Category: Class 12 Mathematics Objective Questions

Mathematics Assessment Questions and Answers on “Determinant – 2”.

1. Evaluate (begin{vmatrix}3&-1&3\6&-5&4\3&-2&3end{vmatrix})
a) 100
b) 223
c) 240
d) 230
Answer: c
Clarification: Expanding along R₁, we get
Δ=(begin{vmatrix}3&-1&3\6&-5&4\3&-2&3end{vmatrix})
Δ=3(begin{vmatrix}-5&45\-2&3end{vmatrix})-(-1)(begin{vmatrix}6&4\3&3end{vmatrix})+3(begin{vmatrix}6&-5\3&-2end{vmatrix})
Δ=3(-15+90)+(18-12)+3(-12+15)
Δ=3(75)+6+9=240.

2. Evaluate (begin{vmatrix}1&0&1\0&0&1\1&0&1end{vmatrix}).
a) 2
b) 0
c) 1
d) -1
Answer: b
Clarification: Δ=(begin{vmatrix}1&0&1\0&0&1\1&0&1end{vmatrix})
Δ=1(begin{vmatrix}0&1\0&1end{vmatrix})-0(begin{vmatrix}0&1\1&1end{vmatrix})+1(begin{vmatrix}0&0\1&0end{vmatrix})
Δ=1(0-0)-0(0-1)+1(0-0)
Δ=0-0+0=0.

3. Evaluate |A|²-5|A|+1, if A=(begin{bmatrix}7&4\5&5end{bmatrix})
a) 161
b) 251
c) 150
d) 151
Answer: d
Clarification: Given that, A=(begin{bmatrix}7&4\5&5end{bmatrix})
|A|=(7(5)-5(4))=35-20=15
|A|²-5|A|+1=(15)²-5(15)+1=225-75+1=151.

4. Evaluate (begin{vmatrix}sin⁡ ,y&0&sin⁡ ,y\cos⁡ ,y&1&cos⁡ ,x\sin⁡ ,y&1&sin ⁡,y end{vmatrix})
a) sin⁡ y (cos⁡ y-cos⁡ x)
b) sin⁡ x (cos⁡ y-cos⁡ x)
c) sin ⁡x (cos⁡ x-cos⁡ y)
d) sin ⁡y (cos⁡ 2y-cos⁡ x)
Answer: a
Clarification: Δ=(begin{vmatrix}sin⁡ ,y&0&sin⁡ ,y\cos⁡ ,y&1&cos⁡ ,x\sin⁡ ,y&1&sin⁡ ,y end{vmatrix})
Δ=sin⁡ y (begin{vmatrix}1&cos⁡ ,x\1&sin ,⁡y end{vmatrix})-0(begin{vmatrix}cos ⁡,y&cos⁡ ,x \sin⁡ ,y&sin ,y end{vmatrix})+sin⁡ y (begin{vmatrix}cos ⁡,y&1\sin ,⁡y&1end{vmatrix})
Δ=sin ⁡y (sin⁡ y-cos⁡ x)-0+sin ⁡y (cos⁡ y-sin ⁡y)
Δ=sin²⁡y-sin ⁡y cos⁡ x+sin⁡ y cos ⁡y-sin²⁡y=sin ⁡y (cos⁡ y-cos⁡ x)

5. Find the value of x, if (begin{vmatrix}2&5\3&xend{vmatrix})=(begin{vmatrix}x&-1\5&3end{vmatrix}).
a) 20
b) -20
c) 30
d) -30
Answer: b
Clarification: (begin{vmatrix}2&5\3&xend{vmatrix})=(begin{vmatrix}x&-1\5&3end{vmatrix})
⇒2x-15=3x+5
⇒x=-20

6. Find the value of x, if (begin{vmatrix}1&-1\3&-5end{vmatrix})=(begin{vmatrix}x&x^2\3&5end{vmatrix}).
a) x=2, –(frac{1}{3})
b) x=-1, –(frac{1}{3})
c) x=-2, –(frac{1}{3})
d) x=0, –(frac{1}{3})
Answer: a
Clarification: Given that, (begin{vmatrix}1&-1\3&-5end{vmatrix})=(begin{vmatrix}x&x^2\3&5end{vmatrix})
-5—(-3)=5x-3x²
-2=5x-3x²
3x²-5x-2=0
Solving for x, we get
x=2, –(frac{1}{3}).

7. Which of the following matrices will not have a determinant?
a) (begin{bmatrix}4&2\5&4end{bmatrix})
b) (begin{bmatrix}1&5&4\3&6&2\4&8&7end{bmatrix})
c) (begin{bmatrix}5&8&9\3&4&6end{bmatrix})
d) (begin{bmatrix}1&2\5&4end{bmatrix})
Answer: c
Clarification: Determinant of the matrix A=(begin{bmatrix}5&8&9\3&4&6end{bmatrix}) is not possible as it is a rectangular matrix and not a square matrix. Determinants can be calculated only if the matrix is a square matrix.

8. Find the determinant of the matrix A=(begin{bmatrix}9&8\7&6end{bmatrix})
a) -1
b) 1
c) 2
d) -2
Answer: d
Clarification: Given that, A=(begin{bmatrix}9&8\7&6end{bmatrix})
⇒Δ=(begin{vmatrix}9&8\7&6end{vmatrix})=9(6)-7(8)=54-56=-2

9. Find the determinant of the matrix A=(begin{bmatrix}-cos⁡θ&-tan⁡θ\cot⁡θ &cos⁡θ end{bmatrix}).
a) sin²⁡θ
b) sin⁡θ
c) -sin⁡θ
d) -sin²⁡θ
Answer: a
Clarification: Given that, A=(begin{bmatrix}-cos⁡θ&-tan⁡θ\cot⁡ θ&cos⁡θ end{bmatrix})
|A|=(begin{vmatrix}-cos⁡θ&-tan⁡θ\cot⁡θ&cos⁡θ end{vmatrix})
|A|=-cos⁡θ (cos⁡θ )-cotθ(-tan⁡θ)
|A|=-cos²⁡θ+1=sin²⁡θ.

10. Evaluate (begin{vmatrix}5&0&5\1&4&3\0&8&6end{vmatrix}).
a) 20
b) 0
c) -40
d) 40
Answer: b
Clarification: Δ=(begin{vmatrix}5&0&5\1&4&3\0&8&6end{vmatrix})
Expanding along R₁, we get
Δ=5(begin{vmatrix}4&3\8&6end{vmatrix})-0(begin{vmatrix}1&3\0&6end{vmatrix})+5(begin{vmatrix}1&4\0&8end{vmatrix})
Δ=5(24-24)-0+5(8-0)
Δ=0-0+40=40.

Mathematics Assessment Questions,

Mathematics Multiple Choice Questions on “Application of Derivative”.

