Mathematics Test

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Mathematics Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

If distance between the foci of an ellipse is 6 and the distance between directrices is 12, then the length of its latus rectum is:

A
$3 \sqrt{2}$

B
$2 \sqrt{3}$

C
$3 \sqrt{5}$

D
3

Solution

given
2ae $=6$
$\therefore a e=3$
Also,
$\frac{2 a}{c}=126$
$\therefore \frac{a}{e}=63$
from (1) and (2)
$e=\frac{1}{\sqrt{2}}, a=3 \sqrt{2}$
since, $b^{2}=a^{2}\left(1-e^{2}\right)$
Substitute the values of "e' and 'a' in above equation
$\Rightarrow b^{2}=9$
$\therefore b=\pm 3$
length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 9}{3 \sqrt{2}}=3 \sqrt{2}$

Hence, option, A is correct answer
Question 2
1 / -0

The differential equation of all non - horizontal lines in a plane is

A
d²y/dx² = 0

B
dx/dy = 0

C
dy/dx = 0

D
d²x/dy²= 0

Solution

Let y = mx+c represents all non horizontal line in a plane.

$∴ \frac{d y}{d x} = m a n d \frac{d^{2} y}{d x^{2}} = 0$

Hence, option, A is correct answer
Question 3
1 / -0

If three natural numbers between 1 and 100 are selected randomly, then the probability that all are divisible by both 2 and 3 is

A
4/105

B
4/33

C
4/35

D
4/1155

Solution

Let E=Events of numbers divisible by 2 and 3 (i.e., divisible by 6)

=(6,12,....,96)

$∴ R e q u i r e d P r o p o b a b i l i t y = \frac{{}^{16}C_{3}}{100_{C_{3}}}$

$= \frac{\frac{\frac{16 \times 15 \times 14}{3 \times 2 \times 1}}{100 \times 99 \times 98}}{3 \times 2 \times 1}$ $= \frac{4}{1155}$

Hence, option, D is correct answer
Question 4
1 / -0

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 bisector of the angel between the lines $x^{2} - 2 m x y - y^{2} = 0 a r e g i v e n b y$

$\frac{x^{2} - y^{2}}{1 - (- 1)} = \frac{x y}{- m} \Rightarrow x^{2} + \frac{2}{m} x y - y^{2} = 0 . . . (i)$

Since, (i) and $x^{2} - 2 n x y - y^{2} = 0 r e r e p r e s e n t s t h e s a m e p a i r o f l i n e s .$

$∴ \frac{1}{1} = \frac{\frac{2}{m}}{- 2 n}$

$\Rightarrow m n = - 1 \Rightarrow m n + 1 = 0$

Hence, option, A is correct answer
Question 5
1 / -0

A man of 2 m height walks at a uniform speed of 6 km/h away from a lamp post of 6 m height. The rate at which the length of his shadow increases is

A
2 km/h

B
1 km/h

C
3 km/h

D
6 km/h

Solution

and in

On differentiating w.r.t. , we get

given

Hence, option, C is correct answer
Question 6
1 / -0

In how many ways can 8 students be arranged in a row?

A
8!

B
7!

C
8

D
7

Solution

Required number of ways = 8!

Hence, option, A is correct answer
Question 7
1 / -0

If f(x) = | x |³, then f′(0) =

A
0

B
1/2

C
-1

D
None of these

Solution

\begin{equation}
\begin{aligned}
&R f^{\prime}(0)=_{1 \rightarrow 0}^{10} \frac{f(0+h)-f(0)}{h}\\
&=\lim _{h \rightarrow 0} \frac{|h|^{3}-0}{h}=\lim _{h \rightarrow 0} \frac{h^{3}}{h}=\lim _{h \rightarrow 0} h^{2}=0\\
&L f^{\prime}(0)=h^{\lim }_{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=h_{h \rightarrow 0}^{\lim } \frac{1+h^{3}}{-h}\\
&R f^{\prime}(0)=L f^{\prime}(0)=0, \text { Hence, } f(0)=0
\end{aligned}
\end{equation}

Hence, option, A is correct answer
Question 8
1 / -0

Let f(x)= ${(x + 1)}^{2}$ −1,(x≥−1). Then the set S={x:f(x)= $f^{- 1}$ (x)}is

A
1

B
$\pm \sqrt{x + 1} - 1$

C
Does not exist

D
None of these

Solution

\begin{equation}
\begin{aligned}
&\text { Let, }\\
&f(x)=(x+1)^{2}-1\\
&y=(x+1)^{2}-1\\
&(x+1)^{2}=y+1\\
&(x+1)=\pm \sqrt{y+1}\\
&x=\pm \sqrt{y+1}-1\\
&f^{-1}(x)=\pm \sqrt{x+1}-1
\end{aligned}
\end{equation}

Hence, option, C is correct answer
Question 9
1 / -0

Let f(x) = x^3/2, then f′(0) =

A
3/2

B
1/2

C
Does not exist

D
None of these

Solution

$R f(0)=\lim _{h \rightarrow 0} \frac{t(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{h^{3 / 2}-0}{h}=\lim _{h \rightarrow 0} n^{1 / 2}=0$
$L f(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{(-h)^{3 / 2}-0}{(-h)} \cdot \lim _{h \rightarrow 0}(-h)^{1 / 2}$
which is imaginary
So, $R f(0)$ Lf (0) . Hence, $f(0)$ does not exist.

