NCERT EXERCISE 1.2

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    Exercise 1.2

    The Exercise 1.2 deals with functions, sets and relation , injection , surjection and certain other properties of sets.

    The exercise 1.2 also include one-one and onto functions. The Exercise 1.2 provides a lot of working problems to make you more familiar and through with the concepts..

    CLICK HERE TO READ EXERCISE 1.2 ONLINE.

    QUESTIONS:

    Question 1: Show that the function f: 𝐑∗ → 𝐑∗ defined by 𝑓(𝑥) = 1/𝑥 is one-one and onto, where 𝐑∗ is the set of all non-zero real numbers. Is the result true, if the domain 𝐑∗ is replaced by N with co-domain being same as 𝐑∗? 
    Solution 1:
    It is given that f: R* → 𝑅∗ is defined by f(x) = 1/𝑥
    For one – one:
    Let x, y ∈ R∗ such that f(x) = f(y)
    ⇒ 1/𝑋=1/𝑌
    ⇒ x = y
    ∴ f is one – one.
    For onto:
    It is clear that for y ∈ R∗ , there exists x = 1/𝑦 ∈ R∗ [as y ≠ 0] such that
    f(x) = 1/(1/𝑦) = y
    ∴ f is onto.
    Thus, the given function f is one – one and onto.
    Now, consider function g: N → R∗ defined by g(x) = 1/x
    We have, 𝑔(𝑥1) = 𝑔(𝑥2) ⇒ =1/X1=1/X2 ⇒ 𝑥1 = 𝑥2
    ∴ g is one – one.
    Further, it is clear that g is not onto as for 1.2 ∈ = R∗ there does not exit any x in N such that g(x) = 1/1.2 .
    Hence, function g is one-one but not onto. 
    Question 2:
    Check the injectivity and surjectivity of the following functions: 
    (i) f: N → N given by f(x) = x²                                                                            Sol.  It is seen that for x, y ∈ N, f(x) = f(y) ⇒ x² = y² ⇒ x = y.                                     ∴ f is injective.
    Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x² = 2.
    ∴ f is not surjective.
    Hence, function f is injective but not surjective.                                                                                                                                                                                          (ii) f: Z → Z is given by f(x) = x²
    It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
    ∴ f is not injective.
    Now, −2 ∈ Z. But, there does not exist any element x ∈ Z such that
    f(x) = −2 or x² = −2.
    ∴ f is not surjective.
    Hence, function f is neither injective nor surjective.
    (iii)f: R → R is given by f(x) = x²
    It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
    ∴ f is not injective.
    Now, −2 ∈ R. But, there does not exist any element x ∈ R such that f(x) = −2
    or x² = −2.
    ∴ f is not surjective.
    Hence, function f is neither injective nor surjective.                                                       (iv) f: N → N given by f(x) = x³
    It is seen that for x, y ∈ N, f(x) = f(y) ⇒ x³ = y³ ⇒ x = y.
    ∴ f is injective.
    Now, 2 ∈ N. But, there does not exist any element x ∈ N such that
    f(x) = 2 or x³ = 2.
    ∴ f is not surjective
    Hence, function f is injective but not surjective.                                                           (v) f: Z → Z is given by f(x) = x³
    It is seen that for x, y ∈ Z, f(x) = f(y) ⇒ x³ = y³ ⇒ x = y.
    ∴ f is injective.
    Now, 2 ∈ Z. But, there does not exist any element x ∈ Z such that
    f(x) = 2 or x³ = 2.
    ∴ f is not surjective.
    Hence, function f is injective but not surjective.
    Question 3:
    Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one – one nor onto, where [x] denotes the greatest integer less than or equal to x.
    Solution :
    f: R → R is given by, f(x) = [x]
    It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1.
    ∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9.
    ∴ f is not one – one.
    Now, consider 0.7 ∈ R.
    It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈ R such that f(x) = 0.7.
    ∴ f is not onto.
    Hence, the greatest integer function is neither one – one nor onto. 
    Question 4:
    In Show that the Modulus Function f: R → R given by 𝑓(𝑥) = |𝑥|, is neither one – one nor onto, where |𝑥| is x, if x is positive or 0 and |X| is − x, if x is negative.
    Solution :
    f: R → R is given by f(x) = |x| = { {X if X≥0, −X if x<0
    It is clear that f(-1) = |-1| = 1 and f(1) = |1| = 1
    ∴ f(−1) = f(1), but −1 ≠ 1.
    ∴ f is not one – one.
    Now, consider −1 ∈ R.
    It is known that f(x) = |𝑥| is always non-negative. Thus, there does not exist any element x in domain R such that f(x) = |𝑥| = −1.
    ∴ f is not onto.
    Hence, the modulus function is neither one-one nor onto.
    Question 5:
    Show that the Signum Function f: R → R, given by 𝑓(𝑥) = {1 if X>0    0, if X=0    −1, if X<0
    is neither one-one nor onto.
    Solution:  f: R → R is given by f(x) == {1 if X>0     0, if X=0       −1, if X<0
    It is seen that f(1) = f(2) = 1, but 1 ≠ 2.
    ∴ f is not one – one.
    Now, as f(x) takes only 3 values (1, 0, or −1) for the element −2 in co-domain
    R, there does not exist any x in domain R such that f(x) = −2.
    ∴ f is not onto.
    Hence, the Signum function is neither one – one nor onto

