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Let R be the set of real numbers. If f : R → R is a function defined by f (x) = x^{2}, then f is :
f (x) = x^{2} is many one as f (1) = f (–1) = 1 Also f is into as – ve real number have no preimage.
∴ F is neither injective nor surjective.
The entire graphs of the equation y = x^{2} + kx – x + 9 is strictly above the xaxis if and only if
For entire graph to be above xaxis, we should have
∴ Put x = 2
LHS = f (2^{2}) = 4 – 1 = 3 and RHS = ( f (2))2 = 1
∴ (a) is not correct
Consider f (x + y) = f (x) + f (y)
Put x = 2, y = 5 we get
f (7) = 6; f (2) + f (5) = 1 + 4 = 5
∴ (b) is not correct
Consider f ( x ) =  f (x) 
Put x = – 5 then f ( –5 ) = f (5) = 4
 f (– 5)  =  – 5 – 1 = 6
∴ (c) is not correct.
Hence (d) is the correct alternative.
If x satisfies x  1 + x  2 + x  3 > 6 , then
Graph of f (x) shows f (x) > 6 for x < 0 or x > 4
If f ( x) = cos(ln x), then the value
The domain of definition of the function
Which of the following functions is periodic?
Clearly it is a periodic function with period 1.
∴ (a) is the correct alternative.
Let f(x) = sin x and g(x) = ln  x . If th e ran ges of th e composition functions fog and gof are R_{1} and R_{2 }respectively, th en
Let f ( x) = ( x + 1)^{2}  1,x > 1 . Th en th e set {x : f (x) = f ^{1} (x)} is
Th e fun ction f(x) = px – q + r  x , x ∈ (∞,∞) where p > 0, q > 0, r > 0 assumes its minimum value only on one point if
From graph (i) infinite many points for min value of f (x)
From graph (ii) only pt. of min of f (x) at x = q/p
From graph (iii) only one pt. of min of f (x) at x = 0
Let f(x) be defined for all x > 0 and be continuous. Let f(x) for all x, y and f(e) = 1. Then
f (x) is continuous and defined for all x > 0 and
⇒ Clearly f (x) = ln x which satisfies all these properties
∴ f (x) = ℓnx
If the function f: [1, ∞) → [1, ∞) is defined by f(x) = 2^{x (x1)}, then f^{–1} (x) is
Let f : R → R be any function. Define g : R → R by g(x) = f(x) for all x. Then g is
Let h (x) = x then g (x) = f (x) = h (f (x))
Since composition of two continuous functions is continuous, therefore g is continuous if f is continuous.
The domain of definition of the function f(x) given by the equation 2^{x} + 2^{y} = 2 is
Let g(x) = 1 + x  [x] and Then for allx, f(g(x)) is equal to
For integral values of x; g (x) = 1
For x < 0; (but not integral value) x – [x] > 0 ⇒ g (x) > 1
For x > 0;(but not integral value) x – [x] > 0 ⇒ g (x) >1
If f:[1, ∞) → [2, ∞) is given by equals
Let E = {1, 2, 3, 4} and F = {1, 2}. Then the number of onto functions from E to F is
From E to F we can define, in all, 2 × 2 × 2 × 2 = 16 functions (2 options for each element of E) out of which 2 are into, when all the elements of E either map to 1 or to 2.
∴ No. of onto functions = 16 – 2 = 14
Then, for what value of a is f (f(x)) = x ?
Suppose f(x) = (x + 1)^{2} for x > 1. If g(x) is the function whose graph is the reflection of the graph of f (x) with respect to the line y = x, then g(x) equals
Given that f (x) = (x + 1)2, x > – 1
Now if g (x) is the reflection of f (x) in the line y = x then it can be obtained by interchanging x and y in f (x) i.e., y = (x + 1)^{2} changes to x = (y + 1)^{2}
Let f unction f : R → R be defined by f(x) = 2x + sin x for x ∈ R , then f is
∴ f is an increasing function ⇒ f is oneone.
Domain of definition of the function
to be defined and real
...(1) But we know that – π/2 < sin^{–1} 2x < π/2 ...(2) Combining (1) and (2), we get
We have
We can see here that as x → ∞ , f (x) → 1 which is the min value of f (x). i.e. f_{min} = 1. Also f (x) is max when
min which is so when x = – 1/2
If f (x) = x^{2} + 2bx + 2c^{2} and g (x) =  x^{2} 2cx+ b^{2} such that min f (x) > max g (x), then the relation between b and c, is
If f(x) = sin x + cos x, g (x) = x^{2} – 1, then g (f(x)) is invertible in the domain
If the functions f(x) and g(x) are defined on R → R such that
Since f – g : R → R for any x there is only one value of (f (x) – g(x)) whether x is rational or irrational.
Moreover as x∈R, f (x) – g (x) also belongs to R.
Therefore, (f – g) is oneone onto.
X an d Y are two sets and f : X → Y. If {f(c) = y; c ⊂ X, y ⊂ Y} and {f–1(d) = x; d ⊂ Y, x ⊂ X}, then the true statement is
Given that X and Y are two sets and f : X → Y.
{f (c) = y; c ⊂ X, y ⊂ Y} and
{f ^{–1}(d) = x : d ⊂ Y, x ⊂ X }
The pictorial representation of given information is as shown:
Since f ^{–1} (d) = x ⇒ f (x) = d Now if a ⊂ x
⇒ f (a) ⊂ f (x) = d ⇒ f ^{–1} [f (a)] = a
∴ f^{ –1} (f (a)) = a, a ⊂ x is the correct option.
If g(x) = f '(x) and given that F(5) = 5, then F(10) is equal to
Let f, g and h be realvalued functions defined on the interval and . If a, b and c denote, respectively, the absolute maximum of f, g and h on [0, 1], then
∴ f (x) is an increasing function on [ 0,1]
∴ g (x) is an increasing function on [ 0,1]
∴ h (x) is an increasing function on [ 0,1]
Hence a = b = c.
Let f (x) = x^{2} and g(x) = sin x for all x ∈ R. Then the set of all x satisfying (f o g o g o f) (x) = (g o g o f) (x), where (f o g) (x) = f (g(x)), is
Then (gof) (x) = sin x_{2}
⇒ (gogof) (x) = sin (sin x^{2})
⇒ (fogogof) (x) = sin 2 (sin x^{2}) As given that (fogogof) (x) = (gogof) (x)
⇒ sin^{2} (sin x^{2}) = sin (sin x^{2})
⇒ sin (sin x^{2}) = 0,1
Th e function f : [0, 3] → [1, 29], defi n ed by f(x) = 2x^{3} – 15x^{2} + 36x + 1, is
We have f(x) = 2x^{3} – 15x^{2} + 36x + 1
⇒ f (x') = 6x^{2} – 30x + 36
= 6(x^{2} – 5x + 6)
= 6 (x – 2) (x – 3)
∴ f(x) is in creasing on [0, 2] and decreasing on [2, 3]
∴ f(x) is many one on [0, 3]
Also f(0) = 1, f(2) = 29, f(3) = 28
∴ Global min = 1 and Global max = 29 i.e., Range of f = [1, 29] = codomain
∴ f is onto.
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