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asked ago in Sets, relations and functions by (38.6k points)

Let f :(π/2,π/2) → R be given by f (x) = (log(sec x + x tan))3. Then

(A)  f(x) is an odd function

(B)  f(x) is a one-one function

(C)  f(x) is an onto function

(D)  f(x) is an even function

1 Answer

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answered ago by (37.5k points)
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Best answer

Correct option (a,b,c)

Explanation :

(-π/2,π/2) → R

f(x) = {log(sec x + tan x)}3

But at x = 0,

 log (sec x + tan x) = ∞

⇒ {log(secx + tanx)}2 > 0

Therefore, f(x) is strictly increasing in (-π/2,π/2).

Now when x → π-/2,f(x) → ∞ and π → π+/2,f(x) → -∞

Therefore, f(x) being continuous in its domain, it covers whole co-domain, i.e. R. See Fig.Hence, it is onto

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