Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Application of Derivatives question

2013 · Shift 2 · Q40
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Mathematics
  4. /Application of Derivatives
  5. /2013 · Shift 2 · Q40

Application of Derivatives question

2013 · Shift 2 · Q40

JEE AdvancedMathematicsApplication of DerivativesMultiple correct+4 / −2
The function f(x)=2∣x∣+∣x+2∣−∣∣x+2∣−2∣x∣∣f(x) = 2\left| x \right| + \left| {x + 2} \right| - \left| {\left| {x + 2} \right| - 2\left| x \right|} \right|f(x)=2∣x∣+∣x+2∣−∣∣x+2∣−2∣x∣∣ has a local minimum or a local maximum at x =
  1. A
    −-− 2
  2. B
    −23{{ - 2} \over 3}3−2​
  3. C
    2
  4. D
    23{{ 2} \over 3}32​
View written solutionFree

Correct answer: A, B

  1. Given function

f(x)=2∣x∣+∣x+2∣−∣∣x+2∣−2∣x∣∣f(x)=2|x|+|x+2|-\left||x+2|-2|x|\right|f(x)=2∣x∣+∣x+2∣−∣∣x+2∣−2∣x∣∣

We need to find where this function has a local minimum or local maximum among the given options.

  1. Use the identity

For any real numbers a,b≥0a,b\ge 0a,b≥0,

a+b−∣a−b∣=2min⁡(a,b).a+b-|a-b|=2\min(a,b).a+b−∣a−b∣=2min(a,b).

Here,

a=2∣x∣,b=∣x+2∣,a=2|x|,\qquad b=|x+2|,a=2∣x∣,b=∣x+2∣,

both are non-negative. Hence

f(x)=2min⁡(2∣x∣,∣x+2∣).f(x)=2\min(2|x|,|x+2|).f(x)=2min(2∣x∣,∣x+2∣).

So we now study

\begin{cases} 4|x|,& 2|x|\le |x+2|,\\[4pt] 2|x+2|,& 2|x|\ge |x+2|. \end{cases}$$ Critical behavior can occur where the expressions inside modulus change sign or where $$2|x|=|x+2|.$$ --- 3. **Solve** $2|x|=|x+2|$ We square both sides: $$4x^2=(x+2)^2$$ $$4x^2=x^2+4x+4$$ $$3x^2-4x-4=0$$ So, $$x=\frac{4\pm 8}{6}$$ Thus, $$x=2,\qquad x=-\frac23.$$ Also the modulus break points are at $$x=0,\qquad x=-2.$$ So we check intervals: $$(-\infty,-2),\ (-2,-\tfrac23),\ (-\tfrac23,0),\ (0,2),\ (2,\infty).$$ --- 4. **Find explicit form on each interval** ### (i) $x<-2$ Then $$|x|=-x,\qquad |x+2|=-(x+2)=-x-2.$$ Compare: $$2|x|=-2x,\qquad |x+2|=-x-2.$$ Since $x<-2$, we have $$-2x>-x-2,$$ so minimum is $|x+2|$. Therefore $$f(x)=2|x+2|=2(-x-2)=-2x-4.$$ Slope $=-2$. --- ### (ii) $-2<x<0$ Then $$|x|=-x,\qquad |x+2|=x+2.$$ Compare: $$2|x|=-2x,\qquad |x+2|=x+2.$$ Solve $$-2x\le x+2 \iff -3x\le 2 \iff x\ge -\frac23.$$ So: - for $-2<x< -\frac23$, minimum is $x+2$, hence $$f(x)=2(x+2)=2x+4;$$ - for $-\frac23 <x<0$, minimum is $-2x$, hence $$f(x)=2(-2x)=-4x.$$ --- ### (iii) $x>0$ Then $$|x|=x,\qquad |x+2|=x+2.$$ Compare: $$2x\le x+2 \iff x\le 2.$$ So: - for $0<x<2$, minimum is $2x$, hence $$f(x)=2(2x)=4x;$$ - for $x>2$, minimum is $x+2$, hence $$f(x)=2(x+2)=2x+4.$$ --- 5. **Collect piecewise form** $$f(x)= \begin{cases} -2x-4, & x<-2,\\ 2x+4, & -2<x< -\dfrac23,\\ -4x, & -\dfrac23 <x<0,\\ 4x, & 0<x<2,\\ 2x+4, & x>2. \end{cases}$$ Now inspect slope changes at candidate points. --- 6. **Check each option** ### Option A: $x=-2$ Left slope: $-2$ Right slope: $+2$ Since slope changes from negative to positive, $f$ decreases then increases. So $x=-2$ is a **local minimum**. ✅ **A is correct**. --- ### Option B: $x=-\dfrac23$ Left slope: $+2$ Right slope: $-4$ Since slope changes from positive to negative, $f$ increases then decreases. So $x=-\dfrac23$ is a **local maximum**. ✅ **B is correct**. --- ### Option C: $x=2$ Left slope: $+4$ Right slope: $+2$ Slope is positive on both sides, so function is increasing on both sides. No local max/min at $x=2$. ❌ **C is incorrect**. --- ### Option D: $x=\dfrac23$ This lies in interval $(0,2)$ where $$f(x)=4x,$$ which is strictly increasing. Hence no local extremum. ❌ **D is incorrect**. --- 7. **Final derived answer** The function has a local minimum or maximum at $$x=-2,\quad x=-\frac23.$$ So the correct options are **A, B**. --- 8. **Comparison with stored answer** Stored correct answer: **A, B** This matches our derived answer.
PreviousNext

More from Application of Derivatives

  • Let f:IR→IR be defined as f(x)=∣x∣+​x2−1​. The total number of points at which f attains either a local maximum or a local minimum is2012 · Numerical
  • Let p(x) be a real polynomial of least degree which has a local maximum at x=1 and a local minimum at x=3. If p(1)=6 and p(3)=2, then p′(0) is2012 · Numerical
  • Let f(x)=(1−x)2sin2x+x2 for all x∈IR and let g(x)=1∫x​(t+12(t−1)​−Int)f(t)dt for…2012 · MCQ
  • Let f(x)=(1−x)2sin2x+x2 for all x∈IR and let g(x)=1∫x​(t+12(t−1)​−Int)f(t)dt for…2012 · MCQ
  • If f(x)=∫0x​et2(t−2)(t−3)dt for all x∈(0,∞), then2012 · Multiple correct
  • Let f be a real-valued differentiable function on R(the set of all real numbers) such that f(1)=1. If the y-intercept of the tangent at any point P(x,y) on the curve y=f(x) is equal to the cube of the abscissa of P, then find…2010 · Numerical
  • Let f be a function defined on R(the set of all real numbers) such that f′(x)=2010(x−2009)(x−2010)2(x−2011)3(x−2012)4 for all x∈R If g…2010 · Numerical
  • The maximum value of the function f(x)=2x3−15x2+36x−48 on the set A={x∣x2+20≤9x∣} is ​.2009 · Numerical