First slide

Derivatives

Question

$Let f (x) be a polynomial of degree 4 with f (2) = - 1, f' (2) = 0, f'' (2) = 2, f''' (2) = - 12, f'''' (2) = 24$ . $Then the value of f^{''} (1) is$

Moderate

A

24
B

26
C

28
D

30

Solution

$\begin{array}{l} Let f (x) = a (x - 2)^{4} + b (x - 2)^{3} + c (x - 2)^{2} + d (x - 2) - 1 \\ f^{'} (x) = 4 a (x - 2)^{3} + 3 b (x - 2)^{2} + 2 c (x - 2) + d \\ f^{''} (x) = 12 a (x - 2)^{2} + 6 b (x - 2) + 2 c \\ f^{'''} (x) = 24 a (x - 2) + 6 b \\ f^{''''} (x) = 24 a \\ Since f (2) = - 1 \\ And f^{'} (2) = 0 \\ \Rightarrow 0 + 0 + 0 + d = 0 \\ \Rightarrow d = 0 \\ and f " (2) = 2 \\ \Rightarrow 0 + 0 + 2 c = 2 \\ \Rightarrow c = 1 \\ And f^{'''} (2) = - 12 \\ \Rightarrow 0 + 6 b = - 12 \end{array}$
$\begin{array}{l} \Rightarrow b = - 2 \\ And f^{''''} (2) = 24 \\ \Rightarrow 24 a = 24 \\ \Rightarrow a = 1 \\ ∴ f^{''} (x) = 12 (1) (x - 2)^{2} + 6 (- 2) (x - 2) + 2 (1) \\ = 12 (x - 2)^{2} - 12 (x - 2) + 2 \\ Hence f^{''} (1) = 12 (1 - 2)^{2} - 12 (1 - 2) + 2 \\ = 26 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App