Calculus
Limits
Learning Objectives:
- Understand concepts of limits- Algebraic, trigonometric, involving infinity, using L-Hôpital's Rule.
A limit is the value that a function or sequence "approaches" as the input or index approaches some value. Sometimes, a function is undefined if it takes some exact value of ’x’. In that case, we can find out where the function was approaching. This can be done by removing the factors causing undefined nature or sometimes by using specific mathematical identities.
L Hôpital’s rule:
If a limit is in indeterminate form (such as \(\dfrac{0}{0}\) or \(\dfrac{\infty }{\infty }\) ) then,
at the first occurance of indeterminancy.
User:HiTe, Public domain, via Wikimedia Commons
Solved Example: 2-1-01
If a function is continuous at a point, then:
A. The limit of function may not exist at that point
B. The function must be derivative at that point.
C. Limit of the function at that point tends to infinity
D. The limit must exist at that point and the value of this limit should be same as value of the function at that point.
A function f(x) is said to be continuous if the limit $\lim_{x\to a} f(x)$ exists and it is matching the actual value of the function irrespective of in what direction (from left or from right) we approach the point along the function. \[\lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = f(a)\]
Correct Answer: D
Solved Example: 2-1-02
Let x denote a real number. Find out the INCORRECT statement.
A. S = {x :x > 3} represents the set of all real numbers greater than 3
B. S = {x : $x^2$ < 0} represents the empty set.
C. S = {x :x $\in$ A and x $\in$ B} represents the union of set A and set B.
D. S = $\{x :a < x < b\}$ represents the set of all real numbers between a and b, where a and b are real numbers.
S = {x :x \(\in\) A AND x \(\in\) B} represents the INTERSECTION of set A and set B.
S = {x :x \(\in\) A OR x \(\in\) B} represents the UNION of set A and set B.
Correct Answer: C
Solved Example: 2-1-03
Consider the function f(x) = $\mid x\mid$ in the interval -1 $\leq$ x $\leq$ 1. At the point x = 0, f (x) is:
A. Continuous and differentiable
B. Non-continuous and differentiable
C. Continuous and non-differentiable
D. Neither continuous nor differentiable
At x= 0, function is continuous but not differentiable because.For x > 0, f'(x)=1, $\lim_{x\to 0^+}$ f'(x)=1
For x < 0, f'(x)=-1, $\lim_{x\to 0^-}$ f'(x)= -1
$\lim_{x\to 0^+}$ f'(x) $\neq$ $\lim_{x\to 0^-}$ f'(x)
Therefore, it is not differentiable.Correct Answer: C
Solved Example: 2-1-04
$\displaystyle \lim_{x\to 0}\left( \dfrac {1-\cos x}{x^{2}}\right)$ is:
A. $\dfrac{1}{4}$
B. $\dfrac{1}{2}$
C. 1
D. 2
\[y= \lim _{x\to 0}\dfrac {\left( 1-\cos x\right) }{x^{2}}\] It forms 0/0 condition. Hence by L'Hopital rule, \[y=\lim _{x\to 0}\dfrac {\dfrac {d}{dx}\left( 1-\cos x\right) }{\dfrac {d}{dx}x^{2}}\] Still this gives $\dfrac{0}{0}$ condition, so again applying L'Hopital rule, \[y =\lim _{x\to 0}\dfrac {\dfrac {d}{dx}\left( \sin x \right) }{\dfrac {d}{dx}\left( 2x\right)} =\lim _{x\to 0}\dfrac {\cos x}{2} =\dfrac{\cos 0}{2} = \dfrac{1}{2}\]
Correct Answer: B
Solved Example: 2-1-05
What is $\displaystyle \lim_{\theta \to 0}\dfrac {\sin \theta }{\theta}$ equal to?
A. $\theta$
B. $\sin \theta$
C. 0
D. 1
This is a standard identity or you may apply L'Hopital's rule, \[y =\lim_{\theta \to 0}\dfrac {\sin \theta }{\theta } =\lim_{\theta \to 0}\dfrac {\dfrac {d}{d\theta }\sin \theta }{\dfrac {d}{d\theta }\theta } =\lim_{\theta \to 0}\dfrac {\cos \theta }{1} =\dfrac {\cos 0}{1} =1\]
Correct Answer: D
Solved Example: 2-1-06
$\displaystyle \lim _{x\rightarrow 0} \dfrac{\sin^2x}{x}$ is equal to:
A. 0
B. 3
C. 1
D. -1
\begin{align*} f\left( x\right) &=\lim _{x\rightarrow 0}\dfrac {\sin ^{2}x}{x}\\ &=\lim _{x\rightarrow 0}\dfrac {\sin ^{2}x}{x^{2}}\times \dfrac {x^{2}}{x}\\ &=\lim _{x\rightarrow 0}\left( \dfrac {\sin x}{x}\right) ^{2}\times x\\ &=\left( 1\right) ^{2}\times 0\\ &=0 \end{align*} Alternative method is to use the L'Hospital's Rule. Take derivative of both numerator as well as denominator. \begin{align*} f\left( x\right) &=\lim _{x\rightarrow 0}\dfrac {\sin ^{2}x}{x}\\ f'\left( x\right)&=\lim _{x\rightarrow 0}\dfrac {2\sin x\cos x}{1}\\ &=\lim _{x\rightarrow 0}\dfrac {\sin 2x}{1}=\dfrac {\sin 0}{1}\\ &=0 \end{align*}
Correct Answer: A
Solved Example: 2-1-07
Which of the following functions is not differentiable in the domain [-1,1]?
A. f(x) = $x^2$
B. f(x) = x - 1
C. f(x) = 2
D. f(x) = maximum (x, - x )
It clearly shows that for the first three functions, f (x) is differentiable at x =- 1, x = 0 and x = 1, i.e. in the domain [- 1, 1]. So, (a), (b) and (c) are differentiable.Correct Answer: D
Solved Example: 2-1-08
At x = 0, the function f (x) = x$^3$+ 1 has:
A. A maximum value
B. A minimum value
C. A singularity
D. A point of inflection
\begin{align*} f(x) &= x^3 + 1\\ f'(x) &= 3x^2,\quad f'(0) = 0\\ f''(x) &= 6x, \quad f''(0) = 0 \end{align*} f''(0) = 0 point of inflection.f''(0) > 0 point of minima.
f''(0) < 0 point of maxima.
Correct Answer: D
Solved Example: 2-1-09
The minimum value of function y = $x^2$ in the interval [1, 5] is:
A. 0
B. 1
C. 25
D. Undefined
Given :
y = x$^2$ ...(i) and interval [1, 5]
At x = 1 and y = 1,
And at x = 5, y = (5)$^2$ = 25
Here the interval is bounded between 1 and 5. So, the minimum value at this interval is 1.
