Subtopic Archive - Celestial Tutors

Test

January 7, 2025/by admin_ct

[latexpage]
At first, we sample $f(x)$ in the $N$ ($N$ is odd) equidistant points around $x^*$:
\[
   f_k = f(x_k),\: x_k = x^*+kh,\: k=-\frac{N-1}{2},\dots,\frac{N-1}{2}
\]
where $h$ is some step.
Then we interpolate points $\{(x_k,f_k)\}$ by polynomial
\begin{equation} \label{eq:poly}
   P_{N-1}(x)=\sum_{j=0}^{N-1}{a_jx^j}
\end{equation}
Its coefficients $\{a_j\}$ are found as a solution of system of linear equations:
\begin{equation} \label{eq:sys}
   \left\{ P_{N-1}(x_k) = f_k\right\},\quad k=-\frac{N-1}{2},\dots,\frac{N-1}{2}
\end{equation}
Here are references to existing equations: (\ref{eq:poly}), (\ref{eq:sys}).
Here is reference to non-existing equation (\ref{eq:unknown}).

dsdsds

Improper integrals

February 26, 2024/by admin_ct

When either one or both limits of integration are infinity, then it is called improper integral. The interval of integration is said to be over an infinite interval. We will call these integrals convergent if the associated limit exists and is a finite number and divergent if the associated limit either doesn’t exist or is (plus or minus) infinity.

Check whether given integral function is convergent or not.

int_1^oo {(4x+1)}/{(x^3-2x^2+x)} dx

Given integral function:

int_1^oo {(4x+1)}/{(x^3-2x^2+x)} dx

Derivative of integral function

December 25, 2023/by admin_ct

The Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration. It says :

If f(x) is continuous over an interval $[a, b]$ and the function F(x) is defined by

$\dpi{120} F(x)=\int_{a}^{x}f(t) dt$

Then $\dpi{120} F'(x)= f(x)$ over closed interval $[a, b]$

Lets work on some examples:

Find derivative of the given function, using the fundamental theorem of Calculus.

$\dpi{120} {\color{Red} g(x) =\int_{0}^{x}\sqrt{1+t^2} dt}$

Taking derivative of both sides,

$\dpi{120} g'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\int_{0}^{x}\sqrt{1+t^2} dt$

Applying fundamental theorem of calculus, integral sign get eliminated with $\dpi{120} \frac{\mathrm{d} }{\mathrm{d} x}$ and we replace variable t with x and we get,

$\dpi{120} g'(x) = \sqrt{1+x^2}$

Derivative of integral function when upper limit is not just x, but it is some function of x.

Then we replace variable t with the upper bound function and take its integral alongwith. Lets look at an example.

2. Use fundamental theorem of calculus to find derivatve of following integral function.

$\dpi{120} {\color{Red} \int_{5}^{x^2 +9}sin(t^3)dt}$

Taking derivative of given integral,

$\dpi{120} \frac{\mathrm{d} }{\mathrm{d} x}\int_{5}^{x^2 +9} sin(t^3)dt=sin(x^2+9)^3 . (x^2+9)'$

$\dpi{120} = sin(x^2+9)^3 . 2x$

$\dpi{120} = 2x sin(x^2 +9)^3$

Derivative of integral function when integrand is a function of two variables.

In such a case we start by taking derivative of whole fucntion using product rule and then apply FDC in next step for second part.

3. Use fundamental theorem of calculus to find derivatve of following integral function.

$\dpi{120} {\color{Red} y=\int_{1}^{tanx}e^x \sqrt{1+t^3} dt}$

First we separate the expressions in x and t.

$\dpi{120} y= e^x \int_{1}^{tanx}\sqrt{1+t^3} dt$

Taking derivative of both sides and using product rule.

$\dpi{120} \frac{\mathrm{d} }{\mathrm{d} x}[y]= \frac{\mathrm{d} }{\mathrm{d} x}[e^x \int_{1}^{tanx}\sqrt{1+t^3} dt]$

$\dpi{120} \frac{\mathrm{d}y }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x}[e^x ] .\int_{1}^{tanx}\sqrt{1+t^3} dt + e^x\frac{\mathrm{d} }{\mathrm{d} x}\int_{1}^{tanx}\sqrt{1+t^3}dt$ [ used product rule here]

$\dpi{120} \frac{\mathrm{d}y }{\mathrm{d} x}= e^x \int_{1}^{tanx}\sqrt{1+t^3} dt + e^x\frac{\mathrm{d} }{\mathrm{d} x}\int_{1}^{tanx}\sqrt{1+t^3}dt$

We observe that first part on right side is just the given original function so we can replace it with y. For second part on right side, we can apply fundamental theorem of Calculus. Replace t with tanx and take derivative of tanx alongwith.

$\dpi{120} \frac{\mathrm{d}y }{\mathrm{d} x}= y + e^x\sqrt{1+(tanx)^3} . (tanx)'$

$\dpi{120} \frac{\mathrm{d}y }{\mathrm{d} x}= y + e^x\sqrt{1+(tanx)^3} . sec^2(x)$

$\dpi{120} \frac{\mathrm{d}y }{\mathrm{d} x}= y + e^x sec^2(x)\sqrt{1+(tanx)^3}$

Derivative of an integral function when both the bounds are functions of x.

When lower bound is no more a constant, instead, both upper and lower bounds are functions of x, then we break the integral into two parts in such a way that lower bound is a constant in both parts. We can break the integral at 0 and manipulate it further to get desired form. Here is an example :

4. Find derivative of the following integral function using fundamental theorem of Calculus.

${\color{Red} y=\int_{x}^{2x}sin(t^2)dt}$

Here we can break the integral at any constant ‘c’. We can also use c as 0.

$y=\int_{x}^{c}sin(t^2)dt +\int_{c}^{2x}sin(t^2)dt$

Presently, it is not in right format. We need to interchange the bounds of first integral. Using property of definite integral we can interchange the bounds by putting a negative sign in front.

$y=-\int_{c}^{x}sin(t^2)dt +\int_{c}^{2x}sin(t^2)dt$

Taking derivative of both sides,

$\frac{dy}{dx}=-\frac{\mathrm{d} }{\mathrm{d} x}\int_{c}^{x}sin(t^2)dt +\frac{\mathrm{d} }{\mathrm{d} x}\int_{c}^{2x}sin(t^2)dt$

Applying fundamental theorem of calculus, integral sign get eliminated with $\dpi{120} \frac{\mathrm{d} }{\mathrm{d} x}$ , variable t replaced with x in first integral and 2x in second integral. Also derivatives of x and 2x taken alongwith.

