Question

Function Domain Range

(i)sinx (P) R (L) [-1, 1]

(ii) cosx (Q) R - {n $\pi$} (M) R

(iii) tanx (R) R- {(2n + 1})$\frac{\pi }{2}$} (N) $(-\infty$, -1] u [1, $\infty$)

(iv) cosecx

(v) secx

(vi) cotx

How many of the following are matched correct?

(i) -P-L, (ii) -P-L, (iii)-R-M, (iv) -Q-N, (v) -R-N, (vi) -Q-M

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Answer 1

Solution: Domain of a function f(x) is the set of all values of x for which the function f(x) is defined. Sinx and cosx are defined for all values of x $\in$ R. secx and cosecx are the reciprocals of cosx and sinx respectively. They won't be defined when sinx and cosx become zero. So cosecx will be defined at all the points where sinx is defined except when sinx = 0. So the domain of cosec x will be R - {n $\pi$}, because sin n n = 0. Similarly the domain of secx will be R-{(2n +1)$\frac{\pi }{2}$}. tan x won't be defined at all the places where cosx = 0 {tan x - $\frac{sinx}{cosx}$}
So the domain of tan x win be R- {(2n + 1})$\frac{\pi }{2}$}. Similarly for cotx, {cotx = $\frac{cosx}{sinx}$} the domain will be R-{n$\pi$}
R- {n$\pi$}.
$underlineRange$
Range is the values a function f(x) can attain. Maximum value of sinx and cosx is 1 and minimum value is -1.
So the range of sinx and cosx is [-1, 1].
Range of tanx is R.[tanx is a continuous function and it attains a value of -$\infty$ at x= $\frac{3\pi }{2}$ and $\infty$ at x = $\frac{\pi }{2}$,Since it is continuous, it should attain all the values between -$\infty$ at x = $\frac{2\pi }{2}$ and $\infty$ at x = $\frac{\pi }{2}$ Since it is continuous, it should attain all the values between -$\infty$ and $\infty$, which is R].
Sec x and cosecx is inverse of cosx and sinx. So the range of them will be (-$\infty$, -1] u (1, $\infty$).
So all of them are matched correct