Correct Answer: B
Solved Example: 2-1-10
The value of: $\displaystyle \lim _{x\rightarrow 8}\dfrac {x^{\frac{1}{3}}-2}{x-8}$
A. $\dfrac{1}{16}$
B. $\dfrac{1}{12}$
C. $\dfrac{1}{8}$
D. $\dfrac{1}{4}$
Using the formula for, $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
Here, a = x$^{\frac{1}{3}}$
and b = 2
\begin{align*}
\lim _{x\rightarrow 8}\frac {x^{\frac{1}{3}}-2}{x-8}
&= \dfrac {(x^{\frac{1}{3}}-2)}{(x^{\frac{1}{3}}-2)(x^{\frac{2}{3}} + {x^{\frac{1}{3}}(2) +4)}}\\
&= \dfrac {1}{(4)+ 2(2)+4}\\
&= \dfrac {1}{12}
\end{align*}
Another option would be to use L'Hopital's rule:
\[\lim _{x\rightarrow 8}\dfrac {x^{\frac{1}{3}}-2}{x-8}\]
Take derivative of numerator and denominator.
\[\lim _{x\rightarrow 8}\dfrac {\dfrac{1}{3}x^{\frac{-2}{3}}}{1}\]
\[= \dfrac{1}{3}8^{\frac{-2}{3}} = \dfrac{1}{12}\]
Correct Answer: B
Solved Example: 2-1-11
If $f(x) = \dfrac{2x^2 -7x +3}{5x^2 -12x -9}$ then $\displaystyle \lim_{x\rightarrow 3} f(x)$ will be:
A. $\dfrac{-1}{3}$
B. $\dfrac{5}{18}$
C. 0
D. $\dfrac{2}{5}$
METHOD I: First we will factorize numerator and denominator separately.
\begin{align*}
2x^2 -7x +3 &=2x^{2}-6x-x+3\\
&=2x\left( x-3\right) -1\left( x-3\right)\\
&=\left( x-3\right) \left( 2x-1\right)\\
\end{align*}
\begin{align*}
5x^{2}-12x-9 &=5x^{2}-15x+3x-9\\
&=5x\left( x-3\right) +3\left( x-3\right)\\
&=\left( x-3\right) \left( 5x+3\right)
\end{align*}
After cancelling the common factor from numerator and denominator, $f\left( x\right) =\dfrac {2x-1}{5x+3}$
Now substitute the value of x = 3, $ f(x)=\dfrac {5}{18}$
METHOD II: Use L'Hospital's Rule and take derivative of both numerator and denominator.
The value of the limit will be: \begin{align*} &= \dfrac{4x - 7}{10x - 12}\\ &= \dfrac{12 - 7}{30 - 12}\\ &= \dfrac{5}{18} \end{align*}
Correct Answer: B
Solved Example: 2-1-12
$\displaystyle \lim_{x \to 1}(1-x)\tan \left( \dfrac{\pi x}{2} \right)$=?
A. $\dfrac{\pi}{2}$
B. $\pi$
C. $\dfrac{2}{\pi}$
D. 0
\[\mathrm{Let\ } L = \lim_{x \to 1} \left( 1-x\right) \tan \left( \dfrac {\pi x}{2}\right)\] \[\mathrm{Put}\ 1-x = \theta, \quad x=1-\theta, \quad \mathrm{As\ } x \rightarrow 1,\ \theta \rightarrow 0\] \begin{align*} L &= \lim_{\theta \to 0} \theta \tan \left[\dfrac {\pi }{2}\left( 1-\theta \right)\right] \\ &= \lim_{\theta \to 0}\theta \tan \left( \dfrac {\pi }{2}-\dfrac {\pi }{2}\theta \right)\\ &= \lim_{\theta \to 0}\theta \left[ \cot \left( \dfrac {\pi }{2}\theta \right) \right]\\ &= \lim_{\theta \to 0}\dfrac {\theta }{\tan \left( \dfrac {\pi }{2}\theta \right) }\\ &=\lim _{\theta \rightarrow 0}\dfrac {{\theta }}{\left[ \dfrac {\tan \left( \dfrac {\pi }{2}\theta \right) }{\left(\dfrac {\pi }{2}\theta \right) }\right]\cdot \dfrac {\pi }{2}{\theta }} =\dfrac{1}{1 \cdot \left(\dfrac {\pi }{2}\right)} =\dfrac {2}{\pi} \end{align*}
Correct Answer: C
Solved Example: 2-1-13
Let f: R $\rightarrow$ R be defined by: $(3x^2 +4) \cos x$ Then \[\lim_{h \to 0} \dfrac{f(h) + f(-h) -8}{h^2}\] is equal to :
A. 0
B. 2
C. $\dfrac{\pi}{2}$
D. $\pi$
\[f\left( h\right) =\left( 3h^{2}+4\right) \cos\ h,\]
\[f\left( -h\right) =\left[ 3\left( -h\right) ^{2}+4\right] \cos \left( -h\right) =\left( 3h^{2}+4\right) \cos\ h\]
Adding, $f\left( h\right) +f\left( -h\right) =\left( 6h^{2}+8\right) \cos\ h$
Subtract 8 from both sides, $f\left( h\right) +f\left( -h\right) -8 =6h^{2}\cos\ h +8\cos\ h -8$
\begin{align*}
& \lim_{h \to 0} \dfrac {f\left( h\right) +f\left( -h\right) -8}{h^{2}}\\ &= \lim_{h \to 0} \dfrac{6h^{2}\cos\ h +8\cos\ h -8}{h^2}\\
&= \lim_{h \to 0} \dfrac {6h^{2}\cos\ h }{h^{2}}+8 \left(\dfrac {\cos\ h -1}{h^{2}}\right)\\
&= \lim_{h \to 0}6\cos\ h + 8\dfrac {\left(- 2\sin ^{2}\dfrac {h}{2}\right) }{h^{2}}\\
&=\lim_{h \to 0} 6\cos\ h - 16\left[ \dfrac{\left(\sin ^{2}\dfrac {h}{2}\right)}{\left( \dfrac {h}{2}\right) ^{2} \times 4}\right]\\
&=\lim_{h \to 0} 6\cos\ h -4 \lim_{\dfrac{h}{2} \to 0} \left[\dfrac {\sin ^{2}\left( \dfrac {h}{2}\right) }{\left( \dfrac {h}{2}\right) ^{2}}\right]\\
&=6(1)-4(1) =2
\end{align*}
Correct Answer: B
Solved Example: 2-1-14
The value of \[\lim_{x \to \infty} \dfrac{x \ln x}{1 + x^2}\] is:
A. 1.0
B. 0.5
C. $\infty$
D. 0
\[\mathrm{Let\ L =} \lim_{x \to \infty} \dfrac{x \ln x}{1 + x^2}\] Applying L'Hopital's rule and taking derivatives of both numerator and denominator: \[= \lim_{x \to \infty} \left[\dfrac{x\left(\dfrac{1}{x}\right) + \ln x}{2x} \right]\] Still this limit is in $\dfrac{\infty}{\infty}$ form, hence applying L'Hopital's rule one more time \begin{align*} L &= \lim_{x \to \infty} \left[\dfrac{0 + \dfrac{1}{x}}{2} \right]\\ &= \lim_{x \to \infty} \left[\dfrac{1}{2x} \right]\\ &= \dfrac{1}{2 \times \infty}\\ &= 0 \end{align*}
Correct Answer: D
Derivatives
Learning Objectives:
- Calculate derivative using basic formulae: Algebraic, trigonometric, inverse trigonometric, logarithmic and exponential.
