Subjective Type Questions: Functions | JEE Advanced | 35 Years Chapter wise Previous Year Solved Papers for JEE PDF Download

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 y² = – x represents upper half of left handed parabola (upper half as y is + ve) and y² = x represents upper half of right handed parabola.
Therefore the resulting graph is as shown below :

Q. 3. If f (x) = x⁹ – 6x⁸ – 2x⁷ + 12x⁶ + x⁴ – 7x³ + 6x² + x – 3, find f (6).

Solution. f (x) = x⁹ – 6x⁸ – 2x⁷ + 12x⁶ + x⁴ – 7x³ + 6x² + x – 3

Then f (6) = 6⁹ – 6 × 6⁸ –  2 × 6⁷ + 12 × 6⁶ + 6⁴ – 7 × 6³ + 6 × 6² + 6 – 3
= 6⁹ – 6⁹ – 2 × 6⁷ + 2 × 6⁷ + 6⁴ – 7 × 6³ + 6³ + 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, x² + y2 < 25] which represents all the points inside and on the circle x² + y² = 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) = 2²

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 eqⁿ becomes
4 {x} = [x] + {x} + [x]  ⇒ 3 {x} = 2 [x]
    ....(1)

    [Using eqⁿ (1)]

   [Using eqⁿ (1)]

   [Using eqⁿ (1)]

    [Using eqⁿ (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) = Ax² + 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) = Ax² + 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) = Ax² + 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.