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**Q. 1. Find the domain and range of the function the function one-to-one?**

**Solution. **Since f (x) is defined and real for all real values of x, therefore domain of f is R.

Also since f (1) = f (–1) = 1/2

∴ f is not one-to-one.

**Q. 2. ****Draw the graph of y = | x | ^{1/2} for – 1 < x < 1.**

**Solution. **

[Here y should be taken always + ve, as by definition y is a + ve square root].

Clearly y^{2} = – x represents upper half of left handed parabola (upper half as y is + ve) and y^{2} = x represents upper half of right handed parabola.

Therefore the resulting graph is as shown below :

**Q. 3. If f (x) = x ^{9} – 6x^{8} – 2x^{7} + 12x^{6} + x^{4} – 7x^{3} + 6x^{2} + x – 3, find f (6).**

**Solution.** f (x) = x^{9} – 6x^{8} – 2x^{7} + 12x^{6} + x^{4} – 7x^{3} + 6x^{2} + x – 3

Then f (6) = 6^{9} – 6 × 6^{8} – 2 × 6^{7} + 12 × 6^{6} + 6^{4} – 7 × 6^{3} + 6 × 6^{2} + 6 – 3

= 6^{9} – 6^{9} – 2 × 6^{7} + 2 × 6^{7 }+ 6^{4}^{ }– 7 × 6^{3} + 6^{3} + 6 – 3 = 3

**Q. 4. **Consider the following relations in the set of real numbers R.

Find the domain and range of R ∩ R'. Is the relation R ∩ R' a function?

**Solution. **R = [(x, y); x _{∈} R, y _{∈} R, x^{2} + y2 __<__ 25] which represents all the points inside and on the circle x^{2} + y^{2} = 52, with centre (0, 0) and radius = 5

which represents all the points inside and on the upward parabola

= The set of all points in shaded region.

For

is not a function because image of an element is not unique , e..g., (0, 1), (0, 2), (0,3) .....

**Q. 5.** **Let A and B be two sets each with a finite number of elements. Assume that there is an injective mapping from A to B and that there is an injective mapping from B to A. Prove that there is a bijective mapping from A to B.**

**Solution. **As there is an injective maping from A to B, each element of A has unique image in B. Similarly as there is an injective mapping from B to A, each element of B has unique image in A. So we can conclude that each element of A has unique image in B and each element of B has unique image in A or in other words there is one to one mapping from A to B. Thus there is bijective mapping from A to B.

**Q.6. Let f be a one-one function with domain {x, y, z} and range {1, 2, 3}. It is given that exactly one of the following statements is true and the remaining two are false f (x) = 1, f (y) ≠ 1, f (z ) ≠ 2 determine f ^{–1}(1).**

**Solution. **f** **is one one function,

Df = {x, y, z}; Rf = {1, 2, 3}

Eaxctly one of the following is true :

f (x) = 1, f (y) ≠ 1, f (z) ≠ 2

To determine f –1 (1):

Case I: f (x) = 1 is tr ue.

⇒ f (y) ≠ 1, f (z) ≠ 2 are false.

⇒ f (y) = 1, f (z) = 2 are true.

But f (x) = 1, f (y) = 1 are true, is not possible as f is one to one.

∴ This case is not possible.

Case II: f (y) ≠ 1 is true.

⇒ f (x) = 1 and f (z) ≠ 2 are false

⇒ f (x) ≠1 and f (z) = 2 are true

Thus, f (x) ≠ 1, f (y) ≠ 1, f (z) = 2

⇒ Either f (x) or f (y) = 2. So, f is not one to one

∴ This case is also not possible.

∴ f (z) ≠ 2 is true

∴ f (x) = 1 and f (y) ≠ 1 are false.

⇒ f (x) ≠ 1 and f (y) = 1 are true.

⇒ f –1 (1) = y

**Q.7. Let R be the set of real numbers and f : R → R be such that for all x and y in R |f (x) – f(y) | < | x – y |^{3} . Prove that f(x) is a constant**

**Solution.**

**Q.8. ****Find the natural number ‘a’ for which where the function ‘f’ satisfies the relation f(x + y) = f (x) f (y) for all natural numbers x, y and further f(1) = 2. **

**Solution. **Given that

To find ‘a’ such that,

...(1)

For this we start with ...(2)

f (1) = 2

Then f (2) = f (1 + 1) = f (1) f (1)

⇒ f (2) = 2^{2}

**Q.9. ****Let {x} and [x] denotes the fractional and integral part of a real number x respectively. Solve 4{x} = x + [x].**

**Solution. **Given that 4 {x} = x + [x]

Where [x] = greatest integer __<__ x

{x} = fractional part of x

∴ x = [x] + {x} for any x ∈ R

∴ Given eq^{n} becomes

4 {x} = [x] + {x} + [x] ⇒ 3 {x} = 2 [x]

....(1)

[Using eq^{n} (1)]

[Using eq^{n} (1)]

[Using eq^{n} (1)]

[Using eq^{n} (1)]

**Q.10. A function f :IR → IR, where IR is the set of real numbers, is defined by ** ** Find the interval of values of α for which f is onto. Is the function one-to-one for α = 3? **

**Solution.**

This shows that

Therefore, f is not one-to-one at α = 3.

**Q.11. Let f(x) = Ax ^{2} + Bx +C where A, B, C are real numbers. Prove that if f(x) is an integer whenever x is an integer, then the numbers 2A, A + B and C are all integers. Conversely, prove that if the numbers 2A, A+B and C are all integers then f(x) is an integer whenever x is an integer.**

**Solution. **Suppose f (x) = Ax^{2} + Bx + C is an integer whenever x is an integer.

∴ f (0), f (1), f (–1) are integers

⇒ C, A + B + C, A – B + C are integers.

⇒ C, A + B, A – B are integers

⇒ C, A + B, (A + B) + (A – B) = 2A are integers.

Conversely suppose 2A, A + B and C are integers.

Let x be any integer.

We have

f (x) = Ax^{2} + Bx + C

Since x is an integer x, x (x – 1)/2 is an integer.

Also 2A, A + B and C are integers.

We get f (x) is an integer for all integer x.

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