Category: Class 12 Mathematics Objective Questions

Mathematics Multiple Choice Questions & Answers (MCQs) on “Properties of Definite Integrals”.

1. What is the difference property of definite integrals?
a) (int_a^b)[-f(x)-g(x)dx
b) (int_a^b)[f(-x)+g(x)dx
c) (int_a^b)[f(x)-g(x)dx
d) (int_a^b)[f(x)+g(x)dx
Answer: c
Clarification: The sum difference property of definite integrals is (int_a^b)[f(x)-g(x)dx
(int_a^b)[f(x)-g(x)dx = (int_a^b)f(x)dx-(int_a^b)g(x)dx

2. The sum property of definite integrals is (int_a^b)[f(x)+g(x)dx?
a) False
b) True
Answer: b
Clarification: The sum property of definite integrals is (int_a^b)[f(x)+g(x)dx
(int_a^b)[f(x)+g(x)dx = (int_a^b)f(x)dx+(int_a^b)g(x)dx
Hence, it is true.

3. What is the constant multiple property of definite integrals?
a) (int_a^b)k⋅f(x)dy
b) (int_a^b)[f(-x)+g(x)dx
c) (int_a^b)k⋅f(x)dx
d) (int_a^b)[f(x)+g(x)dx
Answer: c
Clarification: The constant multiple property of definite integrals is (int_a^b)k⋅f(x)dx
(int_a^b)k⋅f(x)dx = k (int_a^b)f(x)dx

4. What is the reverse integral property of definite integrals?
a) –(int_a^b)f(x)dx=-(int_b^a)g(x)dx
b) –(int_a^b)f(x)dx=-(int_b^a)g(x)dx
c) (int_a^b)f(x)dx=(int_b^a)g(x)dx
d) (int_a^b)f(x)dx=-(int_b^a)f(x)dx
Answer: d
Clarification: In the reverse integral property the upper limits and lower limits are interchanged. The reverse integral property of definite integrals is (int_a^b)f(x)dx=-(int_b^a)f(x)dx.

5. Identify the zero-length interval property.
a) (int_a^b)f(x)dx = -1
b) (int_a^b)f(x)dx = 1
c) (int_a^b)f(x)dx = 0
d) (int_a^b)f(x)dx = 0.1
Answer: c
Clarification: The zero-length interval property is one of the properties used in definite integrals and they are always positive. The zero-length interval property is (int_a^b)f(x)dx = 0.

6. What is adding intervals property?
a) (int_a^c)f(x)dx+(int_b^c)f(x)dx = (int_a^c)f(x) dx
b) (int_a^b)f(x)dx+(int_b^a)f(x)dx = (int_a^c)f(x) dx
c) (int_a^b)f(x)dx+(int_b^c)f(x)dx = (int_a^c)f(x) dx
d) (int_a^b)f(x)dx-(int_b^c)f(x)dx = (int_a^c)f(x) dx
Answer: c
Clarification: The adding intervals property of definite integrals is (int_a^b)f(x)dx+(int_b^c)f(x)dx.
(int_a^b)f(x)dx+(int_b^c)f(x)dx = (int_a^c)f(x) dx

7. What is the name of the property of (int_a^b)f(x)dx+(int_b^c)f(x)dx = (int_a^c)f(x) dx?
a) Zero interval property
b) Adding intervals property
c) Adding integral property
d) Adding integrand property
Answer: b
Clarification: (int_a^b)f(x)dx+(int_b^c)(x)dx = (int_a^c)f(x) dx is a property of definite integrals. (int_a^b)f(x)dx+(int_b^c)f(x)dx = (int_a^c)f(x) dx is called as adding intervals property used to combine a lower limit and upper limit of two different integrals.

8. What is the name of the property (int_a^b)f(x)dx=-(int_b^a)f(x)dx?
a) Reverse integral property
b) Adding intervals property
c) Zero interval property
d) Adding integrand property
Answer: a
Clarification: In the reverse integral property the upper limits and lower limits are interchanged. The reverse integral property of definite integrals is (int_a^b)f(x)dx=-(int_b^a)f(x)dx.

9. What is the name of the property (int_a^b)f(x)dx = 0?
a) Reverse integral property
b) Adding intervals property
c) Zero-length interval property
d) Adding integrand property
Answer: b
Clarification: The zero-length interval property is one of the properties used in definite integrals and they are always positive. The zero-length interval property is (int_a^b)f(x)dx = 0.

10. What property this does this equation come under (int^1_{-1})sin⁡x dx=-(int_1^{-1})sin⁡x dx?
a) Reverse integral property
b) Adding intervals property
c) Zero-length interval property
d) Adding integrand property
Answer: a
Clarification: (int^1_{-1})sin⁡x dx=-(int_1^{-1})sin⁡x dx comes under the reverse integral property.
In the reverse integral property the upper limits and lower limits are interchanged. The reverse integral property of definite integrals is (int_a^b)f(x)dx=-(int_b^a)f(x)dx.

