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

वक्र \(y = x ^{2}-5 x +5\) की स्पर्श रेखा, जो रेखा \(2 y =4 x +1\) के समान्तर है, किस बिन्दु से गुजरेगी

  1. A \(\left( {\frac{7}{2},\frac{1}{4}} \right)\)
  2. B \(\left( {  \frac{1}{8},-7} \right)\)
  3. C \(\left( { - \frac{1}{8},7} \right)\)
  4. D \(\left( {\frac{1}{4},\frac{7}{2}} \right)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\left( {  \frac{1}{8},-7} \right)\)

Step-by-step Solution

Detailed explanation

\(y = {x^2} - 5x + 5\) \(\frac{{dy}}{{dx}} = 2x - 5 = 2 \Rightarrow x = \frac{7}{2}\) at \(x = \frac{7}{2},y = \frac{{ - 1}}{4}\) Equation of tangent at \(\left( {\frac{7}{2},\frac{{ - 1}}{4}} \right)\) is \(\left( {\frac{7}{2},\frac{{ - 1}}{4}} \right)\) Now check options…
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