Category: Class 12 Mathematics Objective Questions

Mathematics MCQs for Engineering Entrance Exams on “Methods of Integration-2”.

1. Integrate 2 sin²⁡x+cos²x.
a) (frac{3x}{2}+frac{sin⁡2x}{4}+C)
b) (frac{3x}{2}-frac{sin⁡2x}{4}+C)
c) (frac{x}{2}+frac{sin⁡2x}{4}+C)
d) (frac{3x}{4}-frac{2sin⁡2x}{2}+C)
Answer: b
Clarification: (int ,2 ,sin^2⁡x +cos^2⁡x=int sin^2⁡x+sin^2⁡x+cos^2⁡x dx)
=(int sin^2⁡x+1 dx)
=(int sin^2⁡x dx+int 1 dx)
=(int frac{1-cos⁡2x}{2} dx+int 1 dx)
=(frac{x}{2}-frac{sin⁡2x}{4}+x)
=(frac{3x}{2}-frac{sin⁡2x}{4}+C)

2. Integrate 8 tan^3⁡x sec^2⁡x.
a) 2 tan⁴⁡x+C
b) 4 cot⁴⁡x+C
c) 2 tan^3⁡x+C
d) tan^4⁡x+C
Answer: a
Clarification: To find: (int 8 ,tan^3⁡x ,sec^2⁡x ,dx)
Let tan⁡x=t
sec^2⁡x dx=dt
∴(int 8 ,tan^3⁡x ,sec^2⁡x ,dx=int 8 ,t^3 ,dt=frac{8t^4}{4}=2t^4)
Replacing t with tan⁡x, we get
(int 8 tan^3⁡x sec^2⁡x dx=2 tan^4⁡x+C)

3. Find the integral of (frac{cos^2⁡x-sin^2⁡x}{7 cos^2⁡x sin^2⁡x}).
a) –(frac{1}{7}) (cot⁡x-tan⁡x)+C
b) –(frac{1}{7}) (cot⁡x-2 tan⁡x)+C
c) –(frac{1}{7}) (cot⁡x+tan⁡x)+C
d) –(frac{1}{7}) (2 cot⁡x+3 tan⁡x)+C
Answer: c
Clarification: To find: (int frac{cos^2⁡x-sin^2⁡x}{7 ,cos^2⁡x ,sin^2⁡x} dx)
(int frac{cos^2⁡x-sin^2⁡x}{7 ,cos^2⁡x ,sin^2⁡x} dx=frac{1}{7} int frac{1}{sin^2⁡x}-frac{1}{cos^2⁡x} dx)
=(frac{1}{7} int cosec^2 x-sec^2⁡x dx)
=(frac{1}{7}) (-cot⁡x-tan⁡x)+C
=-(frac{1}{7}) (cot⁡x+tan⁡x)+C.

4. Find (int sin^2⁡(8x+5) dx)
a) (frac{x}{4}+frac{sin⁡(16x+10)}{32}+C)
b) (frac{x}{2}-frac{cos⁡(16x+10)}{32}+C)
c) (frac{x}{2}-frac{sin⁡(16x+10)}{32}+C)
d) (frac{x}{2}+frac{cos⁡(16x+5)}{32}+C)
Answer: c
Clarification: (int sin^2⁡(8x+5) dx=int frac{1-cos⁡2(8x+5)}{2} dx=int frac{1}{2} dx-frac{1}{2} int cos(16x+10)dx)
=(frac{x}{2}-frac{1}{2} (frac{sin⁡(16x+10)}{16})=frac{x}{2}-frac{sin⁡(16x+10)}{32}+C)

5. Find (int frac{5 cos^2⁡x}{1+sin⁡x} dx).
a) -3(x+cos⁡x)+C
b) 5(x+cos⁡x)+C
c) 5(-x+sin⁡x)+C
d) 5(x-cos⁡x)+C
Answer: b
Clarification: (int frac{5 cos^2⁡x}{1+sin⁡x} dx=int frac{5(1-sin^2⁡x)}{1+sin⁡x}=5int frac{(1+sin⁡x)(1-sin⁡x)}{(1+sin⁡x)} dx)
=5∫ (1-sin⁡x)dx
=5(x-(-cos⁡x))=5(x+cos⁡x)+C

