Let k and m be positive real numbers such that the function f ( x… - Vidyadip

MCQ

Let $k$ and $m$ be positive real numbers such that the function $\quad f ( x )=\left\{\begin{array}{cc}3 x ^2+ k \sqrt{ x +1}, & 0< x <1 \\ mx ^2+ k ^2, & x \geq 1\end{array}\right.$ is differentiable for all $x > 0$. Then $\frac{8 f^{\prime}(8)}{f^{\prime}\left(\frac{1}{8}\right)}$ is equal to $.............$.

✓
$309$
B
$310$
C
$311$
D
$312$

✓

Answer

Correct option: A.

$309$

a
function is differentiable $\forall x < 0$

so $\quad f \left(1^{-}\right)= f (1)$

$3+\sqrt{2} k = m + k ^2......(1)$

and $\quad f _{+}^1\left(1^{-}\right)= f _{-}^1\left(1^{+}\right)$

$\left.2 m x\right|_{x=1}=6 x+\left.\frac{k}{2 \sqrt{x+1}}\right|_{x=1}$

$2 m =6+\frac{ k }{2 \sqrt{2}}$

$m =3+\frac{ k }{4 \sqrt{2}}.......(2)$

$k ^2+3+\frac{ k }{4 \sqrt{2}}=3+\sqrt{2} k$

$k =\frac{7}{4 \sqrt{2}}, 0$

$m =3+\frac{7}{32}$

$m =\frac{103}{32}$

So $\quad \frac{8 f^{\prime}(8)}{f^{\prime}\left(\frac{1}{8}\right)}=8 \times \frac{\left.2 mx \right|_{ x =8}}{6 x +\left.\frac{ k }{2 \sqrt{ x +1}}\right|_{x=\frac{1}{8}}}$

$=\frac{8 \times 2 \times 8 \times \frac{103}{32}}{\frac{16}{12}}$

$=103 \times 3=309$

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