Question:
If a curve passes through the point \(\left(2, \frac{7}{2}\right)\) and has slope \(\left(1-\frac{1}{x^2}\right)\) at any point \((x, y)\) on it, then the ordinate of the point on the curve whose abscissa is \(-2\) is :
If a curve passes through the point \(\left(2, \frac{7}{2}\right)\) and has slope \(\left(1-\frac{1}{x^2}\right)\) at any point \((x, y)\) on it, then the ordinate of the point on the curve whose abscissa is \(-2\) is :
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Integrate the slope and fix the constant using the given point.
Updated On: Oct 31, 2026
- \(-\frac{3}{2}\)
- \(\frac{3}{2}\)
- \(\frac{5}{2}\)
- \(-\frac{5}{2}\)
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The Correct Option is A
Solution and Explanation
\(\dfrac{dy}{dx}=1-\dfrac{1}{x^2} \Rightarrow y=x+\dfrac{1}{x}+C\).
Through \(\left(2,\tfrac72\right)\): \(\tfrac72=2+\tfrac12+C \Rightarrow C=1\).
Curve: \(y=x+\dfrac1x+1\).
At \(x=-2\): \(y=-2-\dfrac12+1=-\dfrac32\).
Answer: (1) \(-\frac{3}{2}\)
Through \(\left(2,\tfrac72\right)\): \(\tfrac72=2+\tfrac12+C \Rightarrow C=1\).
Curve: \(y=x+\dfrac1x+1\).
At \(x=-2\): \(y=-2-\dfrac12+1=-\dfrac32\).
Answer: (1) \(-\frac{3}{2}\)
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