Mathematics Test

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Mathematics Test - 18

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Weekly Quiz Competition

Question 1
4 / -1

On the occasion of Diwali, each student of a class sends greeting cards to every other. If there are 20 students in the class, then the total number of greeting cards exchanged by the students is

A
²⁰C₂

B
2 x ²⁰C₂

C
2 x ²⁰P_{2 -1}

D
None of these

Solution

Required number = 2 x ²⁰C₂ because if A sends to B, then B also sends to A, so it becomes twice for all.
Question 2
4 / -1

If all permutations of the letters of the word AGAIN are arranged as in dictionary, then fiftieth word will be

A
NAAGI

B
NAGAI

C
NAAIG

D
NAIAG

Solution

Starting with the letter A and arranging the other four letters, there are 4 ! = 24 words. These are the first 24 words. Then, starting with G and arranging A, A, I and N in different ways, there are = 4!/2! 1! 1! = 24/2 = 12 words

Next, the 37^th word starts with I. There are 12 words starting with I.

This accounts up to the 48^th word.

The 49^th word is NAAGI.

The 50^th word is NAAIG.
Question 3
4 / -1

\(\int \frac{x^{2}-2}{x^{2} \sqrt{x^{2}-1}} d x\) is equal to

A
0

B
2/3

C
4/3

D
None of these

Solution

Putting \(x^{2}-1=t^{2}\) and \(x d x=t\) dt, we get \(1-\int_{0}^{0} \frac{t^{2}-1}{\left(t^{2}+1\right)^{2}} d t=\int_{0}^{1} \frac{t^{2}-1}{\left(t^{2}+1\right)^{2}} d t+\int_{\left(t^{2}+1\right)^{2}}^{t^{2}-1} d t-1_{1}+1_{2}\)
where \(l_{1}=\int_{0}^{1} \frac{t^{2}-1}{\left(t^{2}+1\right)^{2}} d t\) and \(l_{2}=\int_{\left(t^{2}+1\right)^{2}}^{t^{2}-1} d t\)
Putting \(\mathrm{t}-1 / \mathrm{u}\) in \(\mathrm{I}_{2}\), we get \(I_{2}=-\int_{0}^{1} \frac{u^{2}-1}{\left(u^{2}+1\right)^{2}} d u=-1_{1}\)
\(\Rightarrow 1=1_{1}+1_{2}=1_{1}-1_{1}=0\)
Question 4
4 / -1

The order and degree of differential equation d²s/dt²+ 4 = 0 are

A
2, 2

B
2, 3

C
3, 2

D
None of these

Solution

The order is 2 and degree is 1.

Hence option D is correct.
Question 5
4 / -1

If the mean of a binomial distribution is 25, then its standard deviation lies in the interval

A
[0, 5)

B
(0, 5]

C
[0, 25)

D
(0, 25]

Solution

\(S D . \sigma=\sqrt{n p q} \geq 0\)
Mean, \(n p=25\) and \(q<1\) \(\therefore \sigma=\sqrt{n p q}<\sqrt{n p}\)
\(\sigma<5\)
\(\Rightarrow 0 \leq \sigma<5\)
Question 6
4 / -1

\(\operatorname{Lt}_{x \rightarrow 0} \frac{1-\cos x}{x}\)is equal to

A
1/2

B
0

C
1

D
None of these

Solution

On applying the L' hospital rule, we get Lt_x→0 Sin x = 0.

Hence option B is correct.
Question 7
4 / -1

If the roots of the equation ax² + bx + c = 0 are of the form k+1/k and k+2/k+1 then (a + b + c)² is equal to

