300+ REAL TIME Class 12 Mathematics Objective Questions & Answers 2023

250+ TOP MCQs on Integrals of Some Particular Functions | Class 12 Maths

Mathematics Exam Questions for IIT JEE Exam on “Integrals of Some Particular Functions”.

1. Find (int frac{2 dx}{x^2-64}).
a) –(log⁡left |frac{x+8}{x-8}right |+C)
b) (frac{3}{2} log⁡left |frac{x+8}{x-8}right |+C)
c) (log⁡left |frac{x+8}{x-8}right |+C)
d) (frac{1}{8} log⁡left |frac{x-8}{x+8}right |+C)
Answer: d
Clarification: (int frac{2 dx}{x^2-64}=2int frac{dx}{x^2-8^2})
By using the formula (int frac{dx}{x^2-a^2}=frac{1}{2a} log⁡|frac{x-a}{x+a}|+C)
∴(2int frac{dx}{x^2-8^2}=2(frac{1}{(2(8))} log⁡|frac{x-8}{x+8}|)+2C_1)
=(frac{1}{8} log⁡|frac{x-8}{x+8}|+C)

2. Find (int frac{8 dx}{x^2-16}).
a) (log⁡left |frac{4+x}{4-x}right |+C)
b) –(log⁡left |frac{4+x}{4-x}right |+C)
c) (8 log⁡left |frac{4+x}{4-x}right |+C)
d) (frac{1}{8} log⁡left |frac{4+x}{4-x}right |+C)
Answer: a
Clarification: (int frac{8dx}{16-x^2}=8int frac{dx}{4^2-x^2})
By using the formula (int frac{dx}{a^2-x^2}=frac{1}{2a} left |frac{a+x}{a-x}right |+C)
∴(8int frac{dx}{4^2-x^2}=8(frac{1}{2(4)} log⁡left |frac{4+x}{4-x}right |)+8C_1)
(8int frac{dx}{4^2-x^2}=log⁡left |frac{4+x}{4-x}right |+C)

3. Find (int frac{3dx}{9+x^2}).
a) (tan^{-1}⁡frac{x}{2}+C)
b) (tan^{-1}⁡frac{x}{3}+C)
c) (tan^{-1}frac{x}{5}+C)
d) (tan^{-1}⁡frac{x}{4}+C)
Answer: b
Clarification: (int frac{3dx}{9+x^2}=3int frac{dx}{3^2+x^2})
Using the formula (int frac{dx}{a^2+x^2}=frac{1}{a} tan^{-1}frac{⁡x}{a}+C)
∴(3int frac{dx}{x^2+3^2}=3left (frac{1}{3} tan^{-1}⁡frac{x}{3}right )+3C_1)
(3int frac{dx}{x^2+3^2}=tan^{-1}⁡frac{x}{3}+C).

4. Find (int frac{10 ,dx}{sqrt{x^2-25}}).
a) –(log⁡|x+sqrt{x^2-25}|+C)
b) (log⁡|x+sqrt{x^2-25}|+C)
c) 10 ( log⁡|x+sqrt{x^2-25}|+C)
d) -10 (log⁡|x+sqrt{x^2-25}|+C)
Answer: c
Clarification: (int frac{10 ,dx}{sqrt{x^2-25}}=10int frac{dx}{sqrt{x^2-25}})
By using the formula (int frac{dx}{sqrt{x^2-a^2}}=log⁡|x+sqrt{x^2-a^2}|+C), we get
∴(10 int frac{dx}{sqrt{x^2-25}}=10 log⁡|x+sqrt{x^2-25}|+10C_1)
=10 (log⁡|x+sqrt{x^2-25}|+C)

5. Find (int frac{dx}{sqrt{5-x^2}}).
a) (sin^{-1}⁡frac{x}{sqrt{5}}+C)
b) (2 sin^{-1}⁡frac{x}{sqrt{5}}+C)
c) –(sin^{-1}⁡frac{x}{sqrt{5}}+C)
d) (sin^{-1}⁡frac{x}{5}+C)
Answer: a
Clarification: (int frac{dx}{sqrt{5-x^2}}=int frac{dx}{sqrt{(√5)^2-x^2}})
By using the formula (int frac{dx}{sqrt{a^2-x^2}}=sin^{-1}⁡frac{x}{a}+C)
∴(int frac{dx}{sqrt{x^2-5}}=sin^{-1}⁡frac{x}{sqrt{5}}+C)

6. Integrate (frac{dx}{sqrt{x^2+36}}).
a) –(log⁡|x^2+sqrt{x^2+36}|+C)
b) (log⁡|2x+sqrt{x^2+36}|+C)
c) –(log⁡|x^2+sqrt{x^2+6}|+C)
d) (log⁡|x^2+sqrt{x^2+36}|+C)
Answer: d
Clarification: (int frac{dx}{sqrt{x^2+36}})
By using the formula (int frac{dx}{sqrt{x^2+a^2}}=log⁡|x^2+sqrt{x^2+a^2}|+C)
∴(int frac{dx}{sqrt{x^2+36}}=log⁡|x^2+sqrt{x^2+36}|+C)

7. Find (int frac{dx}{x^2-8x+20}).
a) (frac{1}{2} tan^{-1}⁡frac{x^2-8x}{2}+C)
b) (frac{5}{2} tan^{-1}⁡frac{x-4}{2}+C)
c) (frac{1}{2} tan^{-1}⁡frac{x-4}{2}+C)
d) (x-frac{1}{2} tan^{-1}⁡frac{x-4}{2}+C)
Answer: c
Clarification: (int frac{dx}{x^2-8x+20}=int frac{dx}{(x^2-2(4x)+4^2)+4})
=(int frac{dx}{(x-4)^2+2^2})
Let x-4=t
Differentiating w.r.t x, we get
dx=dt
By using the formula (int frac{dx}{x^2+a^2}=frac{1}{a} tan^{-1}⁡frac{x}{a}+C)
(int frac{dx}{(x-4)^2+2^2}=int frac{dt}{t^2+2^2}=frac{1}{2} tan^{-1}⁡frac{t}{2}+C)
Replacing t with x-4, we get
(int frac{dx}{(x-4)^2+2^2}=frac{1}{2} tan^{-1}⁡frac{x-4}{2}+C)

