There are two columns. Each option on the left column may have one or more than one match on the right column. Answer is accepted as correct only if all matches are correct.
Cracked By:
Sinjan Jana, 12th, Bishop Cotton Boys School, Bangalore.
crack78q.png
The solution below is the edited version of the solution provided by Sinjan:
(a) --> (p) and (s)
(b) --> (p)
(c) --> (q)
(d) --> (q) and (r)
For option (a):
Replacing x by (-x) ---> f(-x) = f(x)
Hence, even function.
Also, the given function is the difference of modulus of two trigonometric functions, which are periodic. So it is periodic too & its period is \(\pi\).
For option (b):
f(-x)=f(x)
For option (c):
It is evident from the curve that f(x)+f(-x)=0
For option (d):
(a) --> (p) and (s)
(b) --> (p)
(c) --> (q)
(d) --> (q) and (r)
For option (a):
Replacing x by (-x) ---> f(-x) = f(x)
Hence, even function.
Also, the given function is the difference of modulus of two trigonometric functions, which are periodic. So it is periodic too & its period is \(\pi\).
For option (b):
f(-x)=f(x)
For option (c):
It is evident from the curve that f(x)+f(-x)=0
For option (d):
arcsin(x) or sin-1x is a strictly increasing function and hence is one-one.
Let y = arcsin(-x)
Therefore, - x = sin (y) => x = sin( -y) => y = -arcsin(x)
PS: arcsinx is one-one, it wouldn't have been invertible hadn't it been so.