Mathematics Test

Result

Mathematics Test - 21

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1

1 / -0

A random variable X has the following probability distribution

X = x	-2	-1	0	1	2	3
P(x)	0.1	0.1	0.2	0.2	0.3	0.1

Then E(x) =

0.8

0.9

0.7

1.1 Solution

$\begin{array}{l}E(x)=\sum x . P(x) \\ X=x \quad-2 \quad-1 \quad 0 \quad 1 \quad 2 \quad 3 \\ P(x) & 0.1 \quad 0.1 \quad 0.20 .2 \quad 0.3 \quad 0.1 \\ x . P(x)-0.2-0.10 & 0.2 \quad 0.6 \quad 0.3 \\ E(x)=\sum x . P(x)=-0.2-0.1+0.2+0.6+0.3 & \\ \text { Thus } E(x)=0.8\end{array}$

Question 2
1 / -0

For every natural number n

A
n> $2^{n}$

B
n< $2^{n}$

C
n≥2n

D
Can't be determined.

Solution

Let n = 1 then option (a) and (d) is eliminated.

Equality can't be attained for any value of n so,

option (b) satisfied.
Question 3
1 / -0

The number of common tangents to the circles x² + y²= 4 and x² + y²− 4x + 2y − 4 = 0 is

A
1

B
2

C
3

D
4

Solution

Given circles are $x^{2}+y^{2}=4$ and $x^{2}+y^{2}-4 x+2 y-4=0$ i.e. $x^{2}+y^{2}=4$ and $(x-2)^{2}-4+(y-1)^{2}-1-4=0$

i.e. $x^{2}+y^{2}=4$ and $(x-2)^{2}+(y-1)^{2}=9$

Center and radii of $(i)$ is $C_{1}=(0,0), r_{1}=2$

Center and radii of $(i i)$ is $C_{2}=(2,-1), r_{2}=3$

Difference between $C_{1}$ and $C_{2}=\sqrt{5}$ Now here, $\mathbf{1}=r_{2}-r_{1}<\sqrt{5}=C_{1} C_{2}<5=r_{1}+r_{2}$

So, there will be 2 common tangents.
Question 4
1 / -0

For positive integers $n_{1}, n_{2}$ the value of the expression $(1+i)^{n_{1}+\left(1+i^{3}\right)^{n_{1}}+}$ $\left(1+i^{5}\right)^{n_{2}}+\left(1+i^{7}\right)^{n_{2}}$ where $i=\sqrt[2]{-1}$ is a real number if and only if

A
n₁=n₂+1

B
n₁=n₂-1

C
n₁=n₂

D
n₁>0,n₂>0

Solution

Using $i^{3}=-i, i^{5}=i$ and $i^{7}=-i,$ we can write the given expression as
\[
\begin{array}{l}
(1+i)^{n_{1}}+\left(1+i^{3}\right)^{n_{1}}+\left(1+i^{5}\right)^{n_{2}}+\left(1+i^{7}\right)^{n_{2}} \text { where } i=\sqrt[2]{1} \\
=2\left[^{n_{1}} C_{0}+^{n_{1}} C_{2}(i)^{2}+^{n_{1}} C_{4}(i)^{4}+^{n_{1}} C_{6}(i)^{6}+\cdots \cdots\right] \\
+2\left[^{n_{2}} C_{0}+^{n_{2}} C_{2}(i)^{2}+^{n_{2}} C_{4}(i)^{4}+^{n_{2}} C_{6}(i)^{6}+\cdots \cdots\right] \\
=2\left[^{n_{1}} C_{0}-^{n_{1}} C_{2}+^{n_{1}} C_{4}-^{n_{1}} C_{6}+\cdots \cdots\right]+2\left[^{n_{2}} C_{0}-^{n_{2}} C_{2}+^{n_{2}} C_{4}-^{n_{2}} C_{6}+\cdots \cdots\right]
\end{array}
\]
This is a real number irrespective of values of $n_{1}$ and $n_{2}$
Question 5
1 / -0

Let P(n) denote the statement that $n^{2}$ + n is odd. It

is seem that P(n) ⇒ P(n + 1), $P_{n}$ is true for all

A
n > 1

B
n

C
n > 2

D
None of these

Solution

$P(n)=n^{2}+n .$ It is always odd (statement) but square of any odd number is always odd and also, sum of odd number is always even. So for no any 'n' for which this statement is true.
Question 6
1 / -0

If the line px + qy = 0 coincides with one of the lines given by ax² + 2hxy + by² = 0 then

A
ap² + 2hpq + bq² = 0

B
aq² + 2hpq + bp² = 0

C
aq²− 2hpq + bp² = 0

D
None of these

Solution

The equation $a x^{2}+2 h x y+b y^{2}=0$ can be written as $a+\frac{2 h y}{x}+b\left(\frac{y}{x}\right)^{2}=0$
Consider $p x+q y=0 .$ or $\frac{y}{x}=\frac{-p}{q}$
Substituting in equation (i), $a+2 h\left(\frac{-p}{q}\right)+b\left(\frac{-p}{q}\right)^{2}=0$
$a q^{2}-2 h p q+b p^{2}=0$
Question 7
1 / -0

