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Application of Derivatives question

2022 · 26 Jun · Shift 1 · Q27
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  5. /2022 · 26 Jun · Shift 1 · Q27

Application of Derivatives question

2022 · 26 Jun · Shift 1 · Q27

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
The sum of the absolute minimum and the absolute maximum values of the function f(x) = |3x −-− x2 + 2|−-− x in the interval [−-− 1, 2] is :
  1. A
    17+32{{\sqrt {17} + 3} \over 2}217​+3​
  2. B
    17+52{{\sqrt {17} + 5} \over 2}217​+5​
  3. C
    5
  4. D
    9−172{{9 - \sqrt {17} } \over 2}29−17​​
View written solutionFree

Correct answer: A

  1. Given function

We need the absolute minimum and absolute maximum of f(x)=∣3x−x2+2∣−xf(x)=|3x-x^2+2|-xf(x)=∣3x−x2+2∣−x on the interval [−1,2].[-1,2].[−1,2].

Let g(x)=3x−x2+2.g(x)=3x-x^2+2.g(x)=3x−x2+2. Then f(x)=∣g(x)∣−x.f(x)=|g(x)|-x.f(x)=∣g(x)∣−x.

  1. Check the sign of g(x)g(x)g(x) on [−1,2][-1,2][−1,2]

g(x)=−x2+3x+2.g(x)= -x^2+3x+2.g(x)=−x2+3x+2. Its zeros are obtained from −x2+3x+2=0  ⟺  x2−3x−2=0.-x^2+3x+2=0 \iff x^2-3x-2=0.−x2+3x+2=0⟺x2−3x−2=0. So x=3±9+82=3±172.x=\frac{3\pm\sqrt{9+8}}{2}=\frac{3\pm\sqrt{17}}{2}.x=23±9+8​​=23±17​​.

These are approximately 3−172≈−0.56,3+172≈3.56.\frac{3-\sqrt{17}}{2}\approx -0.56, \qquad \frac{3+\sqrt{17}}{2}\approx 3.56.23−17​​≈−0.56,23+17​​≈3.56.

Hence on [−1,2][-1,2][−1,2], the expression g(x)g(x)g(x) changes sign at x0=3−172.x_0=\frac{3-\sqrt{17}}{2}.x0​=23−17​​.

Therefore,

  • for x∈[−1,3−172],x\in\left[-1,\frac{3-\sqrt{17}}{2}\right],x∈[−1,23−17​​], we have g(x)≤0g(x)\le 0g(x)≤0,
  • for x∈[3−172,2],x\in\left[\frac{3-\sqrt{17}}{2},2\right],x∈[23−17​​,2], we have g(x)≥0.g(x)\ge 0.g(x)≥0.

So we write f(x)f(x)f(x) piecewise.

  1. Piecewise form of f(x)f(x)f(x)

Case 1:

If x∈[−1,3−172],x\in\left[-1,\frac{3-\sqrt{17}}{2}\right],x∈[−1,23−17​​], then ∣g(x)∣=−g(x)=x2−3x−2.|g(x)|=-g(x)=x^2-3x-2.∣g(x)∣=−g(x)=x2−3x−2. Thus f(x)=x2−3x−2−x=x2−4x−2.f(x)=x^2-3x-2-x=x^2-4x-2.f(x)=x2−3x−2−x=x2−4x−2.

Case 2:

If x∈[3−172,2],x\in\left[\frac{3-\sqrt{17}}{2},2\right],x∈[23−17​​,2], then ∣g(x)∣=g(x)=−x2+3x+2.|g(x)|=g(x)=-x^2+3x+2.∣g(x)∣=g(x)=−x2+3x+2. Thus f(x)=−x2+3x+2−x=−x2+2x+2.f(x)=-x^2+3x+2-x=-x^2+2x+2.f(x)=−x2+3x+2−x=−x2+2x+2.

So,

\begin{cases} x^2-4x-2, & -1\le x\le \dfrac{3-\sqrt{17}}{2},\\[1ex] -x^2+2x+2, & \dfrac{3-\sqrt{17}}{2}\le x\le 2. \end{cases}$$ --- 4. **Find extrema on each interval** We must check: 1. endpoints: $$x=-1,2$$ 2. joining point: $$x=\dfrac{3-\sqrt{17}}{2}$$ 3. critical points inside each subinterval. --- 5. **First piece:** $$f(x)=x^2-4x-2$$ Differentiate: $$f'(x)=2x-4.$$ Set $$f'(x)=0$$: $$2x-4=0 \implies x=2.$$ But $$x=2$$ is **not** in the first interval. So no interior critical point here. Thus extrema on the first interval occur at its endpoints. - At $$x=-1$$: $$f(-1)=|3(-1)-(-1)^2+2|-(-1)=|-3-1+2|+1=|-2|+1=3.$$ - At $$x=\dfrac{3-\sqrt{17}}{2}$$, since this is where $$g(x)=0$$, $$f\left(\frac{3-\sqrt{17}}{2}\right)=0-\frac{3-\sqrt{17}}{2}=\frac{\sqrt{17}-3}{2}.$$ --- 6. **Second piece:** $$f(x)=-x^2+2x+2$$ Differentiate: $$f'(x)=-2x+2.$$ Set $$f'(x)=0$$: $$-2x+2=0 \implies x=1.$$ This lies in the second interval, so check it. - At $$x=1$$: $$f(1)=|3(1)-1+2|-1=|4|-1=3.$$ - At $$x=2$$: $$f(2)=|6-4+2|-2=|4|-2=2.$$ - At $$x=\dfrac{3-\sqrt{17}}{2}$$: $$f\left(\frac{3-\sqrt{17}}{2}\right)=\frac{\sqrt{17}-3}{2}.$$ --- 7. **Compare all candidate values** The values obtained are: $$f(-1)=3,$$ $$f(1)=3,$$ $$f(2)=2,$$ $$f\left(\frac{3-\sqrt{17}}{2}\right)=\frac{\sqrt{17}-3}{2}.$$ Now, $$\frac{\sqrt{17}-3}{2}\approx \frac{4.123-3}{2}\approx 0.56.$$ So, - **Absolute minimum** is $$\frac{\sqrt{17}-3}{2}$$ - **Absolute maximum** is $$3$$ Their sum is $$3+\frac{\sqrt{17}-3}{2}= rac{6+\sqrt{17}-3}{2}= rac{\sqrt{17}+3}{2}.$$ --- 8. **Final answer** $$\boxed{\frac{\sqrt{17}+3}{2}}$$ This corresponds to **Option A**.
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