11. Evaluate (int_2^3)3f(x)-g(x)dx, if (int_2^3)f(x) = 4 and (int_2^3)g(x)dx = 4.
a) 38
b) 12
c) 8
d) 7
Answer: c
Clarification: (int_2^3)3f(x)-g(x)dx = 3 (int_2^3)f(x) – (int_2^3)g(x)dx
= 3(4) – 4
= 8

12. Compute (int_3^2)f(x) dx if (int_2^3)f(x) = 4.
a) – 4
b) 84
c) 2
d) – 8
Answer: c
Clarification: (int_3^2)f(x)dx = – (int_2^3)f(x)dx
= – 4

13. Compute (int_8^2)2f(x)dx if (int_2^8)f(x) = – 3.
a) – 4
b) 84
c) 2
d) – 8
Answer: c
Clarification: (int_8^2)2f(x)dx = -2 (int_2^8)f(x)dx
= – 2(-3)
= 6

14. Compute (int_2^6)7e^x dx.
a) 30.82
b) 7(e⁶ – e²)
c) 11.23
d) 81(e⁶ – e³)
Answer: b
Clarification: (int_2^6)7e^x dx = 7(e^x)⁶₂ dx
= 7(e⁶ – e²)

15. Evaluate (int_3^7)2f(x)-g(x)dx, if (int_3^7)f(x) = 4 and (int_3^7)g(x)dx = 2.
a) 38
b) 12
c) 6
d) 7
Answer: c
Clarification: (int_3^7)2f(x)-g(x)dx = 2 (int_3^7)f(x) – (int_3^7)g(x)dx
= 2(4) – 2
= 6

Mathematics Multiple Choice Questions on “Three Dimensional Geometry – Angle between Two Lines”.

1. If L₁ and L₂ have the direction ratios (a_1,b_1,c_1 ,and ,a_2,b_2,c_2) respectively then what is the angle between the lines?
a) (θ=tan^{-1}⁡left|frac{a_1 a_2+b_1 b_2+c_1 c_2}{sqrt{a_1^2+b_1^2+c_1^2} sqrt{a_2^2+b_2^2+c_2^2}}right |)
b) (θ=2tan^{-1}⁡left|frac{a_1 a_2+b_1 b_2+c_1 c_2}{sqrt{a_1^2+b_1^2+c_1^2} sqrt{a_2^2+b_2^2+c_2^2}}right |)
c) (θ=cos^{-1}⁡left|frac{a_1 a_2+b_1 b_2+c_1 c_2}{sqrt{a_1^2+b_1^2+c_1^2} sqrt{a_2^2+b_2^2+c_2^2}}right |)
d) (θ=2 ,cos^{-1}⁡left|frac{a_1 a_2+b_1 b_2+c_1 c_2}{sqrt{a_1^2+b_1^2+c_1^2} sqrt{a_2^2+b_2^2+c_2^2}}right |)
Answer: c
Clarification: If L₁ and L₂ have the direction ratios (a_1,b_1,c_1 ,and ,a_2,b_2,c_2) respectively then the angle between the lines is given by
(cos⁡θ=left|frac{a_1 a_2+b_1 b_2+c_1 c_2}{sqrt{a_1^2+b_1^2+c_1^2} sqrt{a_2^2+b_2^2+c_2^2}}right |)
(θ=cos^{-1}⁡left|frac{a_1 a_2+b_1 b_2+c_1 c_2}{sqrt{a_1^2+b_1^2+c_1^2} sqrt{a_2^2+b_2^2+c_2^2}}right |)

2. Find the angle between the lines.
(frac{x+2}{1}=frac{y+5}{6}=frac{z-3}{2})
(frac{x-4}{5}=frac{y-3}{-2}=frac{z+3}{1})
a) (cos^{-1}frac{⁡5}{sqrt{1230}})
b) (cos^{-1}⁡frac{⁡3}{sqrt{3120}})
c) (cos^{-1}⁡frac{⁡7}{sqrt{2310}})
d) (cos^{-1}frac{⁡⁡48}{sqrt{1230}})
Answer: a
Clarification: We know that, the angle between two lines is given by the formula
cos⁡θ=(left |frac{a_1 a_2+b_1 b_2+c_1 c_2}{sqrt{a_1^2+b_1^2+c_1^2} sqrt{a_2^2+b_2^2+c_2^2}}right |)
cos⁡θ=(left |frac{1(5)+6(-2)+2(1)}{sqrt{1^2+6^2+2^2).√(5^2+(-2)^2+1^2}}right |)
=(left |frac{-5}{sqrt{41}.sqrt{30}} right |=frac{5}{sqrt{1230}})
∴(θ=cos^{-1}frac{5}{sqrt{1230}})

3. Find the value of p such that the lines
(frac{x-1}{3}=frac{y+4}{p}=frac{z-9}{1})
(frac{x+2}{1}=frac{y-3}{1}=frac{z-7}{-2})
are at right angles to each other.
a) p=2
b) p=1
c) p=-1
d) p=-2
Answer: c
Clarification: The angle between two lines is given by the equation
(cos⁡θ=left |frac{a_1 a_2+b_1 b_2+c_1 c_2}{sqrt{a_1^2+b_1^2+c_1^2} sqrt{a_2^2+b_2^2+c_2^2}}right |)
cos⁡90°=(left |frac{3(1)+p(1)+1(-2)}{sqrt{3^2+p^2+1^2}.sqrt{1^2+1^2+(-2)^2}}right |)
0=(|frac{p+1}{sqrt{10+p^2}.√6}|)
0=p+1
p=-1

