Mathematics Test

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Mathematics Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

The function $f(x)=\log (x+\sqrt{x^{2}+1})$ is

A
an even function

B
an odd function

C
periodic function

D
None of these

Solution

$f(-x)=\log [-x+\sqrt{1+x^{2}}]$
$f(x)+f(-x)=\log [x+\sqrt{1+x^{2}}]+\log [-x+\sqrt{1+x^{2}}]$
$=\log \left[1+x^{2}-x^{2}\right]=\log 1=0$
$\therefore f(-x)=-f(+x)$
So, $f(x)$ is an odd function of $x$
Question 2
1 / -0

For what value of p are 2p+1, 13, 5P–3 are three consecutive terms of an A.P?

A
P = 4

B
P = 6

C
P = 2

D
P = 1

Solution

$13-2 p+1=5 p-3-13$
$12-2 p=5 p-16$
$28=7 p$
$p=4$
Question 3
1 / -0

What is $\int_{0}^{2} \frac{d x}{x^{2}+4}$ equal to?

A
π/2

B
π/4

C
π/8

D
π/3

Solution

Let $I=\int_{0}^{2} \frac{d x}{x^{2}+4}$
$\left[\because \int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]$
$=\left[\frac{1}{2} \tan ^{-1} \frac{x}{2}\right]_{0}^{2}=\frac{1}{2}\left(\frac{\pi}{4}-0\right)=\frac{\pi}{8}$
Question 4
1 / -0

A (3, 4) and B (5, –2) are two points and P is a point such that PA = PB. If the area of ∆ PAB is 10 sq units, then what are the coordinates of P?

A
Only (1, 0)

B
Only (7, 2)

C
Either (1, 0) or (7, 2)

D
Neither (1, 0) nor (7, 2)

Solution

We have $A(3,4)$ and $B(5,-2)$ Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$
Given that $\mathrm{PA}=\mathrm{PB}$
\[
\Rightarrow P A^{2}=P B^{2}
\]
$\therefore(x-3)^{2}+(y-4)^{2}=(x-5)^{2}+(y+2)^{2}$
\[
\Rightarrow x-3 y=1
\]
Area of $\Delta P A B=10$
\[
\begin{array}{l}
\therefore \frac{1}{2}\left|\begin{array}{ccc}
x & y & 1 \\
3 & 4 & 1 \\
5 & -2 & 1
\end{array}\right|=\pm 10 \\
\Rightarrow \\
\Rightarrow(4+2)-y(3-5)+1(-6-20)=\pm 20 \\
\Rightarrow 6 x+2 y-26=\pm 20 \\
\Rightarrow \quad 6 x+2 y=46 \\
\quad \text { Or } 6 x+2 y=6
\end{array}
\]
On solving eq (1) \& (2), we get
\[
x=7, y=2
\]
Similarly, solving eq(1) \& (3) we get
\[
x=1, y=0
\]
Hence coordinates of $\mathrm{P}$ are (7,2) or (1,0)
Question 5
1 / -0

The degree and order respectively of the differential equation $\frac{d y}{d x}=\frac{1}{x+y+1}$ are

A
1, 1

B
1, 2

C
2, 1

D
2, 2

Solution

Since order of the highest derivative in the given diff. equation is 1 and exponent of the derivative is also is (1,1)
Question 6
1 / -0

If $\left|\begin{array}{ccc}x^{2}+2 x & 2 x+1 & 1 \\ 2 x+1 & x+2 & 1 \\ 3 & 3 & 1\end{array}\right|=(x-1)^{k},$ then $\mathrm{k}$ equals to

A
1

B
2

C
3

D
4

Solution

Applying $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{2} \rightarrow R_{2}-R_{3}$ and taking $(\mathrm{x}-1)$ common from $R_{1} \& R_{2}$ we get $\Delta=(x-1)^{2}\left|\begin{array}{ccc}x+1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1\end{array}\right|=(x-1)^{3}$
$\therefore \mathrm{k}=3$
Question 7
1 / -0

The middle point of the segment of the straight line joining the points (p,q) and (q,–p) is $(\frac{r}{2}, \frac{s}{2})$ What is the length of the segment?

