Mathematics Test

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Mathematics Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

\(\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} \mathrm{d} x=\)

A
π / 2−1

B
π / 2+1

C
π - 1

D
3π/2

Solution

\(\int_{0}^{1} \frac{1-x}{\sqrt{1-x^{2}}} d x=\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} d x+\frac{1}{2} \int_{0}^{1} \frac{-2 x}{\sqrt{1-x^{2}}} d x\)
\(=\left(\sin ^{-1} x\right)_{0}^{1}+\frac{1}{2} \cdot 2(\sqrt{1-x^{2}})_{0}^{1}=\frac{\pi}{2}-1\)
Question 2
1 / -0

For a random variable X,E(X)=3 and E(X)²=11. Then, variance of X is

A
8

B
5

C
2

D
1

Solution

Given that, \(E(X)=3\) and \(E\left(X^{2}\right)=11\)
\(\text { Variance of } X=E\left(X^{2}\right)-[E(X)]^{2} \\\)
\(=11-(3)^{2}=11-9=2\)
Question 3
1 / -0

If the points \(\mathrm{P}_{1}\) and \(\mathrm{P}_{2}\) represent two complex numbers \(\mathrm{z}_{1}\) and \(\mathrm{z}_{2}\), then the point \(\mathrm{P}_{3}\) represents the number:

A
\(\mathrm{z}_{1}+\mathrm{z}_{2}\)

B
\(\mathrm{z}_{1}-\mathrm{z}_{2}\)

C
\(\mathrm{z}_{1} \times \mathrm{z}_{2}\)

D
\(\mathrm{z}_{1} \div \mathrm{z}_{2}\)

Solution

This is a parallelogram \(\mathrm{OP}_{1} \mathrm{P}_{2} \mathrm{P}_{3}\).

Then the midpoint of \(P_{1} P_{2}\) and \(O P_{3}\) are the same.

But midpoint of \(\mathrm{P}_{1} \mathrm{P}_{2}\) is \(\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right)\).

So that the coordinates of \(\mathrm{P}_{3}\) are \(\left(\mathrm{x}_{1}+\mathrm{x}_{2}, \mathrm{y}_{1}+\mathrm{y}_{2}\right)\).

Thus, the point \(\mathrm{P}_{3}\) corresponds to sum of the complex number \(\mathrm{z}_{1}\) and\(\mathrm{z}_{2}\).

\(\overrightarrow{\mathrm{OP}}_{3}=\overrightarrow{\mathrm{OP}}_{1}+\overrightarrow{\mathrm{P}_{1} \mathrm{P}_{3}}=\overrightarrow{\mathrm{OP}}_{1}+\overrightarrow{\mathrm{OP}}_{2}=\mathrm{z}_{1}+\mathrm{z}_{2}\)
Question 4
1 / -0

A
4

B
2

C
1

D
1/4

Solution
Question 5
1 / -0

The locus of the point PP which is equidistant from 3x+4y+5=03x+4y+5=0 and 9x+12y+7=09x+12y+7=0 is

A
A hyperbola

B
An ellipse

C
A parabola

D
A straight line

Solution

\(\left|\frac{3 x+4 y+5}{5}\right|=\left|\frac{9 x+12 y+7}{25}\right|\)
\(15 x+20 y+25-9 x-12 y-7=0\)
\(6 x+8 y+18=0\)
\(3 x+4 y+9=0\)

So, locus is a straight line.
Question 6
1 / -0

Two dice are thrown nn times in succession. The probability of obtaining a double six at least once, is

A
(1 / 36)ⁿ

B
1−(35 / 36)ⁿ

C
(1 / 12)ⁿ

D
None of these

Solution

Here,

\(\mathrm{p}=\) Probability of getting double six in two dice
\(=\frac{1}{6^{2}}=\frac{1}{36}\)
and \(q=\) Probability of not getting double six in two dice \(=1-p=\frac{35}{36}\)
\(\therefore\) Required probability
\(=1-\text { (Probability of not getting double } \operatorname{six})^{n}\)
\(=1-\left(\frac{35}{36}\right)^{n}\)
Question 7
1 / -0

If the sum of the first four terms of an AP is 16 and the sum of their squares is 84, then the terms are

