Sample Solution

MCH-014 Solved Assignment 2024

  1. State whether the following statement are TRUE or FALSE. Give reason in support of your answer.
    a) Derivative of 1 x 1 x (1)/(x)\frac{1}{x}1x with respect to x x xxx is 1 .
    b) If A A AAA is a matrix of order 2 by 3 and B B BBB is a matrix of order 3 by 2 , then order of the matrix A + B A + B A+B\mathrm{A}+\mathrm{B}A+B is 2 by 3 .
    c) If a = 3 i ^ 5 j ^ + 4 k ^ a = 3 i ^ 5 j ^ + 4 k ^ vec(a)=3 hat(i)-5 hat(j)+4 hat(k)\vec{a}=3 \hat{i}-5 \hat{j}+4 \hat{k}a=3i^5j^+4k^ and b = 4 i ^ + 4 j ^ + 2 k ^ b = 4 i ^ + 4 j ^ + 2 k ^ vec(b)=4 hat(i)+4 hat(j)+2 hat(k)\vec{b}=4 \hat{i}+4 \hat{j}+2 \hat{k}b=4i^+4j^+2k^ they are perpendicular to each other.
    d) If probability of an event E is 1 / 2 1 / 2 1//21 / 21/2 and probability of the event E F E F EnnF\mathrm{E} \cap \mathrm{F}EF is 1 / 6 1 / 6 1//61 / 61/6, then probability of the event F F FFF is 1 / 3 1 / 3 1//31 / 31/3, where events E E EEE and F F FFF are independent.
    e) If the first term of an AP is 5 and 101 term of the A P A P APA PAP is 1005 then common difference of the AP will be 105.
  2. Solve the following system of equations using Cramer’s rule.
x + 3 y + 2 z = 6 , x + 4 y + 5 z = 8 , 2 x + 5 y + 3 z = 10 x + 3 y + 2 z = 6 , x + 4 y + 5 z = 8 , 2 x + 5 y + 3 z = 10 x+3y+2z=6,-x+4y+5z=8,2x+5y+3z=10x+3 y+2 z=6,-x+4 y+5 z=8,2 x+5 y+3 z=10x+3y+2z=6,x+4y+5z=8,2x+5y+3z=10
  1. a) Prove that A = [ cos 2 α sin 2 α sin 2 α cos 2 α ] A = cos 2 α sin 2 α sin 2 α cos 2 α A=[[cos 2alpha,-sin 2alpha],[sin 2alpha,cos 2alpha]]A=\left[\begin{array}{cc}\cos 2 \alpha & -\sin 2 \alpha \\ \sin 2 \alpha & \cos 2 \alpha\end{array}\right]A=[cos2αsin2αsin2αcos2α] is an orthogonal matrix.
  2. a) Evaluate 5 5 ( x 3 + x ) d x 5 5 x 3 + x d x int_(-5)^(5)(x^(3)+x)dx\int_{-5}^5\left(x^3+x\right) d x55(x3+x)dx.
b) Evaluate ln x x 2 d x ln x x 2 d x int(ln x)/(x^(2))dx\int \frac{\ln x}{x^2} d xlnxx2dx
  1. a) Solve the differential equation sin x d y d x + y cos x = 1 sin x d y d x + y cos x = 1 sin x(dy)/(dx)+y cos x=1\sin x \frac{d y}{d x}+y \cos x=1sinxdydx+ycosx=1.
    b) If a = 2 i ^ + 3 j ^ + 4 k ^ a = 2 i ^ + 3 j ^ + 4 k ^ vec(a)=2 hat(i)+3 hat(j)+4 hat(k)\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}a=2i^+3j^+4k^ and b = 3 i ^ + 5 j ^ + 2 k ^ b = 3 i ^ + 5 j ^ + 2 k ^ vec(b)=3 hat(i)+5 hat(j)+2 hat(k)\vec{b}=3 \hat{i}+5 \hat{j}+2 \hat{k}b=3i^+5j^+2k^ then find a × b a × b vec(a)xx vec(b)\vec{a} \times \vec{b}a×b.
    c) In an iron determination (taking 1 g sample every time) the following four replicate results were obtained: 24.8 , 25.2 , 23.6 24.8 , 25.2 , 23.6 24.8,25.2,23.624.8,25.2,23.624.8,25.2,23.6 and 24.7 mg iron. Calculate the coefficient of variation and relative standard deviation in ppm of the given data.
  2. a) In a factory there are three machines A, B, C which produce 10 % 10 % 10%10 \%10%, 40 % 40 % 40%40 \%40% and 50 % 50 % 50%50 \%50% items respectively. Past experience shows that percentage of defective items produced by machines A, B, C are 5 % 5 % 5%5 \%5%, 4 % , 2 % 4 % , 2 % 4%,2%4 \%, 2 \%4%,2% respectively. An item from the production of these machines is selected at random and it is found defective. What is the probability that it is produced by machine A?
b) Assume that in a population each person is equally likely to have a particular disease and disease status of each individual is independent of each other, then find the probability that out of the 5 randomly selected individuals who are tested for this particular disease exactly 3 have this disease.
c) A hospital specialising in heart surgery. In 2023 total of 1000 patients were admitted for treatment. The average payment made by a patient was Rs 1 , 00 , 000 1 , 00 , 000 1,00,0001,00,0001,00,000 with a standard deviation of Rs 20000 . Under the assumption that payments follow a normal distribution, find the number of patients who paid between Rs 90,000 and Rs 1,10,000.

