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Functions and Limits | Short Questions and Answers

 

maths functions and limits short questions and answers

Functions and Limits Short Questions and Answers. This post is about 2nd Year | FSc. Part-II Mathematics Chapter No.1: (Functions and Limits) | Short Questions and Answers. The following questions are very important for the final examinations of all Boards of Punjab as well as Pakistan:

Chapter No.1: Functions and Limits

Short Questions and Answers

Q.1: What is a function?
Answer:
A function f from a set X to a set Y is a rule that assigns to each element x in X a unique element y in Y. Symbolically we write its as f:XY and read as “f is a function from X to Y”.

Q.2: If f(x)=2x3-5x2+6x+3, then find f(12x),x0?
Answer:
f(x)=2x3-5x2+6x+3   (1)
Putting x=12x in (1), we have
f(12x)=2(12x)3-5(12x)2+6(12x)+3
=28x3-54x2+62x+3
f(12x)=14x3-54x2+3x+3,x0

Q.3: If f(x)=x2-x, then find f(x-1)? (Lahore Board)
Answer:
f(x)=x2-x         (1)
Replacing x by x-1 in (1), we have
f(x-1)=(x-1)2-(x-1)
=x2-2x+1-x+1
=x2-3x+2

Q.4: If f(x)=x+4, then find f(x2+4)? (Lahore Board)
Answer:
f(x)=x+4                (1)
Putting x=x2+4 in (1), we have
f(x2+4)=x2+4+4=x2+8

Q.5: Without finding the inverse, state the domain and range of f-1(x) when f(x)=x-1x-4,x4? (Mirpur Board)
Answer:
f(x)=x-1x-4
First we find the domain and range of f. Since f is not defined at x=4, so this point must not be in the domain of f.
Thus, Domain of f=R-(-1)
Now      Domain of f-1 = Range of f=R-(-1)
And        Range of f-1 = Domain of f=R-(-4)

Q.6: Without finding the inverse, state the domain and range of f-1(x) when f(x)=(x-5)2,x5? (Gujranwala Board)
Answer:
Since the range of f(x) is x0, so the domain of f-1(x) is x0. Since the domain of f(x) is x5, so the range of f-1(x) is x5.

Q.7: Without finding the inverse, state the domain and range of f-1(x) when f(x)=1x+3? (Rawalpindi Board)
Answer:
f(x)=1x+3
First, we find the domain and range of f. Since f is not defined at x=-3, so this point must not be in the domain of f. Thus,
Domain of f=R-(-3)
Since no value of x can give the value of f zero, so 0 is not in the range of f
Range of f=R-(0)
Now      Domain of f-1 = Range of f=R-(0)
And        Range of f-1= Domain of f=R-(-3)

Q.8: Without finding the inverse, state the domain and range of f-1(x) when f(x)=x+2? (Faisalabad Board)
Answer:
Since the range of f(x)=x+2 is x-2, so the domain of f-1(x) is x-2. Since the domain of f(x) is x0, so the range of f-1(x) is x0.

Q.9: Find f-1(x) for f(x)=1x+3? (Lahore Board)
Answer:
Let y=f(x)=1x+3, then x+3=1y

functions and limits
Q.10: If f(x)=2x+1, then show that f-1(f(x))=x? (Multan and Lahore Board)
Answer:
Let y=f(x), then f-1(y)=x.
Now f(x)=2x+1

functions and limits
Q.11: If f(x)=-2x+8, then find f-1(x)? (Lahore Board)
Answer:
Let y=f(x), then f-1(y)=x

functions and limits
Q.12: If f(x)=(-x+9)3, then find f-1(x)? (Lahore Board)
Answer:
Let y=f(x), then f-1(y)=x

functions and limits

Q.13: Given that f(x)=x3-x2+x-1, then find `f(0)? (Lahore Board)

Answer:

f(x)=x3-x2+x-1           (1)

Putting x=0 in (1), we have

f(0)=03-02+0-1

f(0)=-1


Q.14: What is a vertical line test of a function?

Answer:

If a vertical line intersects a graph in more than one point, it is not the graph of a function. In fact, when the graph intersects at more than one points a vertical line, then a single point will have more than one images (points of intersection), so it will not be graph of a function.


Q.15: What is an algebraic function?

Answer:

The functions defined by algebraic expressions are called algebraic functions, for example f(x)=x+2, f(x)=x2-4x+1 etc. are algebraic functions. But `f(x)=sin x is not an algebraic function, because it is not defined by an algebraic expression.


Q.16: Define a polynomial function of degree n? (Gujranwala Board)

Answer:

A function P(x)=anxn+an-1xn-1+an-2xn-2+... for all x, where the coefficients a_n,a_{n-1},a_{n-2},….,a_2,a_1,a_0 are real numbers with a_n\ne0 and the the exponents are non-negative integers, is called a polynomial function of degree n.


Q.17: Express the perimeter P of a square as function of its area A? (Lahore Board, G-I)

Answer:

If x is the side length of the square, then its perimeter P is

P=4x   (1)

The area A of the square is

A=x^2

\Rightarrow x=\sqrt{A}               (2)

Putting this value of x from (2) in (1), we have

P=4\sqrt{A}

Which is the required perimeter P as a function of area A.


Q.18: Express the volume V of the cube as a function of its base area A(DG Khan Board, G-I)

Answer:

Let x is the side length of the cube, then its volume V is

V=x^3                (1)

The area A of the base of the cube is A=x^2 \Rightarrow x=\pm\sqrt{A}. Since the side length cannot be negative, so x=\sqrt{A}=(A)^\frac{1}{2}. Putting this valued of x in (1), we have

V=(A)^\frac{3}{2}

Which is the required volume V of the cube as a function of its base area A.


Q.19: Express the area A of a circle as a function of its circumference C? (DG Khan Board, G-II)

Answer:

Let x be the radius of the circle, then its area A is given by

A=\pi x^2          (1)

The circumference C of the circle is given by C=2\pi x \Rightarrow x=\frac{C}{2\pi}.

Putting this value of x in (1), we have

A=\pi(\frac{C}{2\pi})^2=\frac{\pi C^2}{4\pi^2}=\frac{1}{4\pi}C^2

Which is the required area A of a circle as a function of its circumference C.

Q.20: Find the domain of g(x)=\sqrt{x^2-4}? (Lahore Board)
Answer:
Domain of g(x)=\sqrt{x^2-4} is x^2-4\geq0
(x+2)(x-2)\geq0\Rightarrow x\geq2,x\leq-2
Therefore, the domain of g(x)=\sqrt{x^2-4} is (-\infty,-2]\cup[2,\infty).


NOTE: 

The remaining short questions and answers of Chapter No.1 (Functions and Limits) will be published soon Insha Allah.


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