4. Find the angle between the two lines if the equations of the lines are
(vec{r}=hat{i}+hat{j}+hat{k}+λ(3hat{i}-hat{j}+hat{k}) ,and ,vec{r}=4hat{i}+hat{j}-2hat{k}+μ(2hat{i}+3hat{j}+hat{k}))
a) (cos^{-1}frac{⁡4}{sqrt{14}})
b) (cos^{-1}⁡frac{7}{sqrt{154}})
c) (cos^{-1}⁡frac{4}{154})
d) (cos^{-1}⁡frac{4}{sqrt{154}})
Answer: d
Clarification: Given that, (vec{r}=hat{i}+hat{j}+hat{k}+λ(3hat{i}-hat{j}+hat{k})) and (vec{r}=4hat{i}+hat{j}-2hat{k}+μ(2hat{i}+3hat{j}+hat{k}))
We know that, if the equations of two lines are of the form (vec{r}=vec{a_1}+λvec{b_1} and ,vec{r}=vec{a_2}+μvec{b_2}) then the angle between the two lines is given by
(cos⁡θ=left|frac{vec{b_1}.vec{b_2}}{|vec{b_1}||vec{b_2}|}right|)
=(left |frac{(3(2)+(-1)3+1(1)}{sqrt{3^2+(-1)^2+(1)^2.} sqrt{2^2+3^2+1^2}}right |=frac{4}{sqrt{11}.sqrt{14}}=frac{4}{sqrt{154}})
θ=(cos^{-1}⁡frac{4}{sqrt{154}}).

5. If two lines L₁ and L₂ with direction ratios (a_1,b_1,c_1 ,and ,a_2,b_2,c_2) respectively are perpendicular to each other then
(a_1 a_2+b_1 b_2+c_1 c_2=0)
a) True
b) False
Answer: a
Clarification: The given statement is true.
We know that the angle between two lines is given by the formula
cos⁡θ=(left |frac{a_1 a_2+b_1 b_2+c_1 c_2}{sqrt{a_1^2+b_1^2+c_1^2} sqrt{a_2^2+b_2^2+c_2^2}}right |)
So, if the lines L₁ and L₂ are perpendicular to each to each other then,
θ=90°
⟹(a_1 ,a_2+b_1 ,b_2+c_1 ,c_2)=0

6. Find the value of p such that the lines (frac{x+11}{4}=frac{y+3}{-2}=frac{z-3}{4} ,and ,frac{x-3}{p}=frac{y+12}{2}=frac{z-3}{-12}) are at right angles to each other.
a) p=11
b) p=12
c) p=13
d) p=4
Answer: c
Clarification: We know that, if two lines are perpendicular to each other then,
(a_1 a_2+b_1 b_2+c_1 c_2=0)
i.e.4(p)+(-2)2+4(-12)=0
4p-4-48=0
4p=52
p=(frac{52}{4})=13.

7. If the equations of two lines L₁ and L₂ are (vec{r}=vec{a_1}+λvec{b_1}) and (vec{r}=vec{a_2}+μvec{b_2}), then which of the following is the correct formula for the angle between the two lines?
a) cos⁡θ=(left |frac{vec{a_1}.vec{a_2}}{|vec{b_1}||vec{a_2}|}right |)
b) cos⁡θ=(left |frac{vec{a_1}.vec{a_2}}{|vec{a_1}||vec{a_2}|}right |)
c) cos⁡θ=(left |frac{vec{b_1}.vec{b_2}}{|vec{b_1}||vec{b_2}|}right |)
d) cos⁡θ=(left |frac{vec{a_1}.vec{b_2}}{|vec{a_1}||vec{b_2}|}right |)
Answer: c
Clarification: Given that the equations of the lines are
(vec{r}=vec{a_1}+λvec{b_1} ,and ,vec{r}=vec{a_2}+μvec{b_2})
∴ the angle between the two lines is given by
cos⁡θ=(left |frac{vec{b_1}.vec{b_2}}{|vec{b_1}||vec{b_2}|}right |).

8. Find the angle between the lines (vec{r}=2hat{i}+6hat{j}-hat{k}+λ(hat{i}-2hat{j}+3hat{k})) and (vec{r}=4hat{i}-7hat{j}+3hat{k}+μ(5hat{i}-3hat{j}+3hat{k})).
a) θ=(cos^{-1}frac{⁡20}{sqrt{602}})
b) θ=(cos^{-1}frac{⁡20}{sqrt{682}})
c) θ=(cos^{-1}frac{⁡8}{sqrt{602}})
d) θ=(cos^{-1}⁡frac{14}{sqrt{598}})
Answer: a
Clarification: If two lines have the equations (vec{r}=vec{a_1}+λvec{b_1} ,and ,vec{r}=vec{a_2}+μvec{b_2})
Then, the angle between the two lines will be given by
cos⁡θ=(left |frac{vec{b_1}.vec{b_2}}{|vec{b_1}||vec{b_2}|}right |)
=(left |frac{(hat{i}-2hat{j}+3hat{k}).(5hat{i}-3hat{j}+3hat{k})}{sqrt{1^2+(-2)^2+(3)^2).√(5^2+(-3)^2+3^2}}right |)
=(frac{5+6+9}{√14.√43}=frac{20}{√602})
θ=(cos^{-1}⁡frac{20}{sqrt{602}})

9. If two lines L₁ and L₂ are having direction cosines (l_1,m_1,n_1 ,and ,l_2,m_2,n_2) respectively, then what is the angle between the two lines?
a) cot⁡θ=(left |l_1 ,l_2+m_1 ,m_2+n_1 ,n_2right |)
b) sin⁡θ=(left |l_1 ,l_2+m_1 ,n_2+n_1 ,m_2right |)
c) tan⁡θ=(left |l_1 ,l_2+m_1 ,m_2+n_1 ,n_2right |)
d) cos⁡θ=(left |l_1 ,l_2+m_1 ,m_2+n_1 ,n_2right |)
Answer: d
Clarification: If two lines L₁ and L₂ are having direction cosines (l_1,m_1,n_1 ,and ,l_2,m_2,n_2) respectively, then the angle between the lines is given by
cos⁡θ=(left |l_1 ,l_2+m_1 ,m_2+n_1 ,n_2right |)

