Ncert Solutions For Class 12 Maths Ex 1.2

Ncert Solutions For Class 12 Maths Chapter 1 Ex 1.2

Q-1: Prove that the function f: \(R_{*}\rightarrow R_{*}\) which is defined by f(a) = \(\frac{1}{a}\) which is one- one and onto, where \(R_{*}\) is the set of all of the non- zero real numbers. Check whether the result is true or not, if the domain, say, \(R_{*}\) is replaced by M having co- domain as same as \(R_{*}\)?

Sol:

As per the data given in the question,

\(R_{*}\rightarrow R_{*}\) which is defined by f(a) = \(\frac{1}{a}\)

For one- one condition:

Let, a and b \(\epsilon R_{*} \) such that, f(a) = f(b)

\(\Rightarrow \frac{1}{a} = \frac{1}{b}\)

\(\boldsymbol{\Rightarrow }\) a = b

Hence, the function f is one- one.

For onto condition:

Clearly, for b\(\epsilon R_{*} \), there must exists a = \(\frac{1}{b} \; \epsilon R_{*} \) [since, b ≠ 0], so that

f(x) = \(\frac{1}{\frac{1}{b}}\) = y

Hence, the function f is onto.

Therefore, the given function f(a) is one- one as well as onto.

Let us consider a another function g: M → \( R_{*} \) which is defined by g(a) = \(\frac{1}{a}\).

g(x1) = g(x2)

\(\Rightarrow \frac{1}{a_{1}} = \frac{1}{a_{2}}\)

\(\boldsymbol{\Rightarrow }\) x1 = x2

Hence, the another function g is also one- one.

It is obvious that g is not onto as for 1.2 \(\epsilon R_{*} \), there doesn’t exist any a in M so that:

g(a) = \(\frac{1}{1.2}\)

Therefore, the function ‘g’ is one- one but, it is not onto.

 

 

Q-2: Check the following functions for their injectivity and surjectivity:

(i) f: N →N which is given by f(a) = a2

(ii) f: Z →Z which is given by f(a) = a2

(iii) f: R → R which is given by f(a) = a2

(iv) f: N →N which is given by f(a) = a3

(v) f: Z → Z which is given by f(a) = a3

Sol:

(i) f: N → N which is given by f(a) = a2

We can see that for a, b ɛ N

f(a) = f(y)

\(\boldsymbol{\Rightarrow }\) a2 = b2

\(\boldsymbol{\Rightarrow }\) a = b

Hence, the function f is injective.

2 ɛ N

But, there doesn’t exist any of the a in N, so that f(a) = b2 = 2

Hence, the function f is not surjective.

Therefore, the function f is not surjective but, it is injective.

 

(ii) f: Z → Z which is given by f(a) = a2

We can see that for a, b ɛ Z

f(-1) = f(1) = 1

But, -1≠ 1.

Hence, the function f is not injective.

-2 ɛ Z

But, there doesn’t exist any of the a ɛ Z, so that f(a) = – 2 or a2 = -2

Hence, the function f is not surjective.

Therefore, the function f is neither surjective nor it is injective.

 

(iii) f: R → R which is given by f(a) = a2

We can see that for a, b ɛ R

f(-1) = f(1) = 1

But, -1≠ 1.

Hence, the function f is not injective.

-2 ɛ R

But, there doesn’t exist any of the a ɛ R, so that f(a) = – 2 or a2 = -2

Hence, the function f is not surjective.

Therefore, the function f is neither surjective nor it is injective.

 

(iv) f: N → N which is given by f(a) = a3

We can see that for a, b ɛ N

f(a) = f(y)

\(\boldsymbol{\Rightarrow }\) a3 = b3

\(\boldsymbol{\Rightarrow }\) a = b

Hence, the function f is injective.

2 ɛ N

But, there doesn’t exist any of the element of a ɛ Z, so that f(a) = 2 or a3 = 2

Hence, the function f is not surjective.

