Mathematics Test

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Mathematics Test - 1

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Weekly Quiz Competition

Question 1
1 / -0

\begin{equation}
\text { Evaluate: } \int^{4} 4[x] d x
\end{equation}

A
24

B
36

C
48

D
12

Solution

\begin{equation}
\begin{aligned}
&\mid-4\left[\int[x] d x+\int[x] d x+\int[x] d x\right]\\
&=4 \cdot\left[\int_{3}^{4} 3 d x+\int_{4}^{5} 4 d x \cdot \int_{5}^{6} 5 d x\right.\\
&-4.13 \times 1]^{4}+\left[4 \times 1^{5}+15 \times 19\right.\\
&=4 .[3+4 \cdot 5]=48
\end{aligned}
\end{equation}
Hence, option, C is correct answer
Question 2
1 / -0

\begin{equation}
\text { Simplify } \int_{0}^{\pi / 2} \cos ^{2} x d x
\end{equation}

A
π

B
π/3

C
π/2

D
π/4

Solution

\begin{equation}
\begin{array}{l}
1=\frac{1}{2} \int_{0}^{\pi / 2}(1+\cos 2 x) d x \\
1=\frac{1}{2}\left[x+\frac{\sin 2 x}{2}\right]_{0}^{\pi / 2}=\frac{\pi}{4}
\end{array}
\end{equation}
Hence, option, D is correct answer
Question 3
1 / -0

\begin{equation}
\int_{0}^{\pi / 2} \sin ^{2} x d x
\end{equation}

A
π/4

B
π/3

C
π/2

D
none of these

Solution

\begin{equation}
\begin{array}{l}
1-\frac{1}{2} \int_{0}^{\pi / 2}(1-\cos 2 x) d x \\
=\frac{1}{2}\left[x-\frac{\sin 2 x}{2}\right]_{0}^{\pi / 2}=\frac{\pi}{4}
\end{array}
\end{equation}
Hence, option, A is correct answer
Question 4
1 / -0

$\int_{0}^{n} \sqrt{1 - \cos x} d x =$

A
2

B
1

C
√2

D
2√2

Solution

\begin{equation}
\begin{array}{l}
1-\int_{0}^{\pi} \sqrt{2 \sin ^{2}(x / 2)} d x \cdot-\sqrt{2} \int_{0}^{\pi} \sin (x / 2) d x=\sqrt{2}\left[\frac{-\cos (x / 2)}{(1 / 2)}\right]_{0}^{\pi} \\
-2 \sqrt{2}\left[-\cos \frac{x}{2} \cdot \cos 0\right]-2 \sqrt{2}
\end{array}
\end{equation}
Hence, option, D is correct answer
Question 5
1 / -0

\begin{equation}
\int_{0}^{\pi} \sqrt{1+\sin x} d x \text { is equal to }
\end{equation}

A
2

B
2√2

C
4

D
None of these

Solution

\begin{equation}
\begin{array}{l}
\text { 1 }=\int_{\pi}^{\pi} \sqrt{\cos ^{2}(x / 2)+\sin ^{2}(x / 2)+2 \sin (x / 2) \cos (x / 2)} d x \\
=\int[\cos (x / 2)+\sin (x / 2)] d x \\
=\left[\frac{\sin (x / 2)}{(1 / 2)}-\frac{\cos (x / 2)}{(1 / 2)}\right]_{0}^{\pi} \\
=2[\sin (x / 2)-\cos (x / 2)]_{0}^{\pi}=4
\end{array}
\end{equation}
Hence, option, C is correct answer
Question 6
1 / -0

\begin{equation}
\text { Evaluate: } \int_{0}^{2 \pi} \sqrt{1+\sin (x / 2)} d x
\end{equation}

A
8

B
2

C
4

D
None of these

Solution

\begin{equation}
\begin{array}{l}
1+\sin \frac{x}{2}=\cos ^{2} \frac{x}{4} \cdot \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
\end{array}
\end{equation}
Hence, option, A is correct answer
Question 7
1 / -0

Let f : R be a differentiable function and f(1) = 4. Find the value of
\begin{equation}
\lim _{x \rightarrow 1} \int_{4}^{f(x)} \frac{2 t}{x-1} d t, \text { if } f^{\prime}(1)=2
\end{equation}

A
16

B
8

C
4

D
2

Solution

\begin{equation}
\begin{aligned}
\lim _{x \rightarrow 1} \frac{\int_{4}^{f(x)} 2 t d t}{x-1} & \\
&=\lim _{x \rightarrow 1} \frac{2 f(x) \cdot f^{\prime}(x)}{1} \\
&=2 f(1) \cdot f^{\prime}(1)=2 \cdot 4 \cdot 2=16
\end{aligned}
\end{equation}
Hence, option, A is correct answer
Question 8
1 / -0

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 siven equation is
$$
x=100 t-\frac{25}{2} t^{2}
$$
On differentiating w.r. $t,$ 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 purting $\frac{d x}{d t}=Q$ we get
$$
100-25 t=0
$$
$t=4$
$\therefore x=100 \times 4-\frac{25 \times 16}{2}=400-200$
$$
=200
$$
Hence, option, A is correct answer
Question 9
1 / -0

What does y = x² represent in a 3-dimensional coordinate system?

A
A parabola

B
A cylinder

C
A paraboloid

D
None of these

Solution

y = $x^{2}$ represents a cylinder in a 3 dimensional coordinate system.
Hence, option, B is correct answer
Question 10
1 / -0

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 parallelepiped is

A
2√3

B
3√2

C
√2

D
√3

Solution

According to question, if we draw the parallelepiped in 3D space.
Points P(5, 8,10) and Q(3, 6, 8) appear to be opposite vertices of parallelepiped.
∴ Length of PQ gives the length of diagonal of parallelepiped.
Using distance formula we can find PQ as-
PQ = $\sqrt{{(5 - 3)}^{2} + {(8 - 6)}^{2} + {(10 - 8)}^{2}}$
= $\sqrt{2^{2} + 2^{2} + 2^{2}}$
= $\sqrt{12}$
= $2 \sqrt{3}$
∴ Length of diagonal of the parallelepiped is 2√3 units.
Hence, option(A) is the only correct choice.

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 1

Result

Mathematics Test - 1

Score

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Rank

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SHARING IS CARING

Question 1

24

36

48

12

Solution

Question 2

π

π/3

π/2

π/4

Solution

Question 3

π/4

π/3

π/2

none of these

Solution

Question 4

2

1

√2

2√2

Solution

Question 5

2

2√2

4

None of these

Solution

Question 6

8

2

4

None of these

Solution

Question 7

16

8

4

2

Solution

Question 8

200 m

125 m

160 m

190 m

Solution

Question 9

A parabola

A cylinder

A paraboloid

None of these

Solution

Question 10

2√3

3√2

√2

√3

Solution

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