- Calculate derivative using rules of derivatives such as: Addition/subtraction, product, quotient and chain rule.
- Apply differentiation to calculate tangent and normal, rate of change, maxima/minima and errors/approximation.
First Principle of Derivative:
\[f'(x) = \lim_{\Delta x \to 0}\frac{f(x+ \Delta x)-f(x)}{ \Delta x}\]
Some Basic Formulae:
\[\dfrac{d}{dx} \ c = 0\]
\[\dfrac{d}{dx} ( u \pm v \pm w \pm ....) = \dfrac{du}{dx} \pm \dfrac{dv}{dx} \pm \dfrac{dw}{dx} \pm .....\]
\[\dfrac{d}{dx} ( cu ) = c \dfrac{du}{dx}\]
\[\dfrac{d}{dx} ( uv ) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}\]
\[\dfrac{d}{dx} ( \dfrac {u}{v}) = \dfrac{v\frac{du}{dx} - u\dfrac{dv}{dx} }{v^{2}}\]
Chain Rule:
\[\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx}\]
List of Derivative Formulae:
Algebraic
\[\dfrac{d}{dx} x^n = n x^{n-1}\]
Trigonometric
\[\dfrac{d}{dx} \sin x = \cos x\] \[\dfrac{d}{dx} \cos x = -\sin x\] \[\dfrac{d}{dx} \tan x = \sec^2 x\] \[\dfrac{d}{dx} \cot x = - \csc^2 x\] \[\dfrac{d}{dx} \sec x = \sec x \tan x\] \[\dfrac{d}{dx} \csc x = - \csc x \cot x\]Log/Exp
\[\dfrac{d}{dx} e^x = e^x\] \[\dfrac{d}{dx} a^x = a^x \log a\] \[\dfrac{d}{dx} \log x = \dfrac{1}{x}\]Inv Trig
\[\dfrac{d}{dx} \sin^{-1} x = \dfrac{1}{\sqrt{1-x^2}}\] \[\dfrac{d}{dx} \cos^{-1} x = \dfrac{-1}{\sqrt{1-x^2}}\] \[\dfrac{d}{dx} \tan^{-1} x = \dfrac{1}{{1+x^2}}\] \[\dfrac{d}{dx} \cot^{-1} x = \dfrac{-1}{{1+x^2}}\] \[\dfrac{d}{dx} \sec^{-1} x = \dfrac{1}{x \sqrt{x^2 - 1}}\] \[\dfrac{d}{dx} \csc^{-1} x = \dfrac{-1}{x \sqrt{x^2 - 1}}\]Solved Example: 2-2-02
If $x = a(\theta + \sin \theta)$ and $y = a(1 - \cos \theta)$ then $\dfrac{dy}{dx}$ will be equal to:
A. $\sin \left(\dfrac{\theta}{2}\right)$
B. $\cos \left(\dfrac{\theta}{2}\right)$
C. $\tan \left(\dfrac{\theta}{2}\right)$
D. $\cot \left(\dfrac{\theta}{2}\right)$
\[\dfrac {dx}{d\theta }=a\left( 1+\cos \theta \right),\ \dfrac {dy}{d\theta }=a\left( 0+\sin \theta \right)\] \[\dfrac {dy}{dx} =\dfrac {\left( \dfrac {dy}{d\theta }\right) }{\left( \dfrac {dx}{d\theta }\right) } =\dfrac {\sin \theta }{1+\cos \theta }\\ =\dfrac {2\sin \dfrac {\theta }{2}\cos \dfrac {\theta }{2}}{2\cos ^{2}\dfrac {\theta }{2}} = \tan\dfrac {\theta }{2}\]
Correct Answer: C
Solved Example: 2-2-03
For the following parametric function, what is the $\dfrac{d^2y}{dt^2}$ at t= 0? \[x = 1 - t^2, \quad y= t- t^3\]
A. $\infty$
B. 0
C. -1
D. $\dfrac{2}{3}$
\[y'=\dfrac {dy}{dx}\] Using chain rule, \[y'=\dfrac {dy}{dt}\dfrac {dt}{dx}\] Since x is given in terms of parameter t, it is easier to find $\dfrac{dx}{dt}$ \[y'=\dfrac {\dfrac {dy}{dt}}{\dfrac {dx}{dt}} =\dfrac {1-3t^{2}}{-2t}\] \[y''=\dfrac {\dfrac {dy'}{dt}}{\dfrac {dx}{dt}} =\dfrac {\dfrac {-2t\left( 0-6t\right) -\left( 1-3t^{2}\right) \left( -2\right) }{\left( -2t\right) ^{2}}}{-2t}\] \[y''=\dfrac {\dfrac {12t^{2}+2-6t^{2}}{4t^{2}}}{-2t} =\dfrac {\dfrac {2+6t^{2}}{4t^{2}}}{-2t} =\dfrac {-2-6t^{2}}{8t^{3}}\] \[y''|_{t=0}=\dfrac {-2}{0}=\infty\]
Correct Answer: A
Solved Example: 2-2-04
If $I_n= \dfrac{d^n}{dx^n}(x^n \log x),$ then: $I_n -n I_{n -1}= ?$
A. n
B. n -1
C. n!
D. (n -1)!
\begin{align*} I_{n} &=\dfrac {d^{n}}{dx^{n}}\left( x^{n}\log x\right)\\ &=\dfrac {d^{n-1}}{dx^{n-1}}\left[ \dfrac {d}{dx}\left(x^{n}\log x\right)\right]\\ &=\dfrac {d^{n-1}}{dx^{n-1}}\left[ x^{n}\dfrac {d}{dx}\left( \log x\right) +\log x\dfrac {d}{dx}\left(x^n\right)\right] \\ &=\dfrac {d^{n-1}}{dx^{n-1}}\left[ x^{n}\left(\dfrac{1}{x}\right)+\log x\left(nx^{n-1}\right)\right] \\ &=\dfrac {d^{n-1}}{dx^{n-1}}\left[ x^{n-1}+n\log x\left(x^{n-1}\right)\right] \\ &=\dfrac {d^{n-1}}{dx^{n-1}}\left[ x^{n-1}\right]+ n\dfrac {d^{n-1}}{dx^{n-1}}\left[\log x\left(x^{n-1}\right)\right] \\ I_n &=\left( n -1\right) !+n I_{n -1}\\ I_n-nI_{n-1}&=\left( n-1\right) ! \end{align*}
Correct Answer: D
Solved Example: 2-2-05
Find the derivative of $e^{x^3}$.