10. Find the angle between the pair of lines (frac{x-3}{5}=frac{y+7}{3}=frac{z-2}{2} ,and ,frac{x+1}{3}=frac{y-5}{4}=frac{z+2}{8}).
a) (cos^{-1}⁡frac{43}{sqrt{3482}})
b) (cos^{-1}⁡⁡frac{43}{sqrt{3382}})
c) (cos^{-1}⁡⁡frac{85}{sqrt{3382}})
d) (cos^{-1}⁡⁡frac{34}{sqrt{3382}})
Answer: b
Clarification: The direction ratios are 5, 3, 2 for L₁ and 3, 4, 8 for L₂
∴ the angle between the two lines is given by
cos⁡θ=(frac{(a_1 a_2+b_1 b_2+c_1 c_2)}{sqrt{a_1^2+b_1^2+c_1^2} sqrt{a_2^2+b_2^2+c_2^2}})
=(frac{15+12+16}{sqrt{5^2+3^2+2^2}.sqrt{3^2+4^2+8^2}})
=(frac{43}{sqrt{38}.sqrt{89}}=frac{43}{sqrt{3382}})
θ=(cos^{-1}⁡frac{43}{sqrt{3382}}).

Mathematics Question Papers for Engineering Entrance Exams on “Calculus Application – Motion in a Straight Line – 2”.

1. A particle moving along a straight line with uniform acceleration, passes successively through three points A, B, C. If t₁ = time taken to go from A to B; t₂ = time taken to go from B to C and AB = a, BC = b. If a = b and the velocities of the particle at A, B, C be p, q, r respectively, then, what is the relation between p², q² and r²?
a) A.P
b) G.P
c) H.P
d) They are in any arbitrary series
Answer: a
Clarification: When a = b, then the equation of the motion of the particle from A to B is,
q² = p² + 2fa
Or q² – p² = 2fa ……….(1)
Again, the equation of motion of the particle from B to C is,
r² = q² + 2fa
Or r² – q² = 2fa ……….(2)
From (1) and (2) we get,
q² – p² = r² – q²
Thus, p², q² and r² are in A.P.

2. Two motor cars on the same line approach each other with velocities u₁ and u₂ respectively. When each is seen from the other, the distance between them is x. If f₁ and f₂ to be the maximum retardation of the two cars then a collision can be just avoided then at which condition collision can just be avoided?
a) (u₁²f₂ – u₂²f₁) = 2f₁f₂(x)
b) (u₁²f₂ + u₂²f₁) = 2f₁f₂(x)
c) (u₁²f₂ + u₂²f₁) = f₁f₂(x)
d) (u₁²f₂ – u₂²f₁) = f₁f₂(x)
Answer: b
Clarification: By question the two motor cars approach each other along the same line.
Let P and Q be the positions of the cars on the line when each is seen from the other, where PQ = x.
It is evident that a collision can be just avoided, if the two cars stop at the point somewhere between P and Q that is if velocities of both the motor cars at O are zero.
Since the initial velocity of the motor car is at P is u₁ and it comes to rest at O, hence, its equation is
0 = u₁² – 2f₁(PO)
Or (PO) = u₁²/2f₁
Again, the initial velocity of the motor car at Q is due to and it comes to rest at O; hence its equation of motion is,
0 = u₂² – 2f₂(QO)
Or (QO) = u₂²/2f₂
Now, x = PQ = PO + OQ = u₁²/2f₁ + u₂²/2f₂
Or u₁²f₂ + u₂²f₁)/2f₁f₂ = x
Thus, (u₁²f₂ + u₂²f₁) = 2f₁f₂(x).

3. One motor car A stands 24m in front of a motorcycle B. Both starts from rest along a straight road in the same direction. If A moves with uniform acceleration of 2 m/sec² and B runs with a uniform velocity of 9 m/sec, is it possible for B to overtake A?
a) No
b) Yes
c) Data not sufficient
d) Answer cannot be determined
Answer: a
Clarification: Let P and Q be the initial positions of the motor car and motorcycle B respectively, where PQ = 24m.
If possible, let us assume that B overtakes A after point R on the straight road after time t seconds from the start.
Then, considering the motion of motor car A, we get
PR = 1/2 (2) (t²) = t²
Again, considering the motion of motorcycle B we get QR = 9t.
Now,QP+PR = QR
Or 24 + t² = 9t
Or t² – 9t + 24 = 0
Or t = [9 ± √(81 – 4*24)]/2
Or t = (9 ± √-15)/2
Clearly, the values of t are imaginary.
Therefore, there is no real positive time when B overtake A.
Hence, the in this case B will never overtake A.

