First slide

Tangents and normals

Question

$If the function f (x) = x^{4} + b x^{2} + 8 x + 1 has a horizontal tangent and a point x for f^{''} (x) = 0$ $then the value of b is equal to$

Moderate

A

-1
B

1
C

6
D

-6

Solution

$f^{'} (x) = 0 and f^{''} (x) = 0 for same x = x_{1} (Say)$
$\begin{array}{l} \Rightarrow f^{'} (x) = 4 x^{3} + 2 b x + 8 \Rightarrow f^{''} (x) = 12 x^{2} + 2 b \\ \Rightarrow f^{'} (x_{1}) = 2 (2 x_{1}^{3} + b x_{1} + 4) = 0 \to (1) \\ \Rightarrow f^{''} (x_{1}) = 2 (6 x_{1}^{2} + b) = 0 \to (2) \\ \Rightarrow b = - 6 x_{1}^{2} \\ Substitute in eqn - > 1 \\ 2 x_{1}^{3} + (- 6 x_{1}^{3}) + 4 = 0 \Rightarrow 4 x_{1}^{3} = 4 \Rightarrow x_{1}^{3} = 1 \\ \Rightarrow x_{1} = 1 \Rightarrow b = - 6 \end{array}$

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