8. Find (int frac{(x+3)}{2x^2+6x+7} dx).
a) (frac{1}{4} log⁡(2x^2+6x+7) + frac{3}{4} left (frac{1}{sqrt{2}} tan^{-1}⁡frac{2x+3}{2sqrt{2}}right )+C)
b) (frac{1}{4} log⁡(2x^2+6x+7) – frac{3}{4} (frac{1}{sqrt{2}} tan^{-1}⁡frac{2x+3}{2sqrt{2}} )+C)
c) (log⁡(2x^2+6x+7) + left (tan^{-1}⁡frac{2x+3}{2√2}right )+C)
d) –(log⁡(2x^2+6x+7) – frac{3}{4} left (frac{1}{√2} tan^{-1}⁡frac{2x+3}{2√2}right )+C)
Answer: a
Clarification: We can express
x+3=A (frac{d}{dx}) (2x²+6x+7)+B
x+3=A(4x+6)+B
x+3=4Ax+(6A+B)
Comparing the coefficients, we get
4A=1 ⇒A=1/4
6A+B=3 ⇒B=3/2
(int frac{x+3}{2x^2+6x+7} dx=frac{1}{4} int frac{4x+6}{2x^2+6x+7} dx+frac{3}{2} int frac{1}{2x^2+6x+7} dx)
Let 2x²+6x+7=t
(4x+6)dx=dt
(frac{1}{4} int frac{4x+6}{2x^2+6x+7} dx=frac{1}{4} int frac{dt}{t}=frac{1}{4} log⁡t)
Replacing t with (2x²+6x+7)
(frac{1}{4} int frac{4x+6}{2x^2+6x+7} dx=frac{1}{4} log⁡(2x^2+6x+7))
(frac{3}{2} int frac{1}{2x^2+6x+7} dx=frac{3}{2} int frac{1}{2(x^2+3x+frac{7}{2})} dx=frac{3}{4} int frac{1}{(x+frac{3}{2})^2+2} dx)
=(frac{3}{4} left (frac{1}{sqrt{2}} tan^{-1}⁡frac{2x+3}{2sqrt{2}} right ))
∴(int frac{x+3}{2x^2+6x+7} dx=frac{1}{4} log⁡(2x^2+6x+7) + frac{3}{4} left (frac{1}{sqrt{2}} tan^{-1}⁡frac{2x+3}{2√2} right )+C)

9. Find (int frac{7dx}{x^2-9}).
a) (frac{7}{6} log⁡|frac{x-9}{x+9}|+C)
b) (frac{7}{9} log⁡|frac{x-3}{x+3}|+C)
c) –(frac{7}{6} log⁡|frac{x+3}{x-3}|+C)
d) (frac{7}{6} log⁡|frac{x-3}{x+3}|+C)
Answer: d
Clarification: (int frac{7dx}{x^2-9}=2int frac{7dx}{x^2-3^2})
By using the formula (int frac{dx}{x^2-a^2}=frac{1}{2a} log⁡|frac{x-a}{x+a}|+C)
∴(7int frac{dx}{x^2-3^2}=7(frac{1}{2(3)} log⁡|frac{x-3}{x+3}|)+7C_1)
=(frac{7}{6} log⁡|frac{x-3}{x+3}|+C)

10. Find (int frac{dx}{x^2+4}).
a) –(tan^{-1}frac{⁡x}{4}+C)
b) (frac{1}{2} tan^{-1}⁡frac{x}{2}+C)
c) (frac{3}{4} tan^{-1}⁡x+C)
d) (frac{3}{4} tan^{-1}frac{⁡3x}{2}+C)
Answer: b
Clarification: (int frac{dx}{x^2+4}=int frac{dx}{x^2+2^2})
Using the formula (int frac{dx}{a^2+x^2}=frac{1}{a} tan^{-1}⁡frac{x}{a}+C)
∴(int frac{dx}{x^2+2^2}=(frac{1}{2} tan^{-1}⁡frac{x}{2})+C)

Mathematics Exam Questions for IIT JEE Exam,

250+ TOP MCQs on Types of Vectors | Class 12 Maths

Mathematics Multiple Choice Questions on “Types of Vectors”.

1. Two vectors having the same initial points are called as ________________
a) collinear vectors
b) unit vectors
c) coinitial vectors
d) equal vectors

Answer: c
Clarification: The vectors which have the same initial points are called coinitial vectors. It is not necessary that they should have the same magnitude and direction.

2. A unit vector is a vector whose magnitude is unity.
a) True
b) False

Answer: a
Clarification: The given statement is true. The vector whose magnitude is unity is called a unit vector. The unit vector in the given direction of given vector (vec{b}) is denoted by (hat{b}).

3. Which of the following vectors are collinear in the figure given below?

a) (vec{a}), (vec{c}) and (vec{d})
b) (vec{a}), (vec{b}) and (vec{d})
c) (vec{a}) and (vec{d})
d) (vec{b}) and (vec{d})

Answer: b
Clarification: Collinear vectors are those vectors which are parallel to the same line irrespective of the magnitude and direction. In the given figure, (vec{a}), (vec{b}) and (vec{d}) are collinear vectors.

4. In the figure given below, which vectors are coinitial but not equal?

a) (vec{a}), (vec{b}) and (vec{c})
b) (vec{b}) and (vec{c})
c) (vec{a}) and (vec{b})
d) (vec{a}) and (vec{c})

Answer: d
Clarification: The vectors which start from the same initial point are called coinitial vectors. In the given figure, vectors (vec{a}) and (vec{c}) start from the same point but are of different magnitude. Therefore, vectors (vec{a}) and (vec{c}) ae coinitial but not equal.

