If f(x)=| array ccc x & x & 1 2 x & x^2 & 2 x x & x & 1 array | , then _ x arrow 0 f^ (x) x

Correct answer: A. Exists and is equal to -2 aligned & f(x)=| array ccc x & x & 1 2 x & x^2 & 2 x x & x & 1 array | =& x(x^2-2 x^2)-x(2 x-2 x x) =&-x^2 x-2 x x+2 x^2 x .+2 x x-x^2 x) =& x^2 x-x^2 x & f^ (x)=2 x( x- x)+x^2( ^2 x+ x) & _ x arrow 0 f^ (x) x =& _ x arrow 0 (2 x( x- x)+x^2( ^2 x+ x))/(x) =& _ x arrow 0 ( x- x)+x( ^2 x+ x) =& 2(0-1)+0=-2 So, _ x arrow 0 f^ (x) x =-2 aligned

JEE Main 201815 AprilMorning ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Morning Shift
Mode: Online
Subject: Mathematics
Chapter: Determinants

JEE Main 2018 · 15 April · Morning Shift · Online · Mathematics · Determinants

If f(x)=| array ccc x & x & 1 2 x & x^2 & 2 x x & x & 1 array | ,...

f(x)=\left|

\begin{array}{ccc}\cos x & x & 1 \\ 2 \sin x & x^2 & 2 x \\ \tan x & x & 1\end{array}

\right|

, then

\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{x}

Options

AExists and is equal to $-2$
BDoes not exist
CExist and is equal to 0
DExists and is equal to 2

Correct answer

A. Exists and is equal to $-2$

Step-by-step solution

\begin{aligned} & f(x)=\left|\begin{array}{ccc}\cos x & x & 1 \\ 2 \sin x & x^2 & 2 x \\ \tan x & x & 1\end{array}\right| \\=& \cos x\left(x^2-2 x^2\right)-x(2 \sin x-2 x \tan x) \\=&-x^2 \cos x-2 x \sin x+2 x^2 \tan x \left.+2 x \sin x-x^2 \tan x\right) \\=& x^2 \tan x-x^2 \tan x \\ \Rightarrow & f^{\prime}(x)=2 x(\tan x-\cos x)+x^2\left(\sec ^2 x+\sin x\right) \\ \therefore & \lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{x} \\=& \lim _{x \rightarrow 0} \frac{2 x(\tan x-\cos x)+x^2\left(\sec ^2 x+\sin x\right)}{x} \\=& \lim _{x \rightarrow 0}(\tan x-\cos x)+x\left(\sec ^2 x+\sin x\right) \\=& 2(0-1)+0=-2 \\ \text{ So, } \lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{x}=-2 \end{aligned}

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