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**Q.1. If g(x) = x ^{2} + x - 1 and (gof) (x) = 4x^{2} - 10x + 5, then f (5/4) is equal to (2020)**

(a) 3/2

(b) - 1/2

(c) 1/2

(d) - 3/2

We have

g(x) = x

Now, g(f(x)) = 4x

⇒ g(f(x)) = (4 - 8x + 4x

⇒ g(f(x)) = (2-2x)

Hence,

Let,

Since f(x) is a decreasing function, then

(a) 1/5

(b) 1

(c) 5

(d) 2/5

Ans.

f and g are differentiable functions on R and fog is the identity function. So,

f(g(x)) = x

⇒ f'(g(x)) . g'(x) = 1 (1)

Substituting x = a in Eq. (1), we get

f'(g(a)). g'(a) = 1

⇒ f'(b) × 5 = 1 ⇒ f'(b) = 1/5

(a) f

(b) 1/x f

(c) f

(d) f

Ans.

The given relation is

(b) 8

(c) 4

(d) 14

Ans.

Solution.

2

= 2

= (2

= 8(2

= 8(1 + 15)

= 8 + 15μ

When 2

Hence, fractional part of the number is 8/15

Therefore, value of k is 8

(a) Not injective

(b) Neither injective nor surjective

(c) Surjective but not injective

(d) Injective but not surjective

As A = {x ∈ R: x is not a positive integer}

A function f: A → R given by f(x) = 2x/x-1

f(x

So, f is

As f(x) ≠ 2 for any x ∈ A ⇒ f is not

Hence f is injective but not subjective.

(a) Onto but not one-one.

(b) One-one but not onto.

(c) Both one-one and onto.

(d) Neither one-one nor onto.

⇒ fog is

(b) R - [-1,1]

(c) R - [- 1/2, 1/2]

(d) (-1,1) - {0}

(b) Injective only

(c) Neither injective nor surjective

(d) Both injective as well as surjective

f: (0, ∞) → (0, ∞)

∵ f(1) = 0 and 1 ∈ domain but 0 ∉ co-domain

Hence, f(x) is not a function.

(b) 2f(x

(c) (f(x))

(d) -2f(x)

Given function can be written as

**Q.13. ****If the function f: R - {1, -1} → A defined by f(x) = x ^{2}/1 - x^{2}, is surjective, then A is equal to: (2019)**

(a) R - {-1}

(b) [0, ∞]

(c) R- [-1, 0]

(d) R - (-1, 0]

∴ f(x) increases in x ∈ (0, ∞)

Also f(0) = 0 and

∴ Set A = R - [-1, 0)

And the graph of function f(x) is

Q.14.

(a) 2

(b) 16

(c) 4

(d) 3

∵ f(x + y) = f(x) x f(y)

⇒ Let f(x) = t

∵ f(1) = 2

∵ t = 2

⇒ f(x) = 2

⇒ a = 3

(b) (-2, -1) ∪ (-1, 0) ∪ (2, ∞)

(c) (-1, 0) ∪ (1, 2) ∪ (2, ∞)

(d) (1, 2) ∪ (2, ∞)

To determine domain,

i.e., 4 - x

and x (x - 1) (x + 1) > 0

x∈(-1, 0)

Hence domain is intersection of (1) & (2).

i.e.,

(b) f(g(S)) = S

f(x) = x

g(A) = {x ∈ R: f(x) ∈ A} S = [ 0, 4]

g(S) = {x ∈ R: f(x) ∈ S}

= {x ∈ R: 0 ≤ x

∴ g(S) ≠ S

∴ f(g (S)) ≠ f(S)

g(f(S)) = {x ∈ R: f(x) ∈ f(S)}

= { x ∈ R : x

= {x ∈ R : -4 ≤ x ≤ 4}

∴ g(f(S)) ≠ g(S)

∴ g(f(S)) = g (S) is incorrect.

(a) aα

(b) aα

(c) aα

(d) aα

f(x) = ln (sin x), g (x) = sin

⇒ f(g(x)) = ln (sin (sin

⇒ f(g(α)) = - α

But given that (fog) (α) = b

∴

∴ aα

⇒ aα

∵ φ(x) = ((hof)og)(x)

∵ φ(π/3) = h(f(g(π/3))) = h(f(√3)) = h(3

= (1 - √3)/(1+√3) = - 1/2 (1 + 3 - 2√3) = √3 - 2 = -(-√3 + 2)

= -tan 15º = tan(180º - 15º) = tan(π - π/12) = tan 11π/12

(b) (1 - 2

(c) (3

(d) (1 - 3

f(g(x)) = x

f(3

= 1/(3

2

x(6

x = (2

(b) One-one and onto

(c) Neither one-one nor onto

(d) Onto but not one-one

f(10) = 10 – 5(2) = 0 which is not in codomain

So, the function is many one + into

(a) Is an empty set

(b) Contains exactly one element

(c) Contains exactly two elements.

(d) Contains more than two elements.

Ans.

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