Calculus MCQ Quiz - Objective Question with Answer for Calculus - Download Free PDF

Explanation:

v . curl v = v . ∇ × v = [v ∇ v] = 0

Option (4) is true. 

Analysis:

Consider \(I = \mathop \smallint \limits_0^{\pi /2} \sqrt {\sin \theta } {\cos ^5}\theta {\rm{ }}d\theta \)

Put sin θ = t

cos θ dθ = dt

If θ = 0 to \(\theta = \frac{\pi }{2}\) then t = 0 to t = 1

Now, \(I = \mathop \smallint \limits_{\theta = 0}^{\pi /2} \sqrt {\sin \theta } \cos \theta \cdot {\left( {1 - {{\sin }^2}\theta } \right)^2}d\theta \)

\(I = \mathop \smallint \limits_{t = 0}^1 \sqrt t {\left( {1 - {t^2}} \right)^2}dt\)

\(I = \mathop \smallint \limits_{t = 0}^1 \sqrt t \left( {1 + {t^4} - 2{t^2}} \right)dt\)

\(I = \left( {\frac{{{t^{\frac{3}{2}}}}}{{\frac{3}{2}}} + \frac{{{t^{\frac{{11}}{2}}}}}{{\frac{{11}}{2}}} - \frac{{2{t^{\frac{7}{2}}}}}{{\frac{7}{2}}}} \right)_0^1\)

\(I = \frac{2}{3} + \frac{2}{{11}} - \frac{4}{7}\)

Concept:

Trigonometric Ratio Fundamental Identities

\(\sec x ={1\over \cos x}\)

Integration by parts

when u and v are functions of x.

\(\smallint u× v~dx = u\smallint vdx - \smallint \left[ {\frac{{du}}{{dx}}\smallint vdx} \right]dx\)

Integral of standard function 

\(\smallint \sec^2(ax +b) dx=\tan (ax+b)\)

\(\smallint\tan x dx= log |\sec x|=-log~cos x\)

The derivative of standard function

\(\frac{{d}}{{dx}}(x^n)=nx^{n-1}\)

\(\frac{{d}}{{dx}}(x)=1\)

Calculation:

Given:

we have \(x\over cos^2 x\)

where u = x and \(v ={1\over cos^2 x}= sec^2 x\)

Integration by parts

when u and v are functions of x.

Integration by parts

when u and v are functions of x.

\(\smallint u× v~dx = u\smallint vdx - \smallint \left[ {\frac{{du}}{{dx}}\smallint vdx} \right]dx\)

\(\smallint x× sec^2 x~dx = x\smallint sec^2 x~dx - \smallint \left[ {\frac{{d}}{{dx}}x\smallint sec^2x~dx} \right]dx\)

\(\smallint x× sec^2 x~dx = x \tan x~dx - \smallint \tan x dx\)

∫x × sec² x dx = x tan x - ∫ tan x dx

∫x × sec2 x dx = x tan x - (-log (cos x))

∫ x × sec2 x dx = x tan x + log (cos x)

Concept:

Arc length or curve length is the distance between two points along the section of the curve. Determining the length of an irregular section of the arc is termed as rectification of the curve.

The length of the curve y = f(x) from x = a to x = b is given as:

\(l= \mathop \smallint \limits_{{\rm{x}} = {\rm{a}}}^{{\rm{x}} = {\rm{b}}} \sqrt {1 + {{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}^2}} {\rm{dx}}\)

or,

If the curve is parametrized in the form x = f(t) and y = g(t) with the parameter t going from a to b then

\(l = \mathop \smallint \limits_{{\rm{t}} = {\rm{a}}}^{{\rm{t}} = {\rm{b}}} \sqrt {{{\left( {\frac{{{\rm{dx}}}}{{{\rm{dt}}}}} \right)}^2} + {{\left( {\frac{{{\rm{dy}}}}{{{\rm{dt}}}}} \right)}^2}} {\rm{dt\;}}\)

Calculation:

\({\rm{y}} = \frac{2}{3}{\rm{\;}}{{\rm{x}}^{3/2}}\)

\(\therefore \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{2}{3} \times \frac{3}{2}{\rm{\;}}{{\rm{x}}^{\frac{1}{2}}}\)

\({\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2} = {\rm{x}}\)

Now, the arc length(l) is

\(l= \mathop \smallint \limits_{{\rm{x}} = 0}^{{\rm{x}} = 1} \sqrt {1 + {\rm{x}}} {\rm{\;dx}}\)

\( = \left| {\frac{{{{\left( {1 + {\bf{x}}} \right)}^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right|_0^1 \)

\(= \left( {\frac{2}{3} \times {2^{\frac{3}{2}}}} \right) - \left( {\frac{2}{3}} \right) = 1.21\)

Explanation:

The given integral is an improper integral of 1^st kind.

