Functions questions are provided here, along with their solutions, based on Class XI and XII syllabi. They are prepared as per the NCERT (CBSE) guidelines. Solving these questions will help students understand the concept well and improve their skills regarding the understanding of functions, which will create the basis of higher Calculus. Learn more about Functions.
Functions come into play whenever one quantity depends on the other. For example, the area of the circle depends on the radius of the circle, so we say the area is a function of the radius of the circle.
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A function f is a rule that assigns to each element x in a domain set D to exactly one unique element, called f(x) in a co-domain set E. The f(x) is called the image of x, while x is the pre-image of f(x). |
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There are various types of functions:
- Many-one function: More than one element in the domain set has the same f-image
- One-one function: Every single element of the domain set has different f-images.
- Into function: When the co-domain of the function is not equal to the range of function, that is, a few elements in the co-domain set does not have a pre-image in the domain set.
- Onto function: Every element of the co-domain set of the function has a pre-image in the domain set.
- Bijective function: A function which is both one-one and onto.
Functions Questions with Solution
Here are given a few solved questions which will help students
1. Which of the following relations are functions? Give reasons and also find the domain and range of the function.
(i) f = {(1, 3), (1, 5), (2, 3), (2, 5)}
(ii) g = {(2, 1), (5, 1), (8, 1), (11, 1)}
Solution: (i) f = {(1, 3), (1, 5), (2, 3), (2, 5)}
Here, the elements 1 and 2 have more than one f-images, namely 3 and 5.
Hence, f is not a function.
(ii) g = {(2, 1), (5, 1), (8, 1), (11, 1)}
Here, each first element of the ordered pair has a unique image which is the second coordinate.
Hence, g is a function.
Domain of g = {2, 5, 8, 11} and range of g = {1}
2. Let A = {1, 2} and B = {3, 6} and f and g be functions from A to B, defined by f(x) = 3x and g(x) = x2 + 2. Show that f = g.
Solution: Since both f and g are defined from set A.
Therefore, dom(f) = dom(g)
Now, for the co-domain of f
f(1) = 3. 1 = 3
f(2) = 3. 2 = 6
Co-domain of f ={3, 6}
And the co-domain of g
g(1) = 12 + 2 = 3
g(2) = 22 + 2 = 6
Co-domain of g ={3, 6}
Co-domain of f = Co-domain of g
Also for each x ∈ A, f(x) = g(x)
Hence, f = g.
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Two functions f and g are said to be equal functions if
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3. Let f: R →R : f(x) = x2 + 1. Find f-1(10).
Solution: Let f-1(10) = x.
⇒ f(x) = 10
⇒ x2 + 1 = 10
⇒ x = ±3
Therefore, f-1(10) = {-3, 3}
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The inverse of a Function Let f: X → Y be a function and let for any x in X, f(x) = y ∈ Y. Then the inverse of f is defined by f-1(y) = {x ∈ X: f(x) = y}. |
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4. Find the domain and the range of the real function, f(x) = 1/(x + 3).
Solution: We have f(x) = 1/(x + 3)
Clearly, f is not defined for x = -3
Therefore, dom(f) = R – {-3}
Let y = f(x). Then,
y = 1/(x + 3) ⇒ x = (1/y) – 3 …….(i)
Clearly, (i) is not defined for y = 0
Therefore, range(f) = R – {0}
5. Let f: N → N: f(x) = 2x for all x in N. Check whether f is a bijection.
Solution: To prove f(x) is a bijection, we need to prove that f is one-one as well as onto.
f is one-one:
Let x1, x2 be any arbitrary element in N
f(x1) = f(x2) ⇒ 2x1 = 2x2 ⇒ x1 = x2
Therefore f is one-one
f is onto:
Let f(x) = y in co-domain of f
Then, 2x = y ⇒ x = y/2
For every y in N, y/2 ∉ N
Therefore, f is not onto
Hence, f is not a bijection.
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Composition of Functions Let f: A → B and g: B → C be two given functions. Then, the composition of f and g is represented by g o f which is the function, defined by (g o f): f: A → C: (g o f)(x) = g{f (x)} for every x in A |
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6. Let R be the set of all real numbers. Let f: R →R : f(x) = cos x and let g: R →R : g(x) = 3x2. Show that (g o f) ≠ (f o g).
Solution: Let x be an arbitrary real number. Then,
(g o f)(x) = g{f(x)} = g{ cos x} = 3 cos2 x
(f o g)(x) = f{g(x)} = f{3x2} = cos 3x2
Clearly, (g o f) ≠ (f o g).
Also Read:
7. Let f: R →R: f(x) = 4x + 3 for all x ∈ R. Show that f is invertible and find f-1.
Solution: Let x1, x2 be in R
f(x1) = f(x2) ⇒ 4x1 + 3 = 4x2 + 3 ⇒ x1 = x2
Therefore, f is one-one.
Let f(x) = y in co-domain R of f. Then,
∃ x = (y – 3)/4 in R, for every y in co-domain R of f
Therefore, f is onto
⇒ f is bijection
⇒ f-1 exists
Now, f(x) = y ⇒ f-1(y) = x = (y – 3)/4
Thus inverse of f is defined by
f-1: R → R: f-1(y) = (y – 3)/4 for all y in R
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A function f is invertible (inverse of f exists) if and only if f is a bijection. |
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8. Let f: R →R: f(x) = x2 and g: R →R: g(x) = 2x + 1. Find (f + g)(x).
Solution: Here, dom(f) = R = dom(g)
Therefore, dom(f) ∩ dom(g) = R
Then, (f + g)(x) = f(x) + g(x) = x2 + 2x + 1
9. Let f(x) = √x and g(x) = x be two functions defined over the set of non-negative real numbers. Find (f/g)(x).
Solution: Both functions are defined on the domain of non-zero real numbers.
Then, dom(f/g) = dom(f) ∩ dom(g) – {x: g(x) = 0} = [0, ∞) ∩ [0, ∞) – {0} = (0, ∞)
So, (f/g)(x): (0, ∞) → R is given by
(f/g)(x) = f(x)/g(x) = √x/x = 1/√x ; x is not equal to zero.
10. Let A = {2, 3, 4, 5} and B = {7, 9, 11, 13} and let f = {(2, 7), (3, 9), (4, 11), (5, 13)}. Show that f is invertible and find f-1.
Solution: We have f is a function such that f: A → B: f(x) = {(2, 7), (3, 9), (4, 11), (5, 13)}.
Clearly, each element in the co-domain set has a unique pre-image in the domain set. Therefore, f is one-one.
Now, range(f) = co-domain(f)
⇒ f is onto
⇒ f is a bijection
⇒ f is invertible
Hence, f-1 = {(7, 2), (9, 3), (11, 4), (13, 5)}
Practice Questions
1. Let f: N → Y: f(x) = 4x2 + 12x + 15. Show that f is invertible and find f-1.
2. Find the domain and range of function f defined by f(x) = x2 + 1
3. Find the domain and range of the function f(x) = (x2 +1)/(x2 – 1).
4. Let f: Z → Z: f(x) = x2 and g: Z → Z: g(x) = |x|2 for all x in Z. Show that f = g
5. Show that f: R – {0} → R – {0}: f(x) = 1/x is a bijection.
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