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This mock test of Test: Problems On Composition for JEE helps you for every JEE entrance exam.
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QUESTION: 1

If f(x) = ax + b and g(x) = cx + d, then f[g(x)] – g[f(x)] is equivalent to

Solution:

We have f(x) = ax+b, g(x) = cx+d

Therefore, f{g(x)} = g{f(x)}

⇔ f(cx+d) = g(ax+b)

⇔ a(cx+d)+b = c(ax+b)+d

⇔ ad+b = cb+d

⇔ f(d) - g(b)

QUESTION: 2

If , then inverse of f is:

Solution:

let y = (4x + 3)/(6x - 4)

⇒ y * (6x - 4) = 4x + 3

⇒ 6xy - 4y = 4x + 3

⇒ 6xy - 4y - 4x = 3

⇒ 6xy - 4x = 3 + 4y

⇒ x(6y - 4) = 4y + 3

⇒ x = (4y + 3)/(6y - 4)

So, f^{-1} (x) = (4x + 3)/(6x - 4)

QUESTION: 3

If f(x) = (3 – x^{2})^{1/2}, then fof is:

Solution:

QUESTION: 4

If f: R → R and g: R → R be defined as f(x) = x + 1 and g(x) = x – 1. Then for all x ∈ R

Solution:

Given f:R→R and g: R→R

=> fog : R→R and gof : R→R

We know that I(R) : R→R

So the domains of gof, fog I(R) are the same.

fog(x) = f(g(x)) = f(x-1) = x-1+1 = I(R)x…..(1)

gof(x) = g(f(x)) = g(x-1) = x-1+1 = I(R)x……(2)

From (1) and (2), we get

fog(x) = gof(x) = I(R) for all x belongs to R

Hence fog = gof

QUESTION: 5

If f and g two functions such that they are one-one then g o f is

Solution:

f:A→B and g:B→C are both one-to-one functions.

Suppose a_{1},a_{2}∈A such that (gof)(a_{1})=(gof)(a_{2})

⇒ g(f(a_{1})) = g(f(a_{2})) (definition of composition)

Since gg is one-to-one, therefore,

f(a_{1}) = f(a_{2})

And since ff is one-to-one, therefore,

a_{1} = a_{2}

Thus, we have shown that if (gof)(a_{1})=(gof)(a_{2} then a_{1}=a_{2}

Hence, gof is one-to-one function.

QUESTION: 6

Let A = {a,b,c} and B = {1,2,3} and f: A→B is defined by f={(a,2), (b,1), (c,3)}. Find f^{-1}

Solution:

QUESTION: 7

If f: A→B and g:B→C are onto, then gof:A→C is:

Solution:

since g : B → C is onto

Suppose z implies C, there exist a pre image in B

Let the pre image be y

Hence y implies B such that g(y) = z

Similarly f : A → B is onto

Suppose y implies B, there exist a pre image in A

Let the pre image be x

Hence y implies A such that f(x) = y

Now, gof : A → C gof = g(f(x))

= g(y)

= z

So, for every x in A, there is an image z in C

Thus, gof is onto.

QUESTION: 8

A function in which second element of order pair are distinct is called

Solution:

QUESTION: 9

If f be a mapping defined by f(x) = x^{2} + 5, then f^{-1 }(x) is:

Solution:

f(x) = x^{2} + 5 = y(let)

⇒ y = x^{2}+5

⇒ x^{2} = y-5

⇒ x = (y-5)^{1/2}

⇒ f-1(x) = (x-5)^{1/2}

QUESTION: 10

Let A = {2 , 3 , 6}. Which of the following relations on A are reflexive?

Solution:

R_{1} is a reflexive on A , because (a,a) ∈ R_{1 }for each a ∈ A

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