JEE Mains · Maths · STD 12 - 6. Application of derivatives
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 \(.......\).
- A \(4\)
- B \(2\)
- C \(6\)
- D \(8\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
\(y=\frac{x-a}{(x+b)(x-2)}\) At point \((1,-3)\), \(-3 =\frac{1-9}{(1+b)(1-2)}\) \(\Rightarrow 1-a =3(1+b)\) Now, \(y=\frac{x-a}{(x+b)(x-2)}\) \(\Rightarrow \frac{d y}{d x}=\frac{(x+b)(x-2) \times(1)-(x-a)(2 x+b-2)}{(x+b)^2(x-2)^2}\) At \((1,-3)\) slope of normal is…
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