$\frac{\mathrm{d} y}{\mathrm{d} x}=-sin(x)^2(x)'+ sin(2x)^2 (2x)'$

$\frac{\mathrm{d} y}{\mathrm{d} x}=-sin(x)^2*1+ sin(4x^2) (2)$

$\frac{\mathrm{d} y}{\mathrm{d} x}=-sin(x^2)+ 2sin(4x^2)$

Practice problems:

Using second fundamental theorem of calculus, find derivative of the following integrals:

1. $\int_{2}^{sinx}e^{t^2}dt$

2. $\int_{5}^{5x+3}\sqrt{1+t^2}dt$

3. $\int_{sinx}^{tanx}\sqrt{1-t^2} dt$

Answers:

1. $cosx. e^{sinx}$

2. $5\sqrt{25x^2 +30x+10}$

3. $-cos^2x +sec^2x\sqrt{1-tan^2x}$

Derivative of Inverse function

December 23, 2023/by admin_ct

If f is a function with inverse function $f^{-1}$ , then derivative of inverse function is given by the following formula :

$\dpi{120} (f^{-1})'(x)= \frac{1}{f'(f^{-1}(x))}$

This implies the derivative of $\dpi{120} f^{-1}$ is the reciprocal of the derivative of f(x), with argument and value reversed.

This formula is an immediate consequence of the definition of an inverse function and the chain rule.

$\dpi{120} f(f^{-1}(x))=x$

Taking the derivative of both sides,

$\dpi{120} \frac{\mathrm{d}[ f(f^{-1}(x))]}{\mathrm{d} x}= \frac{\mathrm{d} [x]}{\mathrm{d} x}$

$\dpi{120} f^{'}(f^{-1}(x))\times (f^{-1})'(x) =1$

$\dpi{120} (f^{-1})'(x) =\frac{1}{f^{'}(f^{-1}(x))}$

Alternatively,

If y= g(x) is the inverse of f(x), then

$\dpi{120} g'(x)= \frac{1}{f'(g(x))}$

Lets look at an example.

1. let $\dpi{120} {\color{Red} f(x) = 5+x^2 +tan(\frac{\pi x}{2})}$ defined on interval $\dpi{120} {\color{Red} -1\leq x\leq 1}$ . Then find $\dpi{120} {\color{Red} f'(f^{-1}(5))}$

First we need to find value of inverse function at 5. We observe that

$\dpi{120} f(0)= 5+0^2+tan(\frac{\pi\times 0}{2} )$

$\dpi{120} f(0) = 5 +tan(0)$

$\dpi{120} f(0) = 5$

$\dpi{120} 0=f^{-1}(5)$

Now we can plugin this value into formula and solve further.

$\dpi{120} (f^{-1})'(x) =\frac{1}{f^{'}(f^{-1}(x))}$

$\dpi{120} (f^{-1})'(5) =\frac{1}{f^{'}(f^{-1}(5))}$

$\dpi{120} (f^{-1})'(5) =\frac{1}{f^{'}(0)}$

Next we find derivative of given function at x=0

$\dpi{120} f(x )= 5+x^2 +tan(\frac{\pi x}{2})$

$\dpi{120} f'(x )= 0+ 2x +sec^2(\frac{\pi x}{2})(\frac{\pi }{2})$ [ used chain rule ]

$\dpi{120} f'(x )= 2x +\frac{\pi }{2}sec^2(\frac{\pi x}{2})$

$\dpi{120} f'(0 )= 2(0) +\frac{\pi }{2}sec^2(\frac{\pi (0)}{2})$

$\dpi{120} f'(0)= 0+ \frac{\pi }{2}sec^2(0)$

$\dpi{120} f(0)= \frac{\pi }{2}$

Finally we get the answer as,

$\dpi{120} (f^{-1})'(5) =\frac{1}{f^{'}(0)}=\frac{1}{\frac{\pi }{2}}$

$\dpi{120} (f^{-1})'(5) ={\frac{2 }{\pi }}$

Lets look at another example through this video lesson.

Now its time for some practice.

Given $\dpi{120} f(x) = 1+ln(x-2)$ . Find $\dpi{120} (f^{-1})'(1)$ .
Given $\dpi{120} f(x) = x^5 +2x^3 +7x+1$ . Find $\dpi{120} (f^{-1})'(1)$

Algebra of functions

October 5, 2023/by admin_ct

Algebra of functions means combining different functions using four major algebraic operations addition, subtraction, multiplication and division.

Suppose we want to find sum of two functions f and g (both as functions of same variable x), then we can write this operation as:

(f+g)(x) = f(x) + g(x)

Same way we can apply other operations too.

What is the domain of combined function?

Domain of combined function would be the intersection of domains of two individual functions. Lets assume domain of function f(x) is A and domain of function g(x) is B then Domain of combined functions would be $\dpi{100} A\cap B$

But we have to be careful while writing the domain of quotient function f/g as denominator of an expression can never have 0.

Lets summarize all the rules of combining functions and writing their domain in a tabular form.

Operation	Formula	Domain
Sum	(f+g)(x) = f(x) + g(x)	$A\cap B$
Difference	(f-g)(x) = f(x) – g(x)	$A\cap B$
Product	(fg)(x) = f(x) g(x)	$A\cap B$
Quotient	(f/g)(x) = f(x)/g(x)	$\left \{ x\in A\cap B\|g(x)\neq 0) \right \}$

Lets work on some examples to understand the concept better,

1. Do the operations as directed for given functions f and g and then find the domain.

$\dpi{100} {\color{Red} f(x) = x^2 +x , g(x)= x^2}$

a. (f+g)(x)

b. (f-g)(x)

c. (f*g)(x)

d. (f/g)(x)

Solution:

Here f(x) and g(x) both functions are polynomials. We know that polynomials are defined for all real numbers.

Domain of f = $\dpi{100} \left ( -\infty ,\infty \right )$

Domain of g = $\dpi{100} \left ( -\infty ,\infty \right )$

Intersection of both domains = $\dpi{100} \left ( -\infty ,\infty \right )$

a. Using the sum formula (f+g)(x) = f(x) +g(x)

$\dpi{100} (f+g)(x)= x^2 +x+x^2 = 2x^2 +x$

Domain : $\dpi{100} \left ( -\infty ,\infty \right )$

b. Using difference formula (f-g)(x)= f(x) – g(x)

$\dpi{100} (f-g)(x)= x^2 +x-x^2 = x$

Domain : $\dpi{100} \left ( -\infty ,\infty \right )$

c. Using product formula (f*g)(x)= f(x)*g(x)

$\dpi{100} (f\ast g)(x)= (x^2 +x)x^2 = x^2.x^2+x.x^2$ Distributed x^2

$\dpi{100} (f*g)(x)= x^4 +x^3$ Combined like terms

Domain : $\dpi{100} \left ( -\infty ,\infty \right )$

d. Using quotient formula

$\dpi{100} \frac{f}{g}(x) =\frac{f(x)}{g(x)}$

$\dpi{100} \frac{f}{g}(x)=\frac{x^2+x}{x^2}$ Plugin values of f(x) and g(x)

Using the fact that any function becomes undefined when its denominator is 0 , x^2 can’t be 0

That means x can’t be 0 here. Therefore,

Domain : $\dpi{100} \left \{ x\in \mathbb{R}: x\neq 0 \right \}$ where R is the set of real numbers.

2. Perform the following operations for given functions and find their domain.

$\dpi{100} {\color{Red} f(x) =\sqrt{16-x^2} , g(x)=\sqrt{x^2-1}}$

a. (f +g)(x)

b. (f*g)(x)

c. (f/g)(x)

Solution:

Lets first find domain of each function.

