Let S = − 1 , ∞ and f : S → ℝ be defined as f x = ∫ − 1 x e t − 1 11 2 t − 1 5 t − 2 7 t − 3 12 2 t − 10 61 d t . Let p = Sum of square of the values of x , where f x attains local maxima on S . and q = Sum of the values of x , where f x attains local minima on S . Then, the value of p 2 + 2 q is…

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Let S = − 1 , ∞ and f : S → ℝ be defined as f x = ∫ − 1 x e t − 1 11 2 t − 1 5 t − 2 7 t − 3 12 2 t − 10 61 d t . Let p = Sum of square of the values of x , where f x attains local maxima on S . and q = Sum of the values of x , where f x attains local minima on S . Then, the value of p 2 + 2 q is ________

Aicharya · Accepted Answer

Given, f x = ∫ − 1 x e t − 1 11 2 t − 1 5 t − 2 7 t − 3 12 2 t − 10 61 d t Now, differentiating the above integral using Newton Leibnitz's theorem, we get, f ' x = e x − 1 11 2 x − 1 5 x − 2 7 x − 3 12 2 x − 10 61 Now, from the above diagram using first derivative test we get, Local minima at x = 1 2 , x = 5 Local maxima at x = 0 , x = 2 So, p = 0 + 4 = 4 , q = 1 2 + 5 = 11 2 Hence, p 2 + 2 q = 16 + 11 = 27

Let S = − 1 , ∞ and f : S → ℝ be defined as f x = ∫ − 1 x e t − 1 11 2 t − 1 5 t − 2 7 t − 3 12 2 t − 10 61…

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