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Sequences and Series question

2021 · 25 Feb · Shift 2 · Q34
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  5. /2021 · 25 Feb · Shift 2 · Q34

Sequences and Series question

2021 · 25 Feb · Shift 2 · Q34

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
The minimum value of f(x)=aax+a1−axf(x) = {a^{{a^x}}} + {a^{1 - {a^x}}}f(x)=aax+a1−ax, where a, x∈Rx \in Rx∈R and a > 0, is equal to :
  1. A
    a+1aa + {1 \over a}a+a1​
  2. B
    2a
  3. C
    a + 1
  4. D
    2a2\sqrt a2a​
View written solutionFree

Correct answer: D

  1. We need to find the minimum value of f(x)=aax+a1−ax,a>0.f(x)=a^{a^x}+a^{1-a^x}, \\ a>0.f(x)=aax+a1−ax,a>0.

  2. Let t=ax.t=a^x.t=ax. Since a>0a>0a>0, we have t>0t>0t>0.

So the function becomes f=at+a1−t.f=a^t+a^{1-t}.f=at+a1−t. Now rewrite the second term: a1−t=a⋅a−t.a^{1-t}=a\cdot a^{-t}.a1−t=a⋅a−t. Thus, f=at+aat.f=a^t+\frac{a}{a^t}.f=at+ata​.

  1. Put y=at.y=a^t.y=at. Since a>0a>0a>0, we again have y>0y>0y>0. Then f=y+ay.f=y+\frac{a}{y}.f=y+ya​.

  2. Now minimize y+ay,y>0.y+\frac{a}{y}, \quad y>0.y+ya​,y>0. By AM-GM, y+ay≥2y⋅ay=2a.y+\frac{a}{y} \ge 2\sqrt{y\cdot \frac{a}{y}}=2\sqrt{a}.y+ya​≥2y⋅ya​​=2a​. Equality holds when y=ay  ⟹  y2=a  ⟹  y=ay=\frac{a}{y} \implies y^2=a \implies y=\sqrt ay=ya​⟹y2=a⟹y=a​ (because y>0y>0y>0).

Hence the minimum value is 2a.\boxed{2\sqrt a}.2a​​.

  1. Therefore, the correct option is: D\boxed{\text{D}}D​

  2. Comparison with stored answer: Stored correct answer is D, which matches our result.

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