Category: Engineering Mathematics Objective Questions

250+ TOP MCQs on Fourier Half Range Series and Answers

Fourier Analysis Multiple Choice Questions on “Fourier Half Range Series”.

1. In half range Fourier series expansion, we know the nature of the function in its full time period.
a) True
b) False
View Answer

Answer: b
Explanation: In half range Fourier series expansion, we know the nature of the function only in its half of the period. It can be either odd or even function in its full range. We assume it to be even function when we find half range cosine series and we assume it to be odd function when we find half range sine series.

2. In half range cosine Fourier series, we assume the function to be _________
a) Odd function
b) Even function
c) Can’t be determined
d) Can be anything
View Answer

Answer: b
Explanation: In half range Fourier series expansion, the nature of the function in half of its period is only known. So when we find half range cosine series, there are only cosine terms which imply that the function is even function. f(x) = f(-x).

3. Find the half range sine series of the function f(x) = x, when 0<x<(frac{pi}{2} ) and (π-x) when (frac{pi}{2} )<x< π.
a) (frac{8}{pi}[frac{sinx}{1^{2}} – sinfrac{(3x)}{3^{2}} + sin frac{(5x)}{5^2} – sin frac{(7x)}{7^{2}} +……] )
b) (frac{4}{pi}[frac{sinx}{1^{2}} + sinfrac{(3x)}{3^{2}} + sin frac{(5x)}{5^2} + sin frac{(7x)}{7^{2}} +……] )
c) (frac{8}{pi}[frac{sinx}{1^{2}} + sinfrac{(3x)}{3^{2}} + sin frac{(5x)}{5^2} + sin frac{(7x)}{7^{2}} +……] )
d) (frac{4}{pi}[frac{sinx}{1^{2}} – sinfrac{(3x)}{3^{2}} + sin frac{(5x)}{5^2} – sin frac{(7x)}{7^{2}} +……] )
View Answer

Answer: d
Explanation: bn = (frac{2}{pi}[int_{0}^{pi⁄2}xsinnxdx+int_{pi⁄2}^{pi}(pi-x)sinnxdx)]
= (frac{4}{pi}frac{sinleft(nfrac{pi}{2}right)}{n^2} ) [this term is zero whenever n is even and when odd, it gives 1 or -1]
= (frac{4}{pi}[frac{sinx}{1^{2}} – sinfrac{(3x)}{3^{2}} + sin frac{(5x)}{5^2} – sin frac{(7x)}{7^{2}} +……]. )

4. Find bn when we have to find the half range sine series of the function x2 in the interval 0 to 3.
a) -18 ( frac{cos(nπ)}{nπ} )
b) 18 ( frac{cos(nπ)}{nπ} )
c) -18 ( frac{cos(n pi⁄2)}{nπ} )
d) 18 ( frac{cos(n pi⁄2)}{nπ} )
View Answer

Answer: a
Explanation: bn = (frac{2}{3} int_0^3 x^{2} sinleft(npifrac{x}{3}right)dx )
= (frac{2}{3} left(-27frac{cos(nπ)}{nπ}right) )
= ( -18frac{cos(nπ)}{nπ}.)

5. What is the formula for Parseval’s relation in Fourier series expansion?
a) ( int_{-l}^l (f(x))^2 dx=l[frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2 ) ] )
b) ( int_{-l}^l (f(x))^2 dx=l[frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2 ) ] )
c) ( int_{-l}^l (f(x))^2 dx=l⁄2 [frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2 ) ] )
d) ( lint_{-l}^l (f(x))^2 dx=[frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2 ) ] )
View Answer

Answer: a
Explanation: The real life significance of Parseval’s relation is to find the energy of the signal in its time period (1 time period). This can be found using the Fourier series coefficients. (int_{-l}^l (f(x))^2 dx=l[frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2 ) ] ) is the realtion between the function and the Fourier series coefficients.

6. In Parseval’s relation of Half range Fourier cosine series expansion, which of the following terms doesn’t appear?
a) a0
b) an
c) bn
d) all terms appear
View Answer

Answer: c
Explanation: In the expansion of a function in half range Fourier cosine series, only an and a0 appear. As the function is considered to be an even function, bn term becomes a null value as the integral becomes zero. So, only bn doesn’t appear in the parseval’s relation of half range Fourier cosine series.

