Mathematics Test

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Mathematics Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

$\int_{0}^{\pi / 2} \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} d x=$

A
ℓn(2)

B
ℓn(3)

C
ℓn(4)

D
ℓn(8)

Solution

$I=\int_{0}^{\pi / 2} \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} \cdot d x$
$=\int_{0}^{\pi / 2} \sqrt{\frac{\cos ^{2} x+\sin ^{2} x-2 \sin x \cos x}{\cos ^{2} x+\sin ^{2} x+2 \sin x \cos x}} \cdot d x$
$=\int_{0}^{\pi / 2} \sqrt{\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)^{2}} \mathrm{dx}$
$=\int_{0}^{\pi / 2}\left|\frac{\cos x-\sin x}{\cos x+\sin x}\right| d x$
$=\int_{0}^{\pi / 4} \frac{\cos x-\sin x}{\cos x+\sin x} d x-\int_{\pi / 4}^{\pi / 2} \frac{\cos x-\sin x}{\cos x+\sin x} d x$
put $\cos x+\sin x=t \Rightarrow(\cos x-\sin x) d x=d t$
$=\int_{1}^{\sqrt{2}} \frac{d t}{t}-\int_{\sqrt{2}}^{1} \frac{d t}{t}=\left.\ell n t\right|_{1} ^{\sqrt{2}}-\left.\ell n t\right|_{\sqrt{2}} ^{1}$
$=\ell n \sqrt{2}+\ell n \sqrt{2}=\ell n 2$
Hence, the correct option is (A)
Question 2
1 / -0

The area of an expanding rectangle is increasing at the rate of 48 cm²/sec. The length of the rectangle is always equal to the square of the breadth. At the instant when the breadth is 4.5 cm, The length is increasing at the rate of

A
7.11

B
8.11

C
6.11

D
5.11

Solution

Let $' l^{\prime}\left(=\mathrm{b}^{2}\right)$ and ${ }^{\prime} b^{\prime}$ be the length and breadth respectively of the rectangle.
$\therefore \mathrm{A},$ the area $=l \times b=b^{2} \times b=b^{3}$
$\therefore \frac{d A}{d t}=3 b^{2} \frac{d b}{d t} \Rightarrow 48=3 b^{2} \frac{d b}{d t}$
(1) [By the question]
But $l=b^{2} \Rightarrow \frac{d l}{d t}=2 b \frac{d b}{d t} \ldots \ldots \ldots .(2)$
$\Rightarrow \frac{d \ell}{d t}=2 b \cdot \frac{1}{3 b^{2}} \frac{d A}{d t}$
$=\frac{2}{3 b} \frac{d A}{d t}$
$\left.\Rightarrow \frac{d l}{d t}\right|_{b=4.5}=\frac{2 \times 48}{3 \times 4.5}=7.11 \mathrm{cm} / \mathrm{sec}$
Hence, the correct option is (A)

Question 3

1 / -0

The following age group are included in the proportion indicated

Age Group	Relative Proportion in Population
12 - 17	0.17
18 - 23	0.31
24 - 29	0.27
30 - 35	0.21
36 +	0.04

Then number of people respectively in each age group should be included in a sample of 3000 people to make the sample representative is

850, 155, 135, 905, 955

510, 930, 810, 630, 120

600, 600, 600, 600, 600

510, 630, 950, 100, 810

Solution

If 0.17 is the population proportion of 'x' people following in the age group $12-17$ then $0.17=\frac{x}{3000}(\because$ the population given is of size 3000 )

$\Rightarrow x=0.17(3000)=510$

$\therefore$ out of 3000 people 510 should fall in the age group $12-17$

Similarly

Age Group	No. of people out of 3000
12 - 17	0.17 * 3000 = 510
18 - 23	0.31 * 3000 = 930
24 - 29	0.27* 3000 = 810
30 - 35	0.21* 3000 = 630
36 +	0.04* 3000 = 120

Hence, the correct option is (B)

Question 4
1 / -0

If ln(x + y) = 2xy, theny′ (0) is equal to

A
1

B
-1

C
2

D
0

Solution

Given equation is $\ln (x+y)=2 x y \ldots(1)$
For $x=0$
$\ln (0+y)=2.0 . y=0$
$\Rightarrow \ln y=0$
$\Rightarrow y=1$
Now, differentiating ( 1 ), we get
$\frac{1}{(x+y)}\left(1+\frac{\mathrm{dy}}{\mathrm{dx}}\right)=2 y+2 x \frac{\mathrm{dy}}{\mathrm{dx}}$
$\Rightarrow$ At point (0,1)
$\left(\frac{1}{0+1}\right)\left(1+\frac{\mathrm{dy}}{\mathrm{dx}}\right)=2(1)+2.0 \cdot\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$
$\Rightarrow 1+\frac{d y}{d x}=2$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=1=y^{\prime}(0)$
Hence, the correct option is (A)
Question 5
1 / -0

ABCD is a square whose vertices A, B, C and D are (0, 0), (2, 0), (2, 2) and (0, 2) respectively. This square is rotated in the X - Y plane with an angle of 30° in anticlockwise direction about the vertex A. Now points B, C, D are converted as B', C' and D' respectively. Then equation of diagonal B' D' and equation of circumcircle AB'E is (E is centre of new square)

A
√3 x + (1−√3)y = √3, x²+ y²= 4

B
(1 + √3) x−(1−√2) y=2, x²+ y²= 9

C
(2−√3)x + y = 2(√3−1), x2+y2−x3–√−y=02-3x+y=23-1, x2+y2-x3-y=0

D
None of these

Solution

Side of the square $=2$ unit Coordinates of $\mathrm{B}^{\prime}, \mathrm{C}^{\prime}$ and $\mathrm{D}^{\prime}$ are $(\sqrt{3}, 1),(\sqrt{3}-1, \sqrt{3}+1)$ and $(-1, \sqrt{3})$ respectively.

