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Application of Derivatives question

2022 · Shift 2 · Q28
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Application of Derivatives question

2022 · Shift 2 · Q28

JEE AdvancedMathematicsApplication of DerivativesMultiple correct+4 / −2
Let α=∑k=1∞sin⁡2k(π6)\alpha=\sum\limits_{k = 1}^\infty {{{\sin }^{2k}}\left( {{\pi \over 6}} \right)}α=k=1∑∞​sin2k(6π​) Let g:[0,1]→Rg:[0,1] \rightarrow \mathbb{R}g:[0,1]→R be the function defined by g(x)=2αx+2α(1−x).g(x)=2^{\alpha x}+2^{\alpha(1-x)} .g(x)=2αx+2α(1−x). Then, which of the following statements is/are TRUE ?
  1. A
    The minimum value of g(x)g(x)g(x) is 2762^{\frac{7}{6}}267​
  2. B
    The maximum value of g(x)g(x)g(x) is 1+2131+2^{\frac{1}{3}}1+231​
  3. C
    The function g(x)g(x)g(x) attains its maximum at more than one point
  4. D
    The function g(x)g(x)g(x) attains its minimum at more than one point
View written solutionFree

Correct answer: A, B, C

  1. Compute α\alphaα

We have

α=∑k=1∞sin⁡2k(π6).\alpha=\sum_{k=1}^{\infty} \sin^{2k}\left(\frac{\pi}{6}\right).α=k=1∑∞​sin2k(6π​).

Since

sin⁡(π6)=12,\sin\left(\frac{\pi}{6}\right)=\frac12,sin(6π​)=21​,

so

sin⁡2k(π6)=(12)2k=(14)k.\sin^{2k}\left(\frac{\pi}{6}\right)=\left(\frac12\right)^{2k}=\left(\frac14\right)^k.sin2k(6π​)=(21​)2k=(41​)k.

Thus α\alphaα is a geometric series:

\alpha=\sum_{k=1}^{\infty} \left(\frac14\right)^k= rac{\frac14}{1-\frac14}= rac{1}{3}.

So the function becomes

g(x)=2x/3+2(1−x)/3,x∈[0,1].g(x)=2^{x/3}+2^{(1-x)/3}, \qquad x\in[0,1].g(x)=2x/3+2(1−x)/3,x∈[0,1].
  1. Observe symmetry

Note that

g(1−x)=2(1−x)/3+2x/3=g(x).g(1-x)=2^{(1-x)/3}+2^{x/3}=g(x).g(1−x)=2(1−x)/3+2x/3=g(x).

So the function is symmetric about x=12x=\tfrac12x=21​.

  1. Find critical points using derivative

Differentiate:

g′(x)=ln⁡232x/3−ln⁡232(1−x)/3=ln⁡23(2x/3−2(1−x)/3).g'(x)=\frac{\ln 2}{3}2^{x/3}-\frac{\ln 2}{3}2^{(1-x)/3} =\frac{\ln 2}{3}\left(2^{x/3}-2^{(1-x)/3}\right).g′(x)=3ln2​2x/3−3ln2​2(1−x)/3=3ln2​(2x/3−2(1−x)/3).

Set g′(x)=0g'(x)=0g′(x)=0:

2x/3=2(1−x)/3.2^{x/3}=2^{(1-x)/3}.2x/3=2(1−x)/3.

Since exponential is one-one,

x3=1−x3  ⟹  x=1−x  ⟹  x=12.\frac{x}{3}=\frac{1-x}{3} \implies x=1-x \implies x=\frac12.3x​=31−x​⟹x=1−x⟹x=21​.

So the only critical point is x=12x=\frac12x=21​.

  1. Determine minimum/maximum

For x<12x<\frac12x<21​, we have x<1−xx<1-xx<1−x, hence

2x/3<2(1−x)/3  ⟹  g′(x)<0.2^{x/3}<2^{(1-x)/3} \implies g'(x)<0.2x/3<2(1−x)/3⟹g′(x)<0.

For x>12x>\frac12x>21​, we have x>1−xx>1-xx>1−x, hence

2x/3>2(1−x)/3  ⟹  g′(x)>0.2^{x/3}>2^{(1-x)/3} \implies g'(x)>0.2x/3>2(1−x)/3⟹g′(x)>0.

Therefore, g(x)g(x)g(x) decreases on [0,12][0,\tfrac12][0,21​] and increases on [12,1][\tfrac12,1][21​,1]. So x=12x=\tfrac12x=21​ gives the minimum.

Now evaluate:

g(12)=21/6+21/6=2⋅21/6=27/6.g\left(\frac12\right)=2^{1/6}+2^{1/6}=2\cdot 2^{1/6}=2^{7/6}.g(21​)=21/6+21/6=2⋅21/6=27/6.

So the minimum value is

27/6.2^{7/6}.27/6.

Hence A is true.

  1. Find maximum value

Since the function decreases then increases, the maximum on [0,1][0,1][0,1] occurs at the endpoints x=0x=0x=0 and x=1x=1x=1.

Evaluate:

g(0)=20+21/3=1+21/3,g(0)=2^0+2^{1/3}=1+2^{1/3},g(0)=20+21/3=1+21/3, g(1)=21/3+20=1+21/3.g(1)=2^{1/3}+2^0=1+2^{1/3}.g(1)=21/3+20=1+21/3.

Thus the maximum value is

1+21/3.1+2^{1/3}.1+21/3.

Hence B is true.

Also, the maximum is attained at two distinct points x=0x=0x=0 and x=1x=1x=1, so C is true.

  1. Check statement about minimum attained at more than one point

The only critical point is x=12x=\frac12x=21​, and the function is strictly decreasing before it and strictly increasing after it. Therefore the minimum is attained at exactly one point.

So D is false.

  1. Final conclusion

The true statements are:

A, B, C\boxed{A,\ B,\ C}A, B, C​

This matches the stored correct answer.

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