Since radical (square root) functions are not defined for negative values, so we set radicand (inner part) ≥0 and solve for x.

$\dpi{100} f(x) = \sqrt{16-x^2}$

16- x² ≥ 0

16 ≥ x²

Domain f : [-4,4]

$\dpi{100} g(x) = \sqrt{x^2 -1}$

x² -1 ≥ 0

x² ≥ 1

Domain g : $\dpi{100} \left ( -\infty , -1 \right ]\cup [1, \infty )$

Intersection of both domains is given as : $\dpi{100} \left [ -4, -1 \right ]\cup \left [ 1,4 \right ]$

a. $\dpi{100} (f+g)(x) = f(x) +g(x)$

$\dpi{100} (f+g)(x)= \sqrt{16-x^2}+\sqrt{x^2-1}$ This expression can’t be simplified more so we leave this answer like this.

Domain: $\dpi{100} \left [ -4, -1 \right ]\cup \left [ 1,4 \right ]$

b. $\dpi{100} (f*g)(x) = f(x)*g(x)$

$\dpi{100} (f*g)(x)= \sqrt{16-x^2}\sqrt{x^2 -1}$

$\dpi{100} (f*g)(x) = \sqrt{(16-x^2)(x^2 -1)}$ If needed , inner expressions can be multiplied and simplified.

Domain: $\dpi{100} \left [ -4, -1 \right ]\cup \left [ 1,4 \right ]$

c. $\dpi{100} \frac{f}{g}(x)=\frac{f(x)}{g(x)}$

$\dpi{100} \frac{f}{g}(x)= \frac{\sqrt{16-x^2}}{\sqrt{x^2 -1}}$ For rational function , denominator can’t be 0.

Since denomaintor √(x² -1) would be 0 at -1 and 1 that means -1 and 1 must be excluded from intersection of domains.

Domain: $\dpi{100} \left [ -4, -1 \right )\cup \left ( 1,4 \right ]$

X and Y Intercepts

September 17, 2023/by admin_ct

Lets first understand what are x and y intercepts.

X intercept :

These are the points, where any line or curve intersects or crosses the x-axis.

How to write x-intercepts?

X intercepts are written in the form of ordered pairs as $\dpi{100} \left ( x_{1},0 \right ),\left ( x_{2},0 \right )$ or can also be written using separated commas as $\dpi{100} x= x_{1},x_{2}$ .

Y Intercept :

These are the points, where any line or curve intersects or crosses the y-axis.

How to write y-intercept?

Y intercept too, can also be written two ways, either ordered pair form (0,y) or using separated commas.

Note : A linear equation ( a line graph) can’t have more than one x and y intercept because a line can intersect x or y axis atmost at one point.

How to find x and y intercept ?

Its easy!

For x intercept, we just replace y with 0 and solve for x.

For y intercept we replace x with 0 and solve for y.

So we just do the opposite. (Its easy for you to memorize it like this:)

Lets work on some examples to understand this process more.

1. Find x and y intercepts of linear equation 2x+3y= 12.

x-intercept :

Plugin y as 0 and solve for x.

2x+3(0) = 12

2x+0= 12

2x=12

x=6 In ordered pair form we can write it as (6,0)

y-intercept :

Plugin x as 0 and solve for y.

2(0) +3y = 12

0+3y =12

3y=12

y=4 In ordered pair form we can write it as (0,4)

2. Find x and y intercepts of quadratic equation $\dpi{100} {\color{Red} y=x^{2}-7x+12}$

y-intercept:

Plugin x as 0 and solve for y.

$\dpi{100} y=0^{2}-7(0)+12$

y= 12 (0,12)

x- intercept :

Plugin y as 0 and solve for x.

$\dpi{100} 0=x^{2}-7x+12$

Factoring the above equation, we get

0 = (x-3)(x-4)

x-3 = 0 x-4 = 0

x =3 x = 4

So this curve has two x intercepts written in ordered pair as (3,0) and (4,0)

Practice problems:

Find x and y intercepts for linear equation x+3y = 15
Find x and y intercepts of quadratic equation $\dpi{100} y=x^{2}-5x+6$

Operations on Scientific Notations worksheet

August 25, 2023/by admin_ct

Addition and subtraction in scientific notation:

While performing basic arithmetic operations like addition, subtraction, multiplication or division on numbers written in scientific notation, we need to be careful and should follow some rules to reach the correct answer.

While adding and subtracting the numbers in scientific notation, exponent part of all the numbers must be same. If it is not same , then first make it same and then combine the numbers.

Don’t forget to write the final answer in scientific notation.

When exponent part is same-

Example1: Combine $\dpi{120} {\color{Red} 4.52\times 10^{3}}$ and $\dpi{120} {\color{Red} 8.3\times 10^{3}}$

Solution: Here we see that exponent part of both numbers is same ( $\dpi{120} 10^{3}$ ). So we just need to combine the coefficients of these two numbers.

$\dpi{120} 4.52\times 10^{3}$ + $\dpi{120} 8.3\times 10^{3}$ = $\dpi{120} (4.52+8.3)\times 10^{3}$

= $\dpi{120} 12.82\times 10^{3}$

As the coefficient part can’t exceed 10, so we need to convert this result into scientific notation.

= $\dpi{120} 1.282\times 10^{4}$

When exponent part is not same-

Example2: Simplify the following expression: $\dpi{120} {\color{Red} 8\times 10^{8}- 5\times 10 ^{6}}$

Solution: As the exponent part is not same, we make them same first.

Again you can do it two ways. Either lower the exponent $\dpi{120} 10^{8}$ to $\dpi{120} 10^{6}$ or you can raise the exponent $\dpi{120} 10^{6}$ to $\dpi{120} 10^{8}$

First way: raise the exponent $\dpi{120} 10^{6}$ to $\dpi{120} 10^{8}$

As we know, when decimal is moved to left, exponent is positive (increased). We need to raise the exponent by 2 so decimal is moved towards left by 2 places and we get $\dpi{120} 5\times 10^{6} = 0.05\times 10^{2}\times 10 ^{6}$

= $\dpi{120} 0.05\times 10^{8}$

$\dpi{120} 8\times 10^{8}- 5\times 10 ^{6}$ = $\dpi{120} 8\times 10^{8}- 0.05\times 10 ^{8}$

= $\dpi{120} (8-0.05)\times 10^{8}$

= $\dpi{120} 7.95\times 10^{8}$

Second way: Either lower the exponent $\dpi{120} 10^{8}$ to $\dpi{120} 10^{6}$

When decimal is moved to right, exponent is negative (decreased). We need to lower the exponent by 2 so decimal is moved towards right by 2 places and we get $\dpi{120} 8\times 10 ^{8} = 8\times 10^{2}\times 10^{6} = 800\times 10^{6}$

$\dpi{120} 8\times 10^{8}- 5\times 10 ^{6}$ = $\dpi{120} 800\times 10^{6}- 5\times 10 ^{6}$

= $\dpi{120} 795\times 10^{6}$

But this result is not in scientific notation. To get it in scientific notation decimal should be placed after first significant digit.

