Mathematics Test

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Mathematics Test - 2

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Weekly Quiz Competition

Question 1
1 / -0.25

Area of the region bounded by y = cosx, x = 0, x = πand X-axis is ...sq. units.

A
A. 3

B
B. 1

C
C. 2

D
D. 4

Solution

Required area =

= 2(sinx)^x/2₀ = 2(1 - 0)

= 2sq. Units
Question 2
1 / -0.25

Let a: (p ∧- r) ∨(-q ∨s) and b: (p ∨s) ↔(-q ∧r). If the truth values of p and q are true and that of r and s are false, then the truth values of a and b are, respectively...

A
T, F

B
T, T

C
F, F

D
F, T

Solution

Let 's break down the truth values of a and b using the given truth values of p, q, r, and s.

Given: p = true, q = true, r = false, s = false.

a: (p ∧- r) ∨(-q ∨s)
a: (true ∧- false) ∨(-true ∨false)
a: (true ∧true) ∨(false ∨false)
a: true ∨false
a: true

b: (p ∨s) ↔(-q ∧r)
b: (true ∨false) ↔(-true ∧false)
b: true ↔(false ∧false)
b: true ↔false
b: false

So, the truth values of a and b are True and False, respectively. The correct answer is:

T, F
Question 3
1 / -0.25

∫logx[log(e^x )] ^-2dx = ?

A
(x/1+logx) + c

B
x(1 - logx) + c

C
x(1 + logx) + c

D
(x/1-logx) + c

Solution

Let I = ∫logx[log(e^x )]^-2 dx

Put logx = t ⇒x = e^f

⇒dx = e^f dt

= (e^f /1+t) + C

= (x/1+logx)+C
Question 4
1 / -0.25

If the scalar triple product of the vectors is 272 then λ=

A
9

B
11

C
8

D
10

Solution

The scalar triple product of the given vectors is 272.

(∵scalar triple product of the vectors a, b and c is [a b d]

⇒-3(21 + 5 λ)-7(-9-7 λ)- 3(-15 + 49)= 272

⇒63 - 15 λ+ 63 + 49 λ- 102 = 272

⇒34 λ- 102 = 272

⇒34 λ= 374

⇒λ= 11
Question 5
1 / -0.25

The joint equation of lines passing through origin and having slopes (1 + √2) and   -1    is
                                                                 1 + √2

A
X² + 2xy - y² = 0

B
x² - 2 √2xy - y² = 0

C
-2 √2xy + y² = 0

D
x² + 2xy + y² = 0

Solution

It is given that stapes of the lines passing through origin are m₁ (let) = 1 + √2) and m₂ (let) =  -1     = -(√2 - 1)
                                                                         1+√2

∴Required joint equation of lines passing through origin is

[y-(1+√2)x][y+(√2-1)x] = 0

⇒y² + (√2 - 1)xy-(1 +√2)xy - (2 - 1)x² = 0

⇒Y² - 2xy - x² = 0

⇒X² + 2xy - y² = 0
Question 6
1 / -0.25

θ.dθ = ...

A
A. -20/21

B
B. -8/21

C
C. 20/21

D
D. 8/21

Solution

Let l =

Put cosθ = t

⇒ -sinθ dθ = dt

⇒ Sinθ dθ = -dt

If θ = 0, t = 1 and 0 = (π/2), t = 0

∴
Question 7
1 / -0.25

If A and B are square matrices of order 3 such that |A| = 2, |B| = 4, then |A(adj B)| = ...

A
A. 16

B
B. 8

C
C. 64

D
D. 32

Solution

We have A and B are square matrics of order 3 such that

|A| = 2, |B| = 4

Now, |A(adj)B| = |A||adj B| (∵|AB| = |A||B|)

|A||B|^3-1

=|A||B|²= (2)(4)²= 32
Question 8
1 / -0.25

The polar coordinatesThus, if of P are(2, π/6). If Q is the image of P about the X-axis, then the polar coordinates of Q are......