5. Which of the following condition is true for equal vectors?
a) They have the same direction but not same magnitude
b) They have the same magnitude and direction
c) They have the same initial point
d) They are parallel to the same line

Answer: b
Clarification: Two vectors are said to be equal, if they have the same magnitude and direction. They will be coinitial if they have the same initial point and they will be collinear if both the vectors are parallel to the same line.

6. The vector whose initial and final points coincide is called ____________
a) unit vector
b) coinitial vectors
c) equal vectors
d) zero vector

Answer: d
Clarification: The vector whose initial and final points coincide is called zero vector or a null vector. It is denoted by (vec{0}). It has zero magnitude and can be considered to have any direction.

7. The vectors which start from the same initial point are called collinear vectors.
a) True
b) False

Answer: b
Clarification: The given statement is false. Collinear vectors are those vectors that are parallel same line irrespective of the direction and magnitude. The vectors which have the same initial point are called coinitial vectors.

8. In the given figure, which of the following vectors are coinitial?

a) (vec{a}), (vec{c}) and (vec{b})
b) (vec{d}), (vec{c}) and (vec{b})
c) (vec{a}) and (vec{c})
d) (vec{b}) and (vec{c})

Answer: a
Clarification: Two vectors are said to coinitial if they have the same initial point. In the given figure, vectors (vec{a}), (vec{c}) and (vec{b}) are starting from the point, hence they are coinitial.

9. Which of the following vectors are equal in the figure given below?

a) (vec{a}) and c
b) (vec{c}) and (vec{b})
c) (vec{a}) and (vec{b})
d) (vec{a}), (vec{b}) and (vec{c})

Answer: c
Clarification: Two vectors are said to be equal, if they have the same magnitude. In the given figure, vectors (vec{a}) and (vec{b}) have the same magnitude. Hence, they are equal.

10. Which of the following is the condition for two vectors to be Collinear?
a) The vectors should be parallel to the same line
b) The vectors should have the same initial point
c) The vectors should have the same magnitude
d) The vectors should have the magnitude 1 and 0 respectively

Answer: a
Clarification: If two or more vectors are parallel to the same line then they are called collinear vectors. It is not necessary that they should have the same magnitude and direction.

250+ TOP MCQs on Bayes Theorem | Class 12 Maths

Mathematics Multiple Choice Questions on “Bayes Theorem”.

1. Method in which the previously calculated probabilities are revised with values of new probability is called __________
a) Revision theorem
b) Bayes theorem
c) Dependent theorem
d) Updation theorem

Answer: b
Clarification: Bayes theorem is the method in which the calculated probabilities are revised with values of new probabilities, whereas Updation theorem, Revision theorem and Dependent theorem are not related to the concept of probability.

2. Formula for Bayes theorem is ________
a) P(A|B) = (frac{P(B│A)P(A)}{P(B)})
b) P(A|B) = (frac{P(A)}{P(B)})
c) P(A|B) = (frac{P(B│A)}{P(B)})
d) P(A|B) = (frac{1}{P(B)})

Answer: a
Clarification: Bayes theorem formula is P(A|B) = (frac{P(B│A)P(A)}{P(B)})
The formula provides relationship between P(A|B) and P(B|A). It is mainly derived from conditional probability formula P(A|B) and P(B|A). Where,
P(A|B) = (frac{P(A∩B)}{P(B)}).
P(B|A) = (frac{P(B∩A)}{P(A)}).

3. Formula for conditional probability P(A|B) is _______
a) P(A|B) = (frac{P(A∩B)}{P(B)})
b) P(A|B) = (frac{P(A∩B)}{P(A)})
c) P(A|B) = (frac{P(A)}{P(B)})
d) P(A|B) = (frac{P(B)}{P(A)})

Answer: a
Clarification: Conditional probability P(A | B) indicates the probability of event ‘A’ happening given that event B has happened.
Which in formula can be written as P(A|B) = (frac{P(A∩B)}{P(B)}).
Whereas formula’s P(A|B) = (frac{P(A∩B)}{P(A)}), P(A|B) = (frac{P(A)}{P(B)}), P(A|B) = (frac{P(B)}{P(A)}) doesn’t satisfies the specified conditions.

4. Previous probabilities in Bayes Theorem that are changed with the new available information are called __________
a) independent probabilities
b) dependent probabilities
c) interior probabilities
d) posterior probabilities

Answer: d
Clarification: In Bayesian statistics, we calculate new probability after information becomes available due to new events and this is known as Posterior Probability. There is no term like Independent probabilities and Dependent probabilities, there are only independent events and dependent events. Interior probabilities represent probabilities of the intersection between two events.

5. A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.
a) 1/8
b) 5/8
c) 2/7
d) 3/8

Answer: d
Clarification: Let E = event that the man reports that six in the throwing of the die and let, S1 = event that six occurs and S2 = event that six does not occur.
P(S1) = Probability that six occurs = 1/6.
P(S2) = Probability that six does not occur = 5/6.
Also, P(E|S1) = P(Probability that man reports six occurs when six actually has occurred on the die) = 3/4.
P(E|S2) = P(Probability that man reports six occurs when six not actually occurred on the die) =
1 – 3/4 = 1/4.
By using Bayes’ theorem,
P(S2|E) = P(S1)P(E|S1)/(P(S1)P(E│S1)+P(S2)P(E|S2))
= (1/6 × 3/4) / ((1/6 × 3/4)) + (5/6 × 1/4)) = 3/8.

