Function $f (x) = x^{3} + 6 x^{2} + (9 + 2 k) x + 1$ is increasing function if-

k \geq \frac{3}{2}

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k > \frac{3}{2}

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k < \frac{3}{2}

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k \leq \frac{3}{2}

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Solution

The correct option is A $k \geq \frac{3}{2}$
$f^{'} (x) \geq 0$ for all x.
Hence
$3 x^{2} + 12 x + (9 + 2 k) \geq 0$
$3 x^{2} + 12 x + 9 + 2 k \geq 0$
$3 (x^{2} + 4 x + 3) + 2 k \geq 0$
$3 (x + 1) (x + 3) + 2 k \geq 0$
Or
$k \geq \frac{- 3 (x + 1) (x + 3)}{2}$
Now
Let $g (x) = x^{2} + 4 x + 3$
$g^{'} (x) = 2 x + 4 = 0$
$x = - 2$
Hence it attains a minimum value at $x = - 2$ .
Hence
$k \geq \frac{- 3 (- 1)}{2}$
Or
$k \geq \frac{3}{2} .$

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Similar questions

^{'} k^{'}

x = 3

(x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{{(x + 3)}^{2} - 36}{x - 3}, & x \neq 3 k, & x = 3 \end{matrix}

T = \frac{5}{9} (K - 32)

T = 290

K =

x = 3

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{(x + 3)^{2} - 36}{x - 3} & , x \neq 3 k & , x = 3 \end{matrix}

\frac{j}{k} = 32

k = \frac{3}{2}

\frac{1}{2} j

3^{2 - {log}_{3} 5} = k

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