7. Find the value of (frac{1}{1^2} +frac{1}{3^2} +frac{1}{5^2} +frac{1}{7^2} ) +….when finding the Half range Fourier sine series of the function f(x) = 1 in 0<x<π.
a) (frac{pi^2}{4} )
b) (frac{pi^2}{8} )
c) (frac{pi^2}{2} )
d) (3frac{pi^2}{8} )
View Answer

Answer: b
Explanation: Using parseval’s relation for half range Fourier sine series,
(int_{-l}^l(f(x))^2 dx=l[frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2 ) ] )
L.H.S. = (int_0^pi(1)^2 dx = pi )
bn = (frac{2}{pi} int_0^π1.sin(nx)dx )
= (frac{2}{pi n} (1-(-1)^n ) )
Using parseval’s formula,
(pi = frac{pi}{2} * frac{4}{pi^2} * 2 (frac{1}{1^2} +frac{1}{3^2} +frac{1}{5^2} +frac{1}{7^2} +….) )
Therefore,
(frac{1}{1^2} +frac{1}{3^2} +frac{1}{5^2} +frac{1}{7^2} +….= frac{pi^2}{8}.)

8. Find the value of( frac{1}{1^4} +frac{1}{3^4} +frac{1}{5^4} +frac{1}{7^4} ) +….by finding the half range Fourier cosine series of the function f(x) = x in the interval 0<x<l.
a) (frac{pi^4}{12} )
b) (frac{pi^4}{48} )
c) (frac{pi^4}{24} )
d) (frac{pi^4}{96} )
View Answer

Answer: d
Explanation: Using Parseval’s relation,
(int_{-l}^l(f(x))^2 dx=l[frac{a_0^2}{2}+sum_{n=1}^{∞}(a_n^2+b_n^2 ) ] )
L.H.S. = (int_0^l x^2 dx= frac{l^3}{3} )
a0 = (frac{2}{l} int_0^l xdx= l )
an = (frac{2}{l} int_0^lxcos(frac{nπx}{l})dx )
= (frac{-4l}{pi^2} (frac{1}{1^2} +frac{1}{3^2} +frac{1}{5^2} +…) )
R.H.S. = (frac{l}{2} (frac{l^2}{2}+left(-4frac{l}{pi}right)^2 (frac{1}{1^4} +frac{1}{3^4} +frac{1}{5^4} +…..)) )
Therefore,
(frac{1}{1^4} +frac{1}{3^4} +frac{1}{5^4} +frac{1}{7^4} +….= (frac{l^3}{3}-frac{l^3}{4})left(2frac{pi^4}{(16l^3 )}right) )
= (frac{pi^4}{96}.)

9. In Parseval’s formula for half range Fourier series, the formula contains l/2 multiplied with the square of individual coefficients.
a) True
b) False
View Answer

Answer: a
Explanation: The Parseval’s formula for half range Fourier cosine series is (int_0^l(f(x))^2 dx=frac{l}{2} [frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2)] )
The Parseval’s formula for half range Fourier sine series is (int_0^l(f(x))^2 dx=frac{l}{2} ∑_{n=1}^∞(b_n^2) .)

10. What is the value of a0 if the function is f(x) = x3 in the interval 0 to 5?
a) 25/4
b) 125/4
c) 625/4
d) 5/4
View Answer

Answer: c
Explanation: (int_0^5 x^3 dx= int_{-l}^l(f(x))^2 dx=l[frac{a_0^2}{2}+∑_{n=1}^∞(a_n^2+b_n^2)] ) (from 0 to 5)
= (frac{5^4}{4}= frac{625}{4}.)

Global Education & Learning Series – Fourier Analysis.

To practice all areas of Fourier Analysis, here is complete set of 1000+ Multiple Choice Questions and Answers.

 

250+ TOP MCQs on Logarithm of Complex Numbers and Answers

Complex Analysis Questions and Answers for Campus interviews focuses on “Logarithm of Complex Numbers”.

1. Find the value of log⁡(-6).
a) log6+2iπ
b) log⁡36+iπ
c) log6+2iπ
d) log6+iπ
Answer: d
Explanation: We know that
(log⁡(x-iy)=frac{1}{2} log⁡(x^2+y^2)+itan^{-1} (frac{y}{x}))
Putting x=-6 and y=0.
(log⁡(-6)=frac{1}{2} log⁡(36)+itan^{-1} (frac{0}{-6}))
(log⁡(-6)=log6+iπ).

2. Find the value of log2(-3).
a) (frac{log⁡_3+i8pi}{log_2})
b) (frac{log⁡_3+3ipi}{log_2})
c) (frac{log⁡_3+ipi}{log_2})
d) (frac{log_⁡2+ipi}{log_3})
Answer: c
Explanation: In this problem, we change the base to e
(log_2(-3)=frac{log_e(-3)}{loge(2)} )
(log_2(-3)=frac{log⁡_3+ipi}{log_2}).

