FLCD 1

1) C
2) A
3) B
4) C
5) A

1.c
3.b
4.b
5.a

in q.2 my answer for b is coming out to be 4

1)(C)

2)(A)

Given expression:

lim_x→0 x^-3[3x-(3x)³3!+.....]+ax²+b

= lim_x→0 1x²(3+a) +(-276+b) +....

The rest of the terms are equal to 0.

Therefore,

3x² +ax²=0

-276+b = 0

3)(A)f(g(x))=x

Differentiating both sides,

f'(g(x))*g'(x)=1
→g'(x)=1f'(g(x))=1+[g(x)]⁵

4)(C)

f'(x)= mx^m-1sin(1x)-x^m-2cos(1x) for x≠0
and
f'(x)= 0 for x = 0

sin and cos of any number lies between 0 and 1.

Hence for lim_x→0f'(x) to be 0 lim_x→0x^m-2 should be 0.
The smallest natural number m satisfying this condition is 3.

5)(A)

arctan(2)+arctan(20k)=arctan(k)

arctan(2+20k1-40k)=arctan(k)

→ 40k²+19k+2=0

Sum of roots = -1940