6. Bag 1 contains 3 red and 5 black balls while another Bag 2 contains 4 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it is drawn from bag 2.
a) 31/62
b) 16/62
c) 16/31
d) 31/32

Answer: c
Clarification: Let E1 = event of choosing the bag 1, E2 = event of choosing the bag 2.
Let A be event of drawing a red ball.
P(E1) = P(E2) = 1/2.
Also, P(A|E1) = P(drawing a red ball from Bag 1) = 3/8.
And P(A|E2) = P(drawing a red ball from Bag 2) = 4/10.
The probability of drawing a ball from bag 2, being given that it is red is P(E2|A).
By using Bayes’ theorem,
P(E2|A) = P(E2)P(A|E2)/( P(E1)P(A│E1)+P(E2)P(A|E2))
= (1/2 × 4/10) / ((1/2 × 3/8)) + (1/2 × 4/10)) = 16/31.

7. Bag 1 contains 4 white and 6 black balls while another Bag 2 contains 4 white and 3 black balls. One ball is drawn at random from one of the bags and it is found to be black. Find the probability that it was drawn from Bag 1.
a) 12/13
b) 5/12
c) 7/11
d) 7/12

Answer: d
Clarification: Let E1 = event of choosing the bag 1, E2 = event of choosing the bag 2.
Let A be event of drawing a black ball.
P(E1) = P(E2) = 1/2.
Also, P(A|E1) = P(drawing a black ball from Bag 1) = 6/10 = 3/5.
P(A|E2) = P(drawing a black ball from Bag 2) = 3/7.
By using Bayes’ theorem, the probability of drawing a black ball from bag 1 out of two bags is-:
P(E1|A) = P(E1)P(A|E1)/( P(E1)P(A│E1)+P(E2)P(A|E2))
= (1/2 × 3/5) / ((1/2 × 3/7)) + (1/2 × 3/5)) = 7/12.

250+ TOP MCQs on Transpose of a Matrix | Class 12 Maths

Mathematics Multiple Choice Questions on “Transpose of a Matrix”.

1. Which of the following is not the property of transpose of a matrix?
a) (A’)’=A
b) (A+B)’=A’+B’
c) (AB)’=(BA)’
d) (kA)’=KA’
Answer: c
Clarification: (AB)’=(BA)’is incorrect. We know that matrix multiplication is not commutative i.e. AB≠BA. Hence, its transpose will also not be commutative.
(AB)’=B’A’

2. Find the transpose of A=(begin{bmatrix}1&-2\-1&5end{bmatrix}).
a) A=(begin{bmatrix}-1&-2\-1&-5end{bmatrix})
b) A=(begin{bmatrix}1&2\1&5end{bmatrix})
c) A=(begin{bmatrix}-1&2\-1&5end{bmatrix})
d) A=(begin{bmatrix}1&-1\-2&5end{bmatrix})
Answer: d
Clarification: A=(begin{bmatrix}1&-2\-1&5end{bmatrix}). To find the transpose of the matrix, interchange the rows with columns and columns with rows.
Hence, A’=(begin{bmatrix}1&-1\-2&5end{bmatrix}).

3. If A=(begin{bmatrix}2\7\8end{bmatrix}), B=(begin{bmatrix}-3&4&1end{bmatrix}), find (AB)’.
a) (AB)’=(begin{bmatrix}-6&-21&-24\8&28&32\2&7&8end{bmatrix})
b) (AB)’=(begin{bmatrix}-6&8&2\-21&-28&7\-24&32&8end{bmatrix})
c) (AB)’=(begin{bmatrix}6&21&24\-8&28&7\-2&7&-8end{bmatrix})
d) (AB)’=(begin{bmatrix}-6&8&-21\8&2&7\-24&8&2end{bmatrix})
Answer: a
Clarification: Given that, A=(begin{bmatrix}2\7\8end{bmatrix}), B=(begin{bmatrix}-3&4&1end{bmatrix})
AB=(begin{bmatrix}2\7\8end{bmatrix})(begin{bmatrix}-3&4&1end{bmatrix})=(begin{bmatrix}-6&8&2\-21&28&7\-24&32&8end{bmatrix})
∴To find (AB)’, interchange the rows with columns and columns with rows of the matrix AB
(AB)’=(begin{bmatrix}-6&-21&-24\8&28&32\2&7&8end{bmatrix})

4. If A’=(begin{bmatrix}8&2\6&4end{bmatrix}) and B’=(begin{bmatrix}9&5\7&3end{bmatrix}). Find (A+2B)’.
a) (begin{bmatrix}26&20\10&12end{bmatrix})
b) (begin{bmatrix}26&12\20&10end{bmatrix})
c) (begin{bmatrix}26&10\20&12end{bmatrix})
d) (begin{bmatrix}26&20\12&10end{bmatrix})
Answer: b
Clarification: Given that A’=(begin{bmatrix}8&2\6&4end{bmatrix}) and B’=(begin{bmatrix}9&5\7&3end{bmatrix})
Calculating the transpose of A’ and B’, we get
A=(begin{bmatrix}8&6\2&4end{bmatrix}) and B=(begin{bmatrix}9&7\5&3end{bmatrix})
⇒(A+2B)=(begin{bmatrix}8&6\2&4end{bmatrix})+2(begin{bmatrix}9&7\5&3end{bmatrix})
=(begin{bmatrix}8+18&6+14\2+10&4+6end{bmatrix})=(begin{bmatrix}26&20\12&10end{bmatrix})
Hence, (A+2B)’=(begin{bmatrix}26&12\20&10end{bmatrix}).