A. $e^{x^3}$
B. $3x$
C. $3e^{x^3}$
D. $3x^2e^{x^3}$
Apply chain rule. \begin{align*} &\dfrac{d}{dx}e^{x^3}\\ &= e^{x^3}\dfrac{d}{dx}({x^3})\\ &=e^{x^3}{3x^2}\\ &={3x^2} e^{x^3} \end{align*}
Correct Answer: D
Solved Example: 2-2-06
Evaluate: \[\dfrac{d}{dx} (\log x \cos x)\]
A. $\log x (-\cos x) + \cos x \left(\dfrac{1}{x}\right)$
B. $\log x (-\sin x) + \cos x \left(\dfrac{1}{x}\right)$
C. $\log x (\cos x) + \cos x \left(\dfrac{1}{x}\right)$
D. $\log x (\sin x) - \cos x \left(\dfrac{1}{x}\right)$
\begin{align*} &\dfrac{d}{dx} (\log x \cos x)\\ &= \log x \dfrac{d}{dx} (\cos x) + \cos x \dfrac{d}{dx} (\log x)\\ &= \log x (-\sin x) + \cos x \left(\dfrac{1}{x}\right) \end{align*}
Correct Answer: B
Solved Example: 2-2-07
What is the value of $\dfrac{d}{dx} (e^x \sin x)$ at x = 0?
A. 0
B. 1
C. -1
D. $\infty$
\begin{align*} \dfrac{d}{dx} (e^x \sin x)\\ &= e^x \dfrac{d}{dx} (\sin x) + \sin x\dfrac{d}{dx} (e^x)\\ &= e^x (\cos x) + \sin x (e^x)\\ &= e^x (\cos x + \sin x) \end{align*} At x = 0, \begin{align*} e^x (\cos x + \sin x)\\ &= e^0 (\cos 0 + \sin 0)\\ &= (1) (1 + 0)\\ &= 1 \end{align*}
Correct Answer: B
Solved Example: 2-2-08
What does derivative of a function represent?
A. The rate of change of a function
B. The area under a curve
C. The maxima and minima points
D. Approximation of a function
Correct Answer: A
Solved Example: 2-2-09
For which function, the derivative is same as its original value?
A. Algebraic (x$^n$)
B. Exponential (e$^x$)
C. Logarithmic ($\ln\ x$)
D. Trigonometric ($\sin\ x$)
Correct Answer: B
Application of Derivatives
Learning Objectives:
- Calculate curvature of any curve.
- Calculate rate of change using derivatives.
- Calculate maxima/mimina and inflection points.
Curvature of Any Curve:
If the equation of the curve is easier to differentiate with respect to x, then:
Radius of Curvature:
The radius of curvature R at any point on a curve is defined as the absolute value of the reciprocal of the curvature K at that point.
Kiatdd, CC BY-SA 3.0, via Wikimedia Commons
Rate of Change:
If y = f(t) then the rate of change is given by \(\dfrac{dy}{dt}\).
Maxima/Minima points:
if \(\dfrac{dy}{dx} = 0\), that point indicates point of direction change. It can be a maxima point or it can be a minima point. Further classification requires calculation of second derivative y”.
For maxima point, y” < 0
For minima point, y” > 0
TD, CC BY-SA 3.0, via Wikimedia Commons
Tangents and Normals:
Derivative of a function gives you slope of the tangent.
Since normal is perpendicular to the tangent, we can use the formula studies in 'straight lines' topic to get slope of normal.
\[\mathrm{Slope\ of\ normal} = \dfrac{-1}{\mathrm{slope\ of\ tangent}}\]If you know point of tangency, then we can use slope-point equation $y - y_1 = m(x - x_1)$ to get the equation of tangent line as well as normal line.
Solved Example: 2-3-01
The minimum point of the function $f(x) = \left(\dfrac{x^3}{3}\right) - x $ is at:
A. x = 1
B. x = - 1
C. x = 0
D. x = $\dfrac{1}{\sqrt{3}}$
$f (x) = \dfrac{x^3}{3} - x, \quad f'(x) = x^2- 1, \quad f''(x) = 2x$
Using the principle of maxima – minima and put $f'(x) = 0, \quad x^2- 1 = 0 \mathrm{\ and\ } x =\pm1$
Hence at x = -1, f''(x) = -2 < 0 (Maxima), at x = 1, f''(x) = 2 > 0 (Minima). So, f (x) is minimum at x = 1
Correct Answer: A
Solved Example: 2-3-02
Consider the functions:
I. $e^{-x}$
II. $x^2 - \sin x$
III. $\sqrt{x^3+ 1}$
Which of the above functions is/are increasing everywhere in [0, 1]?
A. III only
B. II only
C. II and III only
D. I and III only
Let's calculate derivatives for each function:
A function is continuously increasing if its derivative (slope) is throughout positive.
For function I: $f(x) = e^{-x}$
$f'(x) = - e^{-x} < 0 $ in [0, 1]
For function II: $f(x) = x^2 - \sin x$
$f'(x) =2x - \cos x < 0$ in [0, 1]
For function III: $f(x) = \sqrt{x^3+ 1}$
$f'(x) = \dfrac{3x^2}{2\sqrt{x^3+ 1}} >0 $ in [0, 1]
Correct Answer: A
Solved Example: 2-3-03
Each of four particles move along an x-axis. Their coordinates (in meters) as functions of time (in seconds) are given by:
Particle I: $x(t) = 3.5 - 2.7t^3$
Particle II: $x(t) = 3.5 + 2.7t^3$
Particle III: $x(t) = 3.5 - 2.7t^2$
Particle IV: $x(t) = 3.5 - 3.4 t - 2.7t^2$
Which of these particles have constant acceleration?
A. All four
B. Only (I) and (II)
C. Only (II) and (III)
D. Only (III) and (IV)
Calculate second derivative for each function. If it comes out just a number (independent of 't', then that function has constant acceleration.