1. What is the slope of the tangent to the curve y = 2x/(x² + 1) at (0, 0)?
a) 0
b) 1
c) 2
d) 3
Answer: c
Clarification: We have y = 2x/(x² + 1)
Differentiating y with respect to x, we get
dy/dx = d/dx(2x/(x² + 1))
= 2 * [(x² + 1)*1 – x * 2x]/(x² + 1)²
= 2 * [1 – x²]/(x² + 1)²
Thus, the slope of tangent to the curve at (0, 0) is,
[dy/dx]_{(0, 0)} = 2 * [1 – 0]/(0 + 1)²
Thus [dy/dx]_{(0, 0)} = 2.

2. The value of f’(x) is -1 at the point P on a continuous curve y = f(x). What is the angle which the tangent to the curve at P makes with the positive direction of x axis?
a) π/2
b) π/4
c) 3π/4
d) 3π/2
Answer: c
Clarification: Let, Φ be the angle which the tangent to the curve y = f(x) at P makes with the positive direction of the x axis.
Then,
tanΦ = [f’(x)]_p = -1
= -tan(π/4)
So, it is clear that this can be written as,
= tan(π – π/4)
= tan(3π/4)
So, Φ = 3π/4
Therefore, the required angle which the tangent at P to the curve y = f(x) makes with positive direction of x axis is 3π/4.

3. What will be the differential function of √(x² + 2)?
a) x√(x² + 2) dx
b) x/√(x² + 2) dx
c) x/√(x² – 2) dx
d) -x/√(x² + 2) dx
Answer: b
Clarification: Let, y = f(x) = √(x² + 2)
So, f(x) = (x² + 2)^1/2
On differentiating it we get,
f’(x) = d/dx[(x² + 2)^1/2]
f’(x) = 1/2 * 1/√(x² + 2) * 2x
So f’(x) = x/√(x² + 2)
So the differential equation is:
dy = f’(x)dx
Hence, dy = x/√(x² + 2) dx

4. What will be the differential function of log(x² + 4)?
a) 2x/(x² + 4) dx
b) 2x/(x² – 4) dx
c) -2x/(x² + 4) dx
d) -2x/(x² – 4) dx
Answer: a
Clarification: Let, y = f(x) = log(x² + 4)
So f(x) = log(x² + 4)
On differentiating it we get,
f’(x) = d/dx[log(x² + 4)]
So f’(x) = 2x/(x² + 4)
So the differential equation is:
dy = f’(x)dx
Hence, dy = 2x/(x² + 4) dx

5. What will be the average rate of change of the function [y = 16 – x²] between x = 3 and x = 4?
a) 7
b) -7
c) 9
d) -9
Answer: b
Clarification: Let, y = f(x) = 16 – x²
If x changes from 3 to 4, then, δx = 4 – 3 = 1
Again f(4) = 16 – 4² = 0
And f(3) = 16 – 3² = 7
Therefore, δy = f(4) – f(3) = 0 – 7 = -7
Hence, the average rate of change of the function between x = 3 and x = 4 is:
δy/δx = -7/1 = -7.

6. What will be the average rate of change of the function [y = 16 – x²] at x = 4?
a) -8
b) 8
c) -9
d) Depends on the value of x
Answer: a
Clarification: Let, y = f(x) = 16 – x²
dy/dx = -2x
Now, [dy/dx]_{x = 4} = [-2x]_{x = 4}
So, [dy/dx]_{x = 4} = -8

7. What will be the value of the co-ordinate whose position of a particle moving along the parabola y² = 4x at which the rate at of increase of the abscissa is twice the rate of increase of the ordinate?
a) (1, 1)
b) (2, 2)
c) (3, 3)
d) (4, 4)
Answer: d
Clarification: Here, y² = 4x ……….(1)
Let, (x, y) be the position of the particle moving along the parabola (1) at time t.
Now, differentiating both sides of (1) with respect to t, we get:
2y(dy/dt) = 4(dx/dt)
Or, y(dy/dt) = 2(dy/dt) ……….(2)
By question, dx/dt = 2 * dy/dt ……….(3)
From (2) and (3) we get, y(dy/dt) = 2 * 2 dy/dt
Or, y = 4
Putting y = 4 in (1) we get, 4² = 4x
So, x = 4
Thus, the co-ordinate of the particle is (4, 4).

8. The time rate of change of the radius of a sphere is 1/2π. When it’s radius is 5cm, what will be the rate of change of the surface of the sphere with time?
a) 10 sq cm
b) 20 sq cm
c) 30 sq cm
d) 40 sq cm
Answer: b
Clarification: Let, r be the radius and s be the area of the surface of the sphere at time t.
By question, dr/dt = 1/2π
Now, s = 4πr²;
Thus, ds/dt = 4π * 2r(dr/dt) = 8πr(dr/dt)
When r = 5cm and dr/dt = 1/2π
Then ds/dt = 8π*5*(1/2π) = 20
Thus, correct answer is 20 sq cm.

9. A solid cube changes its volume such that its shape remains unchanged. For such a cube of unit volume, what will be the value of rate of change of volume?
a) 3/8*(rate of change of area of any face of the cube)
b) 3/4*(rate of change of area of any face of the cube)
c) 3/10*(rate of change of area of any face of the cube)
d) 3/2*(rate of change of area of any face of the cube)
Answer: d
Clarification: Let x be the length of a side of the cube.
If v be the volume and s the area of any face of the cube, then
v = x³ and s = x²
Thus, dv/dt = dx³/dt = 3x² (dx/dt)
And ds/dt = dx²/dt = 2x(dx/dt)
Now, (dv/dt)/(ds/dt) = 3x/2
Or, dv/dt = (3x/2)(ds/dt)
Now, for a cube of unit volume we have,
v = 1
=>x = 1 [as, x is real]
Therefore, for a cube of unit volume [i.e. for x = 1], we get,
dv/dt = (3/2)(ds/dt)
Thus the rate of change of volume = 3/2*(rate of change of area of any face of the cube)

10. A 5 ft long man walks away from the foot of a 12(½) ft high lamp post at the rate of 3 mph. What will be the rate at which the shadow increases?
a) 0mph
b) 1mph
c) 2mph
d) 3mph
Answer: c
Clarification: Let, AB be the lamp-post whose foot is A, and B is the source of light, and given (AB)’ = 12(½) ft.
Let MN denote the position of the man at time t where (MN)’ = 5ft.
Join BN and produce it to meet AM(produced) at P.
Then the length of man’s shadow= (MP)’
Assume, (AM)’ = x and (MP)’ = y. Then,
(PA)’ = (AM)’ + (MP)’ = x + y
And dx/dt = velocity of the man = 3
Clearly, triangles APB and MPN are similar.
Thus, (PM)’/(MN)’ = (PA)’/(AB)’
Or, y/5 = (x + y)/12(½)
Or, (25/2)y = 5x + 5y
Or, 3y = 2x
Or, y = (2/3)x
Thus, dy/dt = (2/3)(dx/dt)
As, dx/dt = 2,
= 2/3*3 = 2mph