5. If A=(begin{bmatrix}cos⁡x&-sin⁡x&-cos⁡x\sin⁡x&-cos⁡x&sin⁡x end{bmatrix}). Find A’A.
a) (begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
b) (begin{bmatrix}1&0&1\1&0&1\1&0&1end{bmatrix})
c) (begin{bmatrix}1&0&1\0&1&0\1&0&1end{bmatrix})
d) (begin{bmatrix}1&0&0\1&1&0\1&1&1end{bmatrix})
Answer: d
Clarification: Given that, A=(begin{bmatrix}cos⁡x&-sin⁡x&cos⁡x\sin⁡x&cos⁡x&sin⁡x end{bmatrix})
∴ A’=(begin{bmatrix}cos⁡x&sinx\-sin⁡x&cos⁡x\cos⁡x&sin⁡x end{bmatrix})
⇒A’ A=(begin{bmatrix}cos⁡x&sin⁡x\-sin⁡x&cos⁡x\cos⁡x&sin⁡x end{bmatrix})(begin{bmatrix}cos⁡ x &-sin⁡x&cos⁡x\sin⁡x&cos⁡x &sin⁡x end{bmatrix})
=(begin{bmatrix}cos^2⁡x+sin^2⁡x&-sin⁡x cos⁡x+sin⁡x cos⁡x&cos^2⁡x+sin^2⁡x\-sin⁡x cos⁡x+sin⁡x cos⁡x&sin^2⁡x+cos^2⁡x&-sin⁡x cos⁡x+cos⁡x sin⁡x\cos^2⁡x+sin^2 x&-cos⁡x sin⁡x+sin⁡x cos⁡x &cos^2⁡x+sin^2⁡x end{bmatrix})
=(begin{bmatrix}1&0&1\0&1&0\1&0&1end{bmatrix})

6. Find the transpose of the matrix A=(begin{bmatrix}-1&2&sqrt{3}\-4&5&sqrt{6}\-7&8&-9end{bmatrix})
a) (begin{bmatrix}1&-2&-sqrt{3}\4&-5&-sqrt{6}\7&-8&9end{bmatrix})
b) (begin{bmatrix}-1&-4&-7\2&5&8\sqrt{3}&sqrt{6}&-9end{bmatrix})
c) (begin{bmatrix}1&4&7\-2&-5&-8\-sqrt{3}&-sqrt{6}&9end{bmatrix})
d) (begin{bmatrix}1&4&7\-2&5&2\1&8&9end{bmatrix})
Answer: b
Clarification: To find the transpose of the matrix of the given matrix, interchange the rows with columns and columns with rows.
Hence, we get A’=(begin{bmatrix}-1&-4&-7\2&5&8\sqrt{3}&sqrt{6}&-9end{bmatrix})

7. If matrix A=(begin{bmatrix}4&1\6&2end{bmatrix}) and B=(begin{bmatrix}-1&3\2&1\6&6end{bmatrix}), then find A’ B’.
a) (begin{bmatrix}14&14\5&4\6&18end{bmatrix})
b) (begin{bmatrix}14&5\14&4\6&18end{bmatrix})
c) (begin{bmatrix}14&14&60\5&4&18end{bmatrix})
d) (begin{bmatrix}14&14&18\5&4&60end{bmatrix})
Answer: c
Clarification: Given that, A=(begin{bmatrix}4&1\6&2end{bmatrix}) and B=(begin{bmatrix}-1&3\2&1\6&6end{bmatrix})
⇒A’=(begin{bmatrix}4&6\1&2end{bmatrix}) and B’=(begin{bmatrix}-1&2&6\3&1&6end{bmatrix})
⇒A’ B’=(begin{bmatrix}4&6\1&2end{bmatrix})(begin{bmatrix}-1&2&6\3&1&6end{bmatrix})=(begin{bmatrix}14&14&60\5&4&18end{bmatrix}).

8. If P=(begin{bmatrix}-1&5\8&3end{bmatrix}) and Q=(begin{bmatrix}4&2\8&5end{bmatrix}). Find (2P+3Q’)’.
a) (begin{bmatrix}10&22\34&21end{bmatrix})
b) (begin{bmatrix}10&21\34&22end{bmatrix})
c) (begin{bmatrix}10&34\22&21end{bmatrix})
d) (begin{bmatrix}10&22\21&34end{bmatrix})
Answer: a
Clarification: Given that, P=(begin{bmatrix}-1&5\8&3end{bmatrix}) and Q=(begin{bmatrix}4&2\8&5end{bmatrix})
⇒Q’=(begin{bmatrix}4&8\2&5end{bmatrix})
⇒2P+3Q’=2(begin{bmatrix}-1&5\8&3end{bmatrix})+3(begin{bmatrix}4&8\2&5end{bmatrix})=(begin{bmatrix}-2+12&10+24\16+6&6+15end{bmatrix})=(begin{bmatrix}10&34\22&21end{bmatrix})
∴(2P+3Q’ )’=(begin{bmatrix}10&22\34&21end{bmatrix}).

9. Which of the following is the reversal law of transposes?
a) (A-B)’=B’-A’
b) (AB)’=B’A’
c) (AB)’=(BA)’
d) (A+B)’=B’+A’
Answer: b
Clarification: According to the reverse law of transposes the transpose of the product is the product of the transposes taken in the reverse order i.e. (AB)’=B’ A’.

10. If A=(begin{bmatrix}i&1\0&iend{bmatrix}), then the correct relation is ___________
a) A+A’=(begin{bmatrix}1&0\-1&0end{bmatrix})
b) A-A’=(begin{bmatrix}1&0\-1&0end{bmatrix})
c) A+A’=(begin{bmatrix}0&1\-1&0end{bmatrix})
d) A-A’=(begin{bmatrix}0&1\-1&0end{bmatrix})
Answer: d
Clarification: Given that, A=(begin{bmatrix}i&1\0&iend{bmatrix})
⇒A’=(begin{bmatrix}i&0\1&iend{bmatrix})
∴A-A’=(begin{bmatrix}i&1\0&iend{bmatrix})–(begin{bmatrix}i&0\1&iend{bmatrix})=(begin{bmatrix}i-i&1-0\0-1&i-iend{bmatrix})=(begin{bmatrix}0&1\-1&0end{bmatrix}).