6. Find the integral of (frac{e^{-x} (1-x)}{sin^2⁡(xe^{-x})}).
a) cot⁡xe^-x+C
b) -cot⁡xe^-x+C
c) -cot⁡xe^x+C
d) -cos^2⁡xe^-x+C
Answer: b
Clarification: (int frac{e^{-x} (1-x)}{sin^2⁡(xe^{-x})} dx)
Let xe^-x=t
Differentiating w.r.t x, we get
(-xe^{-x}+e^{-x} dx=dt)
e^-x (1-x)dx=dt
(int frac{e^{-x} (1-x)}{sin^2⁡(xe^{-x})} dx=int frac{dt}{sin^2⁡t})
=(int cosec^2 ,t ,dt)
=-cot⁡t+C
Replacing t with xe^-x, we get
(int frac{e^{-x} (1-x)}{sin^2⁡(xe^{-x})} dx=-cot⁡xe^{-x}+C).

7. Integrate (frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2}).
a) -log⁡(1+2sin⁡2x)+C
b) (frac{1}{4}) log⁡(1-sin⁡2x)+C
c) –(frac{1}{4}) log⁡(1+cos⁡2x)+C
d) -log⁡(1-sin⁡2x)+C
Answer: d
Clarification: (int frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2} dx=int frac{2 cos⁡2x}{cos^2⁡x+sin^2⁡x-sin⁡2x} ,dx ,(∵2 cos⁡x sin⁡x=sin⁡2x))
=(int frac{2 cos⁡2x}{1-sin⁡2x} dx)
Let 1-sin⁡2x=t
Differentiating w.r.t x, we get
-2 cos⁡2x dx=dt
2 cos⁡2x dx=-dt
(int frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2} dx=-int frac{dt}{t})
=-log⁡t
Replacing t with 1-sin⁡2x, we get
∴(int frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2}) dx=-log⁡(1-sin⁡2x)+C

8. Integrate sin³⁡(x+2).
a) (frac{3}{4} ,(sin⁡(x+2))+frac{1}{12} ,cos⁡(3x+6)+C)
b) –(frac{3}{4} ,(cos⁡(x+2))-frac{1}{5} ,cos⁡(3x+6)+C)
c) –(frac{3}{4} ,(cos⁡(x+2))+frac{1}{12} ,cos⁡(3x+6)+C)
d) –(frac{3}{4} ,(cos⁡(x+2))+frac{1}{12} ,sin⁡(x+2)+C)
Answer: c
Clarification: To find: ∫ 3 sin³⁡(x+2) dx
We know that, sin⁡3x=3 sin⁡x-4 sin³⁡x
∴sin³⁡⁡x=(frac{3 sin⁡x-sin⁡3x}{4})
sin³⁡(x+2)=(frac{(3 sin⁡(x+2)-sin⁡(3x+6))}{4})
(int sin^3⁡(x+2) ,dx=frac{3}{4} int sin⁡(x+2) ,dx-frac{1}{4} int ,sin⁡(3x+6) ,dx)
=-(frac{3}{4} ,(cos⁡(x+2))+frac{1}{12} ,cos⁡(3x+6)+C)

9. Integrate 2x cos⁡(x²+3).
a) sin⁡(x²+3)+C
b) sin²⁡(x²+3)+C
c) cot⁡(x²+3)+C
d) -sin⁡(x²+3)+C
Answer: a
Clarification: ∫ 2x cos⁡(x²+3) dx
Let x²+3=t
Differentiating w.r.t x, we get
2x dx=dt
∫ 2x cos⁡(x²+3) dx=∫ cos⁡t dt
=sin⁡t+C
Replacing w.r.t x, we get
∴∫ 2x cos⁡(x²+3) dx=sin⁡(x²+3)+C

10. Find ∫ 2 sin³⁡x+1 dx
a) (frac{3}{2}-frac{cos⁡3x}{6}+x+C)
b) –(frac{3}{2} cos⁡x+frac{cos⁡3x}{6}+x+C)
c) –(frac{3}{2} cos⁡x-frac{cos⁡3x}{6}-x+C)
d) –(frac{3}{2} cos⁡x+frac{cos⁡3x}{6}+C)
Answer: b
Clarification: We know that, sin⁡3x=3 sin⁡x-4 sin³⁡x
∴sin³x=(frac{3 sin⁡x-sin⁡3x}{4})
(int 2 ,sin^3⁡x+1 ,dx=int frac{(3 sin⁡x-sin⁡3x)}{2} dx+int dx)
=(frac{3}{2} int sin⁡x dx-frac{1}{2} int sin⁡3x dx+int dx)
=-(frac{3}{2} cos⁡x+frac{cos⁡3x}{6}+x+C)

Mathematics MCQs for Engineering Entrance Exams,

Mathematics Multiple Choice Questions on “Vector Algebra Basics”.