Particle I: $x(t) = 3.5 - 2.7t^3$
$\dot{x} = -8.1 t^2$
$\ddot{x} = -16.2 t$
Particle II: $x(t) = 3.5 + 2.7t^3$
$\dot{x} = 8.1 t^2$
$\ddot{x} = 16.2 t$
Particle III: $x(t) = 3.5 - 2.7t^2$
$\dot{x} = -5.4 t$
$\ddot{x} = -5.4$
Particle IV: $x(t) = 3.5 - 3.4 t - 2.7t^2$
$\dot{x} = -3.4 - 5.4 t$
$\ddot{x} = -5.4$
Correct Answer: D
Solved Example: 2-3-04
Equation of the line normal to function $f(x) = (x - 8)^{\frac{2}{3}} + 1$ at P(0, 5) is:
A. y = 3x - 5
B. y = 3x + 5
C. 3y = x + 15
D. 3y = x - 15
\begin{align*} y &=\left( x-8\right) ^{\frac {2}{3}} + 1 \\ \dfrac {dy}{dx} &= \dfrac {2}{3}\left( x-8\right) ^{-\frac {1}{3}} \end{align*} \begin{equation*} \begin{split} \dfrac {dy}{dx}|_{x=0} &=\left( \dfrac {2}{3}\right) \left( -8\right) ^{-\frac {1}{3}}\\ & =\dfrac {2}{3}\left( \dfrac {1}{\sqrt [3] {-8}}\right) =\dfrac {2}{3}\left( \dfrac {1}{-2}\right) =\dfrac {-1}{3} \end{split} \end{equation*} \[\mathrm{Slope\ of\ normal}=\dfrac {-1}{\left( -1/3\right) }=3\] \begin{align*} y-y_{1} &=m\left( x-x_{1}\right)\\ y-5 &=3\left( x-0\right)\\ y-5 &=3x\\ y &=3x+5 \end{align*}
Correct Answer: B
Solved Example: 2-3-05
Derivatives can be used to find out:
A. Slope of tangents and normals
B. Velocity and acceleration
C. Curvature at some point on a curve
D. All of the above
Correct Answer: D
Solved Example: 2-3-06
A negative value of first derivative indicates:
A. The curve is increasing.
B. The curve is decreasing.
C. The curve is stationary.
D. The elevation of curve is negative.
Correct Answer: B
Partial Derivatives
Learning Objectives:
- Define a partial derivative.
- Compute partial derivatives by viewing certain variables as constant.
- Understand notation and computation for higher-order partial derivatives.
When working with functions of more than one variable, the question in calculus becomes: how can we evaluate the rate of change? The answer is called a partial derivative. Given a function f(x, y) or f(x, y, z), the partial derivative of f with respect to x, \(\partial f\) = \(f_x\) , is found by treating all variables other than x as constants.
Technically, we’re finding \[\lim_{h \rightarrow 0} \dfrac{f (x + h, y)- f (x, y)}{h}\] The partial derivatives \(\partial f = f_y\) and \(\partial f = f_z\) have analogous definitions.
Solved Example: 2-4-01
Let $f = y^x$. What is: $\dfrac {\partial ^{2}f}{\partial x\partial y}$ at x = 2, y = 1?
A. 0
B. $\ln 2$
C. 1
D. $\dfrac{1}{\ln 2}$
\begin{align*} f &=y^{x}\\ \dfrac {\partial f}{\partial x} &= y^{x}\log y\\ \dfrac {\partial ^{2}f}{\partial x\partial y} &= xy^{x-1}\log y+\dfrac {1}{y}y^{x} \end{align*} \begin{align*} \dfrac {\partial ^{2}f}{\partial x\partial y}|_{ \left[ x=2,y=1\right]} &=\left( 2\right) \left( 1\right) ^{1}\log 1+\dfrac {1}{1}\left( 1\right) ^{2} \\ &=0+1 \\ &=1 \end{align*}
Correct Answer: C
Solved Example: 2-4-02
The function $f (x,y) = 2x^2+ 2xy - y^3$ has:
A. Only one stationary point at (0, 0)
B. Two stationary points at (0, 0) and $\left(\dfrac{1}{6},\dfrac{-1}{3}\right)$
C. Two stationary points at (0, 0) and (1, - 1)
D. No stationary point
\[f(x,y) = 2x^2+ 2xy - y^3\] Partially differentiate this function w.r.t x and y. For the stationary point of the function, put $\partial$f/$\partial$x and $\partial$f/$\partial$y equal to zero. \[\dfrac {\partial f}{\partial x}=4x+2y=0, \quad \mathrm{or\ ,} 2x+y=0\] \[\dfrac {\partial f}{\partial y}=2x-3y^{2}=0\] From equation (i), y =- 2x substitute in equation (ii), \begin{align*} 2x - 3(- 2x)^2 &= 0\\ 2x - 3 \times 4x^2 &= 0\\ 6x^2- x &= 0, x = 0, \dfrac{1}{6} \end{align*} From equation (i), For x = 0, y =- 2 $\times$(0) = 0 and for x= $\dfrac{1}{6}$, y= -2 $\times$ $\dfrac{1}{6}$ = $\dfrac{-1}{3}$ So, two stationary points at (0, 0) and ($\dfrac{1}{6}$, $\dfrac{-1}{3}$).
Correct Answer: B
Solved Example: 2-4-03
If u = $\dfrac{x+y}{x-y}$, then $\dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y}$ = ??
A. $\dfrac{1}{x-y}$
B. $\dfrac{2}{x-y}$
C. $\dfrac{1}{{x-y}^2}$
D. $\dfrac{2}{{x-y}^2}$
\[u = \dfrac{x+y}{x-y}\] \[\dfrac{\partial u}{\partial x} = \dfrac{(x-y)(1) - (x+y) (1)}{{(x-y)}^2}= \dfrac{-2y}{{(x-y)}^2}\] \[\dfrac{\partial u}{\partial y} = \dfrac{(x-y)(1) - (x+y) (-1)}{{(x-y)}^2} = \dfrac{2x}{{(x-y)}^2}\] Adding, \[\dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} = \dfrac{2x - 2y}{{(x-y)}^2} = \dfrac{2(x-y)}{(x-y)^2} = \dfrac{2}{(x-y)}\]
Correct Answer: B
Solved Example: 2-4-04
If u = $\log (x^2 + y^2)$, then $\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2}$ =??