250+ TOP MCQs on Exponential and Logarithmic Functions | Class 12 Maths

Mathematics Multiple Choice Questions & Answers (MCQs) on “Exponential and Logarithmic Functions”.

1. Differentiate 8e^-x+2e^x w.r.t x.
a) 2e^-x+8e^x
b) 2e^x+8e^-x
c) 2e^-x-8e^x
d) 2e^x-8e^-x
Answer: d
Clarification: To solve:y=(8e^-x+2e^x)
Differentiating w.r.t x we get,
(frac{dy}{dx})=8(-e^-x+2e^x)
∴(frac{dy}{dx})=2e^x-8e^-x.

2. Differentiate 8e^cos2x w.r.t x.
a) 16 sin⁡2x e^cos2x
b) -16 sin⁡2x e^cos2x
c) -16 sin⁡2x e^-cos⁡2x
d) 16 sin⁡2x e^-cos⁡2x
Answer: b
Clarification: Consider y=8e^cos2x
Differentiating w.r.t x by using chain rule, we get
(frac{dy}{dx}=frac{d}{dx})(8e^cos2x)
=8e^cos2x(frac{d}{dx}) (cos⁡2x)
=8e^cos2x(-sin⁡2x)(frac{d}{dx}) (2x)
=8e^cos2x(-sin⁡2x)(2)
∴(frac{dy}{dx})=-16 sin⁡2x e^cos2x

3. Differentiate 3 sin^-1⁡(e^2x) w.r.t x.
a) (frac{6e^2x}{sqrt{1-e^{4x}}})
b) (frac{2e^2x}{sqrt{1-e^{4x}}})
c) –(frac{6e^2x}{sqrt{1-e^{4x}}})
d) (frac{6e^{-2x}}{sqrt{1-e^{4x}}})
Answer: a
Clarification: Consider y=3 sin^-1⁡(e^2x)
(frac{dy}{dx}=frac{d}{dx})(3 sin^-1⁡(e^2x))
(frac{dy}{dx}=left (frac{3}{sqrt{1-(e^{2x})^2}}right )frac{d}{dx})(e^2x)
(frac{dy}{dx}=left (frac{3}{sqrt{1-(e^{2x})^2}}right ))2e^2x
∴(frac{dy}{dx})=(frac{6e^{2x}}{sqrt{1-e^{4x}}})

4. Differentiate log⁡(log⁡x⁵) w.r.t x.
a) –(frac{5}{x log⁡x^5})
b) (frac{1}{log⁡x^5})
c) (frac{5}{x log⁡x^5})
d) –(frac{1}{x log⁡x^5})
Answer: c
Clarification: Consider y=(log⁡(log⁡(x⁵)))
(frac{dy}{dx}=frac{1}{log⁡x^5} frac{d}{dx} (log⁡x^5))
(frac{dy}{dx}=frac{1}{log⁡x^5}.frac{1}{x^5}.frac{d}{dx} (x^5))
(frac{dy}{dx}=frac{1}{log⁡x^5}.frac{1}{x^5}.5x^4)
∴(frac{dy}{dx}=frac{5}{x log⁡x^5})

5. Differentiate 3e^3x³ w.r.t x.
a) 27x^-2 e^3x³
b) 27x² e^3x³
c) -27x² e^3x³
d) -27x^-2 e^3x³
Answer: b
Clarification: Consider y=3e^3x³
(frac{dy}{dx})=(frac{d}{dx})(3e^3x³)
(frac{dy}{dx})=3e^3x³ (frac{d}{dx})(3x³)
(frac{dy}{dx})=3e^3x³ (3(3x²))
(frac{dy}{dx})=27x² e^3x³.

6. Differentiate 5e^x² tan⁡x w.r.t x.
a) 5e^x² (1+tan⁡x)²
b) -5e^x² (1+tan⁡x)²
c) 5e^x² (1-tan⁡x)²
d) -5e^x² (1-tan⁡x)²
Answer: a
Clarification: Consider y=5e^x² tan⁡x
Differentiating w.r.t x by using chain rule, we get
(frac{dy}{dx})=tan⁡x (frac{d}{dx}) (5e^x²)+5e^x² (frac{d}{dx}) (tan⁡x)
(frac{dy}{dx})=tan⁡x (5e^x².2x)+5e^x² (sec^2⁡x)
(frac{dy}{dx})=5e^x² (2x tan⁡x+sec^2⁡x)
(frac{dy}{dx})=5e^x² (1+tan^2⁡x+2x tan⁡x)
(frac{dy}{dx})=5e^x² (1+tan⁡x)²

7. Differentiate log⁡(e^5x³) w.r.t x.
a) (frac{-15x^2}{e^{5x^3}})
b) (frac{15x^2}{e^{5x^3}})
c) 15x²
d) -15x²
Answer: c
Clarification: Consider y=log⁡(e^5x³)
y=5x³ (∴log⁡e^x=x)
⇒(frac{dy}{dx})=(frac{d}{dx} (5x^3))
∴(frac{dy}{dx}=5(3x^2)=15x^2)

8. Differentiate 7 log⁡(x⁴.5e^x³) w.r.t x.
a) (frac{7(4+3x^3)}{x^2})
b) (frac{7(4-3x^3)}{x})
c) –(frac{7(4+3x^3)}{x})
d) (frac{7(4+3x^3)}{x})
Answer: d
Clarification: Consider y=7 log⁡(x⁴.5e^x³)
y=(7(log⁡x^4 +log⁡5e^{x^3}))
y=(7(4 log⁡x+log⁡5e^{x^3}))
(frac{dy}{dx}=7(4 frac{d}{dx} (log⁡x)+frac{d}{dx} (log⁡5e^{x^3})))
(frac{dy}{dx}=7(frac{4}{x}+frac{1}{5e^{x^3}} frac{d}{dx} (5e^{x^3})))
(frac{dy}{dx}=7(frac{4}{x}+frac{1}{5e^{x^3}}.5e^{x^3}.frac{d}{dx} {x^3}))
(frac{dy}{dx}=7(frac{4}{x}+frac{1}{5e^{x^3}}.5e^{x^3}.3x^2))
(frac{dy}{dx}=7(frac{4}{x}+3x^2))
∴(frac{dy}{dx}=frac{7(4+3x^3)}{x})