So final answer is $\dpi{120} 7.95\times 10^{8}$

Example3. Simplify the expression $\dpi{120} {\color{Red} 9\times 10^{5} + 6\times 10 ^{7}}$

Solution: $\dpi{120} 9\times 10^{5} + 6\times 10 ^{7}$ = $\dpi{120} 9\times 10^{5} + 6\times 10 ^{2}\times 10^{5}$

= $\dpi{120} 9\times 10^{5} + 600\times 10^{5}$

= $\dpi{120} (9+600)\times 10^{5}$

= $\dpi{120} 609 \times 10^{5}$

Then we convert it into scientific notation and get answer as $\dpi{120} 6.09 \times 10^{7}$

Example4. Simplify the expression $\dpi{120} {\color{Red} 9\times 10^{5} - 6\times 10 ^{7}}$

Solution: $\dpi{120} 9\times 10^{5} - 6\times 10 ^{7}$ = $\dpi{120} 9\times 10^{5} - 6\times 10 ^{2}\times 10^{5}$

= $\dpi{120} 9\times 10^{5} - 600\times 10^{5}$

= $\dpi{120} (9-600)\times 10^{5}$

= $\dpi{120} -591 \times 10^{5}$

Rewriting it as scientific notation we get the final answer as $\dpi{120} -5.91\times 10^{7}$

Practice problems:

Simplify each problem and write the answer in scientific notation.

$\dpi{120} 1\times 10^{6} - 8\times 10 ^{4}$
$\dpi{120} 7\times 10^{7} + 3\times 10 ^{4}$
$\dpi{120} 4\times 10^{8} - 9\times 10 ^{7}$
$\dpi{120} 3\times 10^{-2} - 6\times 10 ^{-5}$

Answers:

1) $\dpi{120} 9.2\times 10^{5}$

2) $\dpi{120} 7.003\times 10 ^{7}$

3) $\dpi{120} 3.1\times 10^{8}$

4) $\dpi{120} 2.994\times 10^{-2}$

Scientific and Standard Notations

August 25, 2023/by admin_ct

Scientific and standard Notations

When we think about writing very large or very small numbers in an easy and quicker way, scientific notation comes to our rescue.

Therefore, scientific notation is a way to represent very small or very large numbers. It is also easy to compare many such numbers when written in scientific notations

How scientific notation is written:

Scientific notation has two parts, first one is the coefficient and second part consists of base 10 and its exponent.

Coefficient is always a number between 1 and 10 . It may contain one or more than one significant digits with a decimal.

Base is always 10

Exponents is always an integer.

Converting from standard notation to scientific notation

When we move towards left, exponent is positive.

When we move towards right, exponent is negative.

Example1. Convert 450000 to scientific notation.

Solution: Since the coefficient must lie between 1 and 10 so decimal should be placed after 4. As decimal is moved 5 steps towards left we use positive exponent (5)

So the scientific notation is written as

$\dpi{120} 4.5\times 10^{5}$

Example2. Convert 0.000000215 to scientific notation.

Solution: This is very small number less than 1 so exponent would be negative.

Since coefficient must lie between 1 and 10 so we put decimal after first significant digit which is 2. For that we need to move decimal to 7 places right and whenever we move towards right while converting standard to scientific notation, we use negative exponents.

So the scientific notation is written as,

$\dpi{120} 2.15\times 10^{-7}$

Example3. Express 0.8 in scientific notation.

Solution: Since decimal should be placed after first significant digit so decimal is moved to right by only one place.

= $\dpi{120} 8.0\times 10^{-1}$

Or it can also be written as $\dpi{120} 8\times 10^{-1}$

Example4. Expression 30 in scientific notation.

Solution: Coefficient is the first significant digit which is 3 here, so we need to move decimal towards left by one place.

= $\dpi{120} 3.0\times 10^{1}$

So 30 can be expressed in scientific notation as $\dpi{120} 3\times 10^{1}$

Converting from scientific notation to standard form

For given negative exponents we move towards left.

For given positive exponents we move towards right.

Example5. Convert $\dpi{120} {\color{Red} 4\times 10^{-5}}$ to standard notation.

Solution: Since decimal is not given here so we assume 4 as 4.0 and because of negative exponent(-5) we move 5 places towards left.

$\dpi{120} 4\times 10^{-5}$ =

So the standard form of given number is 0.00004

Example6. Convert $\dpi{120} {\color{Red} 3.27\times 10^{4}}$ to standard form.

Solution: Since exponents is given positive so we move decimal towards right by 4 places.

$\dpi{120} 3.27\times 10^{4}$ =

So the standard form of given number is 32700.

Example7. Express $\dpi{120} {\color{Red} 5.2\times 10^{0}}$ to standard form.

Solution: Since exponent is 0 that means we don’t need to move decimal. Moreover $\dpi{120} 10^{0} =1$ so the given number remain same without any change in it.

$\dpi{120} 5.2\times 10^{0}= 5.2\times 1= 5.2$

Practice problems:

Express the given numbers in scientific notation.

850,000
0.0000917
5102000

Express the given numbers in standard form.

$\dpi{120} 4.28\times 10^{-2}$
$\dpi{120} 3.162\times 10 ^{4}$
$\dpi{120} 1.2\times 10 ^{-5}$

Answers:

$\dpi{120} 8.5\times 10^{5}$
$\dpi{120} 9.17\times 10^{-5}$
$\dpi{120} 5.102\times 10 ^{6}$
0.0428
31620
0.000012

Worksheet 2 (Differential Calculus)

October 12, 2020/by admin_ct

1. If f(x)=|x-3| then f'(1) is

a) 0

b) 1

c) -1

d) Non existent

Solution: Derivative of this absolute function can be found either by using definition of absolute value function or by using formula.

Lets first use definition of absolute value function.

$\dpi{120} f(x) =\begin{cases} & x-3 \text{ if } x\geq 3 \\ & -(x-3) \text{ if } x < 3 \end{cases}$

Since x=1 lie in second domain (x<3) so we use lower function.

f(x) = -(x-3)

f ‘(x) = -(1-0) =-1

Therefore f ‘(1) = -1. Option c is correct.

Other way: Using formula for derivative of absolute functions.

Let f(x) = |U| where U= g(x)

Then $\dpi{120} f'(x)=\frac{U}{|U|}*U'$ or $\dpi{120} f'(x)=\frac{g(x)}{|g'(x)|}*g'(x)$

$\dpi{120} f'(x)=\frac{(x-3)}{|x-3|}*(x-3)'$

$f'(1)=\frac{(1-3)}{|1-3|}*1 =\frac{-2}{|-2|} =\frac{-2}{2}=-1$

2. Given that $\dpi{120} {\color{Red} f(x) =x^{10}h(x)}$ and h(-1)=4 , h'(-1) = 7 . Then find f ‘(-1).

a) 4

b) -33

c)-40

d)-70

Solution: Since f(x) is a product of two functions x^10 and h(x) so we use product rule to find derivative f'(x)

$f'(x)=(x^{10})'h(x)+x^{10}h'(x)$

$f'(x)=(10x^{9})'h(x)+x^{10}h'(x)$

$f'(-1)=10(-1)^{9}h(-1)+(-1)^{10}h'(-1)$

Plugin given values of h(1) and h'(-1)

$f'(-1)= -10(4)+ (1)(7)$

f ‘(-1) = -33

Option b) -33 is correct answer.