    Question 6:
    Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one – one.
    Solution:
    It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.
    f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.
    ∴ f (1) = 4, f (2) = 5, f (3) = 6
    It is seen that the images of distinct elements of A under f are distinct.
    Hence, function f is one – one.

    Question 7:
    In each of the following cases, state whether the function is one – one, onto or bijective. Justify your answer.
    (i) f: R → R defined by f(x) = 3 − 4x
    (ii) f: R → R defined by f(x) = 1 + x2
    Solution:
    (i) f: R → R is defined as f(x) = 3 − 4x.
    Let x1, x2 ∈ R such that f(x1) = f(x2)
    ⇒ 3-4×1 = 3- 4×2
    ⇒ -4×1 = -4×2
    ⇒ x1 = x2
    ∴ f is one – one.
    For any real number (y) in R, there exists 3−y/ 4 in R such that f (3−y/ 4) = 3-4 (3−y /4) = y
    ∴ f is onto.
    Hence, f is bijective.
    (ii) f: R → R is defined as f(x) = 1 + x²                                                                  Let x1, x2 ∈ R such that f(x1) = f(x2)
    ⇒ 1+ 𝑋1²=1+𝑥2²
    ⇒ + 𝑋12=𝑋22
    ⇒ x1 = ± x2
    ∴ f(x1) = f(x2) does not imply that x1 = x2
    For example f(1) = f(-1) = 2
    ∴ f is not one – one.
    Consider an element −2 in co-domain R.
    It is seen that f(x) = 1 + x² is positive for all x ∈ R.
    Thus, there does not exist any x in domain R such that f(x) = −2.
    ∴ f is not onto.
    Hence, f is neither one – one nor onto.