1. Which of the following holds true for a vector quantity?
a) It has only magnitude
b) It has only direction
c) A vector has both direction and magnitude
d) A vector can never be negative
Answer: c
Clarification: A quantity which has both magnitude and direction is called a vector quantity. The quantity which has only magnitude but no direction is called a scalar quantity.

2. A scalar quantity has only magnitude and no direction.
a) True
b) False
Answer: a
Clarification: The given statement is true. The quantity which has only magnitude and no direction is called scalar quantity.
Example: Time. Time is a scalar quantity which has only magnitude and no direction.

3. Which of the following id the correct symbol for denoting a given vector A?
a) |A|
b) [A]
c) (vec{A})
d) (underset{A}{rightharpoonup })
Answer: c
Clarification: The correct way denoting a vector given vector A is (vec{A}).

4. Which of the following is used to represent the magnitude of vector (vec{A})?
a) (hat{A})
b) |(vec{A})|
c) (underset{A}{leftrightarrow})
d) (vec{A})
Answer: b
Clarification: The correct way of representing the magnitude of a given vector A is|(vec{A})|. The magnitude of a vector is the distance between the initial and the terminal points of a vector.

5. For a given vector A, |(vec{A})| can be less than zero.
a) True
b) False
Answer: b
Clarification: The given statement is false.|(vec{A})| denotes the magnitude of the vector and since length of the line of the line can never be negative,|(vec{A})|<0 has no meaning.

6. Which of the given qualities is a vector?
a) Speed
b) Time
c) Weight
d) Volume
Answer: a
Clarification: Speed is a vector quantity as it has both magnitude and direction. Time, weight, volume have only magnitude and no direction. Hence, they all are scalar quantity.

7. Which of the following is not a vector quantity?
a) Speed
b) Density
c) Force
d) Velocity
Answer: b
Clarification: Density is a scalar quantity as it has only magnitude but no direction. Speed, force, velocity has both magnitude and direction. Therefore, they all are vectors.

8. Which of the below given is a vector quantity?
a) 8 kg
b) 4 seconds
c) 6 Newton
d) 90 cm³
Answer: c
Clarification: 6 Newton is a vector quantity as it is a force. Force is a vector quantity which has both magnitude and direction.

9. Which of the following is a scalar quantity?
a) 80m³
b) 5 Newton
c) 7 m/s towards east
d) 55m/s²
Answer: a
Clarification: 80 cm³ is a scalar quantity as it is denoting volume. Volume is a scalar quantity and has only magnitude but no direction.

10. If l, m, n are the direction cosines of a position vector (vec{a}), then which of the following is true?
a) l²+m²-n²=0
b) lmn=1
c) l²+m²+n²=1
d) l² m²+n²=1
Answer: c
Clarification: Consider (vec{a}) is the position vector of a point M(x,y,z) and α, β, γ are the angles, made by the vector (vec{a}) with the positive directions of x, y and z respectively. The cosines of the angles, cos⁡α, cos⁡β, cos⁡γ are the direction cosines of the vector (vec{a}) denoted by l, m, n, then
cos²⁡α+cos²⁡β+cos²⁡γ=1 i.e.l²+m²+n²=1.

Mathematics Multiple Choice Questions on “Probability – Independent Events”.

1. Events P and Q are independent if P(A∩B) = P(A) P(B).
a) False
b) True
Answer: b
Clarification: According to the definition of independent events,
P(A/B) = P(A)
P(A ∩ B) ⁄ P(B) = P(A) or P(A ∩ B) = P(A) P(B). Here, P(B) ≠ 0.

2. What are independent events?
a) If the outcome of one event does not affect the outcome of another
b) If the outcome of one event affects the outcome of another
c) Any one of the outcomes of one event does not affect the outcome of another
d) Any one of the outcomes of one event does affect the outcome of another
Answer: a
Clarification: Independent events refer to the outcome of one event does not affect the outcome of another. For example, if two persons participating in two different races, winning one person doesn’t affect the winning of another person.