11. A ladder 20 ft long leans against a vertical wall. If the top end slides downwards at the rate of 2ft per second, what will be the rate at which the slope of the ladder changes?
a) -19/54
b) -21/54
c) -23/54
d) -25/54
Answer: d
Clarification: Let the height on the wall be x and laser touches the ground at distance y from the wall. The length of the ladder is 20ft.
By Pythagoras theorem:
x² + y² = 400
Differentiating with respect to t:
2x(dx/dt) + 2y(dy/dt) = 0
Dividing throughout by 2:
x (dx/dt) + y (dy/dt) = 0
Now, dx/dt = -2 ft /s. negative because downwards
x(-2) + y (dy/dt) = 0 ………..(1)
When lower end is 12 ft from wall, let us find x:
x² + 12² = 400
x² = 400 – 144= 256
x = 16
x(2) + y (dy/dt) = 0 from (1)
16(-2) + 12 (dy/dt) = 0
-32 + 12(dy/dt) = 0
dy/dt = (32/12) = (8/3)
Thus, lower end moves on a horizontal floor when it is 12 ft from the wall at the rate of 8/3 ft/s
Now, assume that the ladder makes an angle θ with the horizontal plane at time t.
If, m be the slope of the ladder at time t, then,
m = tanθ = x/y
Thus, dm/dt = d/dt(x/y) = [y(dx/dt) – x(dx/dt)]/y²
Therefore, the rate of change of slope of ladder is,
[dm/dt]_{y = 2} = [12*(-2) – 16*(8/3)]/(12)²
Now, putting the value of x = 16, when y = 12 and dx/dt = -2, dy/dt = 8/3
We get, [dm/dt]_{y = 12} = [12(-2) – 16(8/3)]/(12)² = -25/54

12. A particle moving in a straight line covers a distance of x cm in t second, where x = t³ + 6t² – 15t + 18. What will be the velocity of the particle at the end of 2 seconds?
a) 20cm/sec
b) 22cm/sec
c) 21cm/sec
d) 23cm/sec
Answer: c
Clarification: We have, x = t³ + 6t² – 15t + 18
Let, v be the velocity of the particle at the end of t seconds. Then,
v = dx/dt = d/dt(t³ + 6t² – 15t + 18)
So, v = 3t² + 12t – 15
Thus, velocity of the particle at the end of 2 seconds is,
[dx/dt]_{t = 2} = 3(2)² + 12(2) – 15 = 21cm/sec.

13. A particle moving in a straight line covers a distance of x cm in t second, where x = t³ + 6t² – 15t + 18. What will be the acceleration of the particle at the end of 2 seconds?
a) 22cm/sec²
b) 23cm/sec²
c) 24cm/sec²
d) 25cm/sec²
Answer: c
Clarification: We have, x = t³ + 6t² – 15t + 18
Let, ‘v’ be the velocity of the particle and ‘a’ be the acceleration of the particle at the end of t seconds. Then,
v = dx/dt = d/dt(t³ + 6t² – 15t + 18)
So, v = 3t² + 12t – 15
Therefore, a = dv/dt = d/dt(3t² + 12t – 15)
So, a = 6t + 12
Thus, acceleration of the particle at the end of 2 seconds is,
[dv/dt]_{t = 2} = 6(2) + 12 = 24cm/sec².

14. A particle moving in a straight line covers a distance of x cm in t second, where x = t³ + 6t² – 15t + 18. When does the particle stop?
a) 1/4 second
b) 1/3 second
c) 1 second
d) 1/2 second
Answer: c
Clarification: We have, x = t³ + 6t² – 15t + 18
The particle stops when dx/dt = 0
And, dx/dt = d/dt(t³ + 6t² – 15t + 18)
3t² + 12t – 15 = 0
=>t² + 4t –5 = 0
=>(t – 1)(t + 5)= 0
Thus, t = 1 or t = -5
Hence, the particle stops at the end of 1 second.

15. Water is flowing into a right circular conical vessel, 45 cm deep and 27 cm in diameter at the rate of 11 cc per minute. How fast is the water level rising when the water is 30 cm deep?
a) 0.033cm/minute
b) 0.043cm/minute
c) 0.053cm/minute
d) 0.045cm/minute
Answer: b
Clarification: Let ‘r’ be the radius and ‘h’ be the height of the water level at time t.
Then the volume of water level ‘V’ at time t is given by,
V = 1/3 (πr²h) ……….(1)
Given the radius of the base of the cone is = (OA)’ = 27/2 cm and its height = (OB)’ = 45cm.
Again at time t, the radius of the water level = r = (CD)’ and its height = h = (CB)’
Clearly, the triangle OAB and CBD are similar.
Therefore, (CD)’/(CB)’ = (OA)’/(OB)’
Or, r/h = (22/7)/45 = 3/10
or, r = 3h/10
Thus, from (1) we get,
V = 1/3 π (3h/10)² h = (3π/100)h³
Thus, dV/dt = (3π/100)*3h²(dh/dt)
= (9πh²/100)(dh/dt)
When, h = 30cm, then,
11 = (9π/100) (30)² (dh/dt) [as for all the values of t we have, dV/dt = 11]
Or, dh/dt = (11/(9π*9)) = 0.043
Thus, the rising rate of rising is 0.043 cm/minute.

Mathematics Online Quiz for IIT JEE Exam on “Evaluation of Definite Integrals by Substitution”.

1. Evaluate the integral (int_0^{frac{π^2}{4}} frac{9 sin⁡sqrt{x}}{2sqrt{x}} dx).
a) 9
b) -9
c) (frac{9}{2})
d) –(frac{9}{2})
Answer: a
Clarification: I=(int_0^{frac{π^2}{4}} frac{9 sin⁡sqrt{x}}{2sqrt{x}} dx)
Let (sqrt{x})=t
Differentiating both sides w.r.t x, we get
(frac{1}{2sqrt{x}} dx=dt)
The new limits are
When x=0 , t=0
When x=(frac{π^2}{4}, t=frac{π}{2})
∴(int_0^{frac{π^2}{4}} frac{9 sin⁡sqrt{x}}{2sqrt{x}} dx=9int_0^{π/2} sin⁡t ,dt)
=(9[-cos⁡t]_0^{π/2})=-9(cos⁡ π/2-cos⁡0)=-9(0-1)=9

2. Find (int_0^1 20x^3 e^{x^4}) dx.
a) (e-1)
b) 5(e+1)
c) 5e
d) 5(e-1)
Answer: d
Clarification: I=(int_0^1 20x^3 e^{x^4}) dx
Let x⁴=t
Differentiating w.r.t x, we get
4x³ dx=dt
∴The new limits
When x=0, t=0
When x=1,t=1
∴(int_0^1 ,20x^3 ,e^{x^4} ,dx=int_0^1 5e^t dt)
(=5[e^t]_0^1=5(e^1-e^0))=5(e-1).