Hence, option, C is correct answer
Question 10
1 / -0

If f(x)=|x−2|and g(x)=f[f(x)], then g′(x)=

______ for x>20

A
2

B
1

C
0

D
None of these

Solution

g (x) = f (f( x ))

= $f (x) - 2$

$||x - 2| - 2|$

$|x - 2 - 2|$ = $|x - 4|$

g(x) = x - 4

g'x = 1

Hence, option, B is correct answer
Question 11
1 / -0

Let f(x + y) = f(x)f(y), for all x, y ∈ R. If f'(0) = 2 and f(4) = 4, then f'(4) is equal to

A
4

B
1

C
1/2

D
8

Solution

\begin{array}{l}
f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
=\lim _{h \rightarrow 0} \frac{f(x) f\left(1+\frac{h}{x}\right)-f(x)}{h} \\
f^{\prime}(x)=\frac{f(x)}{x}\left(f^{\prime}(1)\right) \\
\int \frac{d f(x)}{f(x)}=\int \frac{2}{x} d x \\
\log f(x)=2 \log x+c \\
\text { Given } f(2)=4 \\
\Rightarrow \log 4=2 \log 2+c \\
\Rightarrow c=0 \\
\Rightarrow \log f(x)=2 \log x \\
\Rightarrow f(x)=x^{2} \\
f^{\prime}(x)=2 x \\
f^{\prime}(4)=8
\end{array}

Hence, option, D is correct answer
Question 12
1 / -0

If $f(x)=\frac{1}{1-x}$, then the set of points of discontinuity of the function $f(f(f(x)))$ is

A
{1}

B
$ϕ$

C
{-1, 1}

D
None of these

Solution

\begin{equation}
\begin{aligned}
&f(x)=\frac{1}{1-x}\\
&f(f(x))=\frac{1}{1-\frac{1}{1-x}}=\frac{1-x}{-x}=\frac{x-1}{x}\\
&f(f(f(x)))=\frac{1}{1-\frac{x-1}{x}}=x\\
&\text { This is continuous everywhere. }
\end{aligned}
\end{equation}

Hence, option, B is correct answer
Question 13
1 / -0

\begin{equation}
\begin{aligned}
&\text { Let } f(x) \text { be a function satisfying } f(x+y)-f(x) f(y) \text { for all } x, y\\
&\in R \& f(x)=1+x g(x), \text { where } \lim _{x \rightarrow 0} g(x)-1 f(x) \text { is equal to }
\end{aligned}
\end{equation}

A
g(x)

B
1 + g(x)

C
1 + xg(x)

D
None of these

Solution

\begin{equation}
\begin{array}{l}
f(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{f(x) f(h)-f(x)}{h}-f(x)_{h \rightarrow 0} \frac{\lim _{h} \frac{f(h)-1}{h}}{h}= \\
f(x)\left(\lim _{h \rightarrow 0} \frac{1+h g(h)-1}{h}\right) \\
\lim _{h \rightarrow 0} g(h)=f(x)=1+x g(x)
\end{array}
\end{equation}

Hence, option, C is correct answer
Question 14
1 / -0

What is $\tan \left\{2 \tan ^{-1}\left(\frac{1}{3}\right)\right\}$ equal to?

A
$\frac 23$

B
$\frac 34$

C
$\frac 38$

D
$\frac 19$

Solution

The given expression is,

$\tan \left\{2 \tan ^{-1}\left(\frac{1}{3}\right)\right\}$

We know that,

$2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$ where $-1

Applying the above mentioned formula in the given expression,

\( \tan \left\{2 \tan ^{-1}\left(\frac{1}{3}\right)\right\}$

$=\tan\left\{\tan ^{-1}\left(\frac{2 \times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\right)\right\}$

$=\tan\left\{\tan ^{-1}\left(\frac{3}{4}\right)\right\}\quad[\because\tan\{\tan^{-1}(x)\}=x]$

$=\frac{3}{4}$
Question 15
1 / -0

\begin{equation}
\text { If } y=\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+-\infty\right) \cdot \text { then } \frac{d y}{d x}
\end{equation}

A
0

B
1

C
x

D
y

Solution

\begin{equation}
\begin{array}{l}
y=\left(1+x+\frac{x^{2}}{2}+\frac{x^{3}}{3 !}+\ldots \infty\right) \\
y=e^{x} \\
\frac{d y}{d x}-e^{x}=y
\end{array}
\end{equation}

Hence, option, D is correct answer

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Mathematics Test - 2

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Mathematics Test - 2

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Question 1

32

23

35

3

Solution

Question 2

d2y/dx2 = 0

dx/dy = 0

dy/dx = 0

d2x/dy2= 0

Solution

Question 3

4/105

4/33

4/35

4/1155

Solution

Question 4

mn + 1 = 0

mn – 1 = 0

1/m + 1/n = 0

1/m - 1/n = 0

Solution

Question 5

2 km/h

1 km/h

3 km/h

6 km/h

Solution

Question 6

8!

7!

8

7

Solution

Question 7

0

1/2

-1

None of these

Solution

Question 8

1

±x+1-1

Does not exist

None of these

Solution

Question 9

3/2

1/2

Does not exist

None of these

Solution

Question 10

2

1

0

None of these

Solution

Question 11

4

1

1/2

8

Solution

Question 12

{1}

ϕ

{-1, 1}

None of these

Solution

$3 \sqrt{2}$

$2 \sqrt{3}$

$3 \sqrt{5}$

d²y/dx² = 0

d²x/dy²= 0

$\pm \sqrt{x + 1} - 1$

$ϕ$