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Let us check each option one by one.
(b) f (1) ≠ 3 as function is not defined for x = 1
∴ (b) is not correct.
Let g (x) be a function defined on [– 1, 1]. If the area of the equilateral triangle with two of its vertices at (0,0) and then the function g(x) is
As (0, 0) and (x, g (x)) are two vertices of an equilateral triangle; therefore, length of the side of D is
∴ (b), (c) are the correct answers as (a) is not a function
(∴ image of x is not unique)
stands for the greatest integer function, then
f(x) = cos [π^{2}] x + cos [– π^{2}] x
We know 9 < π^{2} < 10 and – 10 < – π^{2} < – 9
⇒ [π^{2}] = 9 and [–π^{2}] = – 10
⇒ ∴ f (x) = cos 9x + cos (–10x)
f (x) = cos 9x + cos 10 x
Let us check each option one by one.
Thus (a) and (c) are the correct options.
If f(x) = 3x – 5, then f^{–1}(x)
f (x) = 3x – 5 (given), which is strictly increasing on R,
so f –1 (x) exists.
Let y = f (x) = 3x – 5
...(1)
and y = f (x) ⇒ x = f^{ –1}(y) ...(2)
If g (f(x)) =  sin x  and f (g(x)) = (sin √x)^{2}, then
Let us check each option one by one.
Let f : (0, 1) → R be defined by where b is a constant such that 0 < b < 1. Then
∴ f is neither onto nor invertible
Hence a and b are the correct options.
Let f : (–1, 1) ⇒ IR be such that Then the value (s) of
The function f(x) = 2x + x + 2 –  x + 2 – 2 x has a local minimum or a local maximum at x =
the critical points can be obtained by solving x = 0
The graph of y = f(x) is as follows
From graph f(x) has local minimum at –2 and 0 and
R be given by f (x) = (log(sec x + tan x))^{3}.
Then
∴ f is an odd function.
(a) is correct and (d) is not correct.
Also
We know that strictly increasing function is one one.
∴ f is one one
∴ (b) is correct
∴ Range of f = (–∞, ∞) = R
∴ f is an onto function.
∴ (c) is correct.
Let a ∈ R and let f : R → R be given by f (x) = x^{5} – 5x + a. Then
f (x) = x^{5}  5 x+a
f (x) = 0 ⇒ x^{5}  5 x +a = 0
⇒ a = 5x – x^{5} = g(x)
⇒ g(x) = 0 when x = 0, 5^{1/4}  5^{1/4} and g' (x ) = 0 ⇒ x = 1, – 1
Also g (– 1) = – 4 and g(1) = 4
∴ graph of g(x) will be as shown below.
From graph
if a ∈ ( 4,4)
then g(x) = a or f (x) = 0 has 3 real roots If a > 4 or a < – 4
then f(x) = 0 has only one real root.
∴ (b) and (d) are the correct options.
Let sin x for all x ∈ R. Let (fog)(x) denote f(g(x)) and (gof)(x) denote g(f(x)). Then which of the following is (are) true?
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