Mathematics Test

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Mathematics Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

$\int_{1}^{4} \frac{x^{2}+x}{\sqrt{2 x+1}} d x$ is equal to:

A
$\frac{57-\sqrt{3}}{5}$

B
$\frac{57-\sqrt{3}}{4}$

C
$\frac{57-4 \sqrt{3}}{5}$

D
None of the above

Solution

Given:

$I=\int_{1}^{4} \frac{x^{2}+x}{\sqrt{2 x+1}} d x$

Let $2 x+1=t^{2}$ ..(1)

Differentiaiting with respect to $\mathrm{x}$, we get:

$ 2 \mathrm{dx}=2 \mathrm{tdt}$

$ d x=t d t$

From equation (1), we get:

$x=\frac{t^{2}-1}{2}$

Now,

$I=\int_{\sqrt{3}}^{3} \frac{\left(\frac{t^{2}-1}{2}\right)^{2}+\frac{t^{2}-1}{2}}{\sqrt{t^{2}}} t d t$

$=\int_{\sqrt{3}}^{3}\left(\frac{t^{4}-2 t^{2}+1}{4}+\frac{t^{2}-1}{2}\right) d t$

$=\int_{\sqrt{3}}^{3}\left(\frac{t^{4}-2 t^{2}+1+2 t^{2}-2}{4}\right) d t$

$=\frac{1}{4} \int_{\sqrt{3}}^{3}\left(t^{4}-1\right) \mathrm{d} t$

$=\frac{1}{4}\left[\frac{t^{5}}{5}-t\right]_{\sqrt{3}}^{3}$

$=\frac{57-\sqrt{3}}{5}$
Question 2
1 / -0

What is the solution to differential equation $\frac{d x}{d y}+\frac{x}{y}=0$?

A
$\frac{x}{y}=c$

B
xy=c

C
y=cx

D
None of these

Solution

Given:

$\frac{d x}{d y}+\frac{x}{y}=0$

On solving, we get-

$\Rightarrow \frac{d x}{d y}=-\frac{x}{y}$

$\Rightarrow \frac{d x}{x}=-\frac{d y}{y}$

$\Rightarrow \ln x=-\ln y+k$ or

$\Rightarrow \ln x=\ln \frac{1}{y}+\ln c$

$\Rightarrow \ln x=\ln \frac{c}{y}$

$\Rightarrow x=\frac{c}{y}$

$\Rightarrow x y=c$
Question 3
1 / -0

Find the values of ${k}$ so the line $\frac{2 {x}-2}{2 {k}}=\frac{4-{y}}{3}=\frac{{z}+2}{-1}$ and $\frac{{x}-5}{1}=\frac{{y}}{{k}}=\frac{{z}+6}{4}$ are at right angles

A
$0$

B
$-2$

C
$2$

D
$1$

Solution

Given lines are $\frac{2 {x}-2}{2 {k}}=\frac{4-{y}}{3}=\frac{{z}+2}{-1}$ and $\frac{{x}-5}{1}=\frac{{y}}{{k}}=\frac{{z}+6}{4}$

Write the above equation of a line in the standard form of lines

$\Rightarrow \frac{2(x-1)}{2 k}=\frac{-(y-4)}{3}=\frac{z+2}{-1} \Leftrightarrow \frac{(x-1)}{k}=\frac{y-4}{-3}=\frac{z+2}{-1}$

So, the direction ratio of the first line is $({k},-3,-1)$

$\frac{{x}-5}{1}=\frac{{y}}{{k}}=\frac{{z}+6}{4}$

So, direction ratio of second line is $(1,{k}, 4)$

Lines are perpendicular,

$\therefore({k} \times 1)+(-3 \times {k})+(-1 \times 4)=0$

$\Rightarrow k-3 k-4=0$

$\Rightarrow-2 k-4=0$

$\therefore{k}=-2$
Question 4
1 / -0

Find the value of $\lim _{x \rightarrow \infty} 2 x \sin \left(\frac{4}{x}\right)$.

A
$2$

B
$4$

C
$8$

D
$1$

Solution

$\lim _{x \rightarrow \infty} 2 x \sin \left(\frac{4}{x}\right)$

$=2 \times \lim _{x \rightarrow \infty} \frac{\sin \left(\frac{4}{x}\right)}{\left(\frac{1}{x}\right)}$

$=2 \times \lim _{x \rightarrow \infty} \frac{\sin \left(\frac{4}{x}\right)}{\left(\frac{4}{x}\right)} \times 4$

Let $\frac{4}{x}=t$

If $x \rightarrow \infty$ then $t \rightarrow 0$

$=8 \times \lim _{t \rightarrow 0} \frac{\sin t}{t}$

$=8 \times 1$

$=8$
Question 5
1 / -0

If (x) is an odd periodic function with period 2, then f(4) equal to:

A
-4

B
4

C
2

D
0

Solution

Since f(x) is an odd periodic function with period 2.