    Question 8:
    Let A and B be sets. Show that f: A × B → B × A such that (a, b) = (b, a) is bijective function.
    Solution:
    f: A × B → B × A is defined as f(a, b) = (b, a).
    Let (a1, b1 ), (a2, b2 ) ∈ A × B such that f(a1, b1 ) = f(a2, b2 )
    ⇒ (b1, a1 ) = (b2, a2 )
    ⇒ b1 = b2 and a1 = a2
    ⇒ (a1, b1 ) = (a2, b2 )
    ∴ f is one – one.
    Now, let (b, a) ∈ B × A be any element.
    Then, there exists (a, b) ∈ A × B such that f(a, b) = (b, a). [By definition of f ]
    ∴ f is onto.
    Hence, f is bijective.
    Question 9:
    Let f: N → N be defined by f(n) = {n+1/2 ,if n is odd       n/2,if n is even     for all n ∈ N
    State whether the function f is bijective. Justify your answer.
    Solution:
    f: N → N is defined as f(n) == {n+1/2 ,if n is odd         n/2,if n is even        for all n ∈ N
    It can be observed that:
    f(1) = 1+1/2 = 1 and f(2) = 2/2 = 1 [By definition of f(n)]
    f(1) = f(2), where 1 ≠ 2
    ∴ f is not one-one.
    Consider a natural number (n) in co-domain N.
    Case I: n is odd
    ∴ n = 2r + 1 for some r ∈ N. Then, there exists 4r + 1∈ N such that
    f(4r + 1) = 4𝑟+1 + 1/2= 2r + 1
    Case II: n is even
    ∴ n = 2r for some r ∈ N. Then, there exists 4r ∈ N such that
    f(4r) = 4𝑟/2 = 2r.
    ∴ f is onto.
    Hence, f is not a bijective function.
    Question 10:
    Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by (𝑥) = (𝑥−2/𝑥−3) . Is f one-one and onto? Justify your answer.
    Solution:
    A = R − {3}, B = R − {1} and f: A → B defined by f(x) = (𝑋−2/𝑋− 3 )
    Let x, y ∈ A such that f(x) = f(y)
    ⇒ X−2/(X− 3) = Y−2/(Y− 3)
    ⇒ (x – 2)(y – 3) = (y – 2)(x – 3)
    ⇒ xy – 3x – 2y + 6 = xy – 2x – 3y + 6
    ⇒ – 3x – 2y = – 2x – 3y ⇒ x = y
    ∴ f is one-one.
    Let y ∈ B = R − {1}. Then, y ≠ 1.
    The function f is onto if there exists x ∈ A such that f(x) = y.
    Now, f(x) = y
    ⇒ X−2/(X− 3) = y
    ⇒ x – 2 = xy – 3y ⇒ x(1 – y) = – 3y + 2
    ⇒ x = 2− 3y/1−y ∈ A           [y ≠ 1]
    Thus, for any y ∈ B, there exists 2− 3y1−y ∈ A such that
    f (2− 3y/(1−y)) = (2−3𝑦/(1−𝑦))−2(2−3𝑦/(1−𝑦))−3 = 2−3y−2 + 2y/(2−3y−3 + 3y )= −𝑦/−1=𝑦                                                                                                                   ∴ f is onto.
    Hence, function f is one – one and onto.
     Question 11:
    Let f: R → R be defined as f(x) = x4 . Choose the correct answer.
    (A) f is one-one onto
    (B) f is many-one onto
    (C) f is one-one but not onto
    (D) f is neither one-one nor onto
    Solution :
    f: R → R is defined as f(x) = xˆ4.
    Let x, y ∈ R such that f(x) = f(y).
    ⇒ xˆ4 = yˆ4
    ⇒ x = ± y
    ∴ f(x) = f(y) does not imply that x = y.
    For example f(1) = f(–1) = 1
    ∴ f is not one-one.
    Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x) = 2.
    ∴ f is not onto.
    Hence, function f is neither one – one nor onto.
    The correct answer is D.
    Question 12:
    Let f: R → R be defined as f(x) = 3x. Choose the correct answer.
    (A) f is one – one onto
    (B) f is many – one onto
    (C) f is one – one but not onto
    (D) f is neither one – one nor onto
    Solution:
    f: R → R is defined as f(x) = 3x.
    Let x, y ∈ R such that f(x) = f(y).
    ⇒ 3x = 3y
    ⇒ x = y
    ∴f is one-one.                                                                                                             Also, for any real number (y) in co-domain R, there exists 𝑦/3 in R such that f (y/3) = 3          (y/3) = y
    ∴ f is onto.
    Hence, function f is one – one and onto.
    The correct answer is A.
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