Category: Class 12 Mathematics Objective Questions

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.

Mathematics Quiz for Class 12 on “Determinant – 3”.

1. What will be the value of (begin{vmatrix}0 & p-q & a – b\q – p & 0 & x – y\b – a & y – x & 0 end {vmatrix})?
a) 0
b) a + b
c) x + y
d) p + q
Answer: a
Clarification: The above matrix is a skew symmetric matrix and its order is odd
And we know that for any skew symmetric matrix with odd order has determinant = 0
Therefore, the value of the given determinant = 0.

2. What will be the value of f(x) if (begin{vmatrix}x & b & c\a & y & c\a & b & z end {vmatrix})?
a) (x – a)(y – b)(z – c)((frac{x}{x-a} + frac{b}{y – b} – frac{c}{z-c}) – 2)
b) (x – a)(y – b)(z – c)((frac{x}{x-a} – frac{b}{y – b} – frac{c}{z-c}) – 2)
c) (x – a)(y – b)(z – c)((frac{x}{x-a} + frac{b}{y – b} + frac{c}{z-c}) – 2)
d) (x – a)(y – b)(z – c)((frac{x}{x-a} + frac{b}{y – b} + frac{c}{z-c}) + 2)
Answer: c
Clarification: Given, (begin{vmatrix}x & b & c\a & y & c\a & b & z end {vmatrix}) = (begin{vmatrix}x & b & c\a – x & y – b & 0\0 & b – y & z – c end {vmatrix})

Applying the operation R₂ = R₂ – R₁ and R₃ = R₃ – R₂
= (x – a)(y – b)(z – c)(begin{vmatrix}x/(x – a) & b/(y – b) & c/(z – c)\-1 & y 1 & 0\0 & -1 & 1 end {vmatrix})
Now, expanding the determinant we get,
= (x – a)(y – b)(z – c)((frac{x}{x-a} + frac{b}{y – b} + frac{c}{z-c}))
= (x – a)(y – b)(z – c)((frac{x}{x-a} + frac{b}{y – b} + frac{c}{z-c}) – 2)
This is because,
(frac{b}{y – b} + frac{c}{z-c} = frac{y-(y-b)}{y-b} + frac{z-(z-c)}{z-c} = frac{y}{y-b}) – 1 + (frac{z}{z-c}) – 1 = (frac{y}{y-b} + frac{z}{z-c}) – 2