If a circle with the point (−1,1) as its center touches the straight line x + 2y + 9 = 0 then the coordinates of the points of contact is

A
(-3, 3)

B
(-3,-3)

C
(0,0)

D
(7/3, -17/3)

Solution

The normal to the tangent will pass through the centre. Also, the intersection of the normal and tangent will give the point of contact. Hence consider the equation of the tangent. $y=\frac{-x}{2}-4.5$
Hence $m=\frac{-1}{2}$
Therefore the slope of the normal is 2 since slope of normal and tangent are mutually perpendicular. Therefore equation of the normal at (1,-1) is $\frac{y-1}{x+1}=2$ or $y-1=2 x+2$ or $2 x-y=-3$
Therefore, we have two equations $2 x-y=-3$ and $x+2 y+9=0$
Solving these two equations give us $(x, y)=(-3,-3)$
Hence, the point of contact is (-3,-3)
Question 8
1 / -0

The relation R defined on the set of natural numbers as {(a, b) : a differs from b by 3}, is given by

A
{(1, 4, (2, 5), (3, 6),.....}

B
{(4, 1), (5, 2), (6, 3),.....}

C
{(1, 3), (2, 6), (3, 9),..}

D
None of these

Solution

$R = {(a, b) : a, b \in N, a - b = 3} = {((n + 3), n) : n \in N}$

={(4,1),(5,2),(6,3), .......} .
Question 9
1 / -0

Two players A and B throw a die alternately for a prize of Rs 11, which is to be won by a player who first throws a six. If A starts the game, their respective expectations are

A
Rs 6; Rs 5

B
Rs 7; Rs 4

C
Rs 5.50; Rs 5.50

D
Rs 5.57; Rs 5.25

Solution

ANSWER
In his first throw $A^{\prime} s$ chance is $\frac{1}{6},$ in his second its is $\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}$
because each players must have failed once before. $A$ can have a second throw, in his third throw his chance is $\left(\frac{5}{6}\right)^{4} \times \frac{1}{6}$ because each players must have failed twice, and so on.
Thus $A^{\prime} s$ chance is the sum of the infinite series
\[
\frac{1}{6}\left(1+\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)^{4}+\ldots\right)
\]
Similarly $B^{\prime}$ s chance is the sum of the infinite series $\frac{5}{6} \cdot \frac{1}{6}\left(1+\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)^{4}\right)$ Therefore $A^{\prime} s$ chance is to $B^{\prime} s$ as 6 is to 5
their respective chance are therefore $\frac{6}{11}$ and $\frac{5}{11}$ and their expectation are $r s 6$ and $r s 5$ respectively.
Question 10
1 / -0

The coefficient of xin the equation $x^{2}$ + p $x$ + q = 0was taken as 17 in place of 13, its roots were found to be -2 and -15, the roots of the original equation are

A
3, 10

B
- 3, - 10

C
- 5, - 18

D
18,5

Solution

Let the equation (inwritten form) be $x^{2}$ + 17 $x$ + q = 0.
Roots are -2, -15.
So q = 30,
And correct equation is $x^{2}$ + 13 $x$ + 30= 0.
Hence roots are -3, -10.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Mathematics Test - 21

Result

Mathematics Test - 21

Score

-

Rank

-

SHARING IS CARING

Question 1

0.8

0.9

0.7

1.1

Solution

Question 2

n>2n

n<2n

n≥2n

Can't be determined.

Solution

Question 3

1

2

3

4

Solution

Question 4

n1=n2+1

n1=n2-1

n1=n2

n1>0,n2>0

Solution

Question 5

n > 1

n

n > 2

None of these

Solution

Question 6

ap2 + 2hpq + bq2 = 0

aq2 + 2hpq + bp2 = 0

aq2 − 2hpq + bp2 = 0

None of these

Solution

Question 7

(-3, 3)

(-3,-3)

(0,0)

(7/3, -17/3)

Solution

Question 8

{(1, 4, (2, 5), (3, 6),.....}

{(4, 1), (5, 2), (6, 3),.....}

{(1, 3), (2, 6), (3, 9),..}

None of these

Solution

Question 9

Rs 6; Rs 5

Rs 7; Rs 4

Rs 5.50; Rs 5.50

Rs 5.57; Rs 5.25

Solution

Question 10

3, 10

- 3, - 10

- 5, - 18

18,5

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

n> $2^{n}$

n< $2^{n}$

n₁=n₂+1

n₁=n₂-1

n₁=n₂

n₁>0,n₂>0

ap² + 2hpq + bq² = 0

aq² + 2hpq + bp² = 0

aq²− 2hpq + bp² = 0