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Application of Derivatives question

2021 · 20 Jul · Shift 1 · Q30
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  5. /2021 · 20 Jul · Shift 1 · Q30

Application of Derivatives question

2021 · 20 Jul · Shift 1 · Q30

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
Let A=[aij]A = [{a_{ij}}]A=[aij​] be a 3 ×\times× 3 matrix, where aij={1,if i=j−x,if ∣i−j∣=12x+1,otherwise.{a_{ij}} = \left\{ {\begin{matrix} 1 & , & {if\,i = j} \\ { - x} & , & {if\,\left| {i - j} \right| = 1} \\ {2x + 1} & , & {otherwise.} \\ \end{matrix} } \right.aij​=⎩⎨⎧​1−x2x+1​,,,​ifi=jif∣i−j∣=1otherwise.​ Let a function f : R →\to→ R be defined as f(x) = det(A). Then the sum of maximum and minimum values of f on R is equal to:
  1. A
    −2027- {{20} \over {27}}−2720​
  2. B
    8827{{88} \over {27}}2788​
  3. C
    2027{{20} \over {27}}2720​
  4. D
    −8827- {{88} \over {27}}−2788​
View written solutionFree

Correct answer: D

  1. Form the matrix AAA explicitly

Given

aij={1,i=j−x,∣i−j∣=12x+1,otherwisea_{ij}= \begin{cases} 1, & i=j \\ -x, & |i-j|=1 \\ 2x+1, & \text{otherwise} \end{cases}aij​=⎩⎨⎧​1,−x,2x+1,​i=j∣i−j∣=1otherwise​

for a 3×33\times 33×3 matrix.

So,

A=(1−x2x+1−x1−x2x+1−x1).A= \begin{pmatrix} 1 & -x & 2x+1\\ -x & 1 & -x\\ 2x+1 & -x & 1 \end{pmatrix}.A=​1−x2x+1​−x1−x​2x+1−x1​​.

We need

f(x)=det⁡(A).f(x)=\det(A).f(x)=det(A).
  1. Compute the determinant

Expand along the first row:

f(x)=1∣1−x−x1∣−(−x)∣−x−x2x+11∣+(2x+1)∣−x12x+1−x∣.f(x)=1\begin{vmatrix}1 & -x\\ -x & 1\end{vmatrix} -(-x)\begin{vmatrix}-x & -x\\ 2x+1 & 1\end{vmatrix} +(2x+1)\begin{vmatrix}-x & 1\\ 2x+1 & -x\end{vmatrix}.f(x)=1​1−x​−x1​​−(−x)​−x2x+1​−x1​​+(2x+1)​−x2x+1​1−x​​.

Now compute each minor:

  • First minor:
∣1−x−x1∣=1−x2.\begin{vmatrix}1 & -x\\ -x & 1\end{vmatrix}=1-x^2.​1−x​−x1​​=1−x2.
  • Second minor:
∣−x−x2x+11∣=(−x)(1)−(−x)(2x+1)=−x+x(2x+1)=2x2.\begin{vmatrix}-x & -x\\ 2x+1 & 1\end{vmatrix} =(-x)(1)-(-x)(2x+1) =-x+x(2x+1)=2x^2.​−x2x+1​−x1​​=(−x)(1)−(−x)(2x+1)=−x+x(2x+1)=2x2.

So the second term is

−(−x)(2x2)=x(2x2)=2x3.-(-x)(2x^2)=x(2x^2)=2x^3.−(−x)(2x2)=x(2x2)=2x3.
  • Third minor:
∣−x12x+1−x∣=(−x)(−x)−1(2x+1)=x2−2x−1.\begin{vmatrix}-x & 1\\ 2x+1 & -x\end{vmatrix} =(-x)(-x)-1(2x+1)=x^2-2x-1.​−x2x+1​1−x​​=(−x)(−x)−1(2x+1)=x2−2x−1.

So the third term is

(2x+1)(x2−2x−1).(2x+1)(x^2-2x-1).(2x+1)(x2−2x−1).

