Mathematics Test

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Mathematics Test - 9

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Weekly Quiz Competition

Question 1
4 / -1

The area of the region bounded by the curve y² = 4x, y-axis and the line y = 3 is

A
2 sq. units

B
9/4 sq. units

C
6 √3sq. units

D
None of these

Solution

Required area $=\int_{0}^{x} d y$
$=\int_{0}^{3} \frac{y^{2}}{4} d y=\left[\frac{y^{3}}{12}\right]_{0}^{3}=\frac{27}{12}=\frac{9}{4}$
Question 2
4 / -1

Evaluate: $\int_{0}^{2 \pi} \sqrt{1+\sin (x / 2)} d x$

A
8

B
2

C
4

D
None of these

Solution

$1+\sin \frac{x}{2}=\cos ^{2} \frac{x}{4}+\sin ^{2} \frac{x}{4}+2 \sin \frac{x}{4} \cos \frac{x}{4}=\left(\cos \frac{x}{4}+\sin \frac{x}{4}\right)^{2}$
$\mid=\int_{0}^{2 \pi}\left(\cos \frac{x}{4}+\sin \frac{x}{4}\right) d x=4\left[\sin \frac{x}{4}-\cos \frac{x}{4}\right]_{0}^{2 \pi}=8$
Question 3
4 / -1

(tan 3x - tan 2x - tan x) is equal to

A
tan x tan 2x tan 3x

B
-tan x + tan 2x tan 3x

C
tan x tan x tan 3x - tan 2x tan 3x

D
None of these

Solution

$\tan 3 x=\tan (2 x+x)$
$\Rightarrow \tan 3 x=\frac{\tan 2 x+\tan x}{1-\tan 2 x \tan x}$
$\Rightarrow \tan 3 x-\tan 3 x \tan 2 x \tan x=\tan 2 x+\tan x$
$\Rightarrow \tan 3 x-\tan 2 x-\tan x=\tan 3 x \tan 2 x \tan x$
Question 4
4 / -1

Evaluate $\int \frac{x^{4}-\sqrt[3]{x}}{6 \sqrt{x}} d x$

A

$\frac{1}{72} x^{\frac{8}{2}}-\frac{1}{5} x^{\frac{4}{6}}+c$

B

$\frac{1}{20} x^{\frac{6}{4}}-\frac{1}{2} x^{\frac{7}{6}}+c$

C

$\frac{1}{4} x^{\frac{3}{4}}-\frac{1}{7} x^{\frac{4}{5}}+c$

D

$\frac{1}{27} x^{\frac{9}{2}}-\frac{1}{5} x^{\frac{5}{6}}+c$

Solution

since there is no "Quotient Rule" for integrals we'll need to break up the integrand and simplify a little prior to integration.
\[
\int \frac{x^{4}-\sqrt[3]{x}}{6 \sqrt{x}} d x=\int \frac{x^{4}}{6 x^{\frac{1}{2}}}-\frac{x^{\frac{1}{3}}}{6 x^{\frac{1}{2}}} d x=\int \frac{1}{6} x^{\frac{7}{2}}-\frac{1}{6} x^{-\frac{1}{6}} d x
\]
Don't forget to convert the roots to fractional exponents!
At this point there really isn't too much to do other than to evaluate the integral.
\[
\int \frac{x^{4}-\sqrt[3]{x}}{6 \sqrt{x}} d x=\int \frac{1}{6} x^{\frac{7}{2}}-\frac{1}{6} x^{-\frac{1}{6}} d x=\frac{1}{27} x^{\frac{9}{2}}-\frac{1}{5} x^{\frac{5}{6}}+c
\]
Don't forget to add on the "+c" since we know that we are asking what function did we differentiate to get the integrand and the derivative of a constant is zero and so we do need to add that onto the answer.
Question 5
4 / -1

If a parallelepiped is formed by planes drawn through the points (5, 8, 10) and (3, 6, 8) parallel to the coordinate planes, then the length of diagonal of the parallelopiped is

A
2√3

B
3√2

C
√2

D
√3

Solution

A = (x1, y1, z1)

B = (x2, y2, z2)

AB = $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$

= $\sqrt{{(3 - 5)}^{2} + {(6 - 8)}^{2} + {(8 - 10)}^{2}}$

= $\sqrt{4 + 4 + 4} = \sqrt{3 \times 4} = 2 \sqrt{3}$
Question 6
4 / -1

Three identical dice are rolled. The probability that the same number will appear on each of them is

A
1/6

B
1/216

C
1/36

D
None of these

Solution

Exhaustive number of cases = 6³ = 216.