3. Find (int_{-1}^1 frac{5x^4}{sqrt{x^5+3}} dx).
a) 4-(sqrt{2})
b) 4+2(sqrt{2})
c) 4-2(sqrt{2})
d) 1-2(sqrt{2})
Answer: c
Clarification: I=(int_{-1}^1 frac{5x^4}{sqrt{x^5+3}} dx)
Let x⁵+3=t
Differentiating w.r.t x, we get
5x⁴ dx=dt
The new limits
when x=-1,t=2
when x=1,t=4
∴(int_{-1}^1 frac{5x^4}{sqrt{x^5+3}} dx=int_2^4 frac{dt}{sqrt{t}})
=([2sqrt{t}]_2^4=2(sqrt{4}-sqrt{2})=4-2sqrt{2})

4. Find (int_0^{frac{sqrt{π}}{2}} 2x ,cos⁡ x^2 ,dx).
a) 1
b) (frac{1}{sqrt{2}})
c) –(frac{1}{sqrt{2}})
d) (sqrt{2})
Answer: b
Clarification: I=(int_0^{frac{sqrt{π}}{2}} ,2x ,cos⁡ x^2 ,dx)
Let x²=t
Differentiating w.r.t x, we get
2x dx=dt
The new limits
When x=0,t=0
When (x={frac{sqrt{π}}{2}}, t=frac{π}{4})
∴(int_0^{frac{sqrt{π}}{2}} ,2x ,cos⁡ x^2 ,dx=int_0^{frac{π}{4}} ,cos⁡t ,dt)
(I =[sin⁡t]_0^{frac{π}{4}}=sin⁡ frac{π}{4}-sin⁡0=1/sqrt{2}).

5. Evaluate the integral (int_1^6 frac{sqrt{x}+3}{sqrt{x}} ,dx).
a) 9
b) (frac{9}{2})
c) –(frac{9}{2})
d) (frac{4}{5})
Answer: b
Clarification: I=(int_1^4 frac{sqrt{x}+3}{sqrt{x}} ,dx)
Let (sqrt{x}+3=t)
Differentiating w.r.t x, we get
(frac{1}{2sqrt{x}} ,dx=dt)
(frac{1}{sqrt{x}} ,dx=2 ,dt)
The new limits
When x=1,t=4
When x=4,t=5
∴(int_1^4 frac{sqrt{x}+3}{sqrt{x}} dx=int_4^5 ,t ,dt)
=([frac{t^2}{2}]_4^5=frac{5^2-4^2}{2}=frac{9}{2})

6. Find (int_1^2 frac{12 ,log⁡x}{x} ,dx).
a) -12 log⁡2
b) 24 log⁡2
c) 12 log⁡2
d) 24 log⁡4
Answer: b
Clarification: I=(int_1^2 frac{12 log⁡x}{x} ,dx)
Let log⁡x=t
Differentiating w.r.t x, we get
(frac{1}{x} ,dx=dt)
The new limits
When x=1,t=0
When x=2,t=log⁡2
(int_1^2 frac{12 log⁡x}{x} dx=12int_0^{log⁡2} ,t ,dt)
=(12[t^2]_0^{log⁡2}=12((log⁡2)^2-0))
=12 log⁡4=24 log⁡2(∵(log⁡2)²=log⁡2.log⁡2=log⁡4=2 log⁡2)

7. Find (int_0^{π/4} frac{5 ,sin⁡(tan^{-1}⁡x)}{1+x^2} ,dx).
a) 5-(frac{1}{sqrt{2}})
b) 5+(frac{5}{sqrt{2}})
c) -5+(frac{5}{sqrt{2}})
d) 5-(frac{5}{sqrt{2}})
Answer: d
Clarification: I=(int_0^1 frac{5 ,sin⁡(tan^{-1)}x}{1+x^2} ,dx)
Let tan^-1⁡x=t
Differentiating w.r.t x, we get
(frac{1}{1+x^2} ,dx=dt)
The new limits
When x=0, t=tan^-1⁡0=0
When x=1, t=tan^-1)1=π/4
∴(int_0^1 frac{5 ,sin⁡(tan^{-1}⁡x)}{1+x^2} ,dx=int_0^{π/4} ,5 ,sin⁡t ,dt)
=(5[-cos⁡t]_0^{π/4}=-5[cos⁡t]_0^{π/4})
(=-5(cos⁡ frac{π}{4}-cos⁡0)=-5(frac{1}{sqrt{2}-1})=5-frac{5}{sqrt{2}})

8. Find (int_{-1}^1 ,7x^6 ,(x^7+8)dx)
a) -386
b) –(frac{386}{3})
c) (frac{386}{3})
d) 386
Answer: c
Clarification: I=(int_{-1}^1 ,7x^6 ,(x^7+8)dx)
Let x⁷+8=t
Differentiating w.r.t x, we get
7x⁶ dx=dt
The new limits
When x=-1,t=7
When x=1,t=9
∴(int_{-1}^1 ,7x^6 ,(x^7+8)dx=int_7^9 ,t^2 ,dt)
=([frac{t^3}{3}]_7^9=frac{1}{3} (9^3-7^3)=frac{386}{3}).

9. Evaluate (int_{sqrt{2}}^2 ,14x ,log⁡ x^2 ,dx)
a) 14(3 log⁡2-1)
b) 14(3 log⁡2+1)
c) log⁡2-1
d) 3 log⁡2-1
Answer: a
Clarification: I=(int_{sqrt{2}}^2 ,14x ,log⁡ x^2 ,dx)
Let x²=t
Differentiating w.r.t x, we get
2x dx=dt
The new limits
When x=(sqrt{2}), t=2
When x=2, t=4
∴(int_{sqrt{2}}^2 ,14x ,log⁡ x^2 ,dx =int_2^4 ,7 ,log⁡ t ,dt)
Using integration by parts, we get
(int_2^4 ,7 ,log⁡ t ,dt=7(log⁡ tint dt-int (log⁡t)’ int ,dt))
=7 (t log⁡t-t)₂⁴
=7(4 log⁡4-4-2 log⁡2+2)
=7(6 log⁡2-2)=14(3 log⁡2-1)

10. Find (int_2^3 ,2x^2 ,e^{x^3} ,dx).
a) (e^{27}-e^8)
b) (frac{2}{3} (e^{27}-e^8))
c) (frac{2}{3} (e^8-e^{27}))
d) (frac{2}{3} (e^{27}+e^8))
Answer: b
Clarification: I=(int_2^3 ,2x^2 ,e^{x^3} ,dx)
Let x³=t
Differentiating w.r.t x, we get
3x² dx=dt
x² dx=(frac{dt}{3})
The new limits
When x=2, t=8
When x=3, t=27
∴(int_2^3 ,2x^2 ,e^{x^3} ,dx=frac{2}{3} int_8^{27} ,e^t ,dt)
=(frac{2}{3} [e^t]_8^{27}=frac{2}{3} (e^{27}-e^8).)