3. If ${\color{Red} f'(x) =\frac{(x+2)^2)(x^2 -1)}{3}}$ and ${\color{Red} g(x)= f(\sqrt{x+1})}$ what is the value of g'(3) ?

a) 16

b) 1/4

c) 4

d) 16

Solution : First we find derivative of g(x) using chain rule.

$g'(x)= f'(\sqrt{x+1})* (\sqrt{x+1})'$

$g'(x)= f'(\sqrt{x+1})* \frac{1}{2\sqrt{x+1}}$

$g'(3)= f'(\sqrt{3+1})* \frac{1}{2\sqrt{3+1}}$

$g'(3)= f'(2)* \frac{1}{2*2}$

To get f'(2) we plugin x=2 into given expression for f'(x)

$f'(2) =\frac{(2+2)^2(2^2 -1)}{3}= 16$

Using f'(2)= 16 we get g'(3) as

g'(3) = 16 *(1/4) = 4

g'(3)= 4

So option c is the answer.

4. Functions f, g and h are twice differentiable functions with g(2)= h(2)= 4. The line ${\color{Red} y= 4+\frac{2}{3}(x-2)}$ is tangent to both the graphs of g at x=2 and graph of h at x=2.

a) find h'(2)

b) Let a be a function given as

${\color{Red} a(x)= 3x^3 h(x)}$ . Write an expression for a'(x) and a'(2).

c) The function h satisfies that ${\color{Red} h(x)= \frac{x^2-4}{1-(f(x))^3} ; x \neq 2}$ . It is known that ${\color{Red} \lim_{x\rightarrow 2}h(x)}$ can be evaluated using L Hospital’s rule. Use this limit to find f(2) and f ‘(2).

d) It is known that g(x) ≤ h(x) for 1<x<3. Let k be a function satisfying g(x)≤k(x)≤h(x) for 1<x<3. Is k continuous at x=2? Justify your answer.

Solution: It is known that slope of tangent line to the curve at given point is same the slope of curve at that point.

Slope of tangent line at x=2 is 2/3

and that would be the slope of curve h(x) at x=2. so h'(2) = 2/3.

b) $a(x)= 3x^3 h(x)$ . Since given function is product of two functions so using product rule,

$a'(x)= 3[(x^3)' h(x) + x^3 h'(x)]$

$a'(x)= 3[(3x^2)' h(x) + x^3 h'(x)]$

$a'(2)= 3[(3(2)^2)' h(2) + 2^3 h'(2)]$

$a'(2)= 3[12*4 + 8*2/3]$

a'(2) = 160

$\lim_{x\rightarrow 2}h(x)= \lim_{x\rightarrow 2} \frac{x^2-4}{1-(f(x))^3}$

It is given that this limit can be evaluated using L Hospital’s rule. Which means it would have intermediate form(0/0) when x approaches to 2. so we have

$\frac{2^2-4}{1-(f(2))^3} = \frac{0}{0}$

$1-(f(2))^3=0$

$(f(2))^3= 1 \Rightarrow f(2)= 1$

Since h(x) is continuous so we have,

$\lim_{x\rightarrow 2}h(x)=h(2)$

Using L Hospital’s rule to find its limit,

$\lim_{x\rightarrow 2}h(x)= \lim_{x\rightarrow 2} \frac{(x^2-4)'}{(1-(f(x))^3)'} = \frac{2x-0}{0-3(f(x))^2 f'(x)}$ = 4

$\frac{2x}{-3(f(x))^2 f'(x)} =4$

$\frac{2*2}{-3(f(2))^2 f'(2)} =4$

$\frac{4}{-3(1)^2 f'(2)} =4$

$\frac{1}{-3f'(2)} =1$

$-\frac{1}{3} = f'(2)$

d) Given that

$g(x)\leq k(x)\leq h(x) on 1<x<3$

Since g(x) and h(x) are given continuous and g(2)= h(2) = 4

$\lim_{x\rightarrow 2}g(x) =4$ and $\lim_{x\rightarrow 2}h(x) =4$ . Using sandwich theorem we can conclude that,

$\lim_{x\rightarrow 2}k(x) =4$ and k(2)= 4

Therefore K(x) is continuous at x=2.

5. Let ${\color{Red} g(x)= \int_{2}^{x}f(t) dt}$ for 0≤t≤7 where the graph of differentiable function f, is shown below.

a) Find g(3) , g'(3) and g”(3)

Using the given definition of g(x) we have,

$g(3)= \int_{2}^{3}f(t) dt$ which means the area under the curve f(t) from x=2 to x=3. This area can be found using geometric formulas.

Area under the curve for given x values is a trapezoid, so we can use trapezoid’s area formula.

A = 1/2 (sum of parallel sides)* distance between parallel sides.

g(3) = A= 1/2(2+4)*1 = 6/2 = 3

Using Fundamental Theorem of Calculus

g'(3) = f(3) = 2

And g”(3)= f ‘(3) which is slope of tangent line at x=3

slope = (4-0)/(2-4) = -2

so g”(3) = -2

b) Find average rate of change of g(x) on 0≤c≤3

Average rate = $\frac{g(3)-g(0)}{3-0}$

For that first we need to find g(0) , same way as we had found g(3).

$g(0)= \int_{2}^{0}f(t) dt$

Which can be rewritten using property of definite integral.

$g(0)= -\int_{0}^{2}f(t) dt$ . which represent the area under the curve from 0 to 2.

g(0) = – [(1/2)*2*4] = -4

Average rate = $\frac{g(3)-g(0)}{3-0} = \frac{3-(-4)}{3} = \frac{7}{3}$

c) For how many values of c, where 0<c<3, is g'(c) equal to average rate found in part b? Justify your answer.

We know that ,

$g(c)= \int_{2}^{c}f(t) dt$

then g'(c) = f(c) = 7/3

And the function f(t) gets the value 7/3 twice on the interval 0<c<3.

d) Find the Graph of gx coordinate of each point of inflection of the graph of g on the interval 0<t<7. Justify your answer.

We know that g'(x) = f(x) and g”(x)= f'(x).

Graph of g will have inflection points where f(x) changes its direction. We can observe from given graph of f that it changes direction from increasing to decreasing at x=2 and again from decreasing to increasing at x=5. Also slopes f ”(2) and f”(5) are undefined being sharp corners. So the two inflections points are at x=2,5.

6. For what values of a and c , is the piecewise function differentiable.

${\color{Red} f(x)=\begin{cases} & ax^2+sinx, \text{ if } x\leq \pi \\ & 2x-c , \text{ if } x > \pi \end{cases} }$

We know that functions $ax^2+sinx$ and 2x-c are continuous and differentiable in their respective domains.

