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JEE Mains · Maths · STD 12 - 6. Application of derivatives

यदि वक्र \(\mathrm{y}=\frac{\mathrm{x}-\mathrm{a}}{(\mathrm{x}+\mathrm{b})(\mathrm{x}-2)}\) के बिन्दु \((1,-3)\) पर अभिलंब का समीकरण \(\mathrm{x}-4 \mathrm{y}=13\), है तो \(\mathrm{a}+\mathrm{b}\) का मान बराबर है_____________.

  1. A \(4\)
  2. B \(2\)
  3. C \(6\)
  4. D \(8\)
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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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