Expert Answer

MCH-014 Solved Assignment 2024

Question:-1

State whether the following statement are TRUE or FALSE. Give reason in support of your answer.

Question:-1(a)

Derivative of 1 x 1 x (1)/(x)\frac{1}{x}1x with respect to x x xxx is 1.

Answer:

The statement "The derivative of 1 x 1 x (1)/(x)\frac{1}{x}1x with respect to x x xxx is 1" is false.
To justify this, we can compute the derivative of 1 x 1 x (1)/(x)\frac{1}{x}1x using basic differentiation rules.
Given the function f ( x ) = 1 x f ( x ) = 1 x f(x)=(1)/(x)f(x) = \frac{1}{x}f(x)=1x, we apply the power rule of differentiation. Rewriting f ( x ) f ( x ) f(x)f(x)f(x) in terms of exponents, we have:
f ( x ) = x 1 f ( x ) = x 1 f(x)=x^(-1)f(x) = x^{-1}f(x)=x1
The power rule states that the derivative of x n x n x^(n)x^nxn is n x n 1 n x n 1 nx^(n-1)nx^{n-1}nxn1. Applying this rule:
f ( x ) = d d x ( x 1 ) = 1 x 2 = 1 x 2 f ( x ) = d d x ( x 1 ) = 1 x 2 = 1 x 2 f^(‘)(x)=(d)/(dx)(x^(-1))=-1*x^(-2)=-(1)/(x^(2))f'(x) = \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2} = -\frac{1}{x^2}f(x)=ddx(x1)=1x2=1x2
Therefore, the derivative of 1 x 1 x (1)/(x)\frac{1}{x}1x with respect to x x xxx is 1 x 2 1 x 2 -(1)/(x^(2))-\frac{1}{x^2}1x2, not 1.

Question:-1(b)

If A A AAA is a matrix of order 2 by 3 and B B BBB is a matrix of order 3 by 2, then order of the matrix A + B A + B A+B\mathrm{A}+\mathrm{B}A+B is 2 by 3.

Answer:

The statement "If A A AAA is a matrix of order 2 by 3 and B B BBB is a matrix of order 3 by 2, then the order of the matrix A + B A + B A+BA + BA+B is 2 by 3" is false.
To justify this, we need to understand the rules for matrix addition. Two matrices can only be added if they have the same dimensions.
  • Matrix A A AAA has dimensions 2 by 3.
  • Matrix B B BBB has dimensions 3 by 2.
Since A A AAA and B B BBB do not have the same dimensions, their addition A + B A + B A+BA + BA+B is not defined. Therefore, it is impossible to form the matrix A + B A + B A+BA + BA+B, and thus the statement about the order of A + B A + B A+BA + BA+B is meaningless.