A

$\frac{\left[\left(s^{2}+r^{2}\right)^{\frac{1}{2}}\right]}{2}$

B

$\frac{\left(s^{2}+r^{2}\right)^{1 / 2}}{4}$

C

$\left(s^{2}+r^{2}\right)^{1 / 2}$

D

$s+r$

Solution

Mid point of $(p, q)$ and $(q,-p)$ is $\left(\frac{p+q}{2}, \frac{q-p}{2}\right) .$ Which is given $\left(\frac{r}{2}, \frac{s}{2}\right)$
\[
\therefore \frac{p+q}{2}=\frac{r}{2} \Rightarrow p+q=r
\]
And $\frac{q-p}{2}=\frac{s}{2} \Rightarrow q-p=s$
Now length of segment
\[
=\sqrt{(p-q)^{2}+(q+p)^{2}}=\sqrt{s^{2}+r^{2}}
\]
Question 8
1 / -0

If $\int_{-2}^{5} f(x) d x=4$ and $\int_{0}^{5}\{1+f(x)\} d x=7,$ then what is $\int_{-2}^{0} f(x) d(x)$ equal to?

A
–3

B
2

C
3

D
5

Solution

Given, $\int_{-2}^{5} \mathrm{f}(\mathrm{x}) \mathrm{dx}=4, \int_{0}^{5}\{1+\mathrm{f}(\mathrm{x})\} \mathrm{d} \mathrm{x}=7$
Let $\mathrm{I}=\int_{-2}^{5} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{-2}^{0} \mathrm{f}(\mathrm{x}) \mathrm{dx}+\int_{0}^{5} f(x) \mathrm{dx}$
$\Rightarrow 4=\int_{-2}^{0} \mathrm{f}(\mathrm{x}) \mathrm{dx}+\int_{0}^{5}[1+\mathrm{f}(\mathrm{x})-1] \mathrm{d} \mathrm{x}$
$\Rightarrow 4=\int_{-2}^{0} f(x) d x+\int_{0}^{5}[1+\mathrm{f}(\mathrm{x})] \mathrm{d} \mathrm{x}-\int_{0}^{5} 1 \mathrm{dx}$
$\Rightarrow 4=\int_{-2}^{0} \mathrm{f}(\mathrm{x}) \mathrm{dx}+7-[x]_{0}^{5}$
$\Rightarrow \int_{-2}^{0} f(x) \mathrm{dx}=2$
Question 9
1 / -0

Equation of the curve passing through (3, 9) which satisfies the differential equation

$\frac{d y}{d x}=x+\frac{1}{x^{2}}$ is

A

$6 x y=3 x^{2}-6 x+29$

B

$6 x y=3 x^{2}+29 x-6$

C

$6 x y=3 x^{3}+29 x-6$

D

None of these

Solution

Given $\frac{d y}{d x}=x+\frac{1}{x^{2}}$
$y=\frac{x^{2}}{2}-\frac{1}{x}+c$
(I) [integrating]
where $c=\frac{29}{6}$ as it passes through the point (3,9)
Put $c=\frac{29}{6}$ in eq

$Y=\frac{x^{2}}{2}-\frac{1}{x}+\frac{29}{6}$
$\Rightarrow 6 x y=3 x^{2}-6+29 x$
$6 x y=3 x^{2}-6+29 x$
$6 x y=3 x^{2}+29 x-6$
Question 10
1 / -0

A man goes 10m due east and then 24m due north. Find the distance from the starting point.

A
62 m

B
35 m

C
21 m

D
26 m

Solution

Let the initial position of the man be O and his final position be B. Since the man goes 10m due east and then 24m due north. Therefore, ∆AOB is a light angled at A s.t OA = 10m & AB = 24m.

By Pythagoras theorem we have,\

$\mathrm{OB}^{2}=\mathrm{OA}^{2}+\mathrm{AB}^{2}$
$\Rightarrow \mathrm{OB}^{2}=10^{2}+24^{2}=100+576=676$
$\Rightarrow \mathrm{OB}=\sqrt{676}=26 \mathrm{m}$

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 29

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Mathematics Test - 29

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SHARING IS CARING

Question 1

an even function

an odd function

periodic function

None of these

Solution

Question 2

P = 4

P = 6

P = 2

P = 1

Solution

Question 3

π/2

π/4

π/8

π/3

Solution

Question 4

Only (1, 0)

Only (7, 2)

Either (1, 0) or (7, 2)

Neither (1, 0) nor (7, 2)

Solution

Question 5

1, 1

1, 2

2, 1

2, 2

Solution

Question 6

1

2

3

4

Solution

Question 7

\(\frac{\left[\left(s^{2}+r^{2}\right)^{\frac{1}{2}}\right]}{2}\)

\(\frac{\left(s^{2}+r^{2}\right)^{1 / 2}}{4}\)

\(\left(s^{2}+r^{2}\right)^{1 / 2}\)

\(s+r\)

Solution

Question 8

–3

2

3

5

Solution

Question 9

\(6 x y=3 x^{2}-6 x+29\)

\(6 x y=3 x^{2}+29 x-6\)

\(6 x y=3 x^{3}+29 x-6\)

None of these

Solution

Question 10

62 m

35 m

21 m

26 m

Solution

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