3. A dice is thrown twice, what is the probability of getting two 3’s?
a) (frac {1}{66})
b) (frac {1}{16})
c) (frac {1}{36})
d) (frac {1}{36})
Answer: d
Clarification: Probability of getting 3 in the first throw = (frac {1}{6})
Probability of getting of 3 in the second throw = (frac {1}{6})
Total probability = (frac {1}{6} times frac {1}{6} = frac {1}{36})

4. What is the formula for independent events?
a) P(AB) = P(A) P(B)
b) P(A∩B) = P(A) P(B)
c) P(A+B) = P(A) P(B)
d) P(A-B) = P(A) P(B)
Answer: b
Clarification: Independent events refer to the outcome of one event does not affect the outcome of another. The formula for independent events are P(A∩B) = P(A) P(B).

5. What is the probability of obtaining 4 heads in a row when a coin is tossed?
a) (frac {5}{8})
b) (frac {6}{19})
c) (frac {1}{16})
d) (frac {4}{7})
Answer: c
Clarification: Probability of getting 4 heads in a row = (frac {1}{2} times frac {1}{2} times frac {1}{2} times frac {1}{2} = frac {1}{16})
hence, the probability of getting 4 heads in a row = (frac {1}{16})

6. A bag contains a pair of gloves in colours blue, red, yellow and pink. You reach into the
bag and choose a pair of gloves without looking. You replace this pair and then choose another pair of gloves. What is the probability that you will choose the pink pair of gloves both times?
a) (frac {3}{7})
b) (frac {4}{7})
c) (frac {1}{16})
d) (frac {1}{7})
Answer: c
Clarification: Probability of pink = (frac {1}{4})
Probability of choosing pink pair of gloves twice = (frac {1}{4} times frac {1}{4} = frac {1}{16})

7. A box contains a pair of socks in colours blue, red, yellow, green and pink. You reach into the box and choose a pair of socks without looking. You replace this pair and then choose another pair of socks. What is the probability that you will choose the yellow pair of socks both times?
a) (frac {8}{11})
b) (frac {1}{25})
c) (frac {4}{11})
d) (frac {7}{4})
Answer: b
Clarification: Probability of yellow = (frac {1}{5})
Probability of choosing pink pair of gloves twice = (frac {1}{5} times frac {1}{5} = frac {1}{25})

8. What is the probability of a coin landing on the tail and the dice showing 2 when a coin is tossed and dice is thrown?
a) (frac {1}{12})
b) (frac {8}{10})
c) (frac {7}{1})
d) (frac {4}{11})
Answer: a
Clarification: The probability of coin landing on the tail = (frac {1}{2})
The probability of the dice showing 2 = (frac {1}{6})
Total probability = (frac {1}{2} times frac {1}{6} = frac {1}{12})

9. A card is chosen at random from a deck of cards and then replaced, a second card is chosen. What is the probability of choosing a four and then a queen?
a) (frac {7}{11})
b) (frac {7}{100})
c) (frac {1}{169})
d) (frac {9}{11})
Answer: c
Clarification: The probability of drawing a four = (frac {4}{52})
The probability of drawing a queen = (frac {4}{52})
Total probability = (frac {4}{52} times frac {4}{52} = frac {1}{169} )

10. A bag contains 4 red, 2 green and 7 blue balls. What is the probability of drawing a red ball and the first ball drawn is blue? The balls drawn are replaced in the bag.
a) (frac {28}{169})
b) (frac {8}{11})
c) (frac {4}{128})
d) (frac {14}{11})
Answer: a
Clarification: Total number of balls = 4 + 2 + 7 = 13
The probability of the first ball to be blue = (frac {7}{13})
The probability of drawing red ball as the second ball = (frac {4}{13})
Total probability = (frac {7}{13} times frac {4}{13} = frac {28}{169})

11. A bag contains 3 red, 2 white and 4 green balls. What is the probability of drawing the second ball to be white and the first ball drawn is green? The balls are replaced in the bag.
a) (frac {1}{9})
b) (frac {2}{9})
c) (frac {8}{81})
d) (frac {2}{81})
Answer: c
Clarification: Total number of balls = 9
The probability of the first ball to be white = (frac {2}{9})
The probability of drawing green ball for the second time with replacement = (frac {4}{9})
Total probability = (frac {2}{9} times frac {4}{9} = frac {8}{81})

12. A bag contains 3 red, 2 white and 4 green balls. What is the probability of drawing the second ball to be green if the first ball drawn is red? The balls are replaced in the bag.
a) (frac {3}{9})
b) (frac {4}{27})
c) (frac {4}{3})
d) (frac {4}{17})
Answer: b
Clarification: Total number of balls = 9
The probability of the first ball to be red = (frac {3}{9})
The probability of the second ball to be green with replacement = (frac {4}{9})
Total probability = (frac {3}{9} times frac {4}{9} = frac {4}{27})

Mathematics Multiple Choice Questions on “Operations on Matrices”.