Question:-1(c)

If a = 3 i ^ 5 j ^ + 4 k ^ a = 3 i ^ 5 j ^ + 4 k ^ vec(a)=3 hat(i)-5 hat(j)+4 hat(k)\vec{a}=3 \hat{i}-5 \hat{j}+4 \hat{k}a=3i^5j^+4k^ and b = 4 i ^ + 4 j ^ + 2 k ^ b = 4 i ^ + 4 j ^ + 2 k ^ vec(b)=4 hat(i)+4 hat(j)+2 hat(k)\vec{b}=4 \hat{i}+4 \hat{j}+2 \hat{k}b=4i^+4j^+2k^ they are perpendicular to each other.

Answer:

The statement "If a = 3 i ^ 5 j ^ + 4 k ^ a = 3 i ^ 5 j ^ + 4 k ^ vec(a)=3 hat(i)-5 hat(j)+4 hat(k)\vec{a}=3 \hat{i}-5 \hat{j}+4 \hat{k}a=3i^5j^+4k^ and b = 4 i ^ + 4 j ^ + 2 k ^ b = 4 i ^ + 4 j ^ + 2 k ^ vec(b)=4 hat(i)+4 hat(j)+2 hat(k)\vec{b}=4 \hat{i}+4 \hat{j}+2 \hat{k}b=4i^+4j^+2k^, they are perpendicular to each other" is false.
To justify this, we need to determine whether the vectors a a vec(a)\vec{a}a and b b vec(b)\vec{b}b are perpendicular by checking if their dot product is zero.
The dot product of a a vec(a)\vec{a}a and b b vec(b)\vec{b}b is calculated as follows:
a b = ( 3 i ^ 5 j ^ + 4 k ^ ) ( 4 i ^ + 4 j ^ + 2 k ^ ) a b = ( 3 i ^ 5 j ^ + 4 k ^ ) ( 4 i ^ + 4 j ^ + 2 k ^ ) vec(a)* vec(b)=(3 hat(i)-5 hat(j)+4 hat(k))*(4 hat(i)+4 hat(j)+2 hat(k))\vec{a} \cdot \vec{b} = (3 \hat{i} – 5 \hat{j} + 4 \hat{k}) \cdot (4 \hat{i} + 4 \hat{j} + 2 \hat{k})ab=(3i^5j^+4k^)(4i^+4j^+2k^)
Using the distributive property of the dot product:
a b = ( 3 4 ) + ( 5 4 ) + ( 4 2 ) a b = ( 3 4 ) + ( 5 4 ) + ( 4 2 ) vec(a)* vec(b)=(3*4)+(-5*4)+(4*2)\vec{a} \cdot \vec{b} = (3 \cdot 4) + (-5 \cdot 4) + (4 \cdot 2)ab=(34)+(54)+(42)
Simplifying each term:
a b = 12 20 + 8 a b = 12 20 + 8 vec(a)* vec(b)=12-20+8\vec{a} \cdot \vec{b} = 12 – 20 + 8ab=1220+8
Adding these values together:
a b = 12 20 + 8 = 0 a b = 12 20 + 8 = 0 vec(a)* vec(b)=12-20+8=0\vec{a} \cdot \vec{b} = 12 – 20 + 8 = 0ab=1220+8=0
Since the dot product a b a b vec(a)* vec(b)\vec{a} \cdot \vec{b}ab is indeed zero, the vectors a a vec(a)\vec{a}a and b b vec(b)\vec{b}b are perpendicular to each other. Therefore, the statement is true.

Question:-1(d)

If probability of an event E is 1 / 2 1 / 2 1//21 / 21/2 and probability of the event E F E F EnnF\mathrm{E} \cap \mathrm{F}EF is 1 / 6 1 / 6 1//61 / 61/6, then probability of the event F F FFF is 1 / 3 1 / 3 1//31 / 31/3, where events E E EEE and F F FFF are independent.

