Mathematics Test

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Mathematics Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

Find the general solution of the Differential equation dy/dx=2xy+12x+5y+30

A

$\log (y+6)=x^{2}+5 x+C$

B

$\log (x+5)=y^{2}+6 y+C$

C

$\log (y-6)=x^{2}-12 x+C$

D

$\log (x+6)=y^{2}+5 y+C$

Solution
Question 2
1 / -0

If y = log sin √x, then dy/dx = ?

A
1/sin√x

B
cot√x/√x

C
12√xsin√x

D
cot√x/2√x

Solution

$Y=\log \sin \sqrt{x}$
$Y=\log \sin x \frac{1}{2}$
$Y=\log (\sin x)^{\frac{1}{2}}$
$Y=\frac{1}{2} \log (\sin )$
$Y=\frac{\cot \sqrt{x}}{2 \sqrt{x}}$

$\left[(\sin x)^{2}=\sin x^{2}\right]$
$\left[\log m^{n}=n \log m\right]$
Question 3
1 / -0

A
21/4

B
27/2

C
29/6

D
13/5

Solution
Question 4
1 / -0

evaluate
$\int \frac{2 \cos^{2 x} - \cos 2 x}{\sin^{2} x} d x .$

A
tanx + c

B
sin2x -cotx +c

C
-cotx + c

D
none of these

Solution

$\cos 2 x=2 \cos ^{2} x-1$
Then $\int \frac{2 \cos ^{2} x-\cos 2 x}{\sin ^{2} x} d x$
Solution: $=\int \frac{2 \cos ^{2} x-\left(2 \cos ^{2} x-1\right)}{\sin ^{2} x} d x$
$=\int \frac{1}{\sin ^{2} x} d x$
$=\int \cos e c^{2} x d x$
$=-\cot x+c$
Question 5
1 / -0

Find the domain of the function f(x)=2x+3/x−7

A
For all the real values except 2

B
For all the real values except 3

C
For all the real values except 7

D
For all the real values except 1

Solution

If the denominator is not equal to zero, then only the function will be defined for the values.
For the function f(x)=2x+3/x−7
x = 7 does not define the function.
Therefore, the domain of the function is all real numbers except 7
Question 6
1 / -0

A
π/3

B
π/2

C
π/1

D
π/4

Solution
Question 7
1 / -0

The solution of given inequality is:
|2x – 3| ≤ 5

A
-1 < x < 4

B
-1 ≤ x ≤ 4

C
-4 ≤ x ≤ 1

D
-4 < x < 1

Solution

Given inequality is equivalent to:
-5 ≤ 2x – 3 ≤ 5
So, 2x - 3 ≥ -5 and 2x - 3 ≤ 5
x ≥ -1 and x ≤ 4
so, -1 ≤ x ≤ 4
Question 8
1 / -0

If the line AB makes an angle of 120° with the positive direction of the y - axis in a clockwise sense, the slope of the line is

A
-√3

B
-1/√3

C
1/√3

D
√3

Solution

Slope = tan θ
As per the question, the line make an angle of 120° with the positive direction of the y - axis in clockwise sense. Hence, the figure can be interpreted as follows:
Slope = tan 150 = -1/√3 (slope = tan θ, where θ is the angle made by the line with the positive x - axis)
Question 9
1 / -0

A ball is thrown straight up, from 3m above the ground with a velocity of 11m/s when does it hit the ground?

A
2.5 sec

B
4 sec

C
5 sec

D
2 sec

Solution

Ignoring air resistance we can work out on these three things
The height starts at 3m
It travels upward at 11 meters per second (11m/s): 11t
Gravity pulls it down, changing its position by about 4.9m per second squared: -4.9t²
Where -4.9t² is simplified from - (½) at² with a = 9.8 m/s²
Add them up and the height h at any time t is:
h = 3 + 11t − 4.9t²
And the ball will hit the ground when the height is zero:
3 + 11t − 4.9t² = 0
From these three we get the quadratic equation
−4.9t² + 11t + 3 = 0
It looks even better when we multiply all terms by −1:
4.9t² − 11t − 3 = 0
Therefore t = -0.25 or t = 2.5
The "t = −0.25" is a negative time, impossible in our case.
The "t = 2.5" is the answer we want
Hence, the ball hits the ground after 2.5 sec.
Question 10
1 / -0

If p = -0.12, q = -0.01 and r = -0.015 then what will be the correct relation between the three?

A
p< q< r

B
r> p> q

C
p< r< q

D
q> p> r

Solution

=-0.12, q=-0.01 , r=-0.015
Their order
p < q < r