1. The addition of matrices is only possible if they are of the same order.
a) True
b) False
Answer: a
Clarification: The given statement is true. Addition of matrices is possible only if the matrices are of the same order. If there are two matrices of different order, then A+B is not defined.

2. If A = (begin{bmatrix}1&2&3\9&10&11end{bmatrix}) and B = (begin{bmatrix}0&5&0\5&0&5end{bmatrix}), then find A+B.
a) A+B = (begin{bmatrix}1&7&3\11&10&16end{bmatrix})
b) A+B = (begin{bmatrix}1&7&3\14&11&13end{bmatrix})
c) A+B = (begin{bmatrix}1&7&3\14&10&16end{bmatrix})
d) A+B = (begin{bmatrix}1&5&3\14&10&16end{bmatrix})
Answer: c
Clarification: Given that, A = (begin{bmatrix}1&2&3\9&10&11end{bmatrix}) and B = (begin{bmatrix}0&5&0\5&0&5end{bmatrix})
Then A+B = (begin{bmatrix}1+0&2+5&3+0\9+5&10+0&11+5end{bmatrix}) = (begin{bmatrix}1&7&3\14&10&16end{bmatrix}).

3. If A = (begin{bmatrix}3&4\1&2end{bmatrix}) and B = (begin{bmatrix}1&5\2&3end{bmatrix}), find 2A-3B.
a) (begin{bmatrix}3&7\-4&5end{bmatrix})
b) (begin{bmatrix}-3&-7\-4&-5end{bmatrix})
c) (begin{bmatrix}3&7\-4&-5end{bmatrix})
d) (begin{bmatrix}3&-7\-4&-5end{bmatrix})
Answer: d
Clarification: Given that, A = (begin{bmatrix}3&4\1&2end{bmatrix}) and B = (begin{bmatrix}1&5\2&3end{bmatrix})
⇒2A=2(begin{bmatrix}3&4\1&2end{bmatrix})=(begin{bmatrix}6&8\2&4end{bmatrix}) and 3B=3(begin{bmatrix}1&5\2&3end{bmatrix})=(begin{bmatrix}3&15\6&9end{bmatrix})
∴2A-3B = (begin{bmatrix}6&8\2&4end{bmatrix})–(begin{bmatrix}3&15\6&9end{bmatrix})=(begin{bmatrix}3&-7\-4&-5end{bmatrix}).

4. If A+B = (begin{bmatrix}6&7\5&0end{bmatrix})and A = (begin{bmatrix}2&5\1&-1end{bmatrix}). Find the matrix B.
a) B = (begin{bmatrix}4&1\2&4end{bmatrix})
b) B = (begin{bmatrix}4&2\4&1end{bmatrix})
c) B = (begin{bmatrix}4&1\4&2end{bmatrix})
d) B = (begin{bmatrix}4&4\4&2end{bmatrix})
Answer: b
Clarification: Given that, A+B = (begin{bmatrix}6&7\5&0end{bmatrix})and A = (begin{bmatrix}2&5\1&-1end{bmatrix})
⇒B=(A+B)-A = (begin{bmatrix}6&7\5&0end{bmatrix})–(begin{bmatrix}2&5\1&-1end{bmatrix})
B = (begin{bmatrix}4&2\4&1end{bmatrix})