\(I = \mathop \smallint \limits_2^∞ \frac{{\left( {1/x} \right)}}{{\log x}}dx\)

\(I = \left[ {\log \left( {\log x} \right)} \right]_2^∞ \)

I = log [log (∞)] – log [log (2)]

∴ I = ∞

Given integral is divergent and diverges to ∞

Additional Information

An improper integral of first kind is when integral limits have -∞ or +∞ or both.

An improper integral of second kind is when integral limits are finite but function is infinite at some value between those limits.

Concept:

We know that,

By Parts method

\(\smallint u \cdot v \cdot dx = u \cdot \smallint v \cdot dx - \smallint \left[ {\frac{{du}}{{dx}}\smallint v \cdot dx} \right]dx\)

Where, u, v should follow the ILATE sequence.[I= Inverse, L= Logarithmic, A= Algebraic, T= Trigonometric, E= Exponential terms]

Calculation:

Given:

From the given Equation, \(\mathop \smallint \limits_1^e √ x \ln \left( x \right)dx\)

u = ln(x), v = √x

Now,

\(\smallint u \cdot v \cdot dx = u \cdot \smallint v \cdot dx - \smallint \left[ {\frac{{du}}{{dx}}\smallint v \cdot dx} \right]dx\)

\(\mathop \smallint \limits_1^{\rm{e}} {\rm{lnx}} \cdot √ {\rm{x}} {\rm{dx}} = {\rm{lnx}} \cdot \mathop \smallint \limits_1^{\rm{e}} √ {\rm{x}} {\rm{dx}} - \mathop \smallint \limits_1^{\rm{e}} \left[ {\frac{{{\rm{du}}}}{{{\rm{dx}}}} \cdot {\rm{\;}}\smallint √ {\rm{x}} {\rm{dx}}} \right]{\rm{dx}}\)  

\(\mathop \smallint \limits_1^{\rm{e}} {\rm{lnx}} \cdot √ {\rm{x}} {\rm{dx}}= \left[ {\ln \left( x \right) \times \frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_1^e - \smallint \left[ {\frac{1}{x} \times \frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]dx\)

 \(\mathop \smallint \limits_1^{\rm{e}} {\rm{lnx}} \cdot √ {\rm{x}} {\rm{dx}}= \left[ {\ln \left( x \right) \times {x^{\frac{3}{2}}} \times \frac{2}{3} - \frac{4}{9} \times {x^{\frac{3}{2}}}} \right]_1^e\)

∴  \(\mathop \smallint \limits_1^e √ x \ln \left( x \right)dx\)  \(= \frac{2}{9}√ {{e^3}} + \frac{4}{9}\)     

Concept:

For limit evaluation of indeterminate forms, i.e. for \(\frac{0}{0}\) or \(\frac{\infty }{\infty }\) form, we apply L’ Hospital’s Rule as:

\(\mathop {\lim }\limits_{x \to a} \left\{ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right\} = \mathop {\lim }\limits_{x \to a} \left\{ {\frac{{f'\left( x \right)}}{{g'\left( x \right)}}} \right\} \)

Calculation:

The given limit is,

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{sinx}}} \right) = \;\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x ~- ~x}}{{xsin\;x}}} \right) \)

By putting a limit we get \(\frac{0}{0}\) form, therefore, by L'Hospital Rule:

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{sin~x}}} \right) = \;\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\cos x~ - ~1}}{{xcos\;x ~+ ~\sin x}}} \right) \)

Again putting the value of limit we get \(\frac{0}{0}\) form, therefore, again using L'Hospital Rule:

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{sinx}}} \right) = \;\mathop {\lim }\limits_{x \to 0} \left( {\frac{{ - \sin x}}{{x\left( { - \sin x} \right)~ +~ \cos x ~+~ \cos x}}} \right) \)

\( = \;\mathop {\lim }\limits_{x \to 0} \left( {\frac{{ - \sin x}}{{2\cos x -~~ xsin\;x}}} \right) \)

Now, by putting the limit, we get:

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{\sin x}}} \right) = \frac{0}{{2 - 0}} = 0 \)

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{sin~x}}} \right)~=~0 \)

Concept:

To solve \(\mathop {\lim }\limits_{x \to ∞ } \frac{{f\left( x \right)}}{{g\left( x \right)}}\) of this type, divide both numerator and denominator by the variable with the highest power.