3. Represent ii in terms of e.
a) (e^{frac{-pi}{3}})
b) (e^{frac{-3pi}{2}})
c) (e^{frac{-pi}{2}})
d) (e^{frac{-pi}{6}})
Answer: c
Explanation: We know that
(a^x=e^{x loga})
(i^i=e^{i log⁡i})
We also know from the definition of logarithm,
(log⁡i=frac{ipi}{2})
(i^i=e^{i(frac{ipi}{2})}=e^{frac{-pi}{2}}).

Global Education & Learning Series – Complex Analysis.

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250+ TOP MCQs on The nth Derivative of Some Elementary Functions and Answers

Engineering Mathematics Multiple Choice Questions on “The nth Derivative of Some Elementary Functions – 1”.

1. The pth derivative of a qth degree monic polynomial, where p, q are positive integers and 2p4 + 3pq3⁄2 = 3q3⁄2 + 2qp3 is given by?
a) Cannot be generally determined
b) (q – 1)!
c) (q)!
d) (q – 1)! * pq
Answer: c
Explanation: First consider the equation
2p4 + 3pq3⁄2 = 3q3⁄2 + 2qp3
After simplification, we get
(2p3 + 3q1⁄2) (p – q) = 0
This gives us two possibilities
2p3 = – 3q1⁄2
OR
p = q
The first possibility can’t be true as we are dealing with positive integers
Hence, we get
p = q
Thus the pth derivative of any monic polynomial of degree p(p = q) is
p! =q!.

2. The first and second derivatives of a quadratic Polynomial at x = 1 are 1 and 2 respectively. Then the value of f(1) – f(0) Is given by?
a) 3⁄2
b) 1⁄2
c) 1
d) 0
Answer: d
Explanation: Let the quadratic polynomial be
f(x) = ax2 + bx + c
The first derivative at x = 1 is given by
2a + b = 1
Now consider the second derivative at x = 1 which is given by
2a = 2
Solving for the coefficients using equations, we get the values as a = 1 and b = -1
Putting these values back in the polynomial yields
f(x) = x2 – x + c
Now the required value can be computed as
f(1) – f(0) = (12 – 1 + c) – (02 – 0 + c)
= (0 + c) – (0 + c) = 0.

3. Let f(x) = (frac{sin(x)}{x – 54}), then the value of f(100)(54) is given by?
a) Undefined
b) 100
c) 10
d) 0
Answer: a
Explanation: The key here is to expand the numerator into a taylor series centered at 54
Doing this gives us the following
Sin(x)=Sin(54)+(frac{(x-54)times Cos(54)}{1!}-frac{(x-54)^2times Sin(54)}{2!}…infty )
Now the function transforms into f(x)=(frac{Sin(54)}{(x-54)}+frac{Cos(54)}{1!}-frac{(x-54)times Sin(54)}{2!}…infty )
Observe the first term in the infinite series there is always (x – 54) in the denominator which goes to 0 when we substitute x = 54
Every derivative also will have this term
Hence any derivative of the given function goes to ∞ as x = 54
Hence, the answer is Undefined.

4. Let f(x) = ln(x2 + 5x + 6) then the value of f(30)(1) is given by?
a) (29!)((frac{1}{3^{30}}+frac{1}{4^{30}}))
b) (-29!)((frac{1}{3^{30}}+frac{1}{4^{30}}))
c) (30!)((frac{1}{3^{30}}+frac{1}{4^{30}}))
d) (-30!)((frac{1}{3^{30}}+frac{1}{4^{30}}))
Answer: b
Explanation: Given function
f(x) = ln(x2 + 5x + 6)
Factorising the inner polynomial we get
f(x) = ln((x + 3) (x + 2))
Now using the rule of logarithms ln (m * n) = ln(m) + ln(n)
we get
f(x) = ln(x + 3) + ln(x + 2)
Now using the nth derivative of logarithmic function (frac{d^n(ln(x+a))}{dx^n} = frac{(-1)^{n+1}times(n-1)!}{(x+a)^n})
We have
(frac{d^n(f(x))}{dx^n}=frac{d^n(ln(x+3))}{dx^n}+frac{d^n(ln(x+2))}{dx^n})
Which simplifies to (frac{d^n(f(x))}{dx^n}=frac{(-1)^{n+1}times(n-1)!}{(x+3)^n}+frac{(-1)^{n+1}times(n-1)!}{(x+2)^n})
Substituting x=1 and n=30
Gives us the answer as (-29!)((frac{1}{3^{30}}+frac{1}{4^{30}}))

5. f(x) = (int_{0}^{pi/2}sin(ax)da) then the value of f(100)(0) is?
a) a(100) sin(a)
b) -a(100) sin(a)
c) a(100) cos(a)
d) 0
Answer: d
Explanation: First solve the integral f(x)=(int_{0}^{pi/2}sin(ax)da)
Which gives =([frac{-cos(ax)}{x}]_{0}^{pi/2}=frac{1-cos(frac{pi x}{2})}{x})
Now expanding into Taylor series yields
(=(frac{(frac{pi x}{2})^2}{x2!}-frac{(frac{pi x}{2})^4}{x4!}….infty))
Observe that every term in the expansion is odd powered
Hence even derivative at x = 0 has to be 0.