Therefore, the function f is not surjective but it is injective.

 

(v) f: Z → Z which is given by f(a) = a3

We can see that for a, b ɛ Z

f(a) = f(y)

\(\boldsymbol{\Rightarrow }\) a3 = b3

\(\boldsymbol{\Rightarrow }\) a = b

Hence, the function f is injective.

2 ɛ Z

But, there doesn’t exist any of the element of a ɛ Z, so that f(a) = 2 or a3 = 2

Hence, the function f is not surjective.

Therefore, the function f is not surjective but it is injective.

 

 

Q-3: Show that the GIF (Greatest Integer Function) f: R →R which is given by f(a) = [a], which is neither one- one nor onto, where [a] notifies the greatest integer which must be less than or equal to a.

Sol:

f: R → R which is given by f(a) = [a]

f(1.4) = [1.4] = 1, f(1.7) = [1.7] = 1

\(\boldsymbol{\Rightarrow }\) f(1.4) = f(1.7) = 1, but 1.4 ≠ 1.7

Hence, the function f is not one- one.

Let us consider that, 0.9 ɛ R.

We know that, f(a) = [a] will always be any integer.

So, there doesn’t exist any of the element a ɛ R, such that f(a) = 0.9

Hence, the function f is not onto.

Therefore, the GIF (Greatest Integer Function) is neither onto nor one- one.

 

 

Q-4: Prove that the Modulus Function f: R → R which is given by f(a) = [a], which is neither one- one nor it is onto, where |a| is a, if a will be positive or 0 and |a| is –a, if a will be negative.

Sol:

f : R → R which is given by

f(a) = |a| = a, if a ≥ 0 and (-a), if a ≤ 0

It is obvious that f(-2) = |-2| = 2 and f(2) = |2| = 2

So, f(-2) = f(2), but -2 ≠ 2

Hence, f is not one- one

Let us consider -2 ɛ R.

We know that, f (a) = |a| will always be non- negative.

Therefore, neither of any element a exist in the domain R, so that

f(a) = |a| = -2.

Hence, f is not onto.

Hence, the modulus function is neither onto nor one- one.

 

 

Q-5: Prove that the Signum function f: R → R which is given by f(a) =

1, if a > 0

0, if a = 0

-1, if x < 0 is neither onto nor one- one.

Sol:

f: R → R which is given by f(a) =

1, if a > 0

0, if a = 0

-1, if x < 0

Since, f(a) can only take 3 values (i.e., 1, 0, -1) for the elements -3 in the co- domain R, there does not exist any of the a in domain R, so that f(a) = -3.

Hence, f is not onto.

Here, we can see that:

f(2) = f(3) = 1, but 2 ≠ 3

Hence, f is not one- one.

Therefore, the Signum function in this case is neither onto nor one- one.

 

 

Q-6: Let us take P as {2, 3, 4}, Q as {5, 6, 7, 8} and let f = {(2, 5), (3, 6), (4, 7)} which are the function from P and Q. Prove that the function f is one- one.

Sol:

As per the data given in the question, we have

P = {2, 3, 4} and Q = {5, 6, 7, 8}

f : P → Q is defines as f = {(2, 5), (3, 6), (4, 7)}

Thus, f (2) = 5, f(3) = 6 and f(4) = 7

Now,

Here we can see that the images of the distinct elements of P under f are also distinct.

Therefore, the function f is not one- one.

Q-7. Check whether the function in each of the following conditions is one- one, onto or bijective. Justify your answer in each cases.

(i) f : R → R which is defines by f(a) = 4 – 5a

(ii) f : R → R which is defined by f(a) = 2 + a2

Sol:

(i) f : R → R which is defined as f(a) = 4 – 5a

Let, a1, a2 ɛ so that, f(a1) = f(a2)

\(\boldsymbol{\Rightarrow }\) 4 – 5a1 = 4 – 5a2

\(\boldsymbol{\Rightarrow }\) – 5a1 = – 5a2

\(\boldsymbol{\Rightarrow }\) a1 = a2

Hence, the function f is one- one.

