A) \[x+5y=2\]
B) \[x-5y=2\]
C) \[5x-y=2\]
D) \[5x+y-2=0\]
Correct Answer: A
Solution :
\[(1+{{x}^{2}})y=2-x\] ......(i) It meets x-axis, where \[y=0\] i.e., \[0=2-x\Rightarrow x=2\] So, (i) meets x-axis at the point (2, 0) Also from (i), \[y=\frac{2-x}{1+{{x}^{2}}}\] Þ \[\frac{dy}{dx}=\frac{(1+{{x}^{2}})\,(-1)-(2-x)\,(2x)}{{{(1+{{x}^{2}})}^{2}}}\]Þ \[\frac{dy}{dx}=\frac{{{x}^{2}}-4x-1}{{{(1+{{x}^{2}})}^{2}}}\] Slope of tangent at (2, 0)is,\[\frac{4-8-1}{{{(1+4)}^{2}}}=\frac{-5}{25}=\frac{-1}{5}\] \ Equation of tangent at (2, 0) is , \[y-0=-\frac{1}{5}(x-2)\Rightarrow x+5y=2\].
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