Mathematics Test 227

Result

Mathematics Test 227

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Weekly Quiz Competition

Question 1
4 / -1

The coefficient of x⁵ in (1+2x+3x²+……….up to infinite term)^-3/2 is

A
21

B
25

C
26

D
None of these

Solution

(1+2x+3x²+………)^–3/2

=((1–x)^–2)^–3/2

= (1 − x)³ = 1 − 3x + 3x² − x³

∴ coefficient of x⁵ = 0.
Question 2
4 / -1

Sum to the infinity of the series

A
7/15

B
2/5

C
1/3

D
None of these

Solution
Question 3
4 / -1

A
1

B
abc

C
(abc)^2/3

D

Solution
Question 4
4 / -1

If the progressions 3, 10, 17, ....... and 63, 65, 67, ....... are such that their n^th terms are equal, then n is equal to

A
13

B
15

C
19

D
18

Solution

n^th term of 1^st series =n^th term of 2^nd series

⇒ 3 + (n − 1) 7 = 63 + (n − 1) 2

⇒ (n − 1)5 = 60

⇒ n -1=12

⇒ n = 13
Question 5
4 / -1

If the latus rectum of a hyperbola through one focus, subtends 60 ° angle at the other focus, then its eccentricity is

A
√2

B
√3

C
√5

D
√6

Solution

Taking only positive value of e as eccentricity cannot be negative.
Question 6
4 / -1

Find the solution of the differential equation

A

B

C

D

Solution
Question 7
4 / -1

If the mean of a binomial distribution is 25, then its standard deviation lies in the interval

A
[0, 5)

B
(0, 5]

C
[0, 25)

D
(0, 25]

Solution
Question 8
4 / -1

The equation of the normal to the ellipse at the positive end of the latus rectum is

A
x + ey + e³a = 0

B
x − ey − e³a = 0

C
x − ey − e²a = 0

D
None of these

Solution
Question 9
4 / -1

The tangent to the circle x² + y² = 5 at (1, − 2) also touches the circle x² + y² − 8x + 6y + 20 = 0. Then the point of contact is

A
(-1, 3)

B
(3, -1)

C
(3, 1)

D
(-3, - 1)

Solution

Tangent at (1, − 2) to x² + y² = 5 is

x − 2y − 5 = 0 ... (i).

Centre and radius of

x² + y² − 8x + 6y + 20 = 0 are

C (4, − 3) and radius r = √5

Perpendicular distance from

C (4, − 3) to (i) is radius.

∴ (i) is also a tangent to the second circle.

Let P (h, k) be the foot of the drawn circle from C (4, − 3) on (i)

∴ (h, k) = (3, −1)
Question 10
4 / -1

The equation x³ – 3x + [a] = 0, will have three real and distinct roots if –

(where [ ] denotes the greatest integer function)

A
a ∈ (–∞, 2)

B
a ∈ (0,2)

C
a ∈ (–∞, -2) U (0, ∞)

D
a ∈ [–1, 2)

Solution

f(x) = x³ – 3x + [a]

Let [a] = t (where t will be an integer)

f(x) = x³ – 3x + t ……….(i)

⇒ f ’(x) = 3x² – 3

⇒ f ‘(x) = 0 has two real and distinct solution which are x = 1 and x = -1

so f(x) = 0 will have three distinct and real solution when f (1). f(-1) < 0

……………. (ii)

Now,

f(1) = (1)³ -3(1) + t = t – 2

f(–1) = (–1)³ – 3 (–1) + t = t + 2

From equation (ii)

(t –2) (t + 2) < 0

⇒ t ∈ (-2, 2)

Now t = [a]

Hence [a] ∈ (-2, 2)

⇒ a ∈ [-1, 2)