Mathematics Test 113

Result

Mathematics Test 113

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Weekly Quiz Competition

Question 1
4 / -1

The statement p → (q →p) is equivalent to

A
p → (p → q)

B
p → (p v q)

C
p → (p ^ q)

D
p → (p ←→ q)

Solution
Question 2
4 / -1

If the straight lines, x = 1 + s, y = –3 – λs, z = 1 + λs and x = t/2, y = 1+t, z = 2 -t with parameters s and t respectively, are co-planar, then λ equals

A
0

B
-1

C
- 1/2

D
-2

Solution
Question 3
4 / -1

A
0

B
1

C
- 1

D
None of these

Solution

(i – j) · [(j – k) × (k – i)]

= (i – j) · (j × k – j × i – k × k + k × i)

= (i – j) · (i + k – 0 + j)

= i · i + i · k + i · j – j · i – j · k – j · j

= i² + 0 + 0 – 0 – 0 – j² = i² – j² = 1 – 1 = 0
Question 4
4 / -1

If the tangents to the parabola y² = 4ax at (x₁, y₁) and (x₂, y₂) intersect at (x₃, y₃), then

A
x₁, x₃, x₂ are in G.P.

B
x₁, x₂, x₃ are in A.P.

C
y₁, y₂, y₃ are in G.P.

D
y₁, y₂, y₃ are in H.P.

Solution
Question 5
4 / -1

The top of two poles of height 20m and 14m are connected by a wire. If the wire makes an angle 30° with the horizontal, then the length of the wire is

A
12m

B
8m

C
10m

D
6m

Solution

Let AB = 20 and CD = 14 be the two poles.

AC = l is the length of the wire.
Question 6
4 / -1

The sum of the series

A

B

C
2¹⁹

D
2²⁰

Solution
Question 7
4 / -1

If f(x) = xe^x(1–x) then f(x) is

A
increasing on [- 1/2, 1]

B
decreasing on R

C
increasing on R

D
decreasing on [- 1/2, 1]

Solution

We have,

f '(x) = e^x(1–x) + x·e^x(1–x)·(1–2x)

= e^{x(1 – x)} [1 + x(1 – 2x)]

= e^x(1–x)·(–2x² + x + 1)

Now, the sign-scheme for –2x² + x + 1 is
Question 8
4 / -1

Solution of the differential equation

(c is constant of integration)

A
2 log |x| - log |y| - 1/xy = c

B
2 log |x| + log |y| - 1/xy = c

C
2 log |x| - log |y| + 1/xy = c

D
None of these

Solution
Question 9
4 / -1

If P and Q are two points on the line 4x + 3y + 30 = 0 such that OP = OQ = 10, where O is the origin, then the area of the ΔOPQ is

A
48

B
16

C
32

D
None of these

Solution
Question 10
4 / -1

A

B

C

D
None of these

Solution