Now,

For any of the real number (b) in R, there exists \(\frac{4 – b}{5}\) in R, so that

f (\(\frac{4 – b}{5}\)) = \( 4 – 5\left(\frac{4 – b}{5} \right) \)

f (\(\frac{4 – b}{5}\)) = 4 – (4 – b)

f (\(\frac{4 – b}{5}\)) = b

Hence, the function f is onto.

Therefore, the function f is bijective.

 

(ii) f : R → R which is defined as f(a) = 2 + a2

Let, a1, a2 ɛ so that, f(a1) = f(a2)

\(\boldsymbol{\Rightarrow }\) 2 + a12 = 2 + a22

\(\boldsymbol{\Rightarrow }\) a12 = a22

\(\boldsymbol{\Rightarrow }\) a1 = ± a2

Thus, f(a1) = f(a2) which doesn’t imply that a1 = a2

Let us illustrate an example, f(2) = f(-2) = 3

Hence, the function f is not one- one.

Now,

Let us take an element -3 in the co- domain R.

We can see that, f(a) = 2 + a2 which is positive for all of the x ɛ R.

So, there doesn’t exist any a in the domain R so that f(a) = -3.

Hence, the function f is not onto.

Therefore, the function f is neither one- one nor onto so, it is not bijective.

 

 

Q-8: Let us consider P and Q be the two sets. Prove that f : P × Q → Q × P so that (x, y) = (y, x) is bijective function.

Sol:

f : P × Q → Q × P which is defined as f(x, y) = y, x.

Let, (x1, y1), (x2, y2) ɛ P × Q so that, f(x1, y1)= f(x2, y2)

\(\boldsymbol{\Rightarrow }\) (y1, x1) = (y2, x2)

\(\boldsymbol{\Rightarrow }\) y1 = y2 and x1 = x2

\(\boldsymbol{\Rightarrow }\) (x1, y1) = (x2, y2)

Hence, the function f is one- one.

Let us consider that (y, x) ɛ Q × P be any element.

Thus,

There will exist (x, y) ɛ P × Q so that,

f(x, y) = (y, x).

Hence, f is onto.

Therefore, the function f is bijective.

 

 

Q-9: Let f : N → N which will be defined by f(p) =

\( \frac{p + 1}{2}\), if p is odd

\( \frac{p}{2}\), if p is even, for all p ɛ N.

Check and justify your answer that whether the function f is bijective or not?

Sol:

f : N → N which is defined as f(p) =

\(\frac{p + 1}{2}\), if p is odd

\(\frac{p}{2}\), if p is even

for all n ɛ N

We can see that:

f(3) = \(\frac{3 + 1}{2}\) = 2 and f(4) = \(\frac{4}{2}\) = 2

\(\boldsymbol{\Rightarrow }\) f(3) = f(4), but 3 ≠ 4

Hence, f is not one- one.

Let us consider any natural number (p) in the co- domain N.

Case-1: [ p is odd ]

\(\boldsymbol{\Rightarrow }\) p = 2r + 1 for few r ɛ N.

Thus, there exists 4r + 1 ɛ N so that,

f(4r + 1) = \(\frac{4r + 1 + 1}{2}\) = \(\frac{4r + 2}{2}\) = 2r + 1

Case-2: [p is even]

\(\boldsymbol{\Rightarrow }\) p = 2r for few r ɛ N.

Thus, there exists 4r ɛ N so that,

f(4r) = \(\frac{4r}{2}\) = 2r

Hence, the function f is onto.

Therefore, the function f is not a bijective function.

 

 

Q-10: Let P = A – {3} and Q = A – {1}. Let us consider the function f : P → Q which defined by f(a) = \(\left(\frac{a – 3}{a – 4} \right)\). Check and Justify your answer, whether the function f is onto and one- one?

