JEE Mains · Maths · STD 12 - 6. Application of derivatives
If the tangent to the curve, \(y = x^3 + ax -b\) at the point \((1, -5)\) is perpendicular to the line, \(-\,x + y + 4 = 0,\) then which one of the following, points lies on the curve
- A \((2, -2)\)
- B \((-2, 2)\)
- C \((-2, 1)\)
- D \((2, -1)\)
Answer & Solution
Correct Answer
(A) \((2, -2)\)
Step-by-step Solution
Detailed explanation
\(y = {x^3} + ax - b\) \(\left( {1, - 5} \right)\) lies on the cure \( \Rightarrow - 5 = 1 + a - b \Rightarrow a - b = - 6\,\,\,\,\,\,\,\,...\left( i \right)\) Also, \(y' = 3{x^2} + a\) \(y'\left( {1, - 5} \right) = 3 + a\) (slope of tangent) \(\therefore \) this tangent is…
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