A
2b² − ac

B
a²

C
b² − 4ac

D
b² − 2ac

Solution

We have \(\frac{k+1}{k}+\frac{k+2}{k+1}=\frac{-b}{a}\)
and \(\frac{k+1}{k} \cdot \frac{k+2}{k+1}=\frac{c}{a}\)
\(\Rightarrow \frac{k+2}{k}=\frac{c}{a}\)
or \(\frac{2}{k}=\frac{c}{a}-1=\frac{c-a}{a}\)
\(\ldots(2)\)
or \(\mathrm{k}=\frac{2 \mathrm{a}}{\mathrm{c}-\mathrm{a}}\)
Now eliminate k putting the value of \(k\) in \(1^{\text {st }}\) relation, we get \(\frac{c+a}{2 a}+\frac{2 c}{c+a}=\frac{-b}{a}\)
\(\Rightarrow(c+a)^{2}+4 a c=-2 b(a+c)\)
\(\Rightarrow(a+c)^{2}+2 b(a+c)=-4 a c\)
Adding \(\mathrm{b}^{2}\) on both sides, \((a+b+c)^{2}=b^{2}-4 a c\)
Question 8
4 / -1

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 9
4 / -1

\(\int_{0}^{\pi / 2} \frac{\sin ^{2} x}{\sin x+\cos x} d x\) is equal to

A
π/2

B
√2 log (√2 + 1)

C
(1/√2) log (√2 + 1)

D
None of these

Solution

\(\operatorname{Let} \mid=\int_{8}^{\pi / 2} \frac{\sin ^{2} x}{\sin x+\cos x} d x\)
Then \(\mid=\int_{0}^{\pi / 2} \frac{\sin ^{2}(\pi / 2-x)}{\sin (\pi / 2-x)+\cos (\pi / 2-x)} d x\)
\(\Rightarrow 21=\int_{0}^{\pi / 2} \frac{\sin ^{2} x+\cos ^{2} x}{\sin x+\cos x} d x=\int_{0}^{\pi / 2} \frac{1}{\sin x+\cos x} d x\)
\(21=(1 / \sqrt{2}) \int_{\pi / 2}^{\pi / 2} \frac{1}{\cos (x-\pi / 4)} d x\)
\(2 \mid-(1 / \sqrt{2}) \int_{0}^{\pi / 2} \sec (x-\pi / 4) d x\)
\(21=(1 / \sqrt{2})[\log (\sec (x-\pi / 4)+\tan (x-\pi / 4))]_{0}^{\pi / 2}\)
\(21=(1 / \sqrt{2})[\log (\sqrt{2}+1)-\log (\sqrt{2}-1)]\)
\(2 \mid=(1 / \sqrt{2}) \log \left(\frac{12+1}{\sqrt{2}-1}\right)-(1 / \sqrt{2}) \log (\sqrt{2}+1)^{2}\)
\(21=\sqrt{2} \log (\sqrt{2}+1)\)
\(\Rightarrow 1=(1 / \sqrt{2}) \log (\sqrt{2}+1)\)
Question 10
4 / -1

The coefficient of x⁵ in (1+2x+3x²+……….up to infinite term)^-3/2 is

A
21

B
25

C
26

D
None of these

Solution

(1+2x+3x²+………)^–3/2

=((1–x)^–2)^–3/2

= (1 − x)³ = 1 − 3x + 3x² − x³

∴coefficient of x⁵ = 0.

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Mathematics Test - 18

Result

Mathematics Test - 18

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SHARING IS CARING

Question 1

20C2

2 x 20C2

2 x 20P2 -1

None of these

Solution

Question 2

NAAGI

NAGAI

NAAIG

NAIAG

Solution

Question 3

0

2/3

4/3

None of these

Solution

Question 4

2, 2

2, 3

3, 2

None of these

Solution

Question 5

[0, 5)

(0, 5]

[0, 25)

(0, 25]

Solution

Question 6

1/2

0

1

None of these

Solution

Question 7

2b2 − ac

a2

b2 − 4ac

b2 − 2ac

Solution

Question 8

69760

30240

99784

none of these

Solution

Question 9

π/2

√2 log (√2 + 1)

(1/√2) log (√2 + 1)

None of these

Solution

Question 10

21

25

26

None of these

Solution

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²⁰C₂

2 x ²⁰C₂

2 x ²⁰P_{2 -1}

2b² − ac

a²

b² − 4ac

b² − 2ac