Functions

1) pi/2
3)it is invertible because it an bijective funⁿ
let e^x=t;
y=t²-1/2t;
so t²-2ty-1=0;
t=(2y±√4y²+4)/2;
replacin t
e^x=(2y±√4y²+4)/2;
take ln on both side
inverted f(x) is
y=ln( (2x±√4x²+4)/2);

i cudnt understand the font of question 2 & 4

For no 1)..My answer is always comin as π(pi).Cud u pls kindly solve it for me.here is a brief explanation of the method dat i'm following.
period of sinx=2π Thus peiod of sin2x i.e sin(2x+0)=2π2=π (if f(x)is a func wid period T then f(ax+b) is a a periodic func with period T/a)Similarly for cos8x priod is comin as π4. Hence L.C.M= π(Ans)

for no(2) if f(x)=log_e(x²+e/x² +1).........in the numerator it is x²+e and in the denominator it is x²+1 the whole expression inside log_e i.e ln.

For no (4): I am jus explaining the equation.Rest of the language i hope u can understand.

f(x)= α(alpha)x²+6x-8 .(This expression is in da numerator.) In the denominator→ α+6x-8x² .[in each of the nmerator & denominator there is a (minus)sign b4 8].Finally it looks like dis:
f(x)=αx²+6x-8/α+6x-8x² .Now pls solve da ques

arpan bannerjee .....
dude period of |sinx| = pi ........... so sin2x = pi/2 .....

similarly for cosx .... its pi/8 .............

take LCM ...... . and then divide by two .....[becuz both are even functions]

2......

(e-y)/(y-1) = x^2 ....

(e-y)(y-1) >0

(y-e)(y-1)<0

therefore ..... y ε (1,e] ...

therefore .. ln y ε (0,1]