A. $\dfrac{1}{x^2 + y^2}$
B. 0
C. $\dfrac{x^2 -y^2}{(x^2 + y^2)^2}$
D. $\dfrac{y^2 -x^2}{(x^2 + y^2)^2}$
\[u = \log (x^2 + y^2)\] Differentiating given equation partially w.r.t. x, \[\dfrac{\partial u}{\partial x} = \dfrac{1}{x^2 + y^2} \cdot 2x = \dfrac{2x}{x^2 + y^2}\] Differentiating partially one more time, \begin{align*} \dfrac{\partial^2 u}{\partial x^2} &= \dfrac{(x^2 + y^2)(2) - (2x) (2x+0)}{(x^2 + y^2)^2}\\ &= \dfrac{2x^2 + 2y^2 - 4x^2}{(x^2 + y^2)^2}\\ &= \dfrac{2y^2 - 2x^2}{(x^2 + y^2)^2} \end{align*} Differentiating given equation partially now w.r.t. y, \[\dfrac{\partial u}{\partial y} = \dfrac{1}{x^2 + y^2} \cdot 2y = \dfrac{2y}{x^2 + y^2}\] Differentiating partially one more time, \begin{align*} \dfrac{\partial^2 u}{\partial y^2} &= \dfrac{(x^2 + y^2)(2) - (2y) (0+2y)}{(x^2 + y^2)^2}\\ &= \dfrac{2x^2 + 2y^2 - 4y^2}{(x^2 + y^2)^2}\\ &= \dfrac{2x^2 - 2y^2}{(x^2 + y^2)^2} \end{align*} Adding, \[\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = \dfrac{2y^2 - 2x^2}{(x^2 + y^2)^2} + \dfrac{2x^2 - 2y^2}{(x^2 + y^2)^2}= 0\]
Correct Answer: B
Solved Example: 2-4-05
What is the difference between total derivative and partial derivative?
A. In total derivative, in which all variables are allowed to vary whereas in partial derivative one variable variable varies and others are held constant.
B. In partial derivative, in which all variables are allowed to vary whereas in total derivative one variable variable varies and others are held constant.
C. Partial derivative is evaluated using partial fractions.
D. Partial derivative involves constant of derivative.
Correct Answer: A
Solved Example: 2-4-06
For which of the following options, partial derivative and total derivative is the same?
A. x$^3$
B. 3$\ln\ x$
C. 3e$^x$
D. 3
Correct Answer: D
Indefinite Integrals
Learning Objectives:
- Calculate integration of a function using basic formulae - algebraic, trigonometric, inverse trigonometric, logarithmic and exponential.
- Calculate integration by substitution, by partial fractions, by trigonometric substitution, by parts.
Integration Formulae:
Algebraic Group:
\[\int x^n dx = \dfrac {x^{n+1}}{n+1} + C\]
Trigonometric Group:
\[\int \cos x \ dx = \sin x + C\]
\[\int \sin x \ dx = -\cos x + C\]
\[\int \sec^2 x \ dx = \tan x + C\]
\[\int \csc^2 x \ dx = - \cot x + C\]
\[\int \sec x \tan x \ dx = \sec x + C\]
\[\int \csc x \cot x \ dx = -\csc x + C\]
Logarithmic and Exponential Group:
\[\int e^x \ dx = e^x + C\]
\[\int a^x \ dx= \dfrac{a^x}{\log a} + C\]
\[\int \dfrac{1}{x} \ dx = \log x + C\]
Inverse Trigonometric Group:
\[\int \dfrac{1}{\sqrt{1-x^2}} \ dx = \sin^{-1} x + C\]
\[\int \dfrac{-1}{\sqrt{1-x^2}} \ dx = \cos^{-1} x + C\]
\[\int \dfrac{1}{{1+x^2}} \ dx = \tan^{-1} x + C \ or -\cot^{-1} x + C\]
\[\int \dfrac{1}{\vert x \vert \sqrt{x^2 - 1}} \ dx = \sec^{-1} x + C \ or -\csc^{-1} x + C\]
Integration by Substitution Method
There are occasions when it is possible to perform an apparently difficult piece of integration by first making a substitution. \[I = \int x \sin (x^{2}+ 1)dx\] Substitute \[\large (x^{2}+ 1) = u\] Then \[\large 2x dx= du\] or \[x dx= \dfrac{1}{2}du\] Then the integral becomes \[\begin{aligned} I &= \dfrac{1}{2} \int \sin u du\\ &= - \dfrac{1}{2}\ \cos u = - \dfrac{1}{2}\ \cos (x^{2}+ 1) + C\end{aligned}\]
Integration by Parts:
\[\begin{aligned} I = \int x\ \cos x \ dx &= \int u \ dv\\ &= uv - \int v \ du\\ &= x\ \sin x - \int \sin x \ dx\\ &= x \ \sin x + \cos x + C\end{aligned}\]
Solved Example: 2-5-01
\[\int \dfrac {\sec x}{\sec x + \tan x}\ dx\]
A. $\tan x + \sec x + C$
B. $\sec x - \tan x + C$
C. $\tan x - \sec x + C$
D. $\tan x - \cot x + C$
\[\mathrm{Let}\ I = \int \dfrac {\sec x}{\sec x + \tan x}\ dx\] Multiply numerator and denominator by $\cos x$. \[=\int \dfrac {1}{1+\sin x}dx\] multiply and divide by 1- $\sin x $ \begin{align*} I &=\int \dfrac {1-\sin x}{1-\sin^{2}x}dx\\ &=\int \dfrac {1-\sin x}{\cos ^{2}x}dx\\ &=\int (\sec^{2}x-\sec x\tan x) dx\\ &=\tan x-\sec x + C \end{align*}
Correct Answer: C
Solved Example: 2-5-02
Evaluate $\int \sqrt {1+\sin 5x}\ dx$
A. $\dfrac {2}{5}\left[ \sin \dfrac {5x}{2}-\cos \dfrac {5x}{2}\right] +C$
B. $\dfrac {2}{5}\left[ \sin \dfrac {5x}{2}+\cos \dfrac {5x}{2}\right] +C$
C. $\dfrac {2}{3}\left[ \sin \dfrac {3x}{2}-\cos \dfrac {3x}{2}\right] +C$
D. $\dfrac {2}{3}\left[ \sin \dfrac {3x}{2}+\cos \dfrac {3x}{2}\right] +C$