Mathematics Online Quiz for IIT JEE Exam,

Mathematics Question Papers for IIT JEE Exam on “Three Dimensional Geometry – Equation of a Line in Space”.

1. Find the vector equation of the line which is passing through the point (2,-3,5) and parallel to the vector (3hat{i}+4hat{j}-2hat{k}).
a) ((2+3λ) hat{i}+(4λ+3) hat{j}+(5-λ)hat{k})
b) ((9+3λ) hat{i}+(λ-3) hat{j}+(5-2λ)hat{k})
c) ((2+3λ) hat{i}+(4λ-3) hat{j}+(5-2λ)hat{k})
d) ((7+λ) hat{i}+(4λ+3) hat{j}+(5-2λ)hat{k})
Answer: c
Clarification: Given that the line is passing through the point (2,-3,5). Therefore, the position vector of the line is (vec{a}=2hat{i}-3hat{j}+5hat{k}).
Also given that, the line is parallel to a vector (vec{b}=3hat{i}+4hat{j}-2hat{k}).
We know that, the equation of line passing through a point and parallel to vector is given by (vec{r}=vec{a}+λvec{b}), where λ is a constant.
∴(vec{r}=2hat{i}-3hat{j}+5hat{k}+λ(3hat{i}+4hat{j}-2hat{k}))
=((2+3λ) hat{i}+(4λ-3) hat{j}+(5-2λ)hat{k})

2. If the line is passing through the points ((x_1, y_1, z_1)) and has direction cosines l, m, n of the line, then which of the following is the cartesian equation of the line?
a) (frac{x-x_1}{l}=frac{y-y_1}{m}=frac{z-z_1}{n})
b) (frac{x-x_1}{n}=frac{y-y_1}{m}=frac{z-z_1}{l})
c) (frac{x+x_1}{n}=frac{y+y_1}{m}=frac{z-z_1}{l})
d) (frac{x+x_1}{l}=frac{y+y_1}{m}=frac{z+z_1}{n})
Answer: a
Clarification: If the line is passing through the points (x₁, y₁, z₁) and has direction cosines l,m,n of the line, then the cartesian equation of the line is given by
(frac{x-x_1}{l}=frac{y-y_1}{m}=frac{z-z_1}{n}).

3. If a line is passing through two points (A(x_1,y_1,z_1)) and (B(x_2,y_2,z_2)) then which of the following is the vector equation of the line?
a) (vec{r}=vec{a}+λ(vec{b}+vec{a}))
b) (vec{r}=vec{a}+λ(vec{a}-vec{b}))
c) (vec{r}=λvec{a}+(vec{b}-vec{a}))
d) (vec{r}=vec{a}+λ(vec{b}-vec{a}))
Answer: d
Clarification: Let (vec{a} ,and, vec{b}) are the position vectors of two points (A(x_1,y_1,z_1)) and (B(x_2,y_2,z_2))
Then the vector equation of the line is given by the formula passing through two points will be given by
(vec{r}=vec{a}+λ(vec{b}-vec{a})), λ∈R

4. Find the vector equation of a line passing through two points (-5,3,1) and (4,-3,2).
a) ((-5+λ) hat{i}+(3+λ)hat{j}+(1-λ) hat{k})
b) ((-5+λ) hat{i}+(3+6λ)hat{j}+(1+λ) hat{k})
c) ((5+7λ) hat{i}+(8+6λ)hat{j}+(3-5λ) hat{k})
d) ((-5+9λ) hat{i}+(3-6λ)hat{j}+(1+λ) hat{k})
Answer: d
Clarification: Consider the points A(-5,3,1) and B(4,-3,2)
Let (vec{a} ,and, vec{b}) be the position vectors of the points A and B.
∴(vec{a}=-5hat{i}+3hat{j}+hat{k})
(vec{b}=4hat{i}-3hat{j}+2hat{k})
∴(vec{r}=-5hat{i}+3hat{j}+hat{k}+λ(4hat{i}-3hat{j}+2hat{k}-(-5hat{i}+3hat{j}+hat{k})))
=-(5hat{i}+3hat{j}+hat{k}+λ(9hat{i}-6hat{j}+hat{k}))
=((-5+9λ) hat{i}+(3-6λ)hat{j}+(1+λ) hat{k})

5. Find the cartesian equation of a line passing through two points (1,-9,8) and (4,-1,6).
a) (frac{x+1}{3}=frac{y-9}{8}=frac{-z-8}{2})
b) (frac{x-1}{3}=frac{y+9}{8}=frac{z-8}{-2})
c) (frac{x-1}{7}=frac{y+9}{-2}=frac{z-8}{5})
d) (frac{2x-1}{3}=frac{6y+9}{8}=frac{4z-8}{-2})
Answer: b
Clarification: The position vector for the point A(1,-9,8) and B(4,-1,6)
(vec{a}=hat{i}-9hat{j}+8hat{k})
(vec{b}=4hat{i}-hat{j}+6hat{k})
∴(vec{r}=vec{a}+λ(vec{b}-vec{a}))
The above vector equation can be expressed in cartesian form as:
(frac{x-x_1}{x_2-x_1}=frac{y-y_1}{y_2-y_1}=frac{z-z_1}{z_2-z_1})
∴ The cartesian equation for the given line is (frac{x-1}{3}=frac{y+9}{8}=frac{z-8}{-2})

6. Find the cartesian equation of the line which is passing through the point (5,-6,1) and parallel to the vector (5hat{i}+2hat{j}-3hat{k}).
a) (frac{x-5}{5}=frac{y-4}{6}=frac{z+1}{-3})
b) (frac{x-5}{5}=frac{z+6}{2}=frac{y-1}{3})
c) (frac{x+5}{4}=frac{y-8}{2}=frac{z-1}{-3})
d) (frac{x-5}{5}=frac{y+6}{2}=frac{z-1}{-3})
Answer: d
Clarification: The equation of a line passing through a point and parallel to a vector is given by
(vec{r}=vec{a}+λvec{b})
(vec{a}) is the position vector of the given point ∴(vec{a}=5hat{i}-6hat{j}+hat{k})
(vec{b}=5hat{i}+2hat{j}-3hat{k}).
(vec{r}=5hat{i}-6hat{j}+hat{k}+λ(5hat{i}+2hat{j}-3hat{k}))
(xhat{i}+yhat{j}+zhat{k}=(5+5λ) hat{i}+(2λ-6) hat{j}+(1-3λ) hat{k})
∴(frac{x-5}{5}=frac{y+6}{2}=frac{z-1}{-3})
is the cartesian equation of the given line.

