Mathematics Test 212

Result

Mathematics Test 212

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Weekly Quiz Competition

Question 1
4 / -1

The fourth term of equal to 200, then the value of x satisfying this is

A
√102

B
10⁴

C
100

D
10

Solution

Since, fourth term of
Question 2
4 / -1

A
n⋅n!

B
n(n+1) !

C
(n+1) !

D
None of these

Solution
Question 3
4 / -1

If ln(x + y) = 2xy, then y'(0) is equal to

A
1

B
-1

C
2

D
0

Solution

Given equation is ln(x + y) = 2xy …(1)

ln x + y = 2xy …1

For x = 0

ln(0 + y) = 2.0.y = 0

⇒ lny = 0

⇒ y = 1
Question 4
4 / -1

A

B

C

D

Solution
Question 5
4 / -1

Two vertices of a triangle are (3,−2) and (−2, 3) and its orthocentre is (−6, 1). The coordinates of its third vertex are-

A
(1, 6)

B
(−1, 6)

C
(1, −6)

D
None of these

Solution

Let the third vertex be A(α,β)

Using the diagram, OA⊥BC

⇒ Slope of OA × Slope BC = −1
Question 6
4 / -1

The equation of a circle C₁ is x²+ y²− 4x − 2y − 11 = 0. Another circle C₂ of radius 1 unit rolls on the outer surface of the circle C₁. Then the equation of the locus of the centre of C₂ is

A
x²+ y² − 4x − 2y − 20 = 0

B
x² + y² + 4x + 2y − 20 = 0

C
x² + y² − 3x − y − 11 = 0

D
None of these

Solution

If P(α, β) be the centre of C₂ of radius r₂ = 1

We know that, if two circles with centres A and P and radii r₁ and r₂ touches each other externally, then distance between their centres AP is equal to the sum of their radii i.e. AP = r₁+ r₂

The locus is obtained by replacing (α, β) by (x, y)

Hence, the locus is x²+ y² − 4x − 2y − 20 = 0
Question 7
4 / -1

The number of real values of k for which the lines and are intersecting is

A
0

B
2

C
1

D
Infinite

Solution

Any point on the first line is (4r+k, 2r+1,r−1), and any point on the second line is (r′+k+1 ,−r′,2r′+1) for some values of r and r'. The lines are intersecting if these two points coincide i.e

4r + k = r′ + k + 1, 2r + 1 = −r′, r − 1 = 2r′ + 1 for some r and r'

⇒ 4r − r′ = 1, 2r + r′ = −1, r − 2r′ = 2

Now, 4r − r′ = 1, 2r + r′ = −1 ⇒ r = 0, r′ = −1 which satisfy r − 2 r′ = 2.

⇒ The given lines are intersecting for all real values of k.
Question 8
4 / -1

Consider set A = {1, 2, 3}. Number of symmetric relations that can be defined on A containing the ordered pair (1, 2) and (2, 1) is

A
18

B
16

C
24

D
32

Solution

n(A × A) = 9

Number of relations containing (1, 2), (2, 1), (a, a) = 2³

Number of relations containing (1, 2), (2, 1), (1, 3), (3, 1), (a, a) = 2³

Number of relations containing (1, 2), (2, 1), (2, 3), (3, 2), (a, a) = 2³

Number of relations containing (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2), (a, a) = 2³

⇒ Total number of symmetric relations = 32.
Question 9
4 / -1

The negation of the statement p→(q∧r) is

A
~p → ~(q ∧ r)

B
~p ∨ (q ∧ r)

C
(q ∧ r) → p

D
p ∧ (~q ∨ ~r)

Solution

As we know, ~(p → q) = p ∧ ~q

Hence, ~(p→(q ∧ r)) = p ∧(~(q ∧ r))

⇒~(p → (q ∧ r)) = p∧(~q ∨ ~r)
Question 10
4 / -1

A

B

C

D

Solution