Chapter No.1: Functions and Limits
Short Questions and Answers
Answer:
Answer:
`f(x)=2x^3-5x^2+6x+3` (1)
Putting `x=\frac{1}{2x}` in (1), we have
`f(\frac{1}{2x})=2(\frac{1}{2x})^3-5(\frac{1}{2x})^2+6(\frac{1}{2x})+3`
`=\frac{2}{8x^3}-\frac{5}{4x^2}+\frac{6}{2x}+3`
`f(\frac{1}{2x})=\frac{1}{4x^3}-\frac{5}{4x^2}+\frac{3}{x}+3, x\ne0`
Answer:
`f(x)=x^2-x` (1)
Replacing x by x-1 in (1), we have
`f(x-1)=(x-1)^2-(x-1)`
`=x^2-2x+1-x+1`
`=x^2-3x+2`
Answer:
`f(x)=\sqrt{x+4}` (1)
Putting `x=x^2+4` in (1), we have
`f(x^2+4)=\sqrt{x^2+4+4}=\sqrt{x^2+8}`
Answer:
`f(x)=\frac{x-1}{x-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\geq0`, so the domain of `f^{-1}(x)` is `x\geq0`. Since the domain of `f(x)` is `x\geq5`, so the range of `f^{-1}(x)` is `x\geq5`.
Answer:
`f(x)=\frac{1}{x+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)= \sqrt{x+2}` is `x\geq-2`, so the domain of `f^{-1}(x)` is `x\geq-2`. Since the domain of `f(x)` is `x\geq0`, so the range of `f^{-1}(x)` is `x\geq0`.
Answer:
Let `y=f(x)=\frac{1}{x+3}`, then `x+3=\frac{1}{y}`
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)=x^3-x^2+x-1` (1)
Putting x=0 in (1), we have
`f(0)=0^3-0^2+0-1`
`\Rightarrow 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)=x^2-4x+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)= a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_2x^2+a_1x+a_0`
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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