Chapter No.1: Functions and Limits
Short Questions and Answers
Answer:
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,x≠0
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
Answer:
f(x)=√x+4 (1)
Putting x=x2+4 in (1), we have
f(x2+4)=√x2+4+4=√x2+8
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)
Answer:
Since the range of f(x) is x≥0, so the domain of f-1(x) is x≥0. Since the domain of f(x) is x≥5, so the range of f-1(x) is x≥5.
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)
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 x≥0, so the range of f-1(x) is x≥0.
Answer:
Let y=f(x)=1x+3, then x+3=1y
Let y=f(x), then f-1(y)=x.
Now f(x)=2x+1
Let y=f(x), then f-1(y)=x
Let y=f(x), then f-1(y)=x
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).
The remaining short questions and answers of Chapter No.1 (Functions and Limits) will be published soon Insha Allah.
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