NTA Abhyas JEE Main2020MathematicsDifferentiationPractice
If y = t a n - 1 1 x 2 + x + 1 + t a n - 1 1 x 2 + 3 x + 3 + t a n - 1 1 x 2 + 5 x + 7 + t a n - 1 1 x 2 + 7 x + 13 , x > 0 and d y d x x = 0 = - k 1 + k , then the value of k is
Correct answer
16
Step-by-step solution
y = t a n - 1 1 x 2 + x + 1 + t a n - 1 1 x 2 + 3 x + 3 + t a n - 1 ( 1 x 2 + 5 x + 7 ) + t a n - 1 1 x 2 + 7 x + 13 = t a n - 1 1 x x + 1 + 1 + t a n - 1 1 x + 2 x + 1 + 1 + t a n - 1 1 x + 3 x + 2 + 1 c + t a n - 1 1 x + 4 x + 3 + 1 = t a n - 1 x + 1 - x 1 + x + 1 x + t a n - 1 x + 2 - x + 1 1 + x + 2 x + 1 + t a n - 1 x + 3 - x + 2 1 + x + 3 x + 2 + t a n - 1 x + 4 - x + 3 1 + x + 4 x + 3 = t a n - 1 x + 1 - t a n - 1 x + t a n - 1 x + 2 - t a n - 1 x + 1 + t a n - 1 x + 3 - t a n - 1 x + 2 + t a n - 1 x + 4 - t