If $f(x)=\left|\begin{array}{ccc}\sin x & \cos x & \tan x \\ x^3 & x^2 & x \\ 2 x & 1 & 1\end{array}\right|$, then $\lim _{x \rightarrow 0} \frac{f(x)}{x^2}$ is
38 views
0 Votes
0 Votes

If $f(x)=\left|\begin{array}{ccc}\sin x & \cos x & \tan x \\ x^3 & x^2 & x \\ 2 x & 1 & 1\end{array}\right|$, then $\lim _{x \rightarrow 0} \frac{f(x)}{x^2}$ is

(A) 3

(B) -1

(C) 0

(D) 1

1 Answer

0 Votes
0 Votes
 
Best answer

SOLUTION —

$f(x)=x(x-1) \sin x-\left(x^3-2 x^2\right) \cos x-x^3 \tan x$

$=x^2 \sin x-x^3 \cos x-x^3 \tan x+2 x^2 \cos x-x \sin x$ $\therefore \lim _{x \rightarrow 0} \frac{f(x)}{x^2}$

$=\lim _{x \rightarrow 0}\left(\sin x-x \cos x-x \tan x+2 \cos x-\frac{\sin x}{x}\right)$ $=0-0-0+2-1=1$

So, The correct option will be (D).

RELATED DOUBTS

1 Answer
0 Votes
0 Votes
67 Views
1 Answer
0 Votes
0 Votes
58 Views
1 Answer
0 Votes
0 Votes
72 Views
1 Answer
0 Votes
0 Votes
45 Views
1 Answer
0 Votes
0 Votes
51 Views
1 Answer
0 Votes
0 Votes
50 Views
1 Answer
0 Votes
0 Votes
53 Views
1 Answer
0 Votes
0 Votes
55 Views
1 Answer
0 Votes
0 Votes
59 Views
1 Answer
0 Votes
0 Votes
54 Views
1 Answer
0 Votes
0 Votes
48 Views
Peddia is an Online Question and Answer Website, That Helps You To Prepare India's All States Boards & Competitive Exams Like IIT-JEE, NEET, AIIMS, AIPMT, SSC, BANKING, BSEB, UP Board, RBSE, HPBOSE, MPBSE, CBSE & Other General Exams.
If You Have Any Query/Suggestion Regarding This Website or Post, Please Contact Us On : [email protected]

CATEGORIES