\begin{align*} \int \sqrt {1+\sin 5x}\ dx &=\int \sqrt {\left( \cos \dfrac {5x}{2}+\sin \dfrac {5x}{2}\right) ^{2}}dx\\ &=\int \left[ \cos \dfrac {5x}{2}+\sin \dfrac {5x}{2}\right] dx\\ &=\dfrac {\sin \dfrac {5x}{2}}{\dfrac {5}{2}}-\dfrac {\cos \dfrac {5x}{2}}{\dfrac {5}{2}}+C\\ &=\dfrac {2}{5}\left[ \sin \dfrac {5x}{2}-\cos \dfrac {5x}{2}\right] + C \end{align*}
Correct Answer: A
Solved Example: 2-5-03
If \[\int \dfrac{f(x)}{\log \sin x} dx = \log[\log (\sin x)] + c,\] then f(x) is equal to:
A. $\sin x$
B. $\cos x$
C. $\tan x$
D. $\cot x$
\[\int \dfrac{f(x)}{\log \sin x} dx = \log \log \sin x + c,\] Since derivative and integration are opposite of each other: \begin{align*} \dfrac{d}{dx} \left(\log \log \sin x + c\right) &= \dfrac{f(x)}{\log \sin x}\\ \dfrac{1}{\log \sin x} \dfrac{d}{dx} \left(\log \sin x \right) &= \dfrac{f(x)}{\log \sin x}\\ \dfrac{1}{\log \sin x} \dfrac{1}{\sin x}\dfrac{d}{dx} \left( \sin x \right) &= \dfrac{f(x)}{\log \sin x}\\ \dfrac{1}{\log \sin x} \dfrac{1}{\sin x} \left( \cos x \right) &= \dfrac{f(x)}{\log \sin x}\\ \dfrac{1}{\log \sin x} \dfrac{\cos x}{\sin x} &= \dfrac{f(x)}{\log \sin x}\\ \dfrac{\cot x}{\log \sin x} &= \dfrac{f(x)}{\log \sin x} \end{align*} By comparing both sides, \[f(x) = \cot x\]
Correct Answer: D
Solved Example: 2-5-04
Evaluate: \[\int (x - 1) e^{-x}\ dx\]
A. $(x - 2) e^{-x} + C$
B. $xe^{-x} + C$
C. $-xe^{-x} + C$
D. $(x + 1) e^{-x} + C$
\[\int (x - 1) e^{-x}\ dx\] Integrating by parts, \begin{align*} u &= (x - 1)\\ du &= dx \end{align*} \begin{align*} dv &= e^{-x}\\ v &= -e^{-x} \end{align*} \begin{align*} \int (x - 1) e^{-x}\ dx &= uv - \int vdu\\ &= (x - 1)(-e^{-x}) - \int -e^{-x}(dx)\\ &= (x - 1)(-e^{-x}) - e^{-x}\\ &= (-x + 1 - 1)(-e^{-x})\\ &= (-x)(e^{-x}) + C \end{align*}
Correct Answer: C
Partial Fractions
Learning Objectives:
- Distinguish between proper and improper fractions.
- Express an algebraic fraction as the sum of its partial fractions.
For partial fractions, the given fraction must be proper, i.e the highest degree of denominator polynomial must be higher than the highest degree of numerator polynomial. if not, first division must be carried out to make it a proper polynomial.
Rules on how to set Partial Fractions (with examples):
- \(\dfrac{3}{(x+2)(x+3)} = \dfrac{A}{(x+2)} + \dfrac{B}{(x+3)}\)
- \(\dfrac{3}{(x+2)^2} = \dfrac{A}{(x+2)} + \dfrac{B}{(x+2)^2}\)
- \(\dfrac{3}{(x^2+2)(x+3)} = \dfrac{Ax + B}{(x^2+2)} + \dfrac{C}{(x+3)}\)
- \(\dfrac{3}{(x^2+2)(x+3)^2} = \dfrac{Ax + B}{(x^2+2)} + \dfrac{C}{(x+3)} + \dfrac{D}{(x+3)^2}\)
Solved Example: 2-6-01
Evaluate: $\int \dfrac{1}{x^2 - 4} \,dx $
A. $\dfrac{1}{4} \ln \dfrac{ \vert x+2\vert}{\vert x-2\vert } + C $
B. $\dfrac{1}{2} \ln \dfrac{ \vert x-2\vert}{\vert x+2\vert } + C $
C. $\dfrac{1}{4} \ln \dfrac{ \vert x-2\vert}{\vert x+2\vert } + C $
D. $\dfrac{1}{2} \ln \dfrac{ \vert x+2\vert}{\vert x-2\vert } + C $
$ \int \dfrac{1}{x^2 - 4}dx= \int \dfrac{1}{(x+ 2)(x-2)}dx = \int { \left( \dfrac{A}{x+ 2} + \dfrac{B}{x-2}\right)dx} $
(After getting a common denominator, adding fractions, and equating numerators, it follows that
$ \ \ A(x-2) + B(x+2) = 1 $ ;
Let $ x = -2: \ A(-4) + B(0) = 1 \longrightarrow A = -\dfrac{1}{4}$
Let $ x = 2: \ A(0) + B(4) = 1 \longrightarrow B = \dfrac{1}{4}$
\begin{align*} &= \int {( \dfrac {-\dfrac{1}{4}}{x+ 2} + \dfrac{\dfrac{1}{4}}{x-2}})dx\\ &= \int {( -(\dfrac{1}{4})\dfrac{1}{x+2} + (\dfrac{1}{4})\dfrac{1}{x-2} )}dx\\ &= -\dfrac{1}{4} \ln {\vert x+2\vert} + \dfrac{1}{4} \ln {\vert x-2\vert} + C\\ &= \dfrac{1}{4} ( \ln \vert x-2\vert - \ln \vert x+2\vert ) + C\\ &= \dfrac{1}{ 4} \ln \dfrac{ \vert x-2\vert}{\vert x+2\vert } + C \end{align*}(Recall that $ \ \ln m - \ln n = \ln (\dfrac{ m}{n} )$)
Correct Answer: A
Solved Example: 2-6-02
If
$\dfrac{x}{(x+1)(x+2)} = \dfrac{P}{(x+1)} + \dfrac{Q}{(x+2)}$
Then:
A. P = 1, Q = 2
B. P = -1, Q = -2
C. P = 1, Q = -2
D. P = -1, Q = 2
Hint: Use partial fractions.
\[\dfrac{x}{(x+1)(x+2)} = \dfrac{P}{(x+1)} + \dfrac{Q}{(x+2)}\]
Compare only numerators after making denominator common.
\[x = P(x+2) + Q(x+1)\]
Now each factor on RHS make zero.
Step I: Make 'P' factor zero by putting x = -2.
\[-2 = Q(-1)\]
\[Q=2\]
Step II: Make 'Q' factor zero by putting x = -1.
\[-1 = P(1)\]
\[P=-1\]
Correct Answer: D
Definite Integrals
Learning Objectives:
- Find the definite integral of algebraic, trigonometric/inverse trigonometric, exponential and logarithmic functions.
- Use properties of definite integrals.
