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QUESTION: 1

Let R be the set of real numbers. If f : R → R is a function defined by f (x) = x^{2}, then f is :

Solution:

f (x) = x^{2} is many one as f (1) = f (–1) = 1 Also f is into as – ve real number have no pre-image.

∴ F is neither injective nor surjective.

QUESTION: 2

The entire graphs of the equation y = x^{2} + kx – x + 9 is strictly above the x-axis if and only if

Solution:

For entire graph to be above x-axis, we should have

QUESTION: 3

Let f(x) = | x – 1 |. Then

Solution:

∴ 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.

QUESTION: 4

If x satisfies |x - 1| + |x - 2| + |x - 3| __>__ 6 , then

Solution:

Graph of f (x) shows f (x) __>__ 6 for x __<__ 0 or x __>__ 4

QUESTION: 5

If f ( x) = cos(ln x), then the value

Solution:

QUESTION: 6

The domain of definition of the function

Solution:

QUESTION: 7

Which of the following functions is periodic?

Solution:

Clearly it is a periodic function with period 1.

∴ (a) is the correct alternative.

QUESTION: 8

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

Solution:

QUESTION: 9

Let f ( x) = ( x + 1)^{2} - 1,x __>__ -1 . Th en th e set {x : f (x) = f ^{-1} (x)} is

Solution:

QUESTION: 10

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

Solution:

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

QUESTION: 11

Let f(x) be defined for all x > 0 and be continuous. Let f(x) for all x, y and f(e) = 1. Then

Solution:

f (x) is continuous and defined for all x > 0 and

⇒ Clearly f (x) = ln x which satisfies all these properties

∴ f (x) = ℓnx

QUESTION: 12

If the function f: [1, ∞) → [1, ∞) is defined by f(x) = 2^{x (x-1)}, then f^{–1} (x) is

Solution:

QUESTION: 13

Let f : R → R be any function. Define g : R → R by g(x) = |f(x)| for all x. Then g is

Solution:

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.

QUESTION: 14

The domain of definition of the function f(x) given by the equation 2^{x} + 2^{y} = 2 is

Solution:

QUESTION: 15

Let g(x) = 1 + x - [x] and Then for allx, f(g(x)) is equal to

Solution:

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

QUESTION: 16

If f:[1, ∞) → [2, ∞) is given by equals

Solution:

QUESTION: 17

The domain of definition of

Solution:

QUESTION: 18

Let E = {1, 2, 3, 4} and F = {1, 2}. Then the number of onto functions from E to F is

Solution:

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

QUESTION: 19

Then, for what value of a is f (f(x)) = x ?

Solution:

QUESTION: 20

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

Solution:

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}

QUESTION: 21

Let f unction f : R → R be defined by f(x) = 2x + sin x for x ∈ R , then f is

Solution:

QUESTION: 22

Solution:

∴ f is an increasing function ⇒ f is one-one.

QUESTION: 23

Domain of definition of the function

Solution:

to be defined and real

...(1) But we know that – π/2 __<__ sin^{–1} 2x __<__ π/2 ...(2) Combining (1) and (2), we get

QUESTION: 24

Range of the function

Solution:

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

QUESTION: 25

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

Solution:

QUESTION: 26

If f(x) = sin x + cos x, g (x) = x^{2} – 1, then g (f(x)) is invertible in the domain

Solution:

QUESTION: 27

If the functions f(x) and g(x) are defined on R → R such that

Solution:

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 one-one onto.

QUESTION: 28

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

Solution:

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.

QUESTION: 29

If g(x) = f '(x) and given that F(5) = 5, then F(10) is equal to

Solution:

QUESTION: 30

Solution:

QUESTION: 31

Let f, g and h be real-valued 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

Solution:

∴ 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.

QUESTION: 32

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

Solution:

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

QUESTION: 33

Th e function f : [0, 3] → [1, 29], defi n ed by f(x) = 2x^{3} – 15x^{2} + 36x + 1, is

Solution:

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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