6. Let f(x) = x9 ex then the ninth derivative of f(x) at x = 0 is given by?
a) 9!
b) 9! * e9
c) 10!
d) 21!
Answer: a
Explanation: The key here is to expand ex as a mclaurin series and then multiply it by x9
We have ex=(frac{x^0}{0!}+frac{x^1}{1!}+frac{x^2}{2!}…infty)
Now multiplying it by x9 we get
f(x)=(frac{x^9}{0!}+frac{x^{10}}{1!}+frac{x^{11}}{2!}…infty)
This simplifies into polynomial differentiation and the ninth derivative is as follows
f(9)(x)=(9!+frac{10!times x}{1!}+frac{p_9^{11}times x^2}{2!}…infty)
Substituting x = 0 gives us the result
f(9)(0) = 9!.

7. The following moves are performed on g(x).
(i) Pick (x0, y0) on g(x) and travel toward the left/right to reach the y = x line. Now travel above/below to reach g(x). Call this point on g(x) as (x1, y1)
(ii) Let the new position of (x0, y0)be (x0, y1)
This is performed for all points on g(x) and a new function is got. Again these steps may be repeated on new function and another function is obtained. It is observed that, of all the functions got, at a certain point (i.e. after finite number of moves) the nth derivatives of the intermediate function are constant, and the curve passes through the origin. Then which of the following functions could be g(x)
a) y = (sqrt{1 – x^2})
b) xy3⁄2 + y = constant
c) x9 y3⁄2 + y6 x3⁄2 = constant
d) x7 y 8 + 4y = constant
Answer: d
Explanation: The phrasing is a bit convoluted. It is asking us to convert a given function f(x) into its own composition f(f(x))
Now the set of finite number of moves asks us to find the composition of composition i.e. f(f(x)) and f(f(x)) and so on.
It also says that the intermediate composition function has the property of all nth derivatives being constant. This is possible only if the intermediate function is of the form k(x) = ax + b (Linear Function)

Given another condition that it has to pass through origin leads to the conclusion that b = 0
So the intermediate function has the form
k(x) = ax
The first possibility gives rise to
k(x) = x
Now one has to select from the options as to which curve would give k(x) = f(f(….(x)…)) = x when composed with itself a finite number of times.
This can be done by simply interchanging the position of y and x
in the options and check whether it preserves its structure.
For y = (sqrt{1 – x^2}) we have after changing the position
This has the same structure.

8. The first, second and third derivatives of a cubic polynomial f(x) at x = 1, are 13, 23 and 33 respectively. Then the value of f(0) + f(1) – 2f(-1) is?
a) 76
b) 86
c) 126
d) 41.5
Answer: d
Explanation: Assume the polynomial to be of the form f(x) = ax3 + bx2 + cx + d
Now the first derivative at x = 1 yields the following equation
13 = 1 = 3a + 2b + c
The second derivative at x = 1 yields the following expression
23 = 8 = 6a + 2b
The third derivative at x = 1 yields the following equation
33 = 27 = 6a
Solving for a, b and c simultaneously yields
(a, b, c) = (9⁄2, -19⁄2, 13⁄2)
Hence the assumed polynomial is f(x) = 9x3 – 19x2 + 13x ⁄ 2 + d
Now the given expression can be evaluated as
f(0) + f(1) – 2f(-1) = (d) + (3⁄2 + d) – 2(-20 + d)
= 40 + 3⁄2
= 41.5.

9. Let g(x) = ln(x) ⁄ x – 1 Then the hundredth derivative at x = 1 is?
a) 100!⁄101
b) 99!⁄101
c) 101⁄100!
d) 1⁄99!
Answer: a
Explanation: The key here is to again expand the numerator as a Taylor series centered at x = 1
Hence we have the Taylor series as
ln(x)=(frac{(x-1)}{1}-frac{(x-1)^2}{2}+frac{(x-1)^3}{3}…..infty)
Hence our function g(x) transforms into
g(x)=(1-frac{(x-1)}{2}-frac{(x-1)^2}{3}+frac{(x-1)^3}{4}…..infty)
This now simplifies into polynomial differentiation and the hundredth derivative can be written as
g(100)(x)=(frac{100!}{101}-frac{101!times(x-1)}{102}…..infty)
Substituting x = 1 yields
100!⁄101.