7. Find the cartesian equation of a line passing through two points (8,-5,7) and (7,1,4).
a) (frac{x-8}{1}=frac{y+5}{-6}=frac{z-7}{3})
b) (frac{2x-8}{-1}=frac{3y+5}{6}=frac{4z-7}{-3})
c) (frac{x-8}{-1}=frac{y+5}{6}=frac{z-7}{-3})
d) (frac{x-8}{-2}=frac{y+5}{5}=frac{z-7}{-7})
Answer: c
Clarification: Consider A(8,-5,7) and B(7,1,4)
i.e. ((x_1,y_1,z_1))=(8,-5,7) and ((x_2,y_2,z_3))=(7,1,4)
The cartesian equation for a line passing through two points is given by
(frac{x-x_1}{x_2-x_1}=frac{y-y_1}{y_2-y_1}=frac{z-z_1}{z_2-z_1})
∴ the cartesian equation for the given line is (frac{x-8}{-1}=frac{y+5}{6}=frac{z-7}{-3})

8. Find the vector equation of a line passing through two points (1,0,4) and (6,-3,1).
a) ((1+5λ) hat{i}-λhat{j}+(4-3λ) hat{k})
b) ((1+5λ) hat{i}-3λhat{j}+(7-3λ) hat{k})
c) ((1+λ) hat{i}+λhat{j}+(8-3λ) hat{k})
d) ((1+5λ) hat{i}-3λhat{j}+(4-3λ) hat{k})
Answer: d
Clarification: Consider the points A(1,0,4) and B(6,-3,1)
Let (vec{a} ,and ,vec{b}) be the position vectors of the points A and B.
∴(vec{a}=hat{i}+4hat{k})
(vec{b}=6hat{i}-3hat{j}+hat{k})
∴(vec{r}=hat{i}+4hat{k}+λ(6hat{i}-3hat{j}+hat{k}-(hat{i}+4hat{k})))
=(hat{i}+4hat{k}+λ(5hat{i}-3hat{j}-3hat{k}))
=((1+5λ) hat{i}-3λhat{j}+(4-3λ) hat{k})

9. Find the vector equation of the line which is passing through the point (1, -4, 4) and parallel to the vector (2hat{i}-5hat{j}+2hat{k}).
a) ((1+2λ) hat{i}-(4+5λ) hat{j}+(4+2λ) hat{j})
b) ((1+2λ) hat{i}-(4+5λ) hat{j}+2λ hat{j})
c) ((1-2λ) hat{i}+(4+5λ) hat{j}+(4+2λ) hat{j})
d) ((8+λ) hat{i}-(4-5λ) hat{j}+(7-λ) hat{j})
Answer: a
Clarification: We know that, the equation of a vector passing through a point and parallel to another vector is given by (vec{r}=vec{a}+λvec{b}), where λ is a constant.
The position of vector of the point (1,-4,4) is given by (vec{a}=hat{i}-4hat{j}+4hat{k})
And (vec{b}=2hat{i}-5hat{j}+2hat{k})
∴(vec{r}=vec{a}+λvec{b})
=(hat{i}-4hat{j}+4hat{k}+λ(2hat{i}-5hat{j}+2hat{k}))
=((1+2λ) hat{i}-(4+5λ) hat{j}+(4+2λ) hat{j})

10. Find the cartesian equation of the line which is passing through the point (-1, -8, 5) and parallel to the vector (7hat{i}-3hat{j}-3hat{k}).
a) (frac{x+1}{-1}=frac{y+8}{-8}=frac{z-5}{5})
b) (frac{x+1}{7}=frac{y+8}{-3}=frac{z-5}{-3})
c) (frac{x-7}{7}=frac{y+8}{-3}=frac{z-3}{-3})
d) (frac{x+1}{7}=frac{y+8}{-8}=frac{z+5}{3})
Answer: b
Clarification: The position vector of the given point is (vec{a}=-hat{i}-8hat{j}+5hat{k})
The vector which is parallel to the given line is (7hat{i}-3hat{j}-3hat{k})
We know that, (vec{r}=vec{a}+λvec{b})
∴(xhat{i}+yhat{j}+zhat{k}=-hat{i}-8hat{j}+5hat{k}+λ(7hat{i}-3hat{j}-3hat{k}))
=((-1+7λ) hat{i}+(-8-3λ) hat{j}+(5-3λ) hat{j})
⇒(frac{x+1}{7}=frac{y+8}{-3}=frac{z-5}{-3}).

Mathematics Question Papers for IIT JEE Exam,

Mathematics Multiple Choice Questions on “Calculus Application – Motion in a Straight Line – 1”.

1. A particle moving in a straight line with uniform acceleration has a velocity 10 cm/sec and 8 seconds later has a velocity 54 cm/sec. What will be the acceleration?
a) 1 cm/sec²
b) 2 cm/sec²
c) 3 cm/sec²
d) 4 cm/sec²
Answer: c
Clarification: Let, the particle moving with a uniform acceleration of f cm/sec².
By question initial velocity of the particle is u = 10 cm/sec and its velocity after 8 seconds = v = 34 cm/sec.
Thus, using the formula v = u + ft we get
34 = 10 + f*8
Or 8f = 24
Or f = 3
Therefore, the required acceleration of the particle is 3 cm/sec².

2. A particle moving in a straight line with uniform acceleration has a velocity 10 cm/sec and 8 seconds later has a velocity 54 cm/sec. What will be the value of space described?
a) 172 cm
b) 176 cm
c) 178 cm
d) 174 cm
Answer: b
Clarification: Let, the particle moving with a uniform acceleration of f cm/sec².
By question initial velocity of the particle is u = 10 cm/sec and its velocity after 8 seconds = v = 34 cm/sec.
Thus, using the formula v = u + ft we get
34 = 10 + f*8
Or 8f = 24
Or f = 3
Therefore, the required acceleration of the particle is 3 cm/sec².
Thus, the space described by the particle in 8 seconds,
= [10*8 + 1/2(3)(8*8) [using the formula s = ut +1/2(ft²)]
= 80 + 96
= 176 cm.

3. A particle moving in a straight line with uniform acceleration has a velocity 10 cm/sec and 8 seconds later has a velocity 54 cm/sec. What will be the described by the particle during the 10^th second of its motion?
a) 38.5cm
b) 37.5cm
c) 38cm
d) 39.5cm
Answer: a
Clarification: Let, the particle moving with a uniform acceleration of f cm/sec².
By question initial velocity of the particle is u = 10 cm/sec and its velocity after 8 seconds = v = 34 cm/sec.
Thus, using the formula v = u + ft we get
34 = 10 + f*8
Or 8f = 24
Or f = 3
Therefore, the required acceleration of the particle is 3 cm/sec².
The space described during the 10th second of its motion is,
= [10 + 1/2(3)(2*10 – 1)]   [using the formula s_t = ut +1/2(f)(2t – 1)]
= 10 + 28.5
= 38.5cm.