5. Find the matrix M and N, if M+N = (begin{bmatrix}5&6\7&8end{bmatrix}),M-N = (begin{bmatrix}4&5\6&8end{bmatrix}).
a) M=1/2 (begin{bmatrix}9&11\13&16end{bmatrix}), N=1/2 (begin{bmatrix}1&1\1&0end{bmatrix})
b) M=(begin{bmatrix}5&6\7&8end{bmatrix}), N=(begin{bmatrix}4&5\8&6end{bmatrix})
c) M=1/2 (begin{bmatrix}9&2\13&16end{bmatrix}), N=1/2 (begin{bmatrix}1&1\2&5end{bmatrix})
d) M=1/2 (begin{bmatrix}4&5\1&2end{bmatrix}), N=1/2 (begin{bmatrix}1&2\1&2end{bmatrix})
Answer: a
Clarification:M+N = (begin{bmatrix}5&6\7&8end{bmatrix})-(1) and M-N = (begin{bmatrix}4&5\6&8end{bmatrix})-(2)
Adding equation (1) and equation (2), (M+N)+(M-N)=2M=(begin{bmatrix}5&6\7&8end{bmatrix})+(begin{bmatrix}4&5\6&8end{bmatrix})
M=1/2 (begin{bmatrix}9&11\13&16end{bmatrix}).
Subtracting equation (1) and equation (2), (M+N)-(M-N)=2N=(begin{bmatrix}5&6\7&8end{bmatrix})–(begin{bmatrix}4&5\6&8end{bmatrix})
N=1/2 (begin{bmatrix}1&1\1&0end{bmatrix}).

6. Find the value of x and y if 2(begin{bmatrix}5&x\y-4&6end{bmatrix})+(begin{bmatrix}-4&1\3&2end{bmatrix})=(begin{bmatrix}6&3\10&14end{bmatrix})?
a) x=-1, y=9
b) x=-1, y=-9
c) x=1, y=-9
d) x=1, y=9
Answer: d
Clarification: Given that, 2(begin{bmatrix}5&x\y-4&6end{bmatrix})+(begin{bmatrix}-4&1\3&2end{bmatrix})=(begin{bmatrix}6&3\10&14end{bmatrix})
⇒(begin{bmatrix}2(5)-4&2x+1\2(y-4)+3&2(6)+2end{bmatrix})=(begin{bmatrix}6&3\10&14end{bmatrix})
Comparing the two matrices, 2x+1=3, 2y-8=10
Solving the two equations we get, x=1, y=9.

7. Find AB if A = (begin{bmatrix}1&2\3&4end{bmatrix}) and B = (begin{bmatrix}1&5\3&2end{bmatrix}).
a) AB = (begin{bmatrix}15&23\9&7end{bmatrix})
b) AB = (begin{bmatrix}9&7\23&15end{bmatrix})
c) AB = (begin{bmatrix}7&9\15&23end{bmatrix})
d) AB = (begin{bmatrix}7&9\23&15end{bmatrix})
Answer: c
Clarification: Given that, A = (begin{bmatrix}1&2\3&4end{bmatrix}) and B = (begin{bmatrix}1&5\3&2end{bmatrix})
Then, AB = (begin{bmatrix}1&2\3&4end{bmatrix})(begin{bmatrix}1&5\3&2end{bmatrix})
=(begin{bmatrix}1×1+2×3&1×5+2×2\3×1+4×3&3×5+4×2end{bmatrix})=(begin{bmatrix}7&9\15&23end{bmatrix}).

8. Matrix addition and matrix multiplication both are commutative.
a) True
b) False
Answer: b
Clarification: The given statement is false. Matrix addition is commutative i.e. A+B=B+A. But matrix multiplication is not commutative i.e.AB≠BA.

9. Let A=(begin{bmatrix}3&-5&2\-4&-6&2\7&1&5end{bmatrix}). Find the additive inverse of A.
a) (begin{bmatrix}-3&5&-2\-4&6&2\7&1&5end{bmatrix})
b) (begin{bmatrix}3&-5&2\-4&-6&2\7&1&5end{bmatrix})
c) (begin{bmatrix}-3&5&-2\4&6&-2\-7&-1&-5end{bmatrix})
d) (begin{bmatrix}-3&5&2\-4&6&-2\-7&-1&5end{bmatrix})
Answer: c
Clarification: Additive inverse of matrix A is the negative of A i.e. -A.
Therefore, -A=(begin{bmatrix}-3&5&-2\4&6&-2\-7&-1&-5end{bmatrix})

10. Which of the following condition is incorrect for matrix multiplication?
a) A(BC)=(AB)C
b) A(B+C)=AB+AC
c) AB=0 if either A or B is 0
d) AB=BA
Answer: d
Clarification: Matrix multiplication is never commutative i.e. AB≠BA. Therefore, the condition AB=BA is incorrect.