4. One motor car A stands 24m in front of a motorcycle B. Both starts from rest along a straight road in the same direction. If A moves with uniform acceleration of 2 m/sec² and B runs at a uniform velocity of 11 m/sec then when will they meet?
a) After 3 seconds from start when the motorcycle B will overtake the motor car A
b) After 4 seconds from start when the motorcycle B will overtake the motor car A
c) After 2 seconds from start when the motorcycle B will overtake the motor car A
d) After 6 seconds from start when the motorcycle B will overtake the motor car A
Answer: a
Clarification: Let P and Q be the initial positions of the motor car and motorcycle B respectively, where PQ = 24m.
If possible, let us assume that B overtakes A after point R on the straight road after time t seconds from the start.
Then, considering the motion of motor car A, we get
PR = 1/2 (2) (t²) = t²
In this case when B runs ata uniform velocity of 11 m/sec, we shall have, QR=11t.
Therefore, in this case, QP + PR = QR gives,
Or 24 + t² = 11t
Or t² – 11t + 24 = 0
Or (t – 3)(t – 8) = 0
Or t = 3 or t = 8
Clearly, we are getting two real positive values of t.
Therefore, A and B will meet twice during the motion.
They will first meet after 3 seconds from start when the motorcycle B will overtake the motor car A.

5. One motor car A stands 24m in front of a motorcycle B. Both starts from rest along a straight road in the same direction. If A moves with uniform acceleration of 2 m/sec², B runs at a uniform velocity of 11 m/sec how many times will they meet?
a) 3
b) 2
c) 4
d) 1
Answer: b
Clarification: Let P and Q be the initial positions of the motor car and motorcycle B respectively, where PQ = 24m.
If possible, let us assume that B overtakes A after point R on the straight road after time t seconds from the start.
Then, considering the motion of motor car A, we get
PR = 1/2 (2) (t²) = t²
In this case when B runs at a uniform velocity 11 m/sec, we shall have, QR = 11t.
Therefore, in this case, QP + PR = QR gives,
Or 24 + t² = 11t
Or t² – 11t + 24 = 0
Or (t – 3)(t – 8) = 0
Or t = 3 or t = 8
Clearly, we are getting two real positive values of t.
Therefore, A and B will meet twice during the motion.

6. One motor car A stands 24m in front of a motorcycle B. Both starts from rest along a straight road in the same direction. If A moves with uniform acceleration of 2 m/sec², then what will happen if B runs at a uniform velocity of 11 m/sec?
a) A will not overtake B
b) A will again overtake B and they will never meet again and again
c) A will again overtake B and they will meet again
d) A will again overtake B and they will never meet again
Answer: d
Clarification: Let P and Q be the initial positions of the motor car and motorcycle B respectively, where PQ = 24m.
If possible, let us assume that B overtakes A after point R on the straight road after time t seconds from the start.
Then, considering the motion of motor car A, we get
PR = 1/2 (2) (t²) = t²
In this case when B runs at a uniform velocity 11 m/sec, we shall have, QR = 11t.
Therefore, in this case, QP + PR = QR gives,
Or 24 + t² = 11t
Or t² – 11t + 24 = 0
Or (t – 3)(t – 8) = 0
Or t = 3 or t = 8
Clearly, we are getting two real positive values of t.
Therefore, A and B will meet twice during the motion.
They will first meet after 3 seconds from start when the motorcycle B will overtake the motor car A.
But the velocity of A continuously increases, hence after a further period of (8 – 3) = 5 seconds A will again overtake B and they will never meet again.

7. Two particles start from rest from the same point and move along the same straight path; the first starts with a uniform velocity of 20 m/minute and the second with a uniform acceleration of 5 m/ min². Before they meet again, when will their distance be maximum?
a) At t = 8
b) At t = 6
c) At t = 4
d) At t = 2
Answer: c
Clarification: Suppose, the two particle starts from rest at and move along the straight path OA.
Further assume that the distance between the particle is maximum after t minutes from start (before they meet again).
If we be the position of the particle after 30 minutes from the start which moves with uniform acceleration 5 m/min² and C that of the particle moving with uniform velocity 20 m/min, then we shall have,
OB = 1/2(5t²) and OC = 20t
If the distance between the particle after t minutes from start be x m, then,
x = BC = OC – OB = 20t – (5/2)t²
Now, dx/dt = 20 – 5t and d²x/dt² = -5
For maximum or minimum values of x, we have,
dx/dt = 0
Or 20 – 5t = 0
Or t = 4
And [d²y/dx²] = -5 < 0
Thus, x is maximum at t = 4.

8. Two particles start from rest from the same point and move along the same straight path; the first starts with a uniform velocity of 20 m/minute and the second with a uniform acceleration of 5 m/ min². Before they meet again, then what will be the maximum distance?
a) 38m
b) 36m
c) 42m
d) 40m
Answer: d
Clarification: Suppose, the two particle starts from rest at and move along the straight path OA.
Further assume that the distance between the particle is maximum after t minutes from start (before they meet again).
If we be the position of the particle after 30 minutes from the start which moves with uniform acceleration 5 m/min² and C that of the particle moving with uniform velocity 20 m/min, then we shall have,
OB = 1/2(5t²) and OC = 20t
If the distance between the particle after t minutes from start be x m, then,
x = BC = OC – OB = 20t – (5/2)t² ……….(1)
Now, dx/dt = 20 – 5t and d²x/dt² = -5
For maximum or minimum values of x, we have,
dx/dt = 0
Or 20 – 5t = 0
Or t = 4
And [d²y/dx²] = -5 < 0
Thus, x is maximum at t = 4.
Therefore, the maximum value is,
= 20*4 – (5/2)(4²) [putting t = 4 in (1)]
= 40 m

