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CBSE Class 12 Math 2025 All Sets Solved Paper

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Question : 4 of 20
Marks: +1, -0
If ∫  21xx2dx=k⋅21x+C,\int\;\frac{2^{\frac{1}{x}}}{x^{2}} dx = k \cdot 2^{\frac{1}{x}} + C, then k is equal to
Solution:  
∫  21xx2dx=K⋅21x+C\int\;\frac{2^{\frac{1}{x}}}{x^{2}} dx = K \cdot 2^{\frac{1}{x}} + C
Let   1x=t\;\frac{1}{x} = t
Differentiating both sides
  −1x2dx=dt\;\frac{-1}{x^{2}} dx = d t
Putting   1x=t\;\frac{1}{x} = t and   dxx2=−dt\;\frac{d x}{x^{2}} = -d t
⇒∫  21xx2dx=∫2t(−dt)\Rightarrow \int\;\frac{2^{\frac{1}{x}}}{x^{2}} dx = \int 2^{t} (-d t)
=−∫2tdt= -\int 2^{t} dt
=  −2tln⁡2+C=\;\frac{-2^{t}}{\ln 2} + C
=  −21xln⁡2+C=\;\frac{-2^{\frac{1}{x}}}{\ln 2} + C
⇒k=  −1ln⁡2 or   −1log⁡2\Rightarrow k = \;\frac{-1}{\ln 2} \text{ or } \;\frac{-1}{\log 2}
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