Slope of $B ^{\prime} D ^{\prime}=\frac{\sqrt{3}-1}{-1-\sqrt{3}}$
\begin{equation*}
=\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{-2}=\sqrt{3}-2
\end{equation*}$\therefore$ Equation of $BD$ is\begin{equation*}
y-1=(\sqrt{3}-2)(x-\sqrt{3})
\end{equation*}$\Rightarrow \quad(2-\sqrt{3}) x+y=2(\sqrt{3}-1)$
and equation of the circumcircle of the triangle AB'E (Apply diametric form as $AB ^{\prime}$ is diameter)
\begin{equation*}(x-0)(x-\sqrt{3})+(y-0)(y-1)=0
\end{equation*}$\Rightarrow x^{2}+y^{2}-x \sqrt{3}-y=0$
Hence, the correct option is (C)
Question 6
1 / -0

The probability that at least one of the events $A$ and $B$ occurs is $0.6 .$ If $A$ and $B$ occur simultaneously with probability $0.2,$ then $P(\bar{A})+P(\bar{B})$ is: (where $\bar{A}$ and $\bar{B}$ are complements of $A$ and $B$ respectively)

A
0.4

B
0.8

C
1.2

D
1.4

Solution

Here, given that $P(A \cup B)=0.6: P(A \cap B)=0.2$
$\therefore P(\bar{A})+P(\bar{B})=1-P(A)+1-P(B)$
$=2-(P(A)+P(B))$
$=2-[P(A \cup B)+P(A \cup B)] \quad\{$ Using addition rule of probability $P(A \cup B)=P(A)+P(B)-P(A \cap B)\}$
$=2-[0.6+0.2]=2-0.8=1.2$
Hence, the correct option is (C)
Question 7
1 / -0

Consider the straight line ax + by = c, where a, b, c ∈ R⁺this line meets the coordinate axes at A and B respectively. If the area of the ΔOAB, O being origin, is always a constant equal to half, then

A
The three numbers a,c,b are in AP

B
The three numbers a,c,b are in GP

C
The three numbers are a,c,b in HP

D
None of these

Solution

Given line is $a x+b y=c$
$\Rightarrow \frac{x}{\left(\frac{c}{a}\right)}+\frac{y}{\left(\frac{c}{b}\right)}=1$
$\Rightarrow$ This line cut the coordinate axes as points $A\left(\frac{c}{a}, 0\right)$ and $B\left(0, \frac{c}{b}\right)$ respectively.
$$
\begin{aligned}
\therefore \text { Area of } \Delta \mathrm{OAB} &=\frac{1}{2} \cdot \mathrm{OA} \cdot \mathrm{OB} \\
&=\frac{1}{2} \cdot \frac{\mathrm{c}}{\mathrm{a}} \cdot \frac{\mathrm{c}}{\mathrm{b}} \\
&=\frac{\mathrm{c}^{2}}{2 \mathrm{ab}}
\end{aligned}
$$

Equating this area to half, we get,
c^{2}=a b
Therefore, $a, c, b$ are in GP.
Hence, the correct option is (B)
Question 8
1 / -0

The tangent at $P$ on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ meets one of the asymptote at $Q$. Then the locus of the mid point of $\mathrm{PQ}$ is $4\left(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\right)=\mathrm{k},$ where $\mathrm{k}$ is

A
3

B
5

C
11

D
23

Solution

Tangent at point (a sec $\theta, b$ tan $\theta$ ) to the hyperbola
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is
$\frac{x \sec \theta}{a}-y \frac{\tan \theta}{b}=1 \ldots \ldots \ldots$ (i)
equation of one of asymptote $\Rightarrow \frac{x}{a}+\frac{y}{b}=0 \quad \ldots \ldots \ldots$
(ii)
solving (1) and (2)
we get the coordinate of $Q$ on the asymptote.
$\mathrm{Q}:[a(\sec \theta-\tan \theta),-b(\sec \theta-\tan \theta)]$
Let coordinates of mid point of $\mathrm{PQ}$ is $\mathrm{M}$ ( $\mathrm{h}, \mathrm{k}$ )
$\therefore \mathrm{h}=\frac{a \sec \theta+(a \sec \theta-a \tan \theta)}{2}$
or $\frac{\mathrm{h}}{a}=\sec \theta-\frac{\tan \theta}{2} \ldots \ldots .(3)$
similary, $k=\frac{b \tan \theta-b \sec \theta+b \tan \theta}{2}$
$\frac{k}{b}=\tan \theta-\frac{\sec \theta}{2} \quad \ldots \ldots \ldots .(4)$
$(3)+(4) \Rightarrow \frac{\mathrm{h}}{a}+\frac{k}{b}=\frac{\sec \theta+\tan \theta}{2} \quad \cdots \cdots \cdots \cdots$
$(3)-(4) \Rightarrow \frac{h}{a}-\frac{k}{b}=\frac{3}{2}(\sec \theta-\tan \theta) \quad \ldots \ldots \ldots \ldots$
Now, $(