Let g(x) = - f(-1) 2 x 2 (x – 1) – f (0) (x 2 – 1) + f(1) 2 x 2 (x + 1) – f ′ (0) x (x –…
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
Let g(x) = \(-\frac{f(-1)}{2}\) x2 (x – 1) – f (0) (x2 – 1)
\(+\frac{f(1)}{2}\) x2 (x + 1) – f ′ (0) x (x – 1) (x + 1) where f is a thrice differentiable function. Then the correct statements are
\(+\frac{f(1)}{2}\) x2 (x + 1) – f ′ (0) x (x – 1) (x + 1) where f is a thrice differentiable function. Then the correct statements are
- there exists x ∈ (–1, 0) such that f ′ (x) = g′ (x)
- there exists x ∈ (0, 1) such that f ′′ (x) = g′′ (x)
- there exists x ∈ (–1, 1) such that f ′′′ (x) = g′′′ (x)
- there exists x ∈ (–1, 1) such that f ′′′ (x) = 3f (1) – 3f (–1) – 6f ′(0)
Answer
(A) there exists x ∈ (–1, 0) such that f ′ (x) = g ′ (x), (B) there exists x ∈ (0, 1) such that f ′′ (x) = g ′′ (x), (C) there exists x ∈ (–1, 1) such that f ′′′ (x) = g ′′′ (x), (D) there exists x ∈ (–1, 1) such that f ′′′ (x) = 3 f (1) – 3 f (–1) – 6 f ′ (0)
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- _ n a ^ n +b^ n a ^ n -b^ n where a>b>1 is equal to
- The equation e x - x - 1 = 0 has, apart from x = 0
- The sub-tangent at any point of the curve x m y n = a m + n varies as
- The value of _ x (x+1)(3 x+4) x^ 2 (x-8) is equal to
- Let (h, k) be a fixed point, where h > 0, k > 0 . A straight line passing through this point cuts the…
- Given below are two statements. One is labelled as Assertion and the other is labelled as Reason. STATEMENT …
- If a^ 2 +b^ 2 +c^ 2 =1 where a, b, c ∈ R, the the maximum value of (4 a-3 b)^ 2 +(5 b-4 c)^ 2 +(3 c-5 a)^ 2 is
- The shortest distance between the point ( 3 2 , 0 ) and the curve y= x ,(x>0) , is-
More Limit, Continuity and Differentiability questions · Browse all practice questions