$f'(x)=\begin{cases} & 2ax+cosx, \text{ if } x\leq \pi \\ & 2 , \text{ if } x > \pi \end{cases}$

We are given that piecewise function is differentiable at x=π too. That means left and right hand derivatives of given function would be same at x=π. Therefore,

$f'(\pi )=\begin{cases} & 2a \pi +cos(\pi ) , \text{ if } x\leq \pi \\ & 2 , \text{ if } x > \pi \end{cases}$

2aπ +cos(π) = 2

2aπ -1 = 2

2a π = 3

$a = \frac{3}{2\pi }$

Also piecewise function is continuous at pi so left and right hand limits would be same as f(π)

Left limit is,

$\lim_{x\rightarrow \pi^- }(ax^2+sinx)= a\pi ^2+sin(\pi )= a\pi ^2$ which is same as value of function , i.e f(π)

And right limit is ,

$\lim_{x\rightarrow \pi^+ }2x-c =2\pi -c$

For the function to be continuous. we set left and right limits equal ,

$a\pi ^2 =2\pi -c$

$c= 2\pi -a\pi ^2$

Plugin value of a = 3/2π , we get

$c=2\pi -\frac{3}{2\pi }*\pi ^2$

$c=2\pi -\frac{3\pi }{2}$

$c=\frac{\pi }{2}$

So the solution is

$a=\frac{3}{2\pi } , c=\frac{\pi }{2}$

7. When the height of a cylinder is 12 cm and the radius is 4cm, the circumference of the cylinder is increasing at a rate of ${\color{Red} \frac{\pi }{4}}$ cm/min, and the height of the cylinder is increasing four times faster than the radius. How fast is the volume of the cylinder changing?

Given: height of cylinder(h)= 12 cm

radius of cylinder(r) = 4 cm

Rate of increase of circumference (2πr)

$2\pi \frac{dr}{dt}=\frac{\pi }{4}$

$2 \frac{dr}{dt}=\frac{1 }{4} \Rightarrow \frac{dr}{dt}=\frac{1 }{8}$

Height of cylinder is increasing 4 times faster than rate of change of radius.

$\frac{dh}{dt} = 4*\frac{1}{8} =\frac{1}{2}$

To find : Rate of change of volume $\dpi{120} \frac{dV}{dt} =?$

Volume of cylinder is given as,

$\dpi{120} V= \pi r^2 h$

Differentiating both sides using product rule, with respect to time(t) ,

$\dpi{120} \frac{dV}{dt}=\pi \left ( 2r\frac{dr}{dt}*h+r^2\frac{dh}{dt} \right )$

Plugin value into this expression.

$\dpi{120} \frac{dV}{dt}=\pi \left ( 2(4)\frac{1}{8}*12+4^2*\frac{1}{2} \right )$

$\dpi{120} \frac{dV}{dt}=\pi \left ( 1*12+8 \right ) = 20\pi$

So the volume of cylinder is changing at the rate of 20π cubic cm/min.

8. Let f be a function that is differentiable on open interval (1,10), If, f(1) = -6 ,f(5)= 2 , and f(10) = -4. Which of the following must be true. Justify your answer.

i. f has atleast two zeros.

ii. The graph of f has atleast one horizontal tangent.

iii. f ‘(5) =0

iv. For some c, 1<c<10 , f(c) =1

v. For some c, 1<c<10 , f ‘(c)= 2/9

i, ii, iv and v must be true. Here is the reason for each.

i. f has atleast two zeros.

When we move from left to right on interval (1,10) , y values goes from negative to positive (-6 to 2 ) and then again from positive to negative (from 2 to -4) that means function’s graph will cross x axis atleast twice. Which proves f has atleast two zeros.

ii. Since function’s graph cross x axis twice so graph will turn atleast once and there is horizontal tangent at each turn. This proves there is atleast one turn and hence atleast one horizontal tangent.

iv. Since the given function is differentiable and hence continuous. This graph is going from -6 to 2 when x is changing from 1 to 5 , that means continuous graph will pass through y value 1, while going from -6 to 2. So there must be some x value c on given interval such that f(c)=1

v. Since function is given differentiable and hence continuous on interval (1,10) , it satisfy hypothesis of Mean Value Theorem. That means there exist a number c on interval (1,10) which satisfy conclusion of MVT.

Therefoere,

$\dpi{120} f'(c)=\frac{f(10)-f(1)}{10-1}$

$\dpi{120} f'(c)=\frac{-4+6}{9}= \frac{2}{9}$

which proves the statement “ For some c, 1<c<10 , f ‘(c)= 2/9″

9. Graph of velocity function of a particle is shown below, where t is measured in seconds. Find the intervals when particle is speeding up and slowing down.

First we observe the given graph. We see that particle start at t=0 , reach max velocity at t=1 .

We make the following observations by looking at the graph :-

Acceleration is positive as long as velocity graph is increasing and acceleration is negative as long as velocity graph is decreasing because acceleration is nothing but the slope of velocity.

Velocity is positive as long as graph is above x axis and velocity is negative as long as graph is below x axis.

Velocity is 0 at t=0 and t=2 seconds. It reaches to max value at t=1. So we make the following chart.

0<t<1
Velocity (V)
Acceleration (A)

0<t<1
+
+

1<t<2
+
–

2<t<3
–
–

Rules to follow :

Particle is speeding up when velocity and acceleration both have same signs. Particle is slowing down when they have opposite signs.

So the answer is :

Particle is speeding up on intervals : (0,1) and (2,3)

slowing down on interval : (1,2)

10 . A particle moves according to position function ${\color{Red} S(t)= \frac{t}{1+t^2}}$ on the interval t ≥0 where t is measured in seconds and S in feet.

a) Find when particle is at rest.

Particle is at rest when its velocity is 0. So we find velocity function by taking its derivative and then set velocity function =0 and solve for t.

$V(t)=S'(t)= \frac{(1+t^2)t' - t(1+t^2)'}{(1+t^2)^2}$

$V(t)= \frac{(1+t^2)1 - t(2t)}{(1+t^2)^2} =\frac{1+t^2 -2t^2}{(1+t^2)^2}$

$V(t) =\frac{1-t^2}{(1+t^2)^2}$

$V(t) =0 \Rightarrow \frac{1-t^2}{(1+t^2)^2}=0$

This expression would be 0 when its numerator is 0 so,

$1-t^2 =0$

t = -1,1

But t=-1 doesn’t lie in the given domain so we accept only t=1.

That means particle is at rest at t=1 second

b) Find the intervals for which particle is moving in positive(right) and negative(left) directions.

Particle is moving towards right when velocity is positive and moving towards left when velocity is negative. So we need to check the intervals where velocity is positive and where it is negative. Domain is given [0,∞) which is divided into two intervals by t=1 (when velocity is 0) So we get the two intervals as,

[0,1) and (1,∞)

For interval [0,1) We get V(0.5) > 0

Velocity is positive, that means particle is moving towards right on [0,1)

For interval (1,∞) , We get V(2) < 0

Velocity is negative, that means particle is moving towards left on (1,∞)

c) Find the total distance traveled in first 8 seconds.