10. Let f(x) = ln(x3 – 3x2 – 16x – 12) , then the 1729th derivative at x = 234 is?
a) (1728!)(times(frac{1}{228^{1729}}+frac{1}{236^{1729}}+frac{1}{235^{1729}}))
b) (-1728!)(times(frac{1}{228^{1729}}+frac{1}{236^{1729}}+frac{1}{235^{1729}}))
c) (1728!)(times(frac{1}{228^{1729}}+frac{1}{236^{1728}}+frac{1}{235^{1729}}))
d) (-1729!)(times(frac{1}{228^{1729}}+frac{1}{236^{1729}}+frac{1}{235^{1729}}))
Answer: a
Explanation: The function can be written as f(x) = ln((x – 6)(x + 1)(x + 2))
Using property of logarithms we have
f(x) = ln(x + 1) + ln(x + 2) + ln(x – 6)
Using the nth derivative of logarithmic functions
(frac{d^n(ln(x+a))}{dx^n}=frac{(-1)^{n+1}times(n-1)!}{(x+a)^n})
We have the 1729th derivative of f(x) as
(f^{(1729)} (x) = frac{1728!}{(x-6)^{1729}} + frac{1728!}{(x+2)^{1729}} + frac{1728!}{(x+1)^{1729}})
Now substituting x=234 we get our answer as
= (1728!)(times(frac{1}{228^{1729}}+frac{1}{236^{1729}}+frac{1}{235^{1729}})).

11. Find the value of S=(sum_{n=1}^infty frac{(-1)^{n+1} times (2n-1)^3}{(2n-1)!}) using nth derivatives.
a) – 2 * sin(1)
b) 3 * sin(1)
c) 3 * cos(1)
d) – 3 * cos(1)
Answer: a
Explanation: We have to consider the function f(x) = sin(ex) in order to get the series in some way.
Expanding the given function into a Taylor series we have
f(x)=(frac{e^x}{1!}-frac{e^{3x}}{3!}+frac{e^{5x}}{5!}…infty )
Now observe that our series in question doesn’t have the exponential function, this gives us the hint that some derivative of this function has to be taken at x = 0
Observe that the term (2n – 1)3 has exponent equal to 3
Hence we have to take the third derivative of the function to get the required series
Now taking the third derivative yields
f(x)=(frac{e^x}{1!}-frac{3^3e^{3x}}{3!}+frac{5^5e^{5x}}{5!}…infty )
Now substituting x=0 we get
(frac{1^3}{1!}-frac{3^3}{3!}+frac{5^3}{5!}…infty=sum_{n=1}^infty frac{(-1)^{n+1} times (2n-1)^3}{(2n-1)!} )
To find the value of this series we need to take the third derivative of original function at the required point, this is as follows
f(3)(x) = -2exsin(ex)
Substituting x = 0 we get
f(3)(0) = -2sin(1).

12. Let f(x)=(frac{ln(1-x)}{e^x}). Find the third derivative at x = 0.
a) 4
b) 1⁄3
c) Undefined
d) 1⁄4
Answer: b
Explanation: Again the key here is to expand the given function into appropriate Taylor series.
Rewriting the function as f(x) = e-x(ln(1 – x)) and then expanding into Taylor series we have
(f(x)=(1-frac{x}{1!}+frac{x^2}{2!}-frac{x^3}{3!}…infty) times (frac{x}{1}+frac{x^2}{2}+frac{x^3}{3}…infty))
Now the question asks us to find the third derivative at x = 0. Thus, it is enough for us to find the coefficient of the x3 term in the infinite polynomial product above
The third degree terms can be grouped apart as follows
= x3⁄3 – x3⁄2 + x3⁄2
Hence the third derivative at x = 0 is simply the coefficient of the third degree term, which is
coefficient(x3⁄3) = 1⁄3.

250+ TOP MCQs on Indeterminate Forms and Answers

Engineering Mathematics Multiple Choice Questions on “Indeterminate Forms – 2”.