A
1, 5, 7, 9

B
3, 5, 7, 9

C
1, 3, 5, 7

D
4, 6, 8, 10

Solution

The four terms are \((a-3 d)\), \((a-d)\), \((a+d)\) and \((a+3 d)\)
\(a-3 d+a-d+a+d+a+3 d=16\)
\(4 a=16\)
\(a=4\)
\((a-3 d)^{2}+(a-d)^{2}+(a+d)^{2}+(a+3 d)^{2}=84\)
\(4 a^{2}+20 d^{2}=84\)
\(d=1,-1\)
Therefore, the terms are 1,3,5,7 or 7,5 ,3 ,1
Question 8
1 / -0

If A and B are square matrices of order 3, then

A
adj(AB)=adjA+adjB

B
(A + B)⁻¹= A⁻¹⁺ B⁻¹

C
AB=O ⇒|A|=0 or |B|=0

D
AB=O ⇒|A|=0 and |B|=0

Solution

AB=O⇒|AB|=0

⇒|AB|=0

⇒|A|⋅|B|=0

⇒|A|=0or|B|=0
Question 9
1 / -0

\(\text { The domain of definition of } f(x)=\sqrt{\log _{0.4}}\left(\frac{x-1}{x+5}\right) \times \frac{1}{x^{2}-36} \text { is }\)

A
{x : x > 1, x ≠ 6}

B
{x : x > 0, x ≠ 6}

C
{x : x < 0, x ≠ - 6}

D
None of these

Solution

For \(\sqrt{\log _{0.4}}\left(x-\frac{1}{x}+5\right)\) to be defined, we must have \(01\) Moreover, \(\frac{1}{x^{2}}-36\) is
defined for \(x \neq 6,-6\) hence the domain of \(f\) is \([x: x>1\)
\(x \neq 6\)
Question 10
1 / -0

Let S = (x – 1)¹⁰ + (x – 1)⁹ (x + 1) + (x – 1)⁸ (x + 1)² + …….+ (x + 1)¹⁰

Which of the following is a constant term in the expansion of S?

A
2

B
¹¹C₁

C
¹¹C₁₀

D
1

Solution

\(S=(x-1)^{10}+(x-1)^{9}(x+1)+(x-1)^{8}(x+1)^{2}+\ldots+(x+1)^{10}\)
\(=(x-1)^{10}\left\{\frac{\left(\frac{x+1}{x+1}\right)^{11}-1}{\frac{x+1}{x-1}-1}\right)\)
\(=\frac{1}{2}\left\{(x+1)^{11}-(x-1)^{11}\right\}\)
\(=^{11} C_{1} x^{10}+^{11} C_{3} x^{8}+^{11} C_{5} x^{6}+^{11} C_{7} x^{4}+^{11} C_{9} x^{2}+1\)

So, constant term in the expansion of S=1

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Mathematics Test - 17

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Mathematics Test - 17

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SHARING IS CARING

Question 1

π / 2−1

π / 2+1

π - 1

3π/2

Solution

Question 2

8

5

2

1

Solution

Question 3

\(\mathrm{z}_{1}+\mathrm{z}_{2}\)

\(\mathrm{z}_{1}-\mathrm{z}_{2}\)

\(\mathrm{z}_{1} \times \mathrm{z}_{2}\)

\(\mathrm{z}_{1} \div \mathrm{z}_{2}\)

Solution

Question 4

4

2

1

1/4

Solution

Question 5

A hyperbola

An ellipse

A parabola

A straight line

Solution

Question 6

(1 / 36)n

1−(35 / 36)n

(1 / 12)n

None of these

Solution

Question 7

1, 5, 7, 9

3, 5, 7, 9

1, 3, 5, 7

4, 6, 8, 10

Solution

Question 8

adj(AB)=adjA+adjB

(A + B)−1= A−1+ B−1

AB=O ⇒|A|=0 or |B|=0

AB=O ⇒|A|=0 and |B|=0

Solution

Question 9

{x : x > 1, x ≠ 6}

{x : x > 0, x ≠ 6}

{x : x < 0, x ≠ - 6}

None of these

Solution

Question 10

2

11C1

11C10

1

Solution

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(1 / 36)ⁿ

1−(35 / 36)ⁿ

(1 / 12)ⁿ

(A + B)⁻¹= A⁻¹⁺ B⁻¹

¹¹C₁

¹¹C₁₀