9. An express train is running behind a goods train on the same line and in the same direction, their velocities being u₁ and u₂ (u₁ > u₂) respectively. When there is a distance x between them, each is seen from the other. At which point it is just possible to avoid a collision f₁ is the greatest retardation and f₂ is the greatest acceleration which can be produced in the two trains respectively?
a) (u₁ – u₂)² = 2x(f₁ + f₂)
b) (u₁ + u₂)² = 2x(f₁ – f₂)
c) (u₁ – u₂)² = 2x(f₁ – f₂)
d) (u₁ + u₂)² = 2x(f₁ + f₂)
Answer: c
Clarification: Let A be the position of the express train and B, that of the goods train when each is seen from the other, where AB = x.
Evidently, to avoid a collision the express train will apply the greatest possible retardation f₁ and the goods train will apply the greatest possible acceleration f₂.
Now, it is just possible to avoid a collision,if the velocities of the two trains at a point C on the line are equal, when they just touch each other at C(because after the instant the velocity of the express train will decrease due to retardation f₁ and that of the goods train will increase due to acceleration f₂).
Let the two trains reach the point C, at time t after each is seen by the other and their equal velocities at C bev and BC = s.
Then the equation of motion of the express train is,
v = u₁ – f₁t ……….(1)   And x + s = u₁t – (1/2)f₁t² ……….(2)
And the equation of the motion of the goods train is,
v = u₂ + f₂t ……….(3)   And s = u₂t – (1/2)f₂t² ……….(4)
From (1) and (3) we get,
u₁ – f₁t = u₂ + f₂t
Or (f₁+ f₂)t = u₁ – u₂
Or t = (u₁ – u₂)/(f₁ + f₂)
Again, (2)–(4) gives,
x = (u₁ – u₂)t – ½(t²)(f₁ + f₂)
= t[(u₁ – u₂) – (1/2)(t)(f₁ + f₂)]
As, (f₁ + f₂)t = (u₁ – u₂)
So, x = (u₁ – u₂)/(f₁ + f₂)[(u₁ – u₂) – (1/2)(u₁ – u₂)]
So, (u₁ – u₂)² = 2x(f₁ – f₂)

Mathematics Multiple Choice Questions on “Properties of Determinants”.

1. Which of the following is not a property of determinant?
a) The value of determinant changes if all of its rows and columns are interchanged
b) The value of determinant changes if any two rows or columns are interchanged
c) The value of determinant is zero if any two rows and columns are identical
d) The value of determinant gets multiplied by k, if each element of row or column is multiplied by k
Answer: a
Clarification: The value of determinant remains unchanged if all of its rows and columns are interchanged i.e. |A|=|A’|, where A is a square matrix and A’ is the transpose of the matrix A.

2. Find the determinant of the matrix A=(begin{bmatrix}1&x&y\1&x&-y\1&-x^2&y^2end{bmatrix}).
a) (x+1)
b) -2xy(x+1)
c) xy(x+1)
d) 2xy(x+1)
Answer: b
Clarification: Given that, A=(begin{bmatrix}1&x&y\1&x&-y\1&-x^2&y^2end{bmatrix})
Δ=(begin{vmatrix}1&x&y\1&x&-y\1 &-x^2&y^2 end{vmatrix})
Taking x common C₂ and y common from C₃, we get
Δ=xy(begin{vmatrix}1&1&1\1&1&-1\1&-x&yend{vmatrix})
Expanding along R₁, we get
Δ=xy{1(y-x)-1(y+1)+1(-x-1)}
Δ=xy(y-x-y-1-x-1)
Δ=xy(-2x-2)=-2xy(x+1).

3. Evaluate (begin{vmatrix}x^2&x^3&x^4\x&y&z\x^2&x^3&x^4 end{vmatrix}).
a) 0
b) 1
c) xyz
d) x² yz³
Answer: a
Clarification: Δ=(begin{vmatrix}x^2&x^3&x^4\x&y&z\x^2&x^3&x^4 end{vmatrix})
If the elements of any two rows or columns are identical, then the value of determinant is zero. Here, the elements of row 1 and row 3 are identical. Hence, its determinant is 0.

4. Evaluate (begin{vmatrix}cos⁡θ&-cos⁡θ&1\sin^2⁡θ&cos^2⁡θ&1\sin⁡θ&-sin⁡θ&1end{vmatrix}).
a) sin⁡θ+cos²⁡θ
b) -sin⁡θ-cos²⁡⁡θ
c) -sin⁡θ+cos²⁡⁡θ
d) sin⁡θ-cos²⁡⁡θ
Answer: d
Clarification: Δ=(begin{vmatrix}cos⁡θ&-cos⁡θ&1\sin^2⁡θ&cos^2⁡θ&1\sin⁡θ&-sin⁡θ&1end{vmatrix})
Applying C₁→C₁+C₂
Δ=(begin{vmatrix}cos⁡θ-cos⁡θ&-cos⁡θ&1\sin^2⁡θ+cos^2⁡θ&cos^2⁡θ&1\sinθ-sin⁡θ&-sin⁡θ&1end{vmatrix})=(begin{vmatrix}0&-cos⁡θ&1\1&cos^2⁡θ&1\0&-sin⁡θ&1end{vmatrix})
Expanding along C₁, we get
0-1(cos²⁡⁡θ+sinθ)=sin⁡θ-cos²⁡⁡θ.