1. Find (lt_{xrightarrow -2}frac{sin(frac{1+(frac{(x+2)^2(x^2+1)}{x^3+3})}{x+2})}{(x+2)})
a) ∞
b) 0
c) 2
d) -∞
Answer: c
Explanation: First evaluate
(=lt_{xrightarrow -2}frac{ln(1+frac{(x+2)^2(x^2+1)}{x^3+3})}{x+2})
(=lt_{xrightarrow -2}(frac{1}{x+2})times(frac{(x+2)^2(x^2+1)}{(x^3+3)}-frac{(x+2)^3(x^3+1)^2}{2.(x^3+3)^2}+…infty))
(=lt_{xrightarrow -2}times(frac{(x+2)^2(x^2+1)}{(x^3+3)}-frac{(x+2)^3(x^3+1)^2}{2.(x^3+3)^2}+…infty))
We hence the form for the totle limit as
(lt_{xrightarrow a}frac{sin(f(x))}{g(x)})=1
Where f(x)→0 : g(x)→0 as x→a
This is true for the above problem
Thus, we can deduce the limit as
= 1
Hence, 2 is the right answer.

2. Find (lt_{xrightarrow 0}frac{(3e^x-2e^{2x}-e^{3x})}{(e^x+e^{2x}-2e^{3x})})
a) 3⁄2
b) 0
c) 4⁄3
d) –4⁄3
Answer: c
Explanation: Form is 0 / 0
Applying L hospitals rule we have
(=lt_{xrightarrow 0}frac{3e^x-4e^{2x}-3e^{3x}}{e^x+2e^{2x}-6e^{3x}})
(=frac{3-4-3}{1+2-6})
(=frac{4}{3})

3. Find relation between a and b such that the following limit is got after a single application of L hospitals Rule (lt_{xrightarrow 0}frac{ae^x+be^{2x}}{be^x+ae^{2x}})
a) b⁄a = 2
b) a⁄b = 2
c) a = b
d) a = -b
Answer: d
Explanation: Given differentiation is applied once we get
ae0 + be0 = 0 = a + b (numerator → zero)
be0 + ae0 = 0 = a + b (denominator → zero)
Thus the relation between (a, b) and is
a + b = 0
OR
a = -b.

4. Find (lt_{xrightarrow 0}frac{2cos(2x)+3cos(5x)-5cos(19x)}{cos(4x)-cos(3x)})
a) -76
b) -6
c) -7
d) 0
Answer: a
Explanation: Form here 0⁄0
Applying L hospitals rule we have
(=lt_{xrightarrow 0}frac{4cos(2x)+15cos(5x)-95cos(19x)}{4cos(4x)-3cos(3x)})
(frac{4+15-95}{4-3}) = -76.

5. Find how many rounds of differentiation are required to have finite limit for (lt_{xrightarrow 0}frac{cos(ax)+cos(bx)-2cos(cx)}{cos(ax)+2cos(bx)-3cos(cx)}) given that a ≠ b ≠ c
a) 3
b) 0
c) 2
d) 4
Answer: c
Explanation: Applying L hospitals rule
(=lt_{xrightarrow 0}frac{a.cos(ax)+b.cos(bx)-2c.cos(cx)}{a.cos(ax)+2b.cos(bx)-3c.cos(cx)}=lt_{xrightarrow 0}frac{a+b-2c}{a+2b-3c})
Assume now that
a + b + 2c = 0 and a + 2b – 3c = 0
We must have
a = c = b but given a ≠ c ≠ b
Thus, our assumption is false and a finite limit exists after first round of differentiation.
Hence, 2 is the right answer.

6. Find (lt_{prightarrowinfty}frac{p^5.p!}{5.6…(5+p)})
a) 4!
b) 5!
c) 0
d) ∞
Answer: a
Explanation: (=lt_{prightarrowinfty}frac{1.2.3.4}{1.2.3.4}times frac{p^5.p!}{5.6…(5+p)})
(=lt_{prightarrowinfty}frac{4!.p^5.p!}{(p+5)!})
(=lt_{prightarrowinfty}frac{4!.p^5}{(p+5)(p+4)(p+3)(p+2)(p+1)})
(=lt_{prightarrowinfty}(4!)timesfrac{p^5}{(p+5)(p+4)(p+3)(p+2)(p+1)}=4!times(1))
= 4!

7. Find (=lt_{xrightarrow 0}frac{sin(x)}{tan(x)})
a) 0
b) 1
c) ∞
d) 2
Answer: b
Explanation: (=lt_{xrightarrow 0}frac{frac{sin(x)}{x}}{frac{tan(x)}{x}})
(=frac{lt_{xrightarrow 0}frac{sin(x)}{x}}{lt_{xrightarrow 0}frac{tan(x)}{x}})
=1/1=1

8. Find (lt_{xrightarrow 1012345}(frac{[sinh(x)]^2-[cosh(x)]^2}{[sinh(x)]^2+[cosh(x)]^2}))
a) 1⁄cosh(1012345)
b) 90987
c) 1012345
d) ∞
Answer: a
Explanation: (lt_{xrightarrow 1012345}left ( frac{1}{(frac{(e^x+e^{-2x})^2+(e^x-e^{-x})^2}{4})}right ))
(=lt_{xrightarrow 1012345}frac{1}{frac{(e^{2x}+e^{-2x})}{2}}=frac{1}{cosh(1012345)})

9. Let f on (f(x)) denote the composition of f(x) with itself n number of times then the value of ltn → ∞ f on (sin(x)) =
a) -1
b) 2
c) ∞
d) 0
Answer: d
Explanation: Drawing the graph of y = x and y = sin(x) we can write the limit value as 0.