4. A point starts with the velocity 10 cm/sec and moves along a straight line with uniform acceleration 5cm/sec². How much time it takes to describe 80 cm?
a) 4 seconds
b) 2 seconds
c) 8 seconds
d) 6 seconds
Answer: a
Clarification: Let us assume that the point takes t seconds to describe a distance 80 cm.
Then using the formula s = ut +1/2(ft²) we get,
80 = 10*t + 1/2(5)(t²)
Or 5t² + 20t – 160 = 0
Or t² + 4t – 32 = 0
Or (t – 4)(t + 8) = 0
Or t = 4 Or -8
Clearly, t = -8 is inadmissible.
Therefore, the required time = 4 seconds.

5. A motor car travelling at the rate of 40 km/hr is stopped by its brakes in 4 seconds. How long will it go from the point at which the brakes are first applied?
a) 22m
b) 22(2/9)m
c) 22(1/9)m
d) 22(4/9)m
Answer: b
Clarification: Let f be the uniform retardation in m/sec² to the motion of the motor car due to application of brakes.
By question, the car is stopped by its brakes in 4 seconds, hence, the final velocity of the car after 4 seconds = 0.
Therefore, using the formula v = u – ft we get,
0 = ((40*1000)/(60*60) – f(4)) [Since u = initial velocity of the motor car = 40 km/hr = (40*1000)/(60*60) m/sec]
Or f = 25/9
Let the car go through a distance s m from the point at which the brakes are first applied.
Then using the formula s = ut – 1/2(ft²) we get,
s = ((40*1000)/(60*60))*4 – 1/2(25/9)(4*4)
= 200/9
= 22(2/9)
Therefore, the required distance described by the car = 22(2/9)m.

6. A bullet fired into a target loses half of its velocity after penetrating 2.5 cm. How much further will it penetrate?
a) 0.85 cm
b) 0.84 cm
c) 0.83 cm
d) 0.82 cm
Answer: c
Clarification: Let, u cm/sec be the initial velocity of the bullet.
By the question, the velocity of the bullet after pertaining 2.5 cm into the target will be u/2 cm/sec.
Now if the uniform retardation to penetration be f cm/sec²,
Then using the formula, v² = u² – 2fs, we get,
(u/2)² = u² – 2f(2.5)
Or f = 3u²/20
Now let us assume that the bullet can penetrate x cm into the target.
Then the final velocity of the bullet will be zero after penetrating x cm into the target.
Hence, using formula v² = u² – 2fs we get,
0 = u² – 2(3u²/20)(x)
Or 10 – 3x = 0
Or x = 10/3 = 3.33(approx).
Therefore, the required further penetration into the target will be
3.33 – 2.5 = 0.83 cm.

7. A particle moving in a straight line with uniform retardation described 7cm in 5th second and after some time comes to rest. If the particle describes 1/64 part of the total path during the last second of its motion, for how long was the particle in motion?
a) 6 seconds
b) 8 seconds
c) 4 seconds
d) 2 seconds
Answer: b
Clarification: Let the initial velocity of the particle be u cm/second and its uniform retardation be f cm/sec².
Further assume that the particle was in motion for t seconds.
By question, the particle comes to rest after t seconds.
Therefore, using the formula, v = u – ft, we get,
0 = u – ft
Or u = ft
Again, the particle described 7cm in the 5th second. Therefore, using the formula
s_t = ut + 1/2(f)(2t – 1) we get,
7 = u – ½(f)(2.5 – 1)
Or u – 9f/2 = 7
Again, the distance described in the last second (i.e., in the t th second) of its motion
= 1/64 * (distance described by the particle in t seconds)
ut -1/2(f)(2t – 1) = 1/64(ut – ½(ft²))
Putting u = ft, we get,
f/2 = 1/64((ft²)/2)
Or t² = 64
Or t = 8 seconds.

8. A particle moving in a straight line with uniform retardation described 7cm in 5th second and after some time comes to rest. If the particle describes 1/64 part of the total path during the last second of its motion, for how long was the particle in motion?
a) 10 cm/sec
b) 12 cm/sec
c) 14 cm/sec
d) 16 cm/sec
Answer: d
Clarification: Let the initial velocity of the particle be u cm/second and its uniform retardation be f cm/sec².
Further assume that the particle was in motion for t seconds.
By question, the particle comes to rest after t seconds.
Therefore, using the formula, v = u – ft, we get,
0 = u – ft
Or u = ft ……….(1)
Again, the particle described 7cm in the 5th second. Therefore, using the formula
s_t = ut +1/2(f)(2t – 1) we get,
7 = u – ½(f)(2.5 – 1)
Or u – 9f/2 = 7 ……….(2)
Again, the distance described in the last second (i.e., in the t th second) of its motion
= 1/64 * (distance described by the particle in t seconds)
ut – 1/2(f)(2t – 1) = 1/64(ut – ½(ft²))
Putting u = ft, we get,
f/2 = 1/64((ft²)/2)
Or t² = 64
Or t = 8 seconds.
Therefore, from (1) we get, u = 8f
Putting u = 8f in (2) we get,
8f – 9f/2 = 7
Or f = 2
Thus, u = 8f = 8*2 = 16
Therefore, the particle was in motion for 8 seconds and its initial velocity is 16 cm/sec.

9. If a, b, c be the space described in the pth, qth and rth seconds by a particle with a given velocity and moving with uniform acceleration in a straight line then what is the value of a(q – r) + b(r – p) + c(p – q)?
a) 0
b) 1
c) -1
d) Can’t be determined
Answer: a
Clarification: Let, f be the uniform acceleration and u be the given initial velocity of the moving particle.
From the conditions of problem we have the following equation of motion of the particle:
u + ½(f)(2p – 1) = a ……….(1)
u + ½(f)(2q – 1) = b ……….(2)
u + ½(f)(2r – 1) = c ……….(1)
Thus, a(q – r) + b(r – p) + c(p – q) = [u + ½(f)(2p – 1)](q – r) + [u + ½(f)(2q – 1)](r – p) + [u + ½(f)(2r – 1)](p – q)
= u (q – r + r – p + p – q) + f/2[(2p – 1)(q – r) + (2q – 1)(r – p) + (2r – 1)(p – q)]
= u*0 + f/2[2(pq – rp + qr – pq + rp – qr) – q + r – r + p – p + q]
= f/2*0
= 0

10. 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, then what is the value of acceleration?
a) 2(bt₁ – at₂)/t₁t₂(t₁ + t₂)
b) -2(bt₁ – at₂)/t₁t₂(t₁ + t₂)
c) 2(bt₁ + at₂)/t₁t₂(t₁ + t₂)
d) 2(bt₁ – at₂)/t₁t₂(t₁ – t₂)
Answer: a
Clarification: Let the particle beam moving with uniform acceleration f and its velocity at A be u.
Then, the equation of the motion of the particle from A to B is,
ut₁ + ft₁²/2 = a   [as, AB = a] ……….(1)
Again, the equation of motion of the particle from A to C is,
u(t₁ + t₂) + f(t₁ + t₂)²/2 = a + b  [as, AC = AB + BC = a + b] ……….(2)
Multiplying (1) by (t₁ + t₂) and (2) by t₁ we get,
ut₁ (t₁ + t₂) + ft₁²(t₁ + t₂)/2 = a(t₁ + t₂) ……….(3)
And ut₁(t₁ + t₂) + f(t₁ + t₂)²/2 = (a + b)t₁ ……….(4)
Subtracting (3) and (4) we get,
1/2(ft₁)(t₁ + t₂)(t₁ – t₁ – t₂) = at₂ – bt₁
Solving the above equation, we get,
f = 2(bt₁ – at₂)/t₁t₂(t₁ + t₂)

Mathematics Multiple Choice Questions on “Types of Relations”.

