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Application of Derivatives question

2025 · 3 Apr · Shift 2 · Q43
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  5. /2025 · 3 Apr · Shift 2 · Q43

Application of Derivatives question

2025 · 3 Apr · Shift 2 · Q43

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
Let f:R→Rf: \mathrm{R} \rightarrow \mathrm{R}f:R→R be a function defined by f(x)=∣∣x+2∣−2∣x∥f(x)=||x+2|-2| x \|f(x)=∣∣x+2∣−2∣x∥. If mmm is the number of points of local minima and nnn is the number of points of local maxima of fff, then m+nm+nm+n is
  1. A
    3
  2. B
    4
  3. C
    2
  4. D
    5
View written solutionFree

Correct answer: A

We need to analyze f(x)=∣∣x+2∣−2∣ ∣x∣.f(x)=\bigl||x+2|-2\bigr|\,|x|.f(x)=​∣x+2∣−2​∣x∣.

Let g(x)=∣∣x+2∣−2∣.g(x)=\bigl||x+2|-2\bigr|.g(x)=​∣x+2∣−2​. Then f(x)=g(x)∣x∣.f(x)=g(x)|x|.f(x)=g(x)∣x∣.

We will first simplify g(x)g(x)g(x) piecewise.

1. Simplify g(x)=∣∣x+2∣−2∣g(x)=\bigl||x+2|-2\bigr|g(x)=​∣x+2∣−2​

Consider ranges based on x+2x+2x+2.

Case 1: x≥−2x\ge -2x≥−2

Then ∣x+2∣=x+2,|x+2|=x+2,∣x+2∣=x+2, so g(x)=∣x+2−2∣=∣x∣.g(x)=|x+2-2|=|x|.g(x)=∣x+2−2∣=∣x∣.

Case 2: x<−2x<-2x<−2

Then ∣x+2∣=−(x+2)=−x−2,|x+2|=-(x+2)=-x-2,∣x+2∣=−(x+2)=−x−2, so g(x)=∣−x−2−2∣=∣−x−4∣=∣x+4∣.g(x)=|-x-2-2|=|-x-4|=|x+4|.g(x)=∣−x−2−2∣=∣−x−4∣=∣x+4∣.

Hence

\begin{cases} |x+4|, & x<-2,\\ |x|, & x\ge -2. \end{cases}$$ Therefore, $$f(x)= \begin{cases} |x+4|\,|x|, & x<-2,\\ |x|\,|x|=x^2, & x\ge -2. \end{cases}$$ Now simplify the first part further. --- ### 2. Write $f(x)$ completely piecewise For $x<-2$, we split at $x=-4$. #### (i) If $x<-4$ Then $$|x+4|=-(x+4), \qquad |x|=-x.$$ So $$f(x)=(-(x+4))(-x)=x(x+4)=x^2+4x.$$ #### (ii) If $-4\le x<-2$ Then $$|x+4|=x+4, \qquad |x|=-x.$$ So $$f(x)=(x+4)(-x)=-x^2-4x.$$ #### (iii) If $x\ge -2$ We already found $$f(x)=x^2.$$ So, $$f(x)= \begin{cases} x^2+4x, & x<-4,\\ -x^2-4x, & -4\le x<-2,\\ x^2, & x\ge -2. \end{cases}$$ --- ### 3. Find local extrema on each interval Differentiate on each interval. #### For $x<-4$ $$f(x)=x^2+4x \implies f'(x)=2x+4.$$ For $x<-4$, we have $2x+4<0$, so $f$ is decreasing on $(-\infty,-4)$. #### For $-4<x<-2$ $$f(x)=-x^2-4x \implies f'(x)=-2x-4.$$ For $-4<x<-2$, we have $-2x-4>0$, so $f$ is increasing on $(-4,-2)$. #### For $x>-2$ $$f(x)=x^2 \implies f'(x)=2x.$$ So: - decreasing on $(-2,0)$, - increasing on $(0,\infty)$. --- ### 4. Check the junction points The possible local extrema occur at the points where behavior changes: $x=-4,-2,0$. #### At $x=-4$ - To the left, $f$ is decreasing. - To the right, $f$ is increasing. So $x=-4$ is a **local minimum**. Also, $$f(-4)=|-2|-?$$ Using piecewise form, $$f(-4)=(-4)^2+4(-4)=16-16=0.$$ #### At $x=-2$ - On the left, $f$ is increasing. - On the right, for $(-2,0)$, $f$ is decreasing. So $x=-2$ is a **local maximum**. Value: $$f(-2)=(-2)^2=4.$$ #### At $x=0$ - On the left, $f$ is decreasing. - On the right, $f$ is increasing. So $x=0$ is a **local minimum**. Value: $$f(0)=0.$$ --- ### 5. Count minima and maxima - Number of points of local minima: $$m=2$$ at $x=-4,0$. - Number of points of local maxima: $$n=1$$ at $x=-2$. Therefore, $$m+n=2+1=3.$$ --- ### 6. Compare with stored answer Our derived answer is **3**, which matches option **A**. So the stored correct answer is correct.
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