11th NCERT Relatins and functions.Miscellaneous Exercise Questions 12

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Question (1)

The relation f is defined by \[\begin{array}{l}f(x) = {x^2}\quad \;0 \le x \le 3\\\;\quad \quad \; = 3x\;\quad 3 \le x \le 10\end{array}\] The relation g is defined by \[\begin{array}{l}g(x) = {x^2}\quad \;0 \le x \le 2\\\quad \quad = 3x\;\quad 2 \le x \le 10\end{array}\] Show that f is a function and g is not a function.Solution

The relation f is defined by \[\begin{array}{l}f(x) = {x^2}\quad \;0 \le x \le 3\\\;\quad \quad \; = 3x\;\quad 3 \le x \le 10\end{array}\] For 0 ≤ x ≤ 3f(3) = 3

3 ≤ x ≤ 10, then f(x) = 3x.

f(3) = 3(3) = 9

So f(3) has unique value .

So f is a funvction.

The relation g is defined by \[\begin{array}{l}g(x) = {x^2}\quad \;0 \le x \le 2\\\quad \quad = 3x\;\quad 2 \le x \le 10\end{array}\] For 0 ≤ x ≤ 2

g(2) = 2

And for 2 ≤ x ≤ 10, then g(x) = 3x.

g(2) = 3(2) = 6

So g(2) has not unique value .

So g is not a funvction.

Question (2)

If f(x) = xSolution

f(x) = xQuestion (3)

Find the domain of the function \[f(x) = \frac{{{x^2} + 2x + 1}}{{{x^2} - 8x + 12}}\]Solution

\[f(x) = \frac{{{x^2} + 2x + 1}}{{{x^2} - 8x + 12}}\] \[ = \frac{{{{\left( {x + 1} \right)}^2}}}{{\left( {x - 2} \right)\left( {x - 6} \right)}}\] Function will be undefined if denominater becomes zero.The denominator becomes zero.

(x - 2)(x - 6) = 0

x - 2 = 0 or x - 6 = 0

x = 2 or x = 6

So at x = 2 and x = 6 function will not be defined.

So domain of f is R - {2,6}.

Question (4)

Find the domain and the range of the real function f defined by \[f(x) = \sqrt {x - 1} \]Solution

\[f(x) = \sqrt {x - 1} \] We can not find the square root of negative numbers.So x - 1 ≥ 0

So x ≥ 1

So Domain of f is [ 1, ∞ )

Since x - 1 ≥ 0

\[\sqrt {x - 1} \ge 0\] So range of f = [ 0 , ∞)

Question (5)

Find the domain and the range of the real function f defined by f (x) = |x – 1|.Solution

f (x) = |x – 1|.x ∈ R, So domain of function = R

For all values of x , | x - 1 | ≥ 0

So range of function = R

Question (6)

Let \[f = \left\{ {\left( {x,\frac{{{x^2}}}{{1 + {x^2}}}} \right):x \in R} \right\}\] be a function from R into R. Determine the range of f.Solution

Let f : R → RSo x ∈ R,

x

1 + x

\[{\frac{{{x^2}}}{{1 + {x^2}}} \ge 0}\] \[{x^2} \le {x^2} + 1\] \[\frac{{{x^2}}}{{{x^2} + 1}} \le 1\] So range of f is [ 0,1)

Question (7)

Let f, g: R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and.f/g.Solution

f : R → R and g: R → RSo domain of and g are R.

So D

So f+ g , f - g and f/ g can be calculated.

f + g = f(x) + g(x)

= x + 1 + 2x - 3

= 3x - 2

f - g = f(x) - g(x)

= ( x + 1 ) - ( 2x - 3 )

= -x + 4

f/ g is defined for other values of x other than g(x) = 0 .

f/g : R- {3/2) → R

\[\frac{f}{g} = \frac{{f(x)}}{{g(x)}},g(x) \ne 0\] \[ = \frac{{x + 1}}{{2x - 3}},2x - 3 \ne 0\]

Question (8)

Let f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.Solution

f : Z → Zf(x) = ax + b,

f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function

So (0, –1) will saisfy the f

∴ -1 = a(0) + b

∴ -1 = b

So (1, 1) is satisfy the function.

∴ 1 = a(1) + b

Replacing the value of b , we get,

1 = a + (- 1)

1 + 1 = a

2 = a

So the values of a = 2 and b = -1 .

Question (9)

Let R be a relation from N to N defined by R = {(a, b): a, b ∈ N and a = b(i) (a, a) ∈ R, for all a ∈ N

(ii) (a, b) ∈ R, implies (b, a) ∈ R

(iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.

Justify your answer in each case.

Solution

(i) (a, a) ∈ R, for all a ∈ NFor all a ∈ N , a

∴ statement is not true.

(ii) ( a, b) ∈ R,

→ a = b

Let (2, 4 ) ∈ R.

theen ( 4, 2 )∉ R. as 2 is not a square of 4.

So the statement is not true.

(iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.

The statement is not true.

If (2, 4) ∈ R, and (4, 16) ∈ R

but ( 2, 16 ) ∉ R as 16 is not square of 2.

Question (10)

Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are the following true?(i) f is a relation from A to B (ii) f is a function from A to B.

Justify your answer in each case.

Solution

A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}.In f ( x , y ) X ∈ A and y ∈ B.

So it is a relation from A → B.

But as (1, 5) ∈ f and (4, 5) ∈ f

The relation is one to many.

So it is not a function.

Question (11)

Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z: justify your answer.Solution

Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}.Let us consider a = 0, b = 1.

f = ( 0(1), 0 + 1 ) = ( 0, 1 )

If a = 0 and b = 2, then f = (0, 2 )

So for a = 0 and different values of b we get ( ab , a+b) = ( 0, b)

So relation becomes one to many, so it is not function.

Question (12)

Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f(n) = the highest prime factor of n. Find the range of f.Solution

A = {9, 10, 11, 12, 13}f: A → N

f(n) = the highest prime factor of n.

f(9) = 3, f(10) = 5, f(11) = 11, f(12) = 3, f( 13) = 13.

So range of f = { 3, 5, 11, 13}