1. Which of these is not a type of relation?
a) Reflexive
b) Surjective
c) Symmetric
d) Transitive
Answer: b
Clarification: Surjective is not a type of relation. It is a type of function. Reflexive, Symmetric and Transitive are type of relations.

2. An Equivalence relation is always symmetric.
a) True
b) False
Answer: a
Clarification: The given statement is true. A relation R in a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive. Hence, an equivalence relation is always symmetric.

3. Which of the following relations is symmetric but neither reflexive nor transitive for a set A = {1, 2, 3}.
a) R = {(1, 2), (1, 3), (1, 4)}
b) R = {(1, 2), (2, 1)}
c) R = {(1, 1), (2, 2), (3, 3)}
d) R = {(1, 1), (1, 2), (2, 3)}
Answer: b
Clarification: A relation in a set A is said to be symmetric if (a₁, a₂)∈R implies that (a₁, a₂)∈R,for every a₁, a₂∈R.
Hence, for the given set A={1, 2, 3}, R={(1, 2), (2, 1)} is symmetric. It is not reflexive since every element is not related to itself and neither transitive as it does not satisfy the condition that for a given relation R in a set A if (a₁, a₂)∈R and (a₂, a₃)∈R implies that (a₁, a₃)∈ R for every a₁, a₂, a₃∈R.

4. Which of the following relations is transitive but not reflexive for the set S={3, 4, 6}?
a) R = {(3, 4), (4, 6), (3, 6)}
b) R = {(1, 2), (1, 3), (1, 4)}
c) R = {(3, 3), (4, 4), (6, 6)}
d) R = {(3, 4), (4, 3)}
Answer: a
Clarification: For the above given set S = {3, 4, 6}, R = {(3, 4), (4, 6), (3, 6)} is transitive as (3,4)∈R and (4,6) ∈R and (3,6) also belongs to R . It is not a reflexive relation as it does not satisfy the condition (a,a)∈R, for every a∈A for a relation R in the set A.

5. Let R be a relation in the set N given by R={(a,b): a+b=5, b>1}. Which of the following will satisfy the given relation?
a) (2,3) ∈ R
b) (4,2) ∈ R
c) (2,1) ∈ R
d) (5,0) ∈ R
Answer: a
Clarification: (2,3) ∈ R as 2+3 = 5, 3>1, thus satisfying the given condition.
(4,2) doesn’t belong to R as 4+2 ≠ 5.
(2,1) doesn’t belong to R as 2+1 ≠ 5.
(5,0) doesn’tbelong to R as 0⊁1

6. Which of the following relations is reflexive but not transitive for the set T = {7, 8, 9}?
a) R = {(7, 7), (8, 8), (9, 9)}
b) R = {(7, 8), (8, 7), (8, 9)}
c) R = {0}
d) R = {(7, 8), (8, 8), (8, 9)}
Answer: a
Clarification: The relation R= {(7, 7), (8, 8), (9, 9)} is reflexive as every element is related to itself i.e. (a,a) ∈ R, for every a∈A. and it is not transitive as it does not satisfy the condition that for a relation R in a set A if (a₁, a₂)∈R and (a₂, a₃)∈R implies that (a₁, a₃) ∈ R for every a₁, a₂, a₃ ∈ R.

7. Let I be a set of all lines in a XY plane and R be a relation in I defined as R = {(I₁, I₂):I₁ is parallel to I₂}. What is the type of given relation?
a) Reflexive relation
b) Transitive relation
c) Symmetric relation
d) Equivalence relation
Answer: d
Clarification: This is an equivalence relation. A relation R is said to be an equivalence relation when it is reflexive, transitive and symmetric.
Reflexive: We know that a line is always parallel to itself. This implies that I₁ is parallel to I₁ i.e. (I₁, I₂)∈R. Hence, it is a reflexive relation.
Symmetric: Now if a line I₁ || I₂ then the line I₂ || I₁. Therefore, (I₁, I₂)∈R implies that (I₂, I₁)∈R. Hence, it is a symmetric relation.
Transitive: If two lines (I₁, I₃) are parallel to a third line (I₂) then they will be parallel to each other i.e. if (I₁, I₂) ∈R and (I₂, I₃) ∈R implies that (I₁, I₃) ∈R.

8. Which of the following relations is symmetric and transitive but not reflexive for the set I = {4, 5}?
a) R = {(4, 4), (5, 4), (5, 5)}
b) R = {(4, 4), (5, 5)}
c) R = {(4, 5), (5, 4)}
d) R = {(4, 5), (5, 4), (4, 4)}
Answer: d
Clarification: R= {(4, 5), (5, 4), (4, 4)} is symmetric since (4, 5) and (5, 4) are converse of each other thus satisfying the condition for a symmetric relation and it is transitive as (4, 5)∈R and (5, 4)∈R implies that (4, 4) ∈R. It is not reflexive as every element in the set I is not related to itself.

9. (a,a) ∈ R, for every a ∈ A. This condition is for which of the following relations?
a) Reflexive relation
b) Symmetric relation
c) Equivalence relation
d) Transitive relation
Answer: a
Clarification: The above is the condition for a reflexive relation. A relation is said to be reflexive if every element in the set is related to itself.

10. (a₁, a₂) ∈R implies that (a₂, a₁) ∈ R, for all a₁, a₂∈A. This condition is for which of the following relations?
a) Equivalence relation
b) Reflexive relation
c) Symmetric relation
d) Universal relation
Answer: c
Clarification: The above is a condition for a symmetric relation.
For example, a relation R on set A = {1,2,3,4} is given by R={(a,b):a+b=3, a>0, b>0}
1+2 = 3, 1>0 and 2>0 which implies (1,2) ∈ R.
Similarly, 2+1 = 3, 2>0 and 1>0 which implies (2,1)∈R. Therefore both (1, 2) and (2, 1) are converse of each other and is a part of the relation. Hence, they are symmetric.