NCERT Exemplar Class 11 Maths Chapter 2 Relations and Functions Exercise Solutions Long Answer Type

Answer 16 Given: g = {(1, 1), (2, 3), (3, 5), (4, 7)}

Each element of the domain has a unique image.
Therefore, g is a function.

Let
g(x) = αx + β

Using ordered pairs:

For (1, 1):
g(1) = α(1) + β = 1
⇒ α + β = 1 …(i)

For (2, 3):
g(2) = 2α + β = 3
⇒ 2α + β = 3 …(ii)

Subtract (i) from (ii):
α = 2

Substitute α = 2 in (i):
2 + β = 1
⇒ β = −1

Final Answer

• α = 2
• β = −1

Note: Any other two ordered pairs from the given relation can also be used to find α and β.

Answer 17

(i)
Given that:
f(x) = 1 / √(1 − cos x)

We know that:
−1 ≤ cos x ≤ 1

⇒ 1 ≥ −cos x ≥ −1
⇒ 1 + 1 ≥ 1 − cos x ≥ −1 + 1
⇒ 2 ≥ 1 − cos x ≥ 0
⇒ 0 ≤ 1 − cos x ≤ 2

For real value of domain:
1 − cos x ≠ 0 ⇒ cos x ≠ 1

⇒ x ≠ 2nπ, ∀ n ∈ Z

Hence, the domain of f = R − {2nπ, n ∈ Z}

(ii)
Given that:
f(x) = 1 / √(x + |x|)

x + |x| = x + x = 2x, if x ≥ 0
and
x + |x| = x − x = 0, if x < 0

So for x < 0, f is not defined.

Hence, the domain f = R⁺

(iii)
Given that:
f(x) = x|x|

It is clear that f(x) is defined for all x ∈ R.

Hence, the domain of f = R

(iv)
Given that:
f(x) = (x³ − x + 3) / (x² − 1)

Here, f(x) is only defined if:
x² − 1 ≠ 0

(x − 1)(x + 1) ≠ 0
⇒ x ≠ 1, x ≠ −1

Hence, the domain of f = R − {−1, 1}

(v)
Given that:
f(x) = 3x / (28 − x)

Here, f(x) is only defined if:
28 − x ≠ 0 ⇒ x ≠ 28

Hence, the domain = R − {28}

Answer 18.

(i)
Given that:
f(x) = 3 / (2 − x²)

Let y = f(x)
∴ y = 3 / (2 − x²)

⇒ y(2 − x²) = 3
⇒ 2y − yx² = 3
⇒ yx² = 2y − 3
⇒ x² = (2y − 3) / y

Here, x is real if:
2y − 3 ≥ 0 and y ≥ 0

⇒ y ≥ 3/2

Hence, the range of f = [3/2, ∞)

(ii)
Given that:
f(x) = 1 − |x − 2|

We know that:
|x − 2| = −(x − 2), if x < 2
and
|x − 2| = (x − 2), if x ≥ 2

∴ −|x − 2| ≤ 0
⇒ 1 − |x − 2| ≤ 1

Hence, the range of f = (−∞, 1]

(iii)
Given that:
f(x) = |x − 3|

We know that:
|x − 3| ≥ 0 ⇒ f(x) ≥ 0

Hence, the range of f = [0, ∞)

(iv)
Given that:
f(x) = 1 + 3 cos 2x

We know that:
−1 ≤ cos 2x ≤ 1

⇒ −3 ≤ 3 cos 2x ≤ 3
⇒ −3 + 1 ≤ 1 + 3 cos 2x ≤ 3 + 1
⇒ −2 ≤ f(x) ≤ 4

Hence, the range of f = [−2, 4]

Answer 19. Given that:
f(x) = |x − 2| + |2 + x|, −3 ≤ x ≤ 3

Since
|x − 2| = −(x − 2), x < 2
|x − 2| = (x − 2), x ≥ 2

and
|2 + x| = −(2 + x), x < −2
|2 + x| = (2 + x), x ≥ −2

Now,
f(x) = |x − 2| + |2 + x|, −3 ≤ x ≤ 3

=
⎧ −(x − 2) − (2 + x), −3 ≤ x < −2
⎨ −(x − 2) + (2 + x), −2 ≤ x < 2
⎩ (x − 2) + (2 + x), 2 ≤ x ≤ 3

∴

f(x) =
⎧ −2x, −3 ≤ x < −2
⎨ 4, −2 ≤ x < 2
⎩ 2x, 2 ≤ x ≤ 3

Answer 20.

