JEE Mains · Maths · STD 12 - 6. Application of derivatives
The tangent to the curve \(y = {x^2} - 5x + 5,\) parallel to the line \(2y=4x+1,\) also passes through the point
- A \(\left( {\frac{7}{2},\frac{1}{4}} \right)\)
- B \(\left( { \frac{1}{8},-7} \right)\)
- C \(\left( { - \frac{1}{8},7} \right)\)
- D \(\left( {\frac{1}{4},\frac{7}{2}} \right)\)
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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