$\therefore f(-x)=-f(x)$ and $f(x+2)=f(x)$

$\therefore f(2)=f(0+2)=f(0)$

and $f(-2)=f(-2+2)=f(0)$

Now, $f(0)=f(-2)=-f(2)=-f(0)$

$\Rightarrow 2 f(0)=0$, i.e.,$f(0)=0$

$\therefore f(4)=f(2+2)=f(2)=f(0)=0$

Thus, $f(4)=0$
Question 6
1 / -0

Differentiate $x^{x^{x}}$ with respect to 'x'.

A

$x^{x} \cdot\left\{(1+\log x) \log x+\frac{1}{x}\right\}$

B

$x^{x^{x}} \cdot\left\{(1+\log x) \log x+\frac{1}{x}\right\}$

C

$x^{x^{x}} \cdot x^{x} \cdot\left\{(1+\log x) \log x+\frac{1}{x}\right\}$

D
None of these

Solution

Consider, $f(x)=x^{x^{x}}$
Taking log on both side, we get
$\Rightarrow \log f(x)=\log x^{x^{x}}$
$\Rightarrow \log f(x)=x^{x} \log x$
Now, use the Product Rule for Differentiation, we get
$\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{uv})=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}$
$\Rightarrow \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}=\left(\mathrm{x}^{\mathrm{x}}\right) \frac{1}{\mathrm{x}}+\log \mathrm{x} \cdot\left[\mathrm{x}^{\mathrm{x}}(1+\log \mathrm{x})\right]$
$\Rightarrow \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}=\left(\mathrm{x}^{\mathrm{x}}\right)\left[\frac{1}{\mathrm{x}}+\log \mathrm{x} .(1+\log \mathrm{x})\right]$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{f}(\mathrm{x}) \cdot\left(\mathrm{x}^{\mathrm{x}}\right)\left[\frac{1}{\mathrm{x}}+\log \mathrm{x} \cdot(1+\log \mathrm{x})\right]$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{\mathrm{x}^{\mathrm{x}}} \cdot\left(\mathrm{x}^{\mathrm{x}}\right)\left[\frac{1}{\mathrm{x}}+\log \mathrm{x} .(1+\log \mathrm{x})\right]$
Question 7
1 / -0

If the vectors $\vec{a}=2 \hat{i}-3 \hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+4 \hat{j}-2 \hat{k}$ represent the two sides of any triangle, then the area of that triangle is:

A
$\frac{1}{2} \sqrt{232}$ square units

B
$\sqrt{234}$ square units

C
$\sqrt{250}$ square units

D
$\frac{1}{2} \sqrt{230}$ square units

Solution

(1) If $\vec{a}$ and $\vec{b}$ are any two sides of triangle then, area of triangle will be,

$\Delta=\frac{1}{2}|\vec{b} \times \vec{h}|\quad\quad\cdots(1)$

(2) If, $\vec{b}=a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}$ and $\vec{h}=a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{k} \quad\quad$ ....(A)

then,

$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\end{array}\right|\quad\quad\quad\cdots(2)$

$\vec{a} \times \vec{b}=\hat{i}\left(b_{1} c_{2}-b_{2} c_{1}\right)-\hat{j}\left(a_{1} c_{2}-c_{1} a_{2}\right)+\hat{k}\left(a_{1} b_{2}-b_{1} a_{2}\right)$

Given, $\vec{a}=2 \hat{i}-3 \hat{j}-\hat{k}, \vec{b}=\hat{i}+4 \hat{j}-2 \hat{k}$

On comparing with (A) we get,

${a}_{1}=2, {b}_{1}=-3, {c}_{1}=-1 ; {a}_{2}=1, {~b}_{2}=4, {c}_{2}=-2$

Using equation $(2)$,

Then, $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -1 \\ 1 & 4 & -2\end{array}\right|$

$\vec{a} \times \vec{b}=\hat{i}(6+4)-\hat{j}(-4+1)+\hat{k}(8+3)$

$\vec{a} \times \vec{b}=10 \hat{i}+3 \hat{j}+11 \hat{k}$

Area of triangle, using (1),

$\Delta=\frac{1}{2}|\vec{a} \times \vec{b}|=\frac{1}{2}|10 \hat{i}+3 \hat{j}+11 \hat{k}|$

