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If the equation of the normal to the curve y=\frac{x-a}{(x+b)(x-2)} at the point (1,-3)$ is $x-4 y=13, then the value of a+b is equal to

Option: 1

-6


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

(1, –3) is on the curve

\begin{aligned} & \therefore-3=\frac{1-a}{(1+\mathrm{b})(1-2)} \Rightarrow-3=\frac{1-\mathrm{a}}{(-1)(1+\mathrm{b})} \\ & \Rightarrow 3+3 \mathrm{~b}=1-\mathrm{a} \Rightarrow \mathrm{a}+3 \mathrm{~b}=-2 \\ & \mathrm{a}=-2-3 \mathrm{~b} \\ & \ell \mathrm{ny}=\ell \mathrm{n}(\mathrm{x}-\mathrm{a})-\ell \mathrm{n}(\mathrm{x}+\mathrm{b})-\ell \mathrm{n}(\mathrm{x}-2) \\ & \frac{1}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{x}-\mathrm{a}}-\frac{1}{\mathrm{x}+\mathrm{b}}-\frac{1}{\mathrm{x}-2} \\ & \left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{(1,-3)}=-3\left(\frac{1}{1-\mathrm{a}}-\frac{1}{1+\mathrm{b}}-\frac{1}{1-2}\right)=-4 \\ & \Rightarrow\left(\frac{1}{1+2+3 \mathrm{~b}}-\frac{1}{1+\mathrm{b}}+1\right)=\frac{4}{12}=\frac{1}{3} \end{aligned}

\begin{aligned} & \Rightarrow \frac{1}{3(b+1)}-\frac{1}{b+1}=\frac{1}{3}-1 \\ & \Rightarrow \frac{1-3}{3(b+1)}=-\frac{2}{3} \\ & \Rightarrow \mathrm{b}+1=3 \\ & \mathrm{~b}=2 \end{aligned}

a=-2-3 b \, \, \, \, \, \, \, \, \, \quad a+b

\mathrm{a}=-2-3 \times 2 \, \, \, \quad \Rightarrow-8+2

a=-2-6=-8 \quad \Rightarrow-6

Posted by

Ajit Kumar Dubey

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