Sol:

P = A – {3}, Q = A – {1} and f : P → Q which is defined by

f(a) = \(\left(\frac{a – 3}{a – 4} \right)\)

Let, a, b ɛ P so that f(a) = f(b)

\(\boldsymbol{\Rightarrow }\) \( \frac{a – 3}{a – 4} = \frac{b – 3}{b – 4} \)

\(\boldsymbol{\Rightarrow }\) (a – 3)(b – 4) = (b – 3)(a – 4)

\(\boldsymbol{\Rightarrow }\) ab – 4a – 3b + 12 = ab – 4b – 3a + 12

\(\boldsymbol{\Rightarrow }\) a = b

Hence, the function f is one- one.

Let, b ɛ Q = A – {1}.

Thus, b ≠ 1

Then, the function f will be onto if a ɛ P so that, f(a) = b.

f(a) = b

\(\boldsymbol{\Rightarrow }\) \(\frac{a – 3}{a – 4}\) = b

\(\boldsymbol{\Rightarrow }\) (a – 3) = b(a – 4)

\(\boldsymbol{\Rightarrow }\) a – 3 = ab – 4b

\(\boldsymbol{\Rightarrow }\) a(1 – b) = 3 – 4b

\(\boldsymbol{\Rightarrow }\) a = \(\frac{3 – 4b}{1 – b}\) ɛ [b ≠ 1]

Then,

For any value of b ɛ Q, there will exist \(\frac{3 – 4b}{1 – b}\) ɛ P, so that

f(\(\frac{3 – 4b}{1 – b}\)) = \(\frac{\left (\frac{3 – 4b}{1 – b} \right) – 3}{\left (\frac{3 – 4b}{1 – b} \right) – 4}\)

= \(\frac{\left(3 – 4b \right) – 3\left(1 – b \right)}{\left(3 – 4b \right) – 4\left(1 – b \right)}\)

= \(\frac{3 – 4b – 3 + 3b}{3 – 4b – 4 + 4b}\)

= \(\frac{-b}{-1}\) = b

Hence, the function f is onto.

Therefore, the function f is onto as well as it is one- one.

 

 

Q-11: Let us assume, f : R → R will be defined as g(a) = a5. Select the correct answer from the following:

(a) f is many- one onto (b) f is one- one onto

(c) f is neither one- one nor onto (d) f is one- one but, not onto

Sol:

f : R → R which is defined as g(a) = a4.

Let, a, b ɛ R so that f(a) = f(b).

\(\boldsymbol{\Rightarrow }\) a4 = b4

\(\boldsymbol{\Rightarrow }\) a = ± b

Thus, f(a) = f(b) doesn’t imply that a = b.

Example:

f(3) = f(-3) = 3

Hence, the function f is not one- one.

Let us consider an element 3 in the co- domain R. It’s obvious that there doesn’t exist any a in the domain R so that, f(a) = 3.

Hence, the function f is not onto.

Therefore, the function f is neither one- one nor it is onto.

Thus, the correct answer is option (c).

 

 

Q-12: Let us assume, f : R → R will be defined as g(a) = 4a . Select the correct answer from the following:

(a) f is many- one onto (b) f is one- one onto

(c) f is neither one- one nor onto (d) f is one- one but, not onto

Sol:

f : R → R which is defined as g(a) = 4a.

Let, a, b ɛ R so that f(a) = f(b).

\(\boldsymbol{\Rightarrow }\) 4a = 4b

\(\boldsymbol{\Rightarrow }\) a = b

Hence, the function f is one- one.

For any of the real number (b) in the co- domain R, there must exist \(\frac{b}{3}\) in R so that, f(\(\frac{b}{3} = 3\left(\frac{b}{3} \right) = y \))

Hence, the function f is onto.

Therefore, the function f is one- one and also, it is onto.

Thus, the correct answer is option (b).

 

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