Expand:

(2x+1)(x2−2x−1)=2x3−3x2−4x−1.(2x+1)(x^2-2x-1)=2x^3-3x^2-4x-1.(2x+1)(x2−2x−1)=2x3−3x2−4x−1.

Therefore,

f(x)=(1−x2)+2x3+(2x3−3x2−4x−1).f(x)=(1-x^2)+2x^3+(2x^3-3x^2-4x-1).f(x)=(1−x2)+2x3+(2x3−3x2−4x−1).

Hence,

f(x)=4x3−4x2−4x.f(x)=4x^3-4x^2-4x.f(x)=4x3−4x2−4x.

So

f(x)=4x(x2−x−1).f(x)=4x(x^2-x-1).f(x)=4x(x2−x−1).
  1. Check whether maximum and minimum on R\mathbb RR exist

Since

f(x)=4x3−4x2−4xf(x)=4x^3-4x^2-4xf(x)=4x3−4x2−4x

is a cubic polynomial with positive leading coefficient, we have

lim⁡x→+∞f(x)=+∞,lim⁡x→−∞f(x)=−∞.\lim_{x\to +\infty} f(x)=+\infty, \qquad \lim_{x\to -\infty} f(x)=-\infty.x→+∞lim​f(x)=+∞,x→−∞lim​f(x)=−∞.

Therefore, on R\mathbb RR, the function is unbounded above and below. Hence it has no absolute maximum and no absolute minimum on R\mathbb RR.

So, as stated, the phrase "sum of maximum and minimum values of fff on R\mathbb RR" is not meaningful.

  1. Likely intended meaning: sum of local maximum and local minimum values

For such questions, usually they ask for the sum of the local maximum and local minimum values.

Differentiate:

f′(x)=12x2−8x−4=4(3x2−2x−1).f'(x)=12x^2-8x-4=4(3x^2-2x-1).f′(x)=12x2−8x−4=4(3x2−2x−1).

Set f′(x)=0f'(x)=0f′(x)=0:

3x2−2x−1=0.3x^2-2x-1=0.3x2−2x−1=0.

Factor:

3x2−2x−1=(3x+1)(x−1)=0.3x^2-2x-1=(3x+1)(x-1)=0.3x2−2x−1=(3x+1)(x−1)=0.

Thus critical points are

x=1,x=−13.x=1,\qquad x=-\frac13.x=1,x=−31​.

Second derivative:

f′′(x)=24x−8.f''(x)=24x-8.f′′(x)=24x−8.
  • At x=1x=1x=1,
f′′(1)=16>0,f''(1)=16>0,f′′(1)=16>0,

so x=1x=1x=1 gives a local minimum.

  • At x=−13x=-\frac13x=−31​,
f′′(−13)=−16<0,f''\left(-\frac13\right)=-16<0,f′′(−31​)=−16<0,

so x=−13x=-\frac13x=−31​ gives a local maximum.

Now compute the values:

  • Local minimum value:
f(1)=4−4−4=−4.f(1)=4-4-4=-4.f(1)=4−4−4=−4.
  • Local maximum value:
f(−13)=4(−127)−4(19)−4(−13)=−427−1227+3627=2027.f\left(-\frac13\right)=4\left(-\frac1{27}\right)-4\left(\frac19\right)-4\left(-\frac13\right) =-\frac4{27}-\frac{12}{27}+\frac{36}{27} =\frac{20}{27}.f(−31​)=4(−271​)−4(91​)−4(−31​)=−274​−2712​+2736​=2720​.

So the sum is

−4+2027=−108+2027=−8827.-4+\frac{20}{27}=\frac{-108+20}{27}=-\frac{88}{27}.−4+2720​=27−108+20​=−2788​.
  1. Conclusion
  • Strictly speaking, fff has no global maximum/minimum on R\mathbb RR.
  • Interpreting the question in the standard exam sense as the sum of local maximum and minimum values, we get
−8827.-\frac{88}{27}.−2788​.

This matches option D.

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