To find total distance, we find distance traveled in positive direction as well as in negative direction.

Distance traveled in positive direction:

$\left | S(1)-S(0) \right |= \frac{1}{2} -0 = \frac{1}{2}$ ft.

Distance traveled in negative direction:

$\left | S(8)-S(1) \right |= \frac{1}{2} -\frac{8}{65}$

Total Distance :

$\frac{1}{2}+\frac{1}{2}-\frac{8}{65} = 1-\frac{8}{65} =\frac{57}{65}$ ft

d) When the particle is speeding up and slowing down?

To find when the particle is speeding up or slowing down, we need to find acceleration function. We know that acceleration is derivative of velocity function. so,

$a(t)= V'(t)=\frac{(1+t^2)^2(1-t^2)'-((1+t^2)^2)'(1-t^2)}{(1+t^2)^4}$

$a(t)= \frac{(1+t^2)^2(-2t)-2(1+t^2)(2t)(1-t^2)}{(1+t^2)^4}$

$a(t)= \frac{(1+t^2)[(-2t)(1+t^2)-2(2t)(1-t^2)]}{(1+t^2)^4}$

$a(t)= \frac{[(-2t)(1+t^2)-2(2t)(1-t^2)]}{(1+t^2)^3}=\frac{-2t-2t^3-4t+4t^3}{(1+t^2)^3}$

$a(t)= \frac{-6t+2t^3}{(1+t^2)^3}=\frac{2t(t^2-3)}{(1+t)^3}$

Setting a(t)=0 and solving for t, we get

$2t(t^2-3)=0$

$t=0, t=\pm \sqrt{3}$

But -√3 doesn’t lie in the given domain, so we have only t=0 and t=√3 .

Using t=√3 , we get the domain t≥0 divided into two intervals as [0,√3) and (√3,∞). Now we check each interval one by one.

For interval [0,√3) we get ,

$a(1)=\frac{2*1(1^2-3)}{(1+1)^3} = \frac{-1}{2} < 0$

Acceleration is negative during this interval.

For interval (√3,∞)

$a(2)=\frac{2*2(2^2-3)}{(1+2)^3} = \frac{4}{27} > 0$

Acceleration is positive during this interval.

Lets make a sign chart using velocity and acceleration results,

0<t<1
Velocity (v)
Acceleration(a)

0<t<1
+
–

1<t<√3
–
–

t>√3
–
+

Using the fact that particle will be speeding up when both velocity and acceleration have same signs (either both positive or both negative) and speeding down when both have different signs, we get the following results.

Speeding up on 1<t<√3

Speeding down on 0<t<1 and t>√3.

Practice Worksheet

Find derivative of given function: $f(x)= 2sin^{-1}(4x^3)$
Find second derivative of $y^2-x^2=9$
Find x values of the horizontal tangents of the curve $y= x^4 -18x^2-4$
Given the function $f(x)= ax-bx^2$ . Determine the values of a and b such that the line 3y-6x+6=0 is tangent to the function at x=3.
At time t, the position of a body moving along the x-axis is given as $x(t)= t^3 -21t^2 +144t$ . Find the body’s acceleration
each time the velocity is zero.
The graphs of f and g are shown below. Use them to evaluate

a) $\left ( fg \right )'(5)$ b) $\frac{d}{dx}\left ( \frac{g(x)}{f(x)} \right )_{x=5}$

7. A particle moves along a coordinate axis in the positive direction to the right. Its position at time $t$ is given by $s(t)=t^3-4t+2$ . Find $v(1)$ and $a(1)$ and use these values to answer the following questions.

a) Is the particle moving from left to right or from right to left at time $t=1$ ?

b) Is the particle speeding up or slowing down at time

8. The velocity-time graph of an object moving along a straight line is shown below. Find the intervals for which object is slowing down.

9. Find least value of a such that the function $f(x)= x^2+ax+1$ is increasing on [1,2].

10. A man is walking at the rate of 6.5 km/hr towards the foot of a tower 120 m high. At what rate is he approaching the top of tower when he is 50m away from tower?

Answers:

$\frac{24x^2}{\sqrt{1-16x^6}}$
$\frac{9}{y^3}$
x=-3, 0, 3
a= 2/3 , b=-2/9
a(6)= -6m/s^2 and a(8)=6m/s^2
a) 1/6 b)7/24
a) Right to left b) slowing down
(1,3)
-2
2.5 km/hr

Worksheet 1 (Limits)

August 29, 2020/by admin_ct

1.Find the limit and choose the correct option:

$\large {\color{Red} \lim_{x\rightarrow \infty }\sqrt{x^{2}-4x}-x}$

a) -∞

b) -2

c) -4

d) 0

Solution: $\lim_{x\rightarrow \infty }\frac{\sqrt{x^{2}-4x}-x}{1}*\frac{\sqrt{x^{2}-4x}+x}{\sqrt{x^{2}-4x}+x}$

$\lim_{x\rightarrow \infty }\frac{\left (\sqrt{x^{2}-4x} \right )^{2}-x^{2}}{\sqrt{x^{2}-4x}+x}$

$\lim_{x\rightarrow \infty }\frac{x^{2}-4x -x^{2}}{\sqrt{x^{2}-4x}+x}=\frac{-4x}{\sqrt{x^{2}-4x}+x}$

$\lim_{x\rightarrow \infty }\frac{-4x}{x(\sqrt{1-\frac{4}{x}}+1)} = \frac{-4}{\sqrt{1-\frac{4}{x}}+1}$

Applying limit as x tends to infinity , we get

$\frac{-4}{\sqrt{1-0}+1} = \frac{-4}{2} = -2$

Answer is option b) -2

2. $\large {\color{Red} \lim_{x\rightarrow \infty }\frac{sinx}{x}}$

a) 0

b) 1

c) ∞

d) DNE

Solution: we know that sinx ranges from -1 to 1

-1 ≤ sinx ≤ 1

$\frac{-1}{x}\leq \frac{sinx}{x}\leq \frac{1}{x}$

$\lim_{x\rightarrow \infty }\frac{-1}{x}\leq \lim_{x\rightarrow \infty }\frac{sinx}{x}\leq \lim_{x\rightarrow \infty }\frac{1}{x}$

Using Squeeze theorem of limits , we know that

$\lim_{x\rightarrow \infty }\frac{-1}{x} = 0 , \lim_{x\rightarrow \infty }\frac{1}{x}=0$

Therefore ,

$\lim_{x\rightarrow \infty }\frac{sinx}{x}= 0$

So the answer is option a) 0

3. $\large {\color{Red} \lim_{x\rightarrow 2^{-}}\frac{x^2 -2x}{|x^2-x-2|}}$

a) -∞

b) -2/3

c) 2/3

d) 0

Solution: To find limit , we define absolute value function first.