Given that:
f(x) = (x − 1) / (x + 1)

(i)

f(1/x)
= (1/x − 1) / (1/x + 1)
= (1 − x) / (1 + x)
= −(x − 1) / (x + 1)
= −f(x)

Hence,
f(1/x) = −f(x)

(ii)

f(−1/x)
= (−1/x − 1) / (−1/x + 1)
= −(1/x + 1) / −(1/x − 1)
= (1 + x) / (1 − x)
= 1 / [(x − 1)/(x + 1)]
= −1 / f(x)

Hence,
f(−1/x) = −1 / f(x)

Answer 21.

Given that:
f(x) = √x and g(x) = x
be two functions defined in the domain
R⁺ ∪ {0}

(i)
(f + g)(x) = f(x) + g(x)
= √x + x

(ii)
(f − g)(x) = f(x) − g(x)
= √x − x

(iii)
(fg)(x) = f(x) · g(x)
= √x · x
= x³ᐟ²

(iv)
(f / g)(x) = f(x) / g(x)
= √x / x
= 1 / √x

Answer 10.

(i) h is not a function
Since 3 has two images, 9 and 11.

(ii) f is a function
Every element of the domain has a unique image.

(iii) g is a function
For every positive integer n, there is a unique image 1/n.

(iv) s is a function
Square of any integer is unique.

(v) t is a function
Every real number is mapped to the constant value 3.

Answer 11.

Given:
f(x) = x² + 7
g(x) = 3x + 5

(i)
f(3) + g(−5)
= (3² + 7) + [3(−5) + 5]
= (9 + 7) + (−15 + 5)
= 16 − 10
= 6

(ii)
f(1/2) × g(14)
= [(1/2)² + 7] × [3(14) + 5]
= (1/4 + 7) × (42 + 5)
= (29/4) × 47
= 1363/4

(iii)
f(−2) + g(−1)
= [ (−2)² + 7 ] + [ 3(−1) + 5 ]
= (4 + 7) + (−3 + 5)
= 11 + 2
= 13

(iv)
f(t) − f(t − 2)
= (t² + 7) − [ (t − 2)² + 7 ]
= t² − (t² − 4t + 4)
= 4t − 4

(v)
[ f(t) − f(5) ] / (t − 5)

= [ (t² + 7) − (25 + 7) ] / (t − 5)
= (t² − 25) / (t − 5)
= t + 5, t ≠ 5

Answer 12. (x) = 2x + 1
g(x) = 4x − 7

(i) f(x) = g(x)

2x + 1 = 4x − 7
⇒ −2x = −8
⇒ x = 4

Hence, the required real number is x = 4.

(ii) f(x) < g(x)

2x + 1 < 4x − 7
⇒ −2x < −8
⇒ x > 4

Hence, the required real numbers are x > 4.

Answer 13. Given:
f(x) = 2x + 1
g(x) = x² + 1

(i) (f + g)(x)
= f(x) + g(x)
= (2x + 1) + (x² + 1)
= x² + 2x + 2

(ii) (f − g)(x)
= f(x) − g(x)
= (2x + 1) − (x² + 1)
= 2x − x²

(iii) (f·g)(x)
= f(x)g(x)
= (2x + 1)(x² + 1)
= 2x³ + x² + 2x + 1

(iv) (f / g)(x)
= f(x) / g(x)
= (2x + 1) / (x² + 1)

Answer 14. Given:
f(x) = x³ + 1
X = {−1, 0, 3, 9, 7}

Evaluate f(x) for each x ∈ X:

For x = −1
f(−1) = (−1)³ + 1 = −1 + 1 = 0

For x = 0
f(0) = 0³ + 1 = 1

For x = 3
f(3) = 27 + 1 = 28

For x = 9
f(9) = 729 + 1 = 730

For x = 7
f(7) = 343 + 1 = 344

Set of ordered pairs:
{ (−1, 0), (0, 1), (3, 28), (7, 344), (9, 730) }

Range:
{ 0, 1, 28, 344, 730 }

Answer 15. Given:
f(x) = 3x² − 1
g(x) = 3 + x

Since f(x) = g(x),

3x² − 1 = 3 + x

⇒ 3x² − x − 4 = 0

Factorising:
(3x − 4)(x + 1) = 0

⇒ 3x − 4 = 0 or x + 1 = 0

⇒ x = 4/3 or x = −1

Hence, the required values of x are −1 and 4/3.