Calculation:

Given:

\(\mathop {\lim }\limits_{x - ∞ } \frac{{{x^2}\;-\;5x\;+\;4}}{{4{x^2}\;+\;2x}}\)

Replacing x with ∞ in above expression, it is an indeterminate form \(\left( {\frac{∞ }{∞ }} \right).\)

Take out common the highest degree term from numerator and denominator.

\(\therefore \;\mathop {\lim }\limits_{x \to ∞ } \left( {\frac{{{x^2}\left[ {1\;-\;\frac{5}{x}\;+\;\frac{4}{{{x^2}}}} \right]}}{{{x^2}\left[ {4\;+\;\frac{2}{x}} \right]}}} \right)\)

\(\because\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0\)

\(\therefore\;\mathop {{\rm{lim}}}\limits_{x \to ∞ } \left( {\frac{{\left[ {1\;-\;0\;+\;0} \right]}}{{\left[ {4\;+\;0} \right]}}} \right)\)

\(\therefore\frac{1}{4}\)

Explanation

Given

Integral

∫∫∫v 8 xyz dv 

Limits for x, y and z is given as

[2, 3] × [1, 2] × [0, 1]

Volume of the integral

V = ∫∫∫v 8 xyz dv 

i.e. V = ∫ ∫ ∫_V 8 xyz dxdydz

\(V = 8 \times \mathop \smallint \limits_2^3 xdx\mathop \smallint \limits_1^2 ydy\mathop \smallint \limits_0^1 zdz\)

\(V = 8 \times \left[ {\frac{{{x^2}}}{2}} \right]_2^3 \times \left[ {\frac{{{y^2}}}{2}} \right]_1^2 \times \left[ {\frac{{{z^2}}}{2}} \right]_0^1\)

V = 5 × 3 × 1

V = 15 

∴ Volume is 15 

Concept:

A function f(x) is continuous at x = a, if the function is defined at x = a and,

Left limit = Right limit = Function value = Real and finite

A function is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

Analysis:

\(f\left( x \right) = \frac{{{x^2} - 3x - 4}}{{{x^2} + 3x - 4}}\)

This function is not defined for:

x² + 3x – 4 = 0

⇒ (x + 4)(x - 1) = 0 i.e.

At x = 1 and x = - 4

Concept: 

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I.  \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule as:

\(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)

Calculation:

Given:

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{cos\;x\; -\;1}}{{sin\;x\; - \;x}} = \left( {\frac{0}{0}} \right)\) form

Applying L’ Hospital  rule:

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{cos\;x - 1}}{{sin\;x-\;x}} =\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{\frac{d}{dx}({cos\;x\; -\;1})}{{\frac{d}{dx}(sin\;x\;-\;x})} =\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{-sin\;x {\rm{}}}}{{\cos {\rm{x}\;-\;1}}} \)

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{-sin\; {\rm{x}}}}{{\cos {\rm{x}\,-1\;}}} = \left( {\frac{0}{0}} \right){\rm{form}}\)

Once again by L’ Hospital rule,

\({\rm{}}\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{-cos\; {\rm{x}}}}{{{\rm{-sin~x}}}} = \frac{-1}{0} = \infty{\rm{}}\)

Explanation

Given,

 Function f(x) = x e^|x|

 Integral is -1 to 1.

If f(-x) = f(x) then the function is said to be even function

 If f(-x) = - f(x) then the function is said to be odd function.

 f(-x) = -x e^|-x| = -x e|x| = - f(x)

 ∴ The given function is an odd function.