5. Evaluate (begin{vmatrix}b-c&b&c\a&c-a&c\a&b&a-bend{vmatrix}).
a) 2abc
b) 2a{(b-c)(c-a+b)}
c) 2b{(a-c)(a+b+c)}
d) 2c{(b-c)(a-c+b)}
Answer: b
Clarification: Δ=(begin{vmatrix}b-c&b&c\a&c-a&c\a&b&a-bend{vmatrix})
Applying C₂→C₂-C₃
Δ=(begin{vmatrix}b-c&b-c&c\a&-a&c\a&-a&a-bend{vmatrix})
Applying C₁→C₁-C₂
Δ=(begin{vmatrix}0&b-c&c\2a&-a&c\2a&-a&a-bend{vmatrix})
Applying R₂→R₂-R₃
Δ=(begin{vmatrix}0&b-c&c\0&0&c-a+b\2a&-a&a-bend{vmatrix})
Expanding along C₁, we get
Δ=2a{(b-c)(c-a+b)}

6. If A=(begin{bmatrix}1&3\2&1end{bmatrix}), then ________
a) |2A|=4|A|
b) |2A|=2|A|
c) |A|=2|A|
d) |A|=|4A|
Answer: a
Clarification: Given that, A=(begin{bmatrix}1&3\2&1end{bmatrix})
2A=2(begin{bmatrix}1&3\2&1end{bmatrix})=(begin{bmatrix}2&6\4&2end{bmatrix})
|2A|=(begin{vmatrix}2&6\4&2end{vmatrix})=(4-24)=-20
4|A|=4(begin{vmatrix}1&3\2&1end{vmatrix})=4(1-6)=4(-5)=-20
∴|2A|=4|A|.

7. Evaluate (begin{vmatrix}-a&b&c\-2a+4x&2b-4y&2c+4z\x&-y&zend{vmatrix}).
a) 0
b) abc
c) 2abc
d) -1
Answer: a
Clarification: Δ=(begin{vmatrix}-a&b&c\-2a+4x&2b-4y&2c+4z\x&-y&zend{vmatrix})
Using the properties of determinants, the given determinant can be expressed as a sum of two determinants.
Δ=(begin{vmatrix}-a&b&c\-2a&2b&2c\x&-y&zend{vmatrix})+(begin{vmatrix}-a&b&c\4x&-4y&4z\x&-y&zend{vmatrix})
Δ=2(begin{vmatrix}-a&b&c\-a&b&c\x&-y&zend{vmatrix})+4(begin{vmatrix}-a&b&c\x&-y&z\x&-y&zend{vmatrix})
Since two rows are similar in each of the determinants, the determinant is 0.

8. Find the determinant of A=(begin{bmatrix}c^2&cb&ca\ab&a^2&-ac\ab&bc&-b^2end{bmatrix})
a) abc(a³+b³+c³+abc)
b) abc(a³+b³+c³-abc)
c) abc(a³+b³+c³+abc)
d) (a³-b³+c³-abc)
Answer: b
Clarification: Given that, A=(begin{bmatrix}c^2&cb&ca\ab&a^2&-ac\ab&bc&-b^2end{bmatrix})
Taking c a, b common from R₁, R₂, R₃ respectively, we get
Δ=abc(begin{bmatrix}c&b&a\b&a&-c\a&c&-bend{bmatrix})
Δ=abc{(c(-ab+c²)-b(-b²+ac)+a(bc-a²)
Δ=abc(-abc+c³+b³-abc+abc-a³)
Δ=abc(a³+b³+c³-abc).

9. Evaluate (begin{vmatrix}1+m&n&q\m&1+n&q\n&m&1+qend{vmatrix}).
a) -1(1+m+n+q)
b) 1+m+n+q
c) 1+2q
d) 1+q
Answer: a
Clarification: Given that, Δ=(begin{vmatrix}1+m&n&q\m&1+n&q\n&m&1+qend{vmatrix})
Applying C₁→C₁+C₂+C₃
Δ=(begin{vmatrix}1+m+n+q&n&q\1+m+n+q&1+n&q\1+m+n+q&m&1+qend{vmatrix})=(1+m+n+q)(begin{vmatrix}1&n&q\1&1+n&q\1&m&1+qend{vmatrix})
Applying R₁→R₂-R₁
Δ=(1+m+n+q)(begin{vmatrix}0&1&0\1&1+n&q\1&m&1+qend{vmatrix})
Expanding along the first row, we get
Δ=(1+m+n+q)(0-1(1+q-q)+0)
Δ=-1(1+m+n+q).

10. Evaluate (begin{vmatrix}4&8&12\6&12&18\7&14&21end{vmatrix}).
a) 168
b) -1
c) -168
d) 0
Answer: d
Clarification: Δ=(begin{vmatrix}4&8&12\6&12&18\7&14&21end{vmatrix})
Taking 4, 6 and 7 from R₁, R₂, R₃ respectively
Δ=4×6×7(begin{vmatrix}1&2&3\1&2&3\1&2&3end{vmatrix})
Since the elements of all rows are identical, the determinant is zero.

Mathematics Interview Questions and Answers on “Derivatives Application – Rate of Change of Quantities”.

1. If the rate of change of radius of a circle is 6 cm/s then find the rate of change of area of the circle when r=2 cm.
a) 74.36 cm²/s
b) 75.36 cm²/s
c) 15.36 cm²/s
d) 65.36 cm²/s
Answer: b
Clarification: The rate of change of radius of the circle is (frac{dr}{dt})=6 cm/s
The area of a circle is A=πr²
Differentiating w.r.t t we get,
(frac{dA}{dt}=frac{d}{dt}) (πr²)=2πr (frac{dr}{dt})=2πr(6)=12πr.
(frac{dA}{dt})|_r=2=24π= 24×3.14=75.36 cm²/s

2. The edge of a cube is increasing at a rate of 7 cm/s. Find the rate of change of area of the cube when x=6 cm.
a) 578 cm²/s
b) 498 cm²/s
c) 504 cm²/s
d) 688 cm²/s
Answer: c
Clarification: Let the edge of the cube be x. The rate of change of edge of the cube is given by (frac{dx}{dt})=7cm/s.
The area of the cube is A=6x²
∴(frac{dA}{dt}=frac{d}{dt} )(6x²)=12x.(frac{dx}{dt})=12x×7=84x
(frac{dA}{dt})|__x=6=84×6=504 cm²/s.

