KCET · Maths · Application of Derivatives
The equation of the curve passing through the point \( (1,1) \) such that the slope of the tangent at
any point \( (x, y) \) is equal to the product of its co-ordinates is
- A \( 2 \log y=x^{2}+1 \)
- B \( 2 \log x=y^{2}+1 \)
- C \( 2 \log x=y^{2}-1 \)
- D \( 2 \log y=x^{2}-1 \)
Answer & Solution
Correct Answer
(D) \( 2 \log y=x^{2}-1 \)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=x y\)
\(\frac{1}{y} d y=x d x\)
Integrate
\(\log y=\frac{x^{2}}{2}+C \ldots(1)\)
Equation \((1)\) passing through \((1,1)\)
\(\log (1)=\frac{1^{2}}{2}+C\)
\(0=\frac{1}{2}+C\)
\(C=-\frac{1}{2}\)
\(\log y=\frac{\chi^{2}}{2}-\frac{1}{2}\)
\(2 \log y=\chi^{2}-1\)
\(\frac{1}{y} d y=x d x\)
Integrate
\(\log y=\frac{x^{2}}{2}+C \ldots(1)\)
Equation \((1)\) passing through \((1,1)\)
\(\log (1)=\frac{1^{2}}{2}+C\)
\(0=\frac{1}{2}+C\)
\(C=-\frac{1}{2}\)
\(\log y=\frac{\chi^{2}}{2}-\frac{1}{2}\)
\(2 \log y=\chi^{2}-1\)
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