Answer

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

Updated On: 25-2-2020

Apne doubts clear karein ab Whatsapp par bhi. Try it now.

CLICK HERE

Loading DoubtNut Solution for you

Watch 1000+ concepts & tricky questions explained!

35.7 K+

17.2 K+

Text Solution

Solution :

Let `p,q in R " and let " f(p)=f(q)` <br> `impliese^(p)-e^(-p)=e^(q)-e^(-q)` <br> `implies e^(p)-e^(q)=e^(-p)-e^(-q)` <br> `implies e^(p)-e^(q)=(1)/(e^(p))-(1)/(e^(q))` <br> `implies e^(p)-e^(q)=(e^(p)-e^(q))/(e^(p)e^(q))` <br> `implies e^(p)-e^(q)=0 " or " e^(p)e^(q)= -1 ` (not possible) <br> `implies e^(p)=e^(q)` <br> `implies p=q` <br> Thus, f(x) is one-one. <br> Lets us find the range of the function. <br> Clearly when x approaches to infinity `(e^(x)-e^ (-x))` approaches to infinity and when x approaches to negative infinity, `(e^(x)-e^ (-x))` approaches to negative infinity. <br> Also, `(e^(x)-e^ (-x))` continuously exists. <br> So, range of function is R. <br> Thus f(x) bijective and so invertible. <br> Let `e^(x)-e^(-x)=y` <br> `implies e^(2x)-ye^(x)-1=0` <br> `impliese^(x)=(y+-sqrt(y^(2)+4))/(2)` <br> `impliese^(x)=(y+sqrt(y^(2)+4))/(2) " " ("as "y-sqrt(y^(2)+4) lt 0` for all y and `e^(x)` is always positive) <br> `implies x=log_(e)((y+-sqrt(y^(2)+4))/(2))` <br> ` implies f^(-1)(x)=log_(e)((x+-sqrt(x^(2)+4))/(2))`