Answer:

The statement "If the probability of an event E E EEE is 1 2 1 2 (1)/(2)\frac{1}{2}12 and the probability of the event E F E F E nn FE \cap FEF is 1 6 1 6 (1)/(6)\frac{1}{6}16, then the probability of the event F F FFF is 1 3 1 3 (1)/(3)\frac{1}{3}13, where events E E EEE and F F FFF are independent" is true.
To justify this, we use the definition of independent events. For two independent events E E EEE and F F FFF, the probability of their intersection is given by the product of their individual probabilities:
P ( E F ) = P ( E ) P ( F ) P ( E F ) = P ( E ) P ( F ) P(E nn F)=P(E)*P(F)P(E \cap F) = P(E) \cdot P(F)P(EF)=P(E)P(F)
Given:
P ( E ) = 1 2 P ( E ) = 1 2 P(E)=(1)/(2)P(E) = \frac{1}{2}P(E)=12
P ( E F ) = 1 6 P ( E F ) = 1 6 P(E nn F)=(1)/(6)P(E \cap F) = \frac{1}{6}P(EF)=16
We need to find P ( F ) P ( F ) P(F)P(F)P(F). Using the formula for independent events:
P ( E F ) = P ( E ) P ( F ) P ( E F ) = P ( E ) P ( F ) P(E nn F)=P(E)*P(F)P(E \cap F) = P(E) \cdot P(F)P(EF)=P(E)P(F)
1 6 = 1 2 P ( F ) 1 6 = 1 2 P ( F ) (1)/(6)=(1)/(2)*P(F)\frac{1}{6} = \frac{1}{2} \cdot P(F)16=12P(F)
Solving for P ( F ) P ( F ) P(F)P(F)P(F):
P ( F ) = 1 6 1 2 = 1 6 × 2 1 = 2 6 = 1 3 P ( F ) = 1 6 1 2 = 1 6 × 2 1 = 2 6 = 1 3 P(F)=((1)/(6))/((1)/(2))=(1)/(6)xx(2)/(1)=(2)/(6)=(1)/(3)P(F) = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{6} \times \frac{2}{1} = \frac{2}{6} = \frac{1}{3}P(F)=1612=16×21=26=13
Therefore, the probability of the event F F FFF is 1 3 1 3 (1)/(3)\frac{1}{3}13, which confirms that the statement is true.

Question:-1(e)

If the first term of an AP is 5 and 101 term of the A P A P APA PAP is 1005 then common difference of the AP will be 105.

Answer:

The statement "If the first term of an AP is 5 and the 101st term of the AP is 1005, then the common difference of the AP will be 105" is false.
To justify this, let’s use the formula for the n n nnn-th term of an arithmetic progression (AP). The n n nnn-th term of an AP is given by:
a n = a + ( n 1 ) d a n = a + ( n 1 ) d a_(n)=a+(n-1)da_n = a + (n-1)dan=a+(n1)d
where a a aaa is the first term, d d ddd is the common difference, and n n nnn is the term number.
Given:
  • The first term a = 5 a = 5 a=5a = 5a=5
  • The 101st term a 101 = 1005 a 101 = 1005 a_(101)=1005a_{101} = 1005a101=1005
We need to find the common difference d d ddd.
Using the formula for the 101st term:
a 101 = a + 100 d a 101 = a + 100 d a_(101)=a+100 da_{101} = a + 100da101=a+100d
Substitute the given values:
1005 = 5 + 100 d 1005 = 5 + 100 d 1005=5+100 d1005 = 5 + 100d1005=5+100d
Solving for d d ddd:
1005 5 = 100 d 1005 5 = 100 d 1005-5=100 d1005 – 5 = 100d10055=100d
1000 = 100 d 1000 = 100 d 1000=100 d1000 = 100d1000=100d
d = 1000 100 d = 1000 100 d=(1000)/(100)d = \frac{1000}{100}d=1000100
d = 10 d = 10 d=10d = 10d=10
The common difference d d ddd is 10, not 105. Therefore, the statement is false.

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