A
(2, 5 π/6)

B
(2, π/6)

C
(2, π/3)

D
(2, 11 π/6)

Solution

We have polar coordinates of P are (2, π/6). If O is the image of P about the X-axis

∴Q = (2, π/6)

⇒Q = (2, 11 π/6)
Question 9
1 / -0.25

Let X be the number of successes in 'n' independent Bernoulli trials with probability of success p = ¾, The least value of ‘n’ so that P(X ≥ 1) ≥ 0.9375 is .......

A
A. 2

B
B. 1

C
C. 4

D
D. 3

Solution

We have, p = ¾, q = 1 - p = ¼

It is given that P (X ≥ 1) ≥ 0.9375

= 1 - P(X = 0) ≥ 0.9375

= 1 - ⁿ C_o (p^o )(g)^n-o ≥ 0.9375

= 1 - (¼)ⁿ ≥ 0.9375

= 1 - 0.9375 ≥ (¼)ⁿ

= 0.0625 ≥ (¼)ⁿ

= 625/10000 ≥ (¼)ⁿ

= 1/16 ≥ (¼)ⁿ

= 16 ≤ 4^th

⇒ n = 2
Question 10
1 / -0.25

Which of the following statement pattern is a tautology?

A
A. (p → q) v q

B
B. p →(q v p)

C
C. (p v g) → q

D
D. p v (q → p)

Solution

Option (a), (p → g) v q

= (∼p v q) v q

= (∼p v q)

It is not a tautology since if p is true and q is false, then(∼p v q) is false.

Option (b),p → (q v p)

= ~p v (q v p)

= (∼p v p) v q

= T v q(∵ ~ p v p = T)

It is a tautology since if q is true or false then T v q must be true.

Similarity, check other options.
Question 11
1 / -0.25

In ΔABC, with the usual notations, if (tan A/2)(tan B/2) = ¾ then a + b = ...

A
A. 4c

B
B. 2c

C
C. 7c

D
D. 3c

Solution

We have, In ΔABC

⇒

⇒ 4a + 4b - 4c = 3a + 3b + 3c

⇒ a + b = 7c
Question 12
1 / -0.25

A
A. 2

B
B. 1

C
C. 1/2

D
D. 3/2

Solution

We have,
Question 13
1 / -0.25

If f(x) is continuous at x = 3, where

f(x) = ax +1, for x ≤3

= bx + 3, for x > 3 then

A
A. a + b = (-⅔)

B
B. a - b = (-⅔)

C
C. a - b = (⅔)

D
D. a + b = (⅔)

Solution

⇒ a - b = 2/3
Question 14
1 / -0.25

For any non zero vector, a,b,c

a.[(b + c) x (a + b + c)] = ...

A
A. 0

B
B. 2(a b c)

C
C. (a b c)

D
D. (a c b)

Solution

We have, a.[(b + c) x (a + b + c)]

= a.[(b x a ) + (b x c) + (c x a) + (c x b)]

= a[(b x a) + ( b x c) + (c x a) - ( b x c)]

= a[(b x a) + (c x a)]

=[a b a] + [a c a] = 0 + 0 = 0
Question 15
1 / -0.25

A
A. -(3/2)

B
B. -(½)

C
C. 3/2

D
D. 1/2

Solution

We have

⇒

⇒

⇒

⇒ y = 3x/2

∴ dy/dx = 3/2
Question 16
1 / -0.25

The direction ratios of the normal to the plane passing through origin and the line of intersection of the planes x + 2y + 3z = 4 and 4x + 3y + 2z = 1 are .......

A
A. 2, 3, 1

B
B. 1, 2, 3

C
C. 3, 1, 2

D
D. 3, 2, 1

Solution

We have, line of intersection of the planes x + 2y+ 3z = 4 and 4x + 3y + 2z = 1

∴ Equation of plane passing through the given planes is (x +2y + 3z - 4) +λ (4x + 3y + 2z - 1) = 0

⇒ (1 + 4λ)x + (2 + 3λ)y + (3 + 2λ) + (-4 - λ) = 0

Since, a plane passing through the origin.

∴ -4 - λ = 0 ⇒ λ = -4

Now, equation of plane is

(1 - 16)x + (2 - 12)y + (3 - 8)z + 0 = 0

⇒ -15x - 10y - 5z = 0

⇒ 3x + 2y + z = 0

∴ Direction ratios of the normal to the plane are 3, 2, 1.