10. Find (lt_{xrightarrow 0}frac{sin(x^2)}{x})
a) ∞
b) -1
c) 0
d) 22
Answer: c
Explanation: Expand into Taylor Series
(=lt_{xrightarrow 0}(frac{1}{x})times(frac{x^2}{1!}-frac{x^6}{3!}+..infty))
(=lt_{xrightarrow 0}times(frac{x}{1!}-frac{x^5}{3!}+..infty))
=0

11. Find (lt_{xrightarrow -33}frac{ln(x^3+68x^2+1222x+2179)-ln(x+1)}{(x^2+66x+1089)})
a) -33
b) 1⁄2
c) 0
d) 31⁄32
Answer: d
Explanation: (=lt_{xrightarrow -33}frac{ln(1+frac{(x+33)^2(x+2)}{(x+1)})}{(x+33)^2})
(=lt_{xrightarrow -33}(frac{1}{(x+33)^2})times(frac{(x+33)^2(x+2)}{(x+1)}-frac{(x+33)^4(x+2)^2}{2(x+1)^2}….infty))
(=lt_{xrightarrow -33}(frac{(x+2)}{(x+1)}-frac{(x+33)^2(x+2)^2}{2(x+1)^2}….infty))
(=lt_{xrightarrow -33}(frac{(x+2)}{(x+1)})=frac{(-33+2)}{(-33+1)})

12. Find (lt_{prightarrowinfty}frac{p^{frac{1}{2}}.p!}{frac{1}{2}.frac{3}{2}…(p+frac{1}{2})})
a) √π
b) ∞
c) √π⁄2
d) 0
Answer: c
Explanation: Using the Gauss definition of the Gamma function we have
(tau(x)=lt_{prightarrowinfty}frac{p^x.p!}{x.(x+1)…(x+p)})
Where τ(x) is the Gamma function Using formula
(tau(x) times tau(1-x)=frac{pi.x}{sin(pi.x)})
put x=1/2 to get
((tau(frac{1}{2}))^2=frac{pi}{2.sin(frac{pi}{2})}=frac{pi}{2})
(tau(frac{1}{2})=sqrt{frac{pi}{2}})

13. Find (lt_{nrightarrowinfty}(1+frac{1}{n})^n)
a) e
b) e – 1
c) 0
d) ∞
Answer: a
Explanation: (lt_{nrightarrowinfty}(1+frac{1}{n})^n=e^{lt_{nrightarrowinfty}frac{n}{n}})
(lt_{nrightarrowinfty}frac{n}{n}) = 1
= e

250+ TOP MCQs on Partial Differentiation and Answers

Engineering Mathematics Multiple Choice Questions on “Partial Differentiation – 1”.

1. f(x, y) = x2 + xyz + z Find fx at (1,1,1)
a) 0
b) 1
c) 3
d) -1
Answer: c
Explanation: fx = 2x + yz
Put (x,y,z) = (1,1,1)
fx = 2 + 1 = 3.

2. f(x, y) = sin(xy) + x2 ln(y) Find fyx at (0, π⁄2)
a) 33
b) 0
c) 3
d) 1
Answer: d
Explanation: fy = xcos(xy) + x2⁄y
fyx = cos(xy) – xysin(xy) + 2x⁄y
Put (x,y) = (0, π⁄2)
= 1.

3. f(x, y) = x2 + y3 ; X = t2 + t3; y = t3 + t9 Find df⁄dt at t=1.
a) 0
b) 1
c)-1
d) 164
Answer: d
Explanation: Using chain rule we have
(frac{df}{dt}=f_x.frac{dx}{dt}+f_y.frac{dy}{dt})
=(2x).(2t + 3t2) + (3y2).(3t2 + 9t8)
Put t = 1; we have x = 2; y = 2
=4.(5) + 12.(12) = 164.

4. f(x, y) = sin(x) + cos(y) + xy2; x = cos(t); y = sin(t) Find df⁄dt at t = π⁄2
a) 2
b)-2
c) 1
d) 0
Answer: b
Explanation:Using chain rule we have
(frac{df}{dt}=f_x.frac{dx}{dt}+f_y.frac{dy}{dt})
= (cos(x) + y2).(-sin(t)) + (-sin(y) + 2xy).(cos(t))
Put t= π⁄2; we have x=0; y=1
=(1 + 1).(-1) + 0 = -2.