$=\frac{1}{2} \sqrt{(10)^{2}+(3)^{2}+(11)^{2}}$

$\Delta=\frac{1}{2} \sqrt{100+9+121}$

$\Delta=\frac{1}{2} \sqrt{230}$ square unit
Question 8
1 / -0

$\underset{{x \rightarrow 1}}\lim \frac{1-\sqrt{x}}{\cos ^{-1} x}$ is equal to:

A
$0$

B
$\frac{1}{2}$

C
$\frac{1}{4}$

D
$1$

Solution

As we know,
$\underset{{{x} \rightarrow {a}}}\lim \frac{{f}({x})}{{g}({x})}=\underset{{{x} \rightarrow {a}}}\lim \frac{{f}^{\prime}({x})}{{g}^{\prime}({x})}$
Given,
$\underset{x \rightarrow 1}{\lim} \frac{1-\sqrt{x}}{\cos ^{-1} x}$
Apply L-hospital rule
$=\underset{x \rightarrow 1}{\lim} \frac{0-\frac{1}{2 \sqrt{x}}}{-\frac{1}{\sqrt{1-x^{2}}}}$
$=\underset{x \rightarrow 1}{\lim}\frac{\sqrt{1-x^{2}}}{2 \sqrt{x}}=0$
Question 9
1 / -0

The differential equation of the family of curves y = c₁e^x + c₂e^-x is:

A
$\frac{d^{2} y}{d x^{2}} + y = 0$

B
$\frac{d^{2} y}{d x^{2}} - y = 0$

C
$\frac{d^{3} y}{d x^{3}} + y = 0$

D
$\frac{d^{4} y}{d x^{4}} - 1 = 0$

Solution

$y=c_{1} e^{x}+c_{2} e^{-x}$

$\Rightarrow \frac{d y}{d x}=\frac{d}{d x} c_{1} e^{x}+\frac{d}{d x} c_{2} e^{-x}$

$\Rightarrow \frac{d y}{d x}=c_{1} e^{x}-c_{2} e^{-x}$

$\Rightarrow \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(c_{1} e^{x}-c_{2} e^{-x}\right)$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=c_{1} e^{x}+c_{2} e^{-x}=y$

$\Rightarrow \frac{d^{2} y}{d x^{2}}-y=0$
Question 10
1 / -0

If $sinθ = \frac{24}{25}$ and $θ$ lies in the second quadrant, the $secθ+tanθ = $?

A
$-3$

B
$-5$

C
$-7$

D
$-9$

Solution

$\sin \theta=\frac{p}{h}=\frac{24}{25}$

$b=\sqrt{h^{2}-p^{2}}=\sqrt{25^{2}-24^{2}}=7$

In second quadrant, both $\sec \theta$ and $\tan \theta$ are negative

$\sec \theta=\frac{h}{b}$

$=\frac{-25}{7}$

$\tan \theta=\frac{p}{b}$

$=\frac{-24}{7}$

$\sec \theta+\tan \theta=\frac{-25}{7}+\frac{-24}{7}$

$=-7$

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 9

Result

Mathematics Test - 9

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Question 1

\(\frac{57-\sqrt{3}}{5}\)

\(\frac{57-\sqrt{3}}{4}\)

\(\frac{57-4 \sqrt{3}}{5}\)

None of the above

Solution

Question 2

\(\frac{x}{y}=c\)

xy=c

y=cx

None of these

Solution

Question 3

\(0\)

\(-2\)

\(2\)

\(1\)

Solution

Question 4

\(2\)

\(4\)

\(8\)

\(1\)

Solution

Question 5

-4

4

2

0

Solution

Question 6

\(x^{x} \cdot\left\{(1+\log x) \log x+\frac{1}{x}\right\}\)

\(x^{x^{x}} \cdot\left\{(1+\log x) \log x+\frac{1}{x}\right\}\)

\(x^{x^{x}} \cdot x^{x} \cdot\left\{(1+\log x) \log x+\frac{1}{x}\right\}\)

None of these

Solution

Question 7

\(\frac{1}{2} \sqrt{232}\) square units

\(\sqrt{234}\) square units

\(\sqrt{250}\) square units

\(\frac{1}{2} \sqrt{230}\) square units

Solution

Question 8

\(0\)

\(\frac{1}{2}\)

\(\frac{1}{4}\)

\(1\)

Solution

Question 9

d2ydx2+y=0

d2ydx2-y=0

d3ydx3+y=0

d4ydx4-1=0

Solution

Question 10

\(-3\)

\(-5\)

\(-7\)

\(-9\)

Solution

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$\frac{d^{2} y}{d x^{2}} + y = 0$

$\frac{d^{2} y}{d x^{2}} - y = 0$

$\frac{d^{3} y}{d x^{3}} + y = 0$

$\frac{d^{4} y}{d x^{4}} - 1 = 0$