fxcos xx12 sin xx22xtan xx1 then Ltx→0f1xx

fx=cos xx12 sin xx22xtan xx1 =cos x(x2-2x2)-2sin x(x-x)+tan x(2x2-x2)=cos x(-x2)+tan x(x2)=x2(tan x-cos x)f1(x)=2x (tan x-cos x)+x2(sec2x+sin x)Ltx→0 f1(x)x=Ltx→0 2x(tan x-cos x)+x2(sec2x+sin x)x=Ltx→0 2(tan x-cos x)+x(sec2x+sin x) =2(0-1)+0=-2

fxcos xx12 sin xx22xtan xx1 then Ltx→0f1xx

fx=cos xx12 sin xx22xtan xx1 =cos x(x2-2x2)-2sin x(x-x)+tan x(2x2-x2)=cos x(-x2)+tan x(x2)=x2(tan x-cos x)f1(x)=2x (tan x-cos x)+x2(sec2x+sin x)Ltx→0 f1(x)x=Ltx→0 2x(tan x-cos x)+x2(sec2x+sin x)x=Ltx→0 2(tan x-cos x)+x(sec2x+sin x) =2(0-1)+0=-2

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$f (x) |\begin{matrix} \cos x & x & 1 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1 \end{matrix}| then \underset{x \to 0}{Lt} \frac{f^{1} (x)}{x}$

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does not exist

$- 2$

answer is D.

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Detailed Solution

$f (x) = |\begin{matrix} \cos x & x & 1 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1 \end{matrix}|$
$= \cos x (x^{2} - 2 x^{2}) - 2 \sin x (x - x) + \tan x (2 x^{2} - x^{2})$
$= \cos x (- x^{2}) + \tan x (x^{2})$
$= x^{2} (\tan x - \cos x)$
$f^{1} (x) = 2 x (\tan x - \cos x) + x^{2} (\sec^{2} x + \sin x)$
$\underset{x \to 0}{Lt} \frac{f^{1} (x)}{x} = \underset{x \to 0}{Lt} \frac{2 x (\tan x - \cos x) + x^{2} (\sec^{2} x + \sin x)}{x}$
$= \underset{x \to 0}{Lt} 2 (\tan x - \cos x) + x (\sec^{2} x + \sin x)$
$= 2 (0 - 1) + 0 = - 2$