5. f(x, y, z, t) = xy + zt + x2 yzt; x = k3 ; y = k2; z = k; t = √k
Find df⁄dt at k = 1
a) 34
b) 16
c) 32
d) 61
Answer: b
Explanation: Using Chain rule we have
(frac{df}{dt}=f_x.frac{dx}{dk}+f_y.frac{dy}{dk}+f_z.frac{dz}{dk}+f_t.frac{dt}{dk})
= (y + 2xyzt).(3k2) + (x + x2zt).(2k) + (t + x2yt).(1) + (z + x2yz).((frac{1}{2sqrt{k}})
Put k=1; we have x=y=z=t=1
9 + 4 + 2 + 1 = 16.

6. The existence of first order partial derivatives implies continuity.
a) True
b) False
Answer: b
Explanation: The mere existence cannot be declared as a condition for contnuity because the second order derivatives should also be continuous.

7. The gradient of a function is parallel to the velocity vector of the level curve.
a) True
b) False
Answer: b
Explanation: The gradient is perpendicular and not parallel to the velocity vector of the level curve.

8. f(x, y) = sin(y + yx2) / 1 + x2 Value of fxy at (0,1) is
a) 0
b) 1
c) 67
d) 90
Answer: a
Explanation: First find
fy = cos(y + yx2)
Hence
fyx = fxy = – (2xy).sin(y + yx2)
Now put (x,y) = (0,1)
= 0.

9. f(x, y) = sin(xy + x3y) / x + x3 Find fxy at (0,1).
a) 2
b) 5
c) 1
d) undefined
Answer: c
Explanation: First find
fy = sin(xy + x3y)
Hence
fyx = fxy = (cos(xy + x3y)) . (y + 3x23y)
Now put (x,y) = (0,1)
= 1.

250+ TOP MCQs on Rectification in Polar and Parametric Forms and Answers

Differential and Integral Calculus Multiple Choice Questions on “Rectification in Polar and Parametric Forms”.

1. Find the length of the curve given by the equation.
(x^{frac{2}{3}}+y^frac{2}{3}=a^frac{2}{3})
a) (frac{3a}{2})
b) (frac{-7a}{2})
c) (frac{-3a}{4})
d) (frac{-3a}{2})
View Answer

Answer: d
Explanation: We know that,
S=(int_{x1}^{x2}sqrt{1+frac{dy}{dx}^2})
(y^frac{2}{3}=a^frac{2}{3}-x^frac{2}{3})
Differentiating on both sides
(frac{2}{3} y^{frac{2}{3}-1}= frac{-2}{3} x^{frac{2}{3}-1})
(frac{dy}{dx} = -frac{y}{x}^{frac{1}{3}})
((frac{dy}{dx})^2 = (frac{y}{x})^{frac{1}{3}})
(1+(frac{dy}{dx})^2=1+(frac{y}{x})^frac{2}{3})
Substituting from the original equation-
(1+(frac{dy}{dx})^2=(frac{a}{x})frac{2}{3})
(sqrt{1+frac{dy^2}{dx}}=(frac{a}{x})^{frac{1}{3}})
(S=int_{a}^{0}(frac{a}{x})^{frac{1}{3}} dx )
(s=frac{-3a}{2})
Thus, length of the given curve is (frac{-3a}{2}).

2. Find the length of one arc of the given cycloid.

x=a(θ-sinθ)
y=a(1+cosθ)

a) a
b) 4a
c) 8a
d) 2a
View Answer

Answer: c
Explanation: We know that
(s=int_{theta1}^{theta2}sqrt{(frac{dx}{dtheta})^2+(frac{dy}{dtheta})^2})
(frac{dx}{dtheta}=a(1-costheta))
(frac{dy}{dtheta}=a(-sintheta))
((frac{dx}{dtheta})^2+(frac{dy}{dtheta})^2=a^2(1-costheta)^2+a^2 sin^2theta)
((frac{dx}{dtheta})^2+(frac{dy}{dtheta})^2=4a^2 sin^2frac{theta}{2})
(s=int_{0}^{2}pisqrt{4a^2 sin^2frac{theta}{2}} dtheta)
On solving the given integral, we get
s=8a
Thus length of one arc of the given cycloid is 8a.

Global Education & Learning Series – Differential and Integral Calculus.

To practice all areas of Differential and Integral Calculus, here is complete set of 1000+ Multiple Choice Questions and Answers.