look im giving u da correct soln

 f(x)dx =g(x)+c , then

 

 

now see

f^-1(x)dx

we put x=f(t)

dx=f`(t).dt

f^-1(x)dx

=

tf ^`(t)dt

use integration by parts

we get

=tf(t)- f(t)dt

  f(t)dx =g(t)+c

=tf(t)-g(t)+C

t=f^-1(x)

so we get

xf^-1(x)-g(f^-1(x))+c

so (c) is correct
please check whether there has been any mistake in the question

in brochure it is written board percentage not PCM

----------------------------------------------------------------------------

Q-is i am eligible

ans-are yes you are eligible

look joy we have to use basics for such questions rather than equations!!

when we take case of slip down friction acts on opposite side i.e up

and when it skids up friction acts at downdard to the banking

------------------------------------------------------------------------------------

the road has been banked so

the angle

tan=v²/rg

v=10m/s r=20m g=10m/s²

from here we get

tan =1/2

-------------

Case 1:-slips down

direction of friction is opposite to motion

we take components of Normal reaction and friction forces along vertical and horizontal axis

N(cos+sin)=mg ----------1

N(sin-cos)=mv²/r-------2

divide eq 1 and 2

and calculate v

Case 2:-skids up

direction of friction is opposite to motion

we take components of Normal reaction and friction forces along vertical and horizontal axis

N(cos-sin)=mg ----------1

N(sin+cos)=mv²/r-------2

divide eq 1 and 2

and get v

 

 

hope it helps
!!