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

જો વક્ર \(y = x^3 + ax -b\) ના બિંદુ \((1, -5)\) આગળનો સ્પર્શકએ રેખા \(-\,x + y + 4 = 0\) ને લંબ હોય તો આપેલ પૈકી વક્ર પરનું બિંદુ મેળવો.

  1. A \((2, -2)\)
  2. B \((-2, 2)\)
  3. C \((-2, 1)\)
  4. D \((2, -1)\)
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Correct Answer

(A) \((2, -2)\)

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\(y = {x^3} + ax - b\) \(\left( {1, - 5} \right)\) 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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