9. Differentiate 2e^x⁴ log⁡x w.r.t x.
a) (frac{2e^{x^4} (4x^4 log⁡x+1)}{x^2})
b) (frac{e^{x^4} (4x^4 log⁡x+1)}{x})
c) (frac{2e^{x^4} (4x^4 log⁡x+1)}{x})
d) –(frac{2e^{x^4} (4x^4 log⁡x+1)}{x})
Answer: c
Clarification: Consider y=2e^x⁴ log⁡x
(frac{dy}{dx})=(frac{d}{dx}) (2e^x⁴ log⁡x)
Differentiating w.r.t x by using chain rule, we get
(frac{dy}{dx})=2(log⁡x (frac{d}{dx} (e^{x^4})+e^{x^4}frac{d}{dx}) (log⁡x))
(frac{dy}{dx})=(2(log⁡x.e^{x^4}frac{d}{dx} {x^4}+e^{x^4}.frac{1}{x}))
(frac{dy}{dx})=(2(log⁡x.e^{x^4}.4x^3+e^{x^4}.frac{1}{x}))
(frac{dy}{dx})=(2e^{x^4} (4x^3 log⁡x+frac{1}{x}))
∴(frac{dy}{dx}=frac{(2e^{x^4} (4x^4 log⁡x+1))}{x})

10. Differentiate (log⁡(cos⁡(sin⁡(e^{x^3})))) w.r.t x.
a) –(3x^2 ,e^{x^3} ,cos⁡e^{x^3} ,tan⁡(sin⁡e^{x^3}))
b) (3x^2 ,e^{x^3} ,cos⁡e^{x^3} ,tan⁡(sin⁡e^{x^3}))
c) –(3e^{x^3} ,cos⁡e^{x^3} ,cos⁡(sin⁡e^{x^3}))
d) –(x^2 e^{x^3} ,cos⁡e^{x^3} ,tan⁡(sin⁡e^{x^3}))
Answer: a
Clarification: Consider y=(log⁡(cos⁡(sin⁡(e^{x^3}))))
Differentiating w.r.t x by using chain rule, we get
(frac{dy}{dx})=(frac{d}{dx} (log⁡(cos⁡(sin⁡(e^{x^3})))))
(frac{dy}{dx})=((frac{1}{cos⁡(sin⁡e^{x^3})} frac{d}{dx} (cos⁡(sin⁡e^{x^3}))))
(frac{dy}{dx})=((frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3}) frac{d}{dx}(sin⁡e^{x^3}))))
(frac{dy}{dx})=((frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3})(cos⁡e^{x^3}) frac{d}{dx} (e^{x^3}))))
(frac{dy}{dx})=((frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3})(cos⁡e^{x^3})(e^{x^3}) frac{d}{dx} {x^3})))
(frac{dy}{dx})=((frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3}).cos⁡e^{x^3} .e^{x^3}.3x^2))
(frac{dy}{dx})=-((frac{3x^2 e^{x^3} cos⁡e^{x^3} sin⁡(sin⁡e^{x^3})}{cos⁡(sin⁡e^{x^3})}))
(frac{dy}{dx})=-(3x^2 e^{x^3} cos⁡e^{x^3} tan⁡(sin⁡e^{x^3}))

250+ TOP MCQs on Integration by Partial Fractions | Class 12 Maths

Mathematics Multiple Choice Questions on “Integration by Partial Fractions”.

1. What form of rational function (frac{px+q}{(x-a)(x-b)}), a≠b represents?
a) (frac{A}{(x-a)})
b) (frac{B}{(x-b)})
c) (frac{A+B}{(x-a)(x-b)})
d) (frac{A}{(x-a)} + frac{B}{(x-b)})
Answer: d
Clarification: The given function (frac{px+q}{(x-a)(x-b)}), a≠b can also be written as
(frac{A}{(x-a)} + frac{B}{(x-b)}) and is further used to solve integration by partial fractions numerical.

2. Find (int frac{x^2+1}{x^2-5x+6} dx).
a) x – 5log|x-2| + 10log|x-3|+C
b) x – 3log|x-2| + 5log|x-3|+C
c) x – 10log|x-2| + 5log|x-3|+C
d) x – 5log|x-5| + 10log|x-10|+C
Answer: a
Clarification: As it is not proper rational function, we divide numerator by denominator and get
(frac{x^2+1}{x^2-5x+6} = 1-frac{5x-5}{x^2-5x+6} = 1+frac{5x-5}{(x-2)(x-3)})
Let (frac{5x-5}{(x-2)(x-3)}=frac{A}{(x-2)} + frac{B}{(x-3)})
So that, 5x–5 = A(x-3) + B(x-2)
Now, equating coefficients of x and constant on both sides, we get A + B = 5 and 3A + 2B = 5. Solving these equations, we get A=-5 and B=10.
Therefore, (frac{x^2+1}{x^2-5x+6} = 1 – frac{5}{(x-2)} + frac{10}{(x-3)}).
(int frac{x^2+1}{x^2-5x+6} dx = int dx – 5int frac{dx}{(x-2)} + 10int frac{dx}{(x-3)}).
= x – 5log|x-2| + 10log|x-3|+C

3. Find (int frac{dx}{(x+1)(x+2)}).
a) (Log left|frac{x+1}{x+2}right|+ C)
b) (Log left|frac{x-1}{x+2}right|+ C)
c) (Log left|frac{x+2}{x+1}right|+ C)
d) (Log left|frac{x+1}{x-2}right|+ C)
Answer: a
Clarification: It is a proper rational function. Therefore,
(frac{1}{(x+1)(x+2)} = frac{A}{(x+1)} + frac{B}{(x+2)})
Where real numbers are determined, 1 = A(x+2) + B(x+1), Equating coefficients of x and the constant term, we get A+B = 0 and 2A+B = 1. Solving it we get A=1, and B=-1.
Thus, it simplifies to, (frac{1}{(x+1)} + frac{-1}{(x+2)} = int frac{dx}{(x+1)} – int frac{dx}{(x+2)}).
= log|x+1| – log|x+2| + C
= (Log left|frac{x+1}{x+2}right|+ C).