Mathematics Multiple Choice Questions on “Differentiability”.

1. Find the derivative of f(x) = sin(x²).
a) -sin(x²)
b) 2xcos(x²)
c) -2xcos(x²)
d) -2xsin(x²)
Answer: b
Clarification: Differentiation of the function f(x) = sin(x²) is done with chain rule. First we differentiate sin function which becomes cos and then differentiate the inner (x²) which becomes 2x, hence it comes out to be 2xcos(x²).

2. What is derivative of xⁿ?
a) n
b) nxⁿ
c) nx^n-1
d) nx^n-2
Answer: c
Clarification: It is a standard rule for derivative of a function of this form in which the original power comes in front and the value in the power is decreased by one. Therefore the only option of this form is nx^n-1 from the given options.

3. Find derivative of tan(x+4).
a) sec²(x+4)
b) 4 sec²(x+4)
c) 4x sec²(x+4)
d) sec²(x)
Answer: a
Clarification: We know that derivative of tanx is sec²(x), now in the above question we get tan(x+4), hence its derivative comes out to be sec²(x+4), as the inside expression (x+4) is differentiated into 1.
Therefore the answer is sec²(x+4).

4. What is value of (frac{dy}{dx}) if x-y = 1?
a) 1
b) 2
c) -1
d) 2
Answer: a
Clarification: We know x-y = 1, hence we differentiate it on both sides-:
We get 1- (frac{dy}{dx}) = 0, (frac{dy}{dx}) = 1, hence the value of (frac{dy}{dx}) comes out to be 1.

5. Value after differentiating cos (sinx) is _________
a) sin (sinx).cosx
b) -sin (sinx).cosx
c) sin (sinx)
d) sin (cosx).cosx
Answer: b
Clarification: We differentiate the given function with the help of chain rule so we first differentiate the outer function which becomes –sin and then we differentiate the inner function sinx which is differentiated and comes out to be cosx, hence the differentiated function comes out to be -sin (sinx).cosx.

6. Value after differentiating cos (x²+5) is ________
a) 5.sin (x²+5)
b) -sin (x²+5).2x
c) sin (x²+5).2x
d) cos (x²+5).2x
Answer: b
Clarification: We differentiate the given function with help of chain rule and hence the outer function becomes –sin and the inner function is differentiated into 2x, therefore the answer comes out to be -sin (x²+5).2x.

7. Find (frac{dy}{dx}) of 2x+3y = sinx.
a) (frac{cosx-2}{3})
b) (frac{cosx-2}{2})
c) (frac{cosx-3}{2})
d) (frac{sinx-2}{3})
Answer: a
Clarification: Differentiating on both sides we get 2 + 3(frac{dy}{dx}) = cosx.
3(frac{dy}{dx}) = cosx-2.
(frac{dy}{dx} = frac{cosx-2}{3}).

8. What is derivative of cotx?
a) tanx
b) –sec²x
c) –cosec²x
d) cosec²x
Answer: c
Clarification: The derivative of cotx is –cosec²x, as this function has a fixed derivative like sinx has its derivative cosx. Therefore the answer to the above question is –cosec²x.

9. If (y = tan^{-1}(frac{3x-x^3}{1-3x^2}), frac{-1}{sqrt{3}} < x < frac{-1}{sqrt{3}})
a) 3
b) (frac{3}{1+x})
c) –(frac{3}{1+x^2})
d) (frac{3}{1+x^2})
Answer: d
Clarification: Given function is (y = tan^{-1}(frac{3x-x^3}{1-3x^2})),
Now taking RHS and substituting x = tang in it and then we get,
y = (tan^{-1}(frac{3tang-tang^3}{1-3tang^2})), Now it becomes the expansion of the function tan3g,
Hence the given function becomes y= tan^-1(tan3g). Which is equal to 3g, now substituting the value of g= tan^-1x, now after differentiating both sides we get the answer (frac{3}{1+x^2}).

10. Find (frac{dy}{dx}) of y = sin (ax + b).
a) a.cos (ax + b)
b) b.sin (ax + b)
c) a.sin (ax + b)
d) a.cos (ax + b)
Answer: a
Clarification: We differentiate the given function with the help of chain rule and hence
the outer function is differentiated into cos, and the inner function comes out to be a and the constant b becomes 0, which is multiplied to the whole function and the answer comes out to be
=> a.cos (ax + b).

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,