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Q1. defined by f(x) = 2x, g(y) = 3y + 4 and h(z) = cosz for all x, y, z in N. Show that ho(gof) = (hog)of.

Solution

Here, we have f(x) = 2x, g(y) = 3y + 4 and h(z) = cosz for all x, y, z in N. L.H.S. = ho(gof)(x) = ho[g(f(x))] = h[gof(x)] = h(g(f(x))) = h(g(2x)) = h(3(2x) + 4) = h(6x + 4) = cos (6x + 4)   R.H.S. = (hog)of (x) = hog(f(x)) = h(g(f(x))) = h(g(2x)) = h(3(2x) + 4) = h(6x + 4) = cos(6x + 4)   Hence, ho(gof) = (hog)of.

Q2. For the binary operation * defined by a * b = a + b + 1. Find the identity element for operation *.

Solution

Let e be the identity element for operation * and we know that a * e = a = e * a.   or a + e + 1 = a   and  e + a +1 = a or e = – 1 and e = – 1. Thus, – 1 is the identity element for operation *.

Q3. Check the function  given by  is one-one or not.

Solution

For x = 1, f(1) = 0, For x = 2, f(2) = 0, For x = 3, f(3) = 0. Since f(1) = f(2) = f(3) = 0, therefore f is not one-one.

Q4. Show that the function  defined by  is onto function.

Solution

Since f(x) = y, therefore f is onto function.

Q5. If  are defined as f(x) = cos x and g(x) = x². Show that .

Solution

We f(x) = cos x and g(x) = x². gof(x) = g(f(x)) = g(cos x) = (cos x)² = cos²x and fog(x) = f(g(x)) = f(x²) = cos(x²). Hence, .

Q6.

Solution

Q7. Show that the function  defined by   is one-one and onto.

Solution

(i) f is one-one: For     for x < 0,     For – 1 < x < 1,  lies between .   So, f is one-one.   (ii) f is onto: For   For     In both cases, each value of y in co-domain of f has a unique value in its domain. Hence, f is onto.

Q8. Let   be defined by Show that the function f is a bijective function.

Solution

Here, be defined by (a) f is one-one:   Therefore, f is one-one, i.e. injective.   (b) f is onto:   For every natural number, f has a unique value in co-domain. So, every member of co-domain has a pre-image has a pre-image in its domain. So, f is onto. f is surjective.   Since f is injective and surjective both, therefore f is a bijective function.

Q9. For the binary operation * defined by a * b = ab/1000. Find the inverse of 0.01.

Solution

Let e be the identity element for operation * and we know that   Also, let i be the inverse of 0.01, then

Q10. If show that fof(x) = x for all . Also find the inverse of f.

Solution

 We have,  

Q11. Show that the modulus function  defined by f(x) = | 2x | is neither one-one nor onto.

Solution

Here, the modulus function  defined by . If f(1) = 2 and f(– 1) = – (– 2) = 2. Since, f(1) = f(– 1) = 2, therefore f is not one-one.   No negative values belonging to co-domain of f has any pre-image in its domain. So, f is not onto function.

Q12. Show the relation R in the set is an equivalence relation.

Solution

Set A = {0,1, 2, 3, 4, 5, ....,10} and or R = {(0, 3), (3, 0), (0, 6), (6, 0), (0, 9), (9, 0), (1, 4), (4, 1),(2, 5), (5, 2), (3, 6), (6, 3), (3, 9),(9, 3),(4, 7), (7, 4),(4, 10), (10, 4), (1, 7), (7, 1),(1, 10), (10, 1), (2, 8), (8, 2),(5, 8), (8, 5),(6, 9), (9, 6),(7, 10), (10, 7), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9), (10, 10)}   (i) a - a = 0 = 3k, where k = 0, . So, R is reflexive. (ii) Here, So, R is symmetric.   (iii) Here,   So, the relation R is also transitive.   Since, relation R is reflexive, symmetric and transitive, therefore the R is an equivalence relation.

Q13. A binary operation defined by a * b = a^b. Is the binary operation associative?

Solution

No. Here, a * b = a^b.

Q14. If P = {5, 6} and Q = {1, 6, 7}, then find P x Q and Q x P.

Solution

P x Q = {(5, 1), (5, 6), (5, 7), (6, 1), (6, 6), (6, 7)} and Q x P = {(1, 5), (1, 6), (6, 5), (6, 6), (7, 5), (7, 6)}

Q15.

Solution

Q16. If L is the set of all lines in the plane and R is the relation in L defined by R = {(l₁, l₂) : l₁ is parallel to l₂}. Show that the relation R is equivalence relation.

Solution

Here, L is the set of all lines in the plane and R = {(l₁, l₂) : l₁ is parallel to l₂}.   (i) Every line in a plane is always parallel to itself, i.e., l₁ || l₂. So, relation R is reflexive.   (ii) If a line l₁ is parallel to line l₂, then l₂ is also parallel to l₁. . So, R is symmetric.   (iii) If a line l₁ is parallel to line l₂ and l₂ is parallel to l₃, then l₁ is also parallel to l₃. . So, R is transitive.   Since, relation R is reflexive, symmetric and transitive, therefore the R is an equivalence relation.

Q17. If  is defined as f(x) = x³ – 2. Find f^–1(62).

Solution

We have f(x) = x³ – 2.

Q18.

Solution

Q19.

Solution

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