4. An improper integration fraction is reduced to proper fraction by _____
a) multiplication
b) division
c) addition
d) subtraction
Answer: b
Clarification: An improper integration factor can be reduced to proper fraction by division, i.e., if the numerator and denominator have same degree, then they must be divided in order to reduce it to proper fraction.

5. (int frac{dx}{x(x^2+1)}) equals ______
a) (log|x| – frac{1}{2} log(x^2+1)) + C
b) (log|x| + frac{1}{2} log(x^2+1)) + C
c) –(log|x| + frac{1}{2} log(x^2+1)) + C
d) (frac{1}{2} log|x| + log(x^2+1)) + C
Answer: a
Clarification: We know that (int frac{dx}{x(x^2+1)} = frac{A}{x} + frac{Bx+C}{x^2+1})
By simplifying it we get, (int frac{dx}{x(x^2+1)}=frac{(A+B) x^2+Cx+A}{x(x^2+1)})
Now equating the coefficients we get A = 0, B = 0, C=1.
(int frac{dx}{x(x^2+1)} = int frac{dx}{x} + int frac{-xdx}{(x^2+1)})
Therefore after integrating we get (log|x| – frac{1}{2} log(x^2+1)) + C.

6. (int frac{dx}{(x^2-9)}) equals ______
a) (frac{1}{6} log frac{x+3}{x-3}) + C
b) (frac{1}{6} log frac{x-3}{x+3}) + C
c) (frac{1}{5} log frac{x+3}{x-3}) + C
d) (frac{1}{3} log frac{x+3}{x-3}) + C
Answer: b
Clarification: (int frac{dx}{(x^2-9)}=frac{A}{(x-3)} + frac{B}{(x+3)})
By simplifying, it we get (frac{A(x+3)+B(x-3)}{(x^2-9)} = frac{(A+B)x+3A-3B}{(x^2-9)})
By solving the equations, we get, A+B=0 and 3A-3B=1
By solving these 2 equations, we get values of A=1/6 and B=-1/6.
Now by putting values in the equation and integrating it we get value,
(frac{1}{6} log (frac{x-3}{x+3})) + C.

7. Which form of rational function (frac{px+q}{(x-a)^2}) represents?
a) (frac{A}{(x-a)} + frac{B}{(x-a)^2})
b) (frac{A}{(x-a)^2} + frac{B}{(x-a)})
c) (frac{A}{(x-a)} – frac{B}{(x-a)^2})
d) (frac{A}{(x-a)} – frac{B}{(x-a)})
Answer: a
Clarification: It is a form of the given partial fraction (frac{px+q}{(x-a)^2}) which can also be written as
(frac{A}{(x-a)} + frac{B}{(x-a)^2}) and is further used to solve integration by partial fractions numerical.

8. (int frac{(x^2+x+1)dx}{(x+2)(x^2+1)}) equals ______
a) (frac{3}{5}log|x+2| + frac{1}{5}log|x^2+1|+frac{1}{5} tan^{-1}x+5C)
b) (frac{3}{5}log|x+2| + frac{1}{5}log|x^2+1|+frac{1}{6} tan^{-1}x+C)
c) (frac{3}{5}log|x+2| + frac{1}{6}log|x^2+1|+frac{1}{6} tan^{-1}x+C)
d) (frac{3}{5}log|x+2| + frac{1}{5}log|x^2+1|+frac{1}{5} tan^{-1}x+C)
Answer: d
Clarification: (int frac{(x^2+x+1)dx}{(x+2)(x^2+1)} = frac{A}{(x+2)} + frac{Bx+C}{(x^2+1)})
Now equating, (x²+x+1) = A (x²+1) + (Bx+C) (x+2)
After equating and solving for coefficient we get values,
A=(frac{3}{5}), B=(frac{2}{5}), C=(frac{1}{5}), now putting these values in the equation we get,
(int frac{(x^2+x+1)dx}{(x+2)(x^2+1)} = frac{3}{5} int frac{dx}{(x+2)} + frac{1}{5} int frac{2xdx}{(x^2+1)} + frac{1}{5} int frac{dx}{(x^2+1)})
Hence it comes, (frac{3}{5} log|x+2| + frac{1}{5} log|x^2+1|+frac{1}{5}tan^{-1}x+C)

9. Identify the type of the equation (x+1)².
a) Linear equation
b) Cubic equation
c) Identity
d) Imaginary
Answer: c
Clarification: As it represents the identity (b+a)² it satisfies the identity (b+a)² = (a² + b² +2ab) and is not linear, cubic or an imaginary equation so the correct option is Identity Equation.

10. For the given equation (x+2) (x+4) = x² + 6x + 8, how many values of x satisfies this equation?
a) Two values of x
b) One value of x
c) All value of x
d) No value of x
Answer: c
Clarification: If we solve the L.H.S. (Left Hand Side) of the equation, we get the following value.
(x+2) (x+4) = x² + 4x + 2x + 8 = x² + 6x + 8.
This value is same as the R.H.S. (Right Hand Side).
So, all the values of x satisfy the equality.