$|x^2 -x-2|=\begin{cases} (x-2)(x+1) & \text{ if } x\leq -1 \\ -(x-2)(x+1)& \text{ if } -1<x<2 \\ (x-2)(x+1))& \text{ if } x\geq 2 \end{cases}$

To find left hand limit at x=2 we use function |x^2 -x -2|= -(x-2)(x+1)

$\lim_{x\rightarrow 2^{-}}\frac{x^2 -2x}{|x^2-x-2|} = \lim_{x\rightarrow 2}\left [\frac{x(x-2)}{-(x-2)(x+1)} \right ]$

$= \lim_{x\rightarrow 2}\left [\frac{x}{-(x+1)} \right ]$

$=\frac{2}{-(2+1)} = -\frac{2}{3}$

Answer is option b) -2/3

4. If $\large {\color{Red} \lim_{x\rightarrow a}\frac{\tan^{-1}(2x)-\tan^{-1}(2a)}{x-a}=\frac{2}{3}}$ then a could equal

a) √2

b) 1

c) √2/2

d) 1/2

Solution: We know that this is limit definition formula for derivative of tan(2x) at x= a. Therefore,

$\frac{\mathrm{d} \tan^{-1}(2a)}{\mathrm{d} x} = \frac{2}{3}$

$\frac{2}{1+4a^{2}} = \frac{2}{3}$

Cross multiplying , we get

$6= 2(1+4a^2)$

$3= 1+4a^2$

$2=4a^2$

$\frac{1}{2}=a^2 => a= \frac{\sqrt{2}}{2}$

Answer is option c) √2/2

Graph of f is shown above. Which of the following is true.

I. $\lim_{x\rightarrow 2^{-}}f(x)=1$

II. $\lim_{x\rightarrow 2^{+}}f(x)= -2$

III. $\lim_{x\rightarrow 2}f(x)= -1$

a) I only

b) II only

c) III only

d) I and II only

e) I,II and III

Solution: From the given graph, we see that graph is approaching to 1 from left at x=2. Therefore left limit is 1.

From right graph is approaching to -2 at x=2 , therefore limit is -2.

So I and II are correct.

But a limit exist only when left and right limits are same. But here left and right limits are different so limit at x=2 doesn’t exist.

Hence III is not correct.

So the correct answer is option d) I and II only.

6. If f(x) is continuous for all x, the maximum number of horizontal asymptotes that the graph of f can have is:

a) 0

b) 1

c) 2

d) None

Solution: Horizontal asymptote is the end behavior of a graph. Since a continuous function have only two ends so we analyze the end behavior of graph by finding limits at – inf and +inf.

Therefore a continuous function can have maximum 2 horizontal asymptotes.

So the correct option is c) 2

7. If f(x) is continuous for all x, the maximum number of vertical asymptotes that the graph of f can have is:

a) 0

b) 1

c) 2

c) No maximum number

Solution: If a function is continuous for all x, then it won’t have any vertical asymptote because a vertical asymptote always breaks the graph in parts and make it discontinuous.

Therefor a function continuous for all x, can’t have any vertical asymptote.

So the answer is option a) 0

8. $\large {\color{Red} f(x)= \frac{x^3 -64}{x-4}}$ has point discontinuity. Assign a value to f(x) that removes discontinuity.

a) 4

b) 64

c) 48

d) -4

Solution: We can see this function has a hole (point discontinuity) at x=4

To remove point discontinuity we can redefine the given function. For that we factor it and simplify.

$\lim_{x\rightarrow 4} \frac{x^3 -64}{x-4}$

$\lim_{x\rightarrow 4} \frac{(x-4)(x^2 +4x+16)}{x-4}$

$\lim_{x\rightarrow 4} (x^2 +4x+16) = 48$

So this function can be redefined as,

$f(x)=\begin{cases} \frac{x^3 -64}{x-4} & \text{ if } x\neq 4\\ 48 & \text{ if } x= 4 \end{cases}$

Therefore correct answer is option c) 48

9. Find $\large {\color{Red} \lim_{x\rightarrow \infty }\frac{1-x}{cosx}}$

a) 2

b) 1

c) 0

d) DNE

Solution;When we plugin limit we get the result as indeterminate form.

Therefore apply L Hospital’s rule,

$\lim_{x\rightarrow \infty }\frac{(1-x)'}{(cosx)'}=\lim_{x\rightarrow \infty }\frac{-1}{-sinx}$

$\lim_{x\rightarrow \infty }\frac{1}{sinx}$

Since sin x is periodic function. It doesn’t approach to any value when x approach to – inf or + inf. Thus the limit does not exist.

So the limit of original function also does not exist.

So answer is option d) DNE

10. Find $\large {\color{Red} \lim_{x\rightarrow \infty }\frac{x^n}{e^x}}$

a) -∞

b) 0

c) ∞

d) DNE

Solution: Let $y = \lim_{x\rightarrow \infty }\frac{x^n}{e^x}$

Taking ln of both sides,

$lny = \lim_{x\rightarrow \infty }ln\left (\frac{x^n}{e^x} \right )$

$lny = \lim_{x\rightarrow \infty }ln\left (x^n \right )-ln\left ( e^x \right )$

$lny = \lim_{x\rightarrow \infty }nln\left (x \right )-x$

$lny = \lim_{x\rightarrow \infty }x\left ( \frac{n lnx}{x} -1\right )$

$lny = \lim_{x\rightarrow \infty }x* \lim_{x\rightarrow \infty } \left ( \frac{n lnx}{x} -1\right )$

$lny = \lim_{x\rightarrow \infty }x* n\lim_{x\rightarrow \infty } \left ( \frac{ lnx}{x} \right )-1$

Using L Hospital’s rule $\lim_{x\rightarrow \infty }\frac{lnx}{x} =0$

ln y= ∞ * (0-1)

lny = -∞

$y= e^{-\infty }= 0$

Therefore limit is 0 and correct option is b)

Practice worksheet

$\dpi{120} \lim_{x\rightarrow 5}\frac{|5-x|}{5-x}$
$\dpi{120} \lim_{x\rightarrow -\infty }\frac{-6+(\frac{2}{x})}{7-(\frac{1}{x^2})}$
$\dpi{120} \lim_{x\rightarrow 0^{-}}\frac{8x}{|x|}$
$\dpi{120} \lim_{x\rightarrow 0}\left [\frac{1}{x^{2}} -\frac{1}{sin^{2}x} \right ]$
$\dpi{150} \lim_{x\rightarrow \infty }e^{\frac{ln2}{1+lnx}}$
If $\dpi{120} \lim_{x\rightarrow a}\frac{x^{5}-a^{5}}{x-a}= 405$ then find all possible values of a.
$\dpi{120} \lim_{x\rightarrow 9}\frac{x^{\frac{3}{2}}- 27}{x-9}$
$\lim_{x\rightarrow 0}\frac{sin2x+sin3x}{2x+sin3x}$
$\dpi{120} \lim_{x\rightarrow \pi }\frac{1+sec^3x}{tan^2x}$ Hint : factor numerator and denominator
$\dpi{120} \lim_{x\rightarrow 0}\frac{3^{2x}-2^{3x}}{x}$