The same number can appear on each of the dice in the following ways: (1, 1, 1), (2, 2, 2).... (6, 6, 6).

So, favourable number of cases = 6.

Hence, required probability = 6/216 = 1/36.
Question 7
4 / -1

If n(U) = 20, n(A) = 12, n(B) = 9, n(AB) = 4, where U is the universal set, A and B are subsets of U, then n | (A U B)^o| is equal to

A
17

B
9

C
11

D
3

Solution

$n(U)=20, n(A)=12, n(B)=9, n(A \cap B)=4$
$\therefore n(A \cup B)=n(A)+n(B)-n(A \cap B)$
$=12+9-4=17$
Hence, $n\left[(A \cup B)^{c}\right]=n(U)-n(A \cup B)$
$=20-17=3$
Question 8
4 / -1

A missile fired from the ground level rises x meters vertically upwards in t seconds, where x = 100t - 25/2 t². The maximum height reached is

A
200 m

B
125 m

C
160 m

D
190 m

Solution

The given equation is
\[
x=100 t-\frac{25}{2} t^{2}
\]
On differentiating w.r.t. $t$, we get
\[
\frac{d x}{d t}=100-\frac{25}{2} \cdot(2 t)=100-25 t
\]
We know that the velocity of missile is zero at maximum height.
$\therefore$ On putting $\frac{d x}{d t}=0,$ we $8 e t$
\[
100-25 t=0
\]
$\Rightarrow \quad t=4$
\[
\begin{aligned}
\therefore x=100 \times 4-\frac{25 \times 16}{2} &=400-200 \\
&=200
\end{aligned}
\]
Question 9
4 / -1

A question paper is divided into two parts A and B and each part contains 5 questions. The number of ways in which a candidate can answer 6 questions by selecting at least two questions from each part is:

A
80

B
100

C
200

D
None

Solution

The number of ways that the candidate may select 2 questions from $A$ and 4 from $B=$ ${ }^{5} \mathrm{C}_{2} \times{ }^{5} \mathrm{C}_{4}$

3 questions from $A$ and 3 from $B={ }^{5} C_{3} \times{ }^{5} C_{3}$

4 questions from $A$ and 2 from $B={ }^{5} C_{4} \times{ }^{5} C_{2}$

So, total number of ways are 200.
Question 10
4 / -1

If the two pairs of lines x²– 2mxy – y² = 0 and x² – 2nxy – y² = 0 are such that one of them represents the bisector of the angles between the other, then

A
mn + 1 = 0

B
mn – 1 = 0

C
1/m + 1/n = 0

D
1/m - 1/n = 0

Solution

Equation of the bisectors of the angle between the lines $x^{2}-2 m x y-y^{2}=0$ are given by
$\frac{x^{2}-y^{2}}{1-(-1)}=\frac{x y}{-m} \Rightarrow x^{2}+\frac{2}{m} x y-y^{2}=0$
since.
(i) and $x^{2}-2 n x y-y^{2}=0$ represents the same pair of lines. $\therefore \quad \frac{1}{1}=\frac{\frac{2}{m}}{-2 n}$
$\Rightarrow \quad m n=-1 \quad \Rightarrow \quad m n+1=0$

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Mathematics Test - 9

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Mathematics Test - 9

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Question 1

2 sq. units

9/4 sq. units

6 √3sq. units

None of these

Solution

Question 2

8

2

4

None of these

Solution

Question 3

tan x tan 2x tan 3x

-tan x + tan 2x tan 3x

tan x tan x tan 3x - tan 2x tan 3x

None of these

Solution

Question 4

\(\frac{1}{72} x^{\frac{8}{2}}-\frac{1}{5} x^{\frac{4}{6}}+c\)

\(\frac{1}{20} x^{\frac{6}{4}}-\frac{1}{2} x^{\frac{7}{6}}+c\)

\(\frac{1}{4} x^{\frac{3}{4}}-\frac{1}{7} x^{\frac{4}{5}}+c\)

\(\frac{1}{27} x^{\frac{9}{2}}-\frac{1}{5} x^{\frac{5}{6}}+c\)

Solution

Question 5

2√3

3√2

√2

√3

Solution

Question 6

1/6

1/216

1/36

None of these

Solution

Question 7

17

9

11

3

Solution

Question 8

200 m

125 m

160 m

190 m

Solution

Question 9

80

100

200

None

Solution

Question 10

mn + 1 = 0

mn – 1 = 0

1/m + 1/n = 0

1/m - 1/n = 0

Solution

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