $$2(x-2)+3(x+3)=2x-4+3x+9=5x+5=5(x+1).$$ Hence, $$f'(x)=5(x+3)(x-2)^2(x+1).$$ 3. Critical points are given by $$f'(x)=0 \implies x=-3,-1,2.$$ All lie in the interval $[-4,4]$. 4. Evaluate $f(x)$ at endpoints and critical points. - At $x=-4$: $$f(-4)=(-4+3)^2(-4-2)^3=( -1)^2(-6)^3=1\cdot (-216)=-216.$$ - At $x=-3$: $$f(-3)=(0)^2(-5)^3=0.$$ - At $x=-1$: $$f(-1)=(2)^2(-3)^3=4\cdot (-27)=-108.$$ - At $x=2$: $$f(2)=(5)^2(0)^3=0.$$ - At $x=4$: $$f(4)=(7)^2(2)^3=49\cdot 8=392.$$ 5. Identify maximum and minimum. From the values: $$f(-4)=-216,\, f(-3)=0,\, f(-1)=-108,\, f(2)=0,\, f(4)=392.$$ So, $$M=392,\qquad m=-216.$$ 6. Compute $M-m$: $$M-m=392-(-216)=392+216=608.$$ 7. Therefore, the correct option is $$\boxed{608}$$ which is option $\boxed{\text{C}}$. 8. Comparison with stored correct answer: Stored correct answer is C, which matches our result.

Application of Derivatives question

Application of Derivatives question

More from Application of Derivatives

Application of Derivatives question

Application of Derivatives question

More from Application of Derivatives