Question

Which function has the range (-infinity,0)u(2,infinity)?

Answer 1

Answer: y= sec(x)+1

Step-by-step explanation:

Answer 2

Answer with explanation:

We have to find that function,which has range ,(¬∞,0)∪(2,∞) out of the following four trigonometric functions.

Range of a function is defined as,those values of function,if the function is →y=f(x), that is those values of y , for which x is defined.

A:→ y =sec x

x=$sec^{-1}y$

x is defined for , y≥1 and y≤-1.

B: →y=cot (2 x)-1

y+1=cot (2 x)

$x=\frac{1}{2}*cot^{-1}(y+1)$

Since cotangent function is defined for all values of y, so, x∈[-∞,∞],that is ,x∈R.

So, Cot(2 x) -1, have the same range as Cot x.

C: y=Cos (x+1)

$x+1=Cos^{-1}y\\\\x=Cos^{-1}y-1$

$-1\leq Cos^{-1}y\leq 1\\\\-1-1\leq Cos^{-1}y-1\leq 1-1\\\\-2\leq Cos^{-1}y-1\leq 0$

D: y=Cosec (x)+1

Range of Cosec x is, (-∞,-1] ∪ [1,∞).

Range of Cosec (x) +1 will be , (-∞,-1+1] ∪ [1+1,∞).

      = (-∞, 0] ∪ [2,∞)

None of the option is true.