Mathematics Test 122

Result

Mathematics Test 122

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Question 1
4 / -1

Let A(h, k), B(1, 1) and C(2, 1) be vertices of a right angled triangle with AC as its hypotenuse. If area of triangle is 2, then value of h² + k² can be

A
25

B
9

C
26

D
11

Solution

k – 1 = ± 4

k = 1 + 4 k = 1 – 4
k = 5 k = –3
∴ values of h² + k² = 1² + 5² = 26
or
h² + k² = 1² + 9 = 10
Question 2
4 / -1

Set of values of 'α' for which the angle bisector of the lines (α + 1)x + 2y + 5 = 0 and 4x + αy – 3 = 0 containing origin is also the obtuse angle bisector, is-

A

B

C

D

Solution

(α + 1)x + 2y + 5 = 0

–4x – αy + 3 = 0

a₁a₂ + b₁b₂ > 0
Question 3
4 / -1

A second degree equation

x² + 2xy + y² – 8x – 8y + 12 = 0 represents

A
A pair of parallel lines

B
A pair of perpendicular lines

C
A point

D
An ellipse

Solution

Method 1 :

x² + 2xy + y² – 8x – 8y + 12 = 0
⇒ (x + y)² – 8(x + y) + 12 = 0
⇒ (x + y)² – 6(x + y) – 2(x + y) + 12 = 0
⇒ (x + y) [x + y– 6] – 2[x + y – 6] = 0
⇒ (x + y – 2) (x + y – 6) = 0
⇒ a pair of parallel lines
Question 4
4 / -1

A

B

C

D

Solution
Question 5
4 / -1

A

B

C

D

Solution
Question 6
4 / -1

Let f(x) is a polynomial function such that and f(5) = 126, then value of is-
(where C is an integertion constant)

A

B

C

D

Solution

f(x) = 1 + x³
Question 7
4 / -1

If = ƒ(x) + C where ƒ(0) = and ƒ (C is constant of integration), then

A
The minimum value of |ƒ(x)| is 1/3

B
The minimum value of |ƒ(x)| is 0

C
The value of ƒ(π) is 1/3

D
option 2 and option 3 are correct

Solution
Question 8
4 / -1

A

B

C

D
All are correct

Solution
Question 9
4 / -1

If ƒ(x) = where ƒ then

A

B

C

D
option 2 and option 3 are correct

Solution
Question 10
4 / -1

Let P and Q be any two points on the lines represented by 2x – 3y = 0 and 2x + 3y = 0 respectively. If area of triangle OPQ, (O is origin) is 5, then equation of locus of mid point of PQ can be :-

A
4x² – 9y² + 30 = 0

B
9x² – 4y² + 30 = 0

C
4x² – 9y² – 30 = 0

D
option 1 and option 3 are correct

Solution

4αβ = ±30
also 2h = α + β, 3k = α – β
(α + β)² – (α – β)² = 4αβ
4x² – 9y²= ± 30