IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

akku

_{[ ]}

^{[ ]} f(x)dx =g(x)+c , then _{[ ]}

^{[ ]} f^-1(x)dx ( where f^-1(x)=f inverse ) is

a) xf^-1(x)+c

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

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

d)g^-1(x)+c

PLS GIVE THE ANS WITH COMPLETE EXPLANATION

aysh

Hi akku!!!
it's really simple...
The answer is (c).

SHORTCUT:
Put f(x)=x.
Then,
g(x)=x²/2.
and
f ^-1(x)=x.

Thus,

f ^-1(x)=(x²)/2+c.

Since only Opt. (c) satisfies it,
the approach is justified...

PLEAZZZZZ RATE MA EFFORT....

aditya_arora04

Hi,

Sorry, i am not able to get either the question or the solution. Technically, i am getting the same as aysh, but i am not able to get that how is it satisfying option c !!!

akku

But the ans is a)

akku

pls reply

shakirshafi12

look im giving u da correct soln

im telling u textbook method

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

if it is (a)

then there is no function of g(X) in it

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