Mathematics Test

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Mathematics Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

If A² - A + I = 0, then inverse of A is

A
I - A

B
A - I

C
A

D
A + I

Solution

When A = A^-1and B is a matrix, then
rank (AB) = rank (B)
Question 2
1 / -0

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)
Question 3
1 / -0

If the progressions 3, 10, 17, ....... and 63, 65, 67, ....... are such that their nth 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 4
1 / -0

Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated is

A
69760

B
30240

C
99784

D
none of these

Solution

The total number of words that can be formed is 10⁵ and number of these words in which no letters are repeated is ¹⁰P₅.
Hence the required number
= 10⁵ − ¹⁰P⁵
= 100000 − 10 × 9 × 8 × 7 × 6
= 69760
Question 5
1 / -0

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-4/1 = k=3/-2 = - [1.4 -2. (-3)-5]
∴ (h, k) = (3, −1)
Question 6
1 / -0

A
14

B
-14

C
12

D
15

Solution
Question 7
1 / -0

If f : R → R is defined by
f (x) = x² – 3x; if x > 2
= x + 2; if x < 2,
then f(4) =

A
4

B
-4

C
8

D
-8

Solution

f (4) = x² - 3x
= (4)² - 3(4)
= 16 - 12 = 4
Question 8
1 / -0

If f : [1,∞) [2,∞) is given by f(x) = x +1/x , then f –1(x) is equal to

A
x + √x² -4/2

B
x/1 -x²

C
x - √x² -4/2

D
1 + √x² -4

Solution
Question 9
1 / -0

Suppose f(x) = (x + 1)² for x - 1. If g(x) is the function whose graph is the reflection of the graph of f(x) with respect to line y = x, then g(x) is equal to

A
-√x - 1, x ≥ 0

B
1/(x+1)²,x > -1

C
√x+1, x≥ -1

D
√x-1, x ≥ -1

Solution

From the first function we can draw the following output chart:
x y
- 1 0
0 1
1 4
2 9
As g(x) is the reflection of f(x) about line y = x, it must satisfy the opposite of the graph, i.e.
x y
0 - 1
1 0
4 1
9 2, which is satisfied only by option 4.
Question 10
1 / -0

Let function f : R → R be defined by f(x) = cos x for x∈ R. Then f is

A
one to one and onto

B
one to one but not onto

C
onto, but not one to one

D
neither one to one nor onto

Solution

f(0) = cos 0 = 1 and f(2π) = cos (2π) = 1
So, f is not one to one and cos(x) lie between - 1 and 1.
Therefore, range is not equal to its codomain.
Hence, it is not onto.
Question 11
1 / -0

If f(x) = x/1+x is defined as [0,∞) → [0, ∞), then f(x) is

∞

A
one-one and onto

B
one-one, but not onto

C
onto, but not one-one

D
neither one-one nor onto

Solution
Question 12
1 / -0

Find the range of f(x) = x² + x + 2/x² + x + 1.

A
(1, ∞)

B
(1, 11/7)

C
(1, 7/3)

D
(1, 7/5)

Solution

We can write as 1+ 1/x² + x + 1. After solving, we get third option is correct.
Question 13
1 / -0

Cos^-1 (cos7π/6) =

A
7π/6

B
5π/6

C
π/6

D
None of these

Solution
Question 14
1 / -0

If tan^-1(x - 1) + tan^-1 x + tan^-1 (x + 1) = tan^-1 3x, then x is equal to

A
± 1/2

B
0, 1/2

C
0, -1/2

D
0, ±1/2

Solution
Question 15
1 / -0

If |x| > 1, then which one of the following is different from the other three?

A
√ x² - 1

B
sec(cosec^-1x)

C
cosec(sec^-1x)

D
|x | / √x² - 1

Solution

sec (cosec^-1x) = cosec (sec^-1x) = |x|/√x² -1.
So options (2) = (3) = (4); Hence option (1) is different from the other three.

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Mathematics Test - 9

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Mathematics Test - 9

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SHARING IS CARING

Question 1

I - A

A - I

A

A + I

Solution

Question 2

a ∈ (–∞, 2)

a ∈ (0,2)

a ∈ (–∞, -2) U (0, ∞)

a ∈ [–1, 2)

Solution

Question 3

13

15

19

18

Solution

Question 4

69760

30240

99784

none of these

Solution

Question 5

(-1, 3)

(3, -1)

(3, 1)

(-3, - 1)

Solution

Question 6

14

-14

12

15

Solution

Question 7

4

-4

8

-8

Solution

Question 8

x + √x2 -4/2

x/1 -x2

x - √x2 -4/2

1 + √x2 -4

Solution

Question 9

-√x - 1, x ≥ 0

1/(x+1)2,x > -1

√x+1, x≥ -1

√x-1, x ≥ -1

Solution

Question 10

one to one and onto

one to one but not onto

onto, but not one to one

neither one to one nor onto

Solution

Question 11

one-one and onto

one-one, but not onto

onto, but not one-one

neither one-one nor onto

Solution

Question 12

(1, ∞)

(1, 11/7)

(1, 7/3)

(1, 7/5)

Solution

x + √x² -4/2

x/1 -x²

x - √x² -4/2

1 + √x² -4

1/(x+1)²,x > -1

√ x² - 1

sec(cosec^-1x)

cosec(sec^-1x)

|x | / √x² - 1