If f ( x ) = | cos x x 1 2 sin x x 2 2 x tan x x 1 | , then lim x 0 f ' ( x ) x is equal to

Question

If $f (x) =$ $| \begin{matrix} \cos x & x & 1 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1 \end{matrix} |$ , then $\lim_{x \to 0} \frac{f' (x)}{x}$ is equal to

Smart Achievers · Accepted Answer

(4)

$Applying R_{1} \to R_{1} - R_{2}$

$f (x) =$ $| \begin{matrix} \cos x - \tan x & 0 & 0 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1 \end{matrix} |$ $= (\cos x - \tan x) (x^{2} - 2 x^{2})$

$= - x^{2} (\cos x - \tan x)$

$∴$ $f' (x) = - 2 x (\cos x - \tan x) - x^{2} (- \sin x - \sec^{2} x)$

$∴$ $\lim_{x \to \infty} \frac{f' (x)}{x} = \lim_{x \to \infty} [- 2 (\cos x - \tan x)] + \lim_{x \to \infty} [x (\sin x + \sec^{2} x)] = - 2 \times 1 = - 2$

Solution

Q.
If $f (x) =$ $| \begin{matrix} \cos x & x & 1 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1 \end{matrix} |$ , then $\lim_{x \to 0} \frac{f' (x)}{x}$ is equal to

1 1

2 - 1

3 2

4 - 2

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Solution

Q. If f(x)=|cosxx12sinxx22xtanxx1|, then limx→0f'(x)x is equal to

1 1

2 - 1

3 2

4 - 2

Ans. (4) Applying R1→R1-R2 f(x)=|cosx-tanx002sinxx22xtanxx1|=(cosx-tanx)(x2-2x2) =-x2(cosx-tanx) ∴ f'(x)=-2x(cosx-tanx)-x2(-sinx-sec2x) ∴ limx→∞f'(x)x=limx→∞[-2(cosx-tanx)]+limx→∞[x(sinx+sec2x)]=-2×1=-2

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Q.
If $f (x) =$ $| \begin{matrix} \cos x & x & 1 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1 \end{matrix} |$ , then $\lim_{x \to 0} \frac{f' (x)}{x}$ is equal to