 For an odd function:

 \(\mathop \smallint \nolimits_{ - a}^a x\;{f(x)}dx\) = 0

For a even function

 \(\mathop \smallint \nolimits_{ - a}^a x\;{f(x)}dx\) = 2 × \(\mathop \smallint \nolimits_{ 0}^a x\;{f(x)}dx\)

 Now, as the function is odd

 \(\mathop \smallint \nolimits_{ - 1}^1 x\;{e^{\left| x \right|}}dx\) = 0

Concept:

If a triangle is formed by three vectors, then the sum of the vectors should be zero.

AB + BC + CA = 0 

Cross product of the vectors:

For two vectors \(\bar a = ai+bj+ck\) and \(\bar b = di+ej+fk\) the cross, the product is given by: \(\bar a\times \bar b = \begin{vmatrix} i & j & k\\ a & b & c\\ d & e & f \end{vmatrix}=pi+qj+rk\)

The magnitude of the cross product is:

\(A = |a\times b| = \sqrt{p^2+q^2+r^2}\)

Area of a triangle:

If the vectors \(\bar a\mbox{ and } \bar b\) form adjacent sides of the triangle then the area of the triangle is given by: \(A = \dfrac{1}{2}|\bar a\times \bar b|\)

Calculation:

Given:

Let, AB = 3i + 4j and CA = 5i +7j + k

If a triangle is formed by three vectors, then the sum of the vectors should be zero.

AB + BC + CA = 0 ⇒ 3i + 4j  + BC + 5i +7j + k = 0

BC = - 8i - 11j - k

Let the adjacent vectors be AB (a), AC (b)  \(\bar a = 3i+4j\) , and \(\bar b = 5i+7j+k\).

First, we will calculate the cross product as follow:

\(\begin{align*} \bar a\times \bar b &= \begin{vmatrix} i & j & k\\ 3 & 4 & 0\\ 5 & 7 & 1 \end{vmatrix}\\ &= i(4-0) - j(3-0) + k(21-20)\\ &= 4i-3j+1k \end{align*}\)

Therefore, the magnitude of the cross product is:

\(\begin{align*} |\bar a\times \bar b| &= \sqrt{16+9+1}\\ &= \sqrt {26} \end{align*}\)

Using the formula for the area of the triangle, the area is given by:

\(A = {1\over 2}|a\times b| = {1\over 2}\times \sqrt {26} = {\sqrt {26}\over 2}\)

Concept:

We know 

⇒ 1 - cos x = 2 sin²(x/2)​

⇒ \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\)

Calculation:

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{1 - {\rm{co}}{{\rm{s}}x}}}{{{x^2}}} = \mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{2{{\sin }^2}\left( {\frac{x}{2}} \right)}}{{{x^2}}}\)

Multiply and divide the denominator by 4

\(\mathop {\lim }\limits_{x \to 0} \frac{{2 \times {{\sin }^2}\left( {\frac{x}{2}} \right)}}{{4 \times \left( {\frac{{{x^2}}}{4}} \right)}}\)= \(\frac{1}{2}\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{\sin \left( {\frac{x}{2}} \right)}}{{\left( {\frac{x}{2}} \right)}}} \right)^2} = \frac{1}{2} \)

Concept:

A function is written in the form of the ratio of two polynomial functions is called a rational function.

Rational functions are continuous at all the points except for the points where the denominator becomes zero.

\(f(x)=\dfrac{P(x)}{Q(x)}\)

where P(x) and Q(x) are polynomials and Q(x) ≠ 0.

f(x) will be discontinuous at points where Q(x) = 0.

Calculation:

Given:

\(f(x)=\dfrac{4-x^2}{4x-x^3}\)

This is a rational function, so it will be discontinuous at points where the denominator becomes zero.

4x - x³ = 0

x(4 - x²) = 0

x(2² - x2) = 0

x(2 + x)(2 - x) = 0

x = 0, x = - 2 and x = 2

Hence the function \(f(x)=\dfrac{4-x^2}{4x-x^3}\) will be discontinuous at exactly three points 0, - 2 and 2.

Mistake Points

There may be a doubt that some factors are eliminating each other so first, we have to simplify this. 

Note that, 

If (4 - x2) = 0 then the f(x) will come indeterminant or 0/0 form.

So, the function will not have any value for x  = ± 2. So, these will also be the points of discontinuity.

Also, if  (4 - x2) ≠ 0

⇒ \(f(x)=\dfrac{4-x^2}{4x-x^3} = \frac{1}{x}\)

Here, x = 0 is also the point of discontinuity.

There will be  exactly three points 0, - 2 and 2.