Properties of Definite Integrals:
For even function, f(-x) = f(x) i.e the graph is symmetrical about y-axis. In such case, \[\int_{-a}^{a} f(x)dx = 2 \int_{0}^{a} f(x)dx\] For odd functions, f(-x) = -f(x) and graph of f(x) is mirrored about x-axis AND y-axis. In such case, \[\int_{-a}^{a} f(x)dx = 0\] This is because the ’negative’ area balances the ’positive’ area after the origin. \[\int_{0}^{2a} f(x)dx = \int_{0}^{a} f(x)dx + \int_{0}^{a} f(2a - x)dx\]
Solved Example: 2-8-01
\[\int_{-a}^{a}(\sin^6 x + \sin^7 x)dx = ?\]
A. 2$\int_{0}^{a}(\sin^6 x)dx$
B. 2$\int_{0}^{a}(\sin^7 x)dx$
C. 2$\int_{0}^{a}(\sin^6 x + \sin^7 x)dx$
D. zero
A function is said to be even function if: $f(-x) = f(x)$
A function is said to be odd function if: $f(-x) = -f(x)$
Here, $\sin(x)$ is an odd function because: $\sin(-x) = - \sin x$
$\sin^6 x$, being an even power of $\sin x$ is an even function.
$\sin^7 x$, being an odd power of $\sin x$ is an odd function.
\begin{align*}
\int_{-a}^{a}(\sin^6 x + \sin^7 x)dx
&= \int_{-a}^{a}(\sin^6 x) dx + \int_{-a}^{a}(\sin^7 x)dx\\
&= \int_{-a}^{a}(\sin^6 x) dx + 0\\
&= 2 \int_{0}^{a}(\sin^6 x) dx
\end{align*}
Correct Answer: A
Solved Example: 2-8-02
The value of the integral: $\int ^{\infty }_{-\infty }\dfrac {dx}{1+x^{2}}$ is:
A. $-\pi$
B. $\dfrac{-\pi}{2}$
C. $\dfrac{\pi}{2}$
D. $\pi$
\begin{align*} I &=\int ^{\infty }_{-\infty }\dfrac {dx}{1+x^{2}}\\ &=\left[ \tan ^{-1}x\right] ^{\infty }_{-\infty }\\ &=\left[ \tan ^{-1}\left(\infty \right) -\tan ^{-1}\left( -\infty \right) \right]\\ &=\dfrac {\pi }{2}-\left( -\dfrac {\pi }{2}\right)\\ &=\pi \end{align*}
Correct Answer: D
Solved Example: 2-8-03
If f (x) is an even function and a is a positive real number, then $\int ^{a}_{-a}f\left( x\right) dx$ equals:
A. $ 2 \int ^{a}_{0}f\left( x\right) dx$
B. a
C. 2a
D. 0
For a function, whose limits bounded between -a to a and a is a positive real number. The solution is given by:
If f(x) is an even function, which means if f(-x) = f(x) $\int ^{a}_{-a}f\left( x\right) dx=2\int ^{a}_{0}f\left( x\right) dx$
If f(x) is an odd function, which means if f(-x) = - f(x) $\int ^{a}_{-a}f\left( x\right) dx=0$
Correct Answer: A
Solved Example: 2-8-04
Let f be a continuous and +ve real valued function on [0, 1]. Then $\int_0^\pi f(\sin x) \cos x\ dx = ?$
A. 0
B. 1
C. -1
D. $\infty$
Let t = $\sin x$, So, dt = $\cos x\ dx$
When x = 0, t = 0, When x = $\pi$, t = 0
$\int_0^\pi f(\sin x) \cos x\ dx$ becomes $ \int_0^0 f(t)\ dt$
Since, both upper and lower limits are identical, the value of the integral is zero.
Correct Answer: A
Solved Example: 2-8-05
The value of: \[\int_{-1}^1 x e^{\lvert x \rvert} dx\] is:
A. $\dfrac{1}{e}$
B. $\dfrac{-1}{e}$
C. $\dfrac{1}{\ln \pi}$
D. 0
\[f(x) = x e^{\lvert x \rvert}\] \[f(-x) = -x e^{\lvert -x \rvert} = -x e^{\lvert x \rvert} = -f(x)\]$x e^{\lvert x \rvert}$ is an odd function because f(-x) = -f(x), also the limits are symmetric about zero. Hence, the value of the integral between -1 to 0 cancels out the one from 0 to +1.
Correct Answer: D
Progressions
Learning Objectives:
- Identify type of a given proression.
- Identify the properties such as common difference or common ratio of a progression.
- Find a formula for the general term of a progression.
- Arithmatic Progressions (AP): Arithmetic progressions are a special case of progressions in which the difference between a term and its predecessor is always constant, known as common difference.
- Geometric Progression (GP): A geometric progression (GP) is a particular type of sequence or number sequence in which each next term in the sequence is produced by multiplying the previous term by a fixed number. Fixed numbers are called common ratios.
- Harmonic Progression (HP): A sequence is called HP in mathematics if the reciprocal of the terms is in AP.
Solved Example: 1-12-01
How many three digit numbers are divisible by 6?
A. 196
B. 149
C. 150
D. 151
Correct Answer: C
Solved Example: 1-12-02
The common difference of AP will be _______ with \[a_{25} - a_{12} = - 52\]
A. -14
B. -4
C. -3
D. -5
Correct Answer: B
Solved Example: 1-12-03
How many numbers between 300 and 1000 are divisible by 7?
A. 101
B. 301
C. 994
D. 100
Correct Answer: D
Solved Example: 1-12-04
For which value of k; the series 2, 3 + k and 6 are in A.P.?
A. 4
B. 3
C. 1
D. 2
Correct Answer: C
Solved Example: 1-12-05
If a, b, c are in arithmetic progression then:
A. 2a = b + c
B. 2c = a + b
C. 3b = 2a + 3c
D. 2b = a + c
Correct Answer: D
Solved Example: 1-12-06
What will be the 10th term of the arithmetic progression 2, 7, 12, _____?
A. 45
B. 43
C. 97
D. 47
Correct Answer: D
Series
Learning Objectives:
- Identify type of a given series.
- Identify the properties such as common difference or common ratio of a series.
- Find sum of first n terms a series.
A sequence is defined as the arrangement of numbers in a particular order. A series, on the other hand, is defined as the sum of the elements of a sequence.
Solved Example: 1-14-01
Use the summation notation to write the series 49 + 54 + 59 + ...for 14 terms.
A. $\displaystyle \sum_{n=1}^{14} (49 + 5n)$
B. $\displaystyle \sum_{n=1}^{14} (44 + 5n)$
C. $\displaystyle \sum_{n=1}^{13} (44 + 5n)$
D. $\displaystyle \sum_{n=1}^{44} (49 + 5n)$
Correct Answer: B
Solved Example: 1-14-02
What will be the sum of 3 + 7 + 11 + 15 + 19 + ... upto 80 terms?
A. 12880
B. 12400
C. 25760
D. 24800
Correct Answer: A