3. The rate of change of area of a square is 40 cm²/s. What will be the rate of change of side if the side is 10 cm.
a) 2 cm/s
b) 4 cm/s
c) 8 cm/s
d) 6 cm/s
Answer: a
Clarification: Let the side of the square be x.
A=x², where A is the area of the square
Given that, (frac{dA}{dt})=2x (frac{dx}{dt})=40 cm²/s.
(frac{dx}{dt}=frac{20}{x} )cm/s
(frac{dx}{dt}=frac{20}{10})=2 cm/s.

4. The total cost P(x) in rupees associated with a product is given by P(x)=0.4x²+2x-10. Find the marginal cost if the no. of units produced is 5.
a) Rs.3
b) Rs.4
c) Rs.5
d) Rs.6
Answer: d
Clarification: The Marginal cost is the rate of change of revenue w.r.t the no. of units produced, we get
(frac{dP(x)}{dt})=0.8x+2
cost(MC)=(frac{dP(x)}{dt}|)_x=5=0.8x+2=0.8(5)+2=4+2=6.

5. At what rate will the lateral surface area of the cylinder increase if the radius is increasing at the rate of 2 cm/s when the radius is 5 cm and height is 10 cm?
a) 40 cm/s
b) 40π cm/s
c) 400π cm/s
d) 20π cm/s
Answer: b
Clarification: Let r be the radius and h be the height of the cylinder. Then,
(frac{dr}{dt})=2 cm/s
The area of the cylinder is given by A=2πrh
(frac{dA}{dt})=2πh((frac{dr}{dt}))=4πh=4π(10)=40π cm/s.

6. If the circumference of the circle is changing at the rate of 5 cm/s then what will be rate of change of area of the circle if the radius is 6cm.
a) 20 cm²/s
b) 40 cm²/s
c) 70 cm²/s
d) 30 cm²/s
Answer: d
Clarification: The circumference of the circle is given by C=2πr, where r is the radius of the circle.
∴(frac{dC}{dt})=2π.(frac{dr}{dt})=5 cm/s
(frac{dr}{dt})=5/2π cm/s
(frac{dA}{dt})|_r=6=2πr.(frac{dr}{dt})=2πr.(frac{5}{2 pi})=5r=5(6)=30 cm²/s.

7. The total cost N(x) in rupees, associated with the production of x units of an item is given by N(x)=0.06x³-0.01x²+10x-43. Find the marginal cost when 5 units are produced.
a) Rs. 1.44
b) Rs. 144.00
c) Rs. 14.4
d) Rs. 56.2
Answer: b
Clarification: The marginal cost is given by the rate of change of revenue.
Hence, (frac{dN(x)}{dt})=0.18x²-0.02x+10.
∴(frac{dN(x)}{dt})|__x=5=0.18(5)²-0.02(5)+10
=4.5-0.1+10
=Rs. 14.4

8. The length of the rectangle is changing at a rate of 4 cm/s and the area is changing at the rate of 8 cm/s. What will be the rate of change of width if the length is 4cm and the width is 1 cm.
a) 5 cm/s
b) 6 cm/s
c) 2 cm/s
d) 1 cm/s
Answer: d
Clarification: Let the length be l, width be b and the area be A.
The Area is given by A=lb
(frac{dA}{dt})=l.(frac{db}{dt})+b.(frac{dl}{dt}) -(1)
Given that, (frac{dl}{dt})=4cm/s and (frac{dA}{dt})=8 cm/s
Substituting in the above equation, we get
8=l.(frac{db}{dt})+4b
Given that, l=4 cm and b=1 cm
∴8=4((frac{db}{dt}))+4(1)
8=4((frac{db}{dt}))+4
(frac{db}{dt})=1 cm/s.

9. For which of the values of x, the rate of increase of the function y=3x²-2x+7 is 4 times the rate of increase of x?
a) -1
b) (frac{1}{3})
c) 1
d) 0
Answer: c
Clarification: Given that, (frac{dy}{dt}=4.frac{dx}{dt})
y=3x²-2x+7
(frac{dy}{dt})=(6x-2) (frac{dx}{dt})
4.(frac{dx}{dt})=(6x-2) (frac{dx}{dt})
4=6x-2
6x=6
⇒x=1

10. The volume of a cube of edge x is increasing at a rate of 12 cm/s. Find the rate of change of edge of the cube when the edge is 6 cm.
a) (frac{1}{8})
b) (frac{2}{9})
c) –(frac{1}{9})
d) (frac{1}{9})
Answer: d
Clarification: Let the volume of cube be V.
V=x³
(frac{dV}{dt})=3x² (frac{dx}{dt})
12=3x² (frac{dx}{dt})
(frac{dx}{dt}=frac{4}{x^2})
(frac{dx}{dt})|__x=6=(frac{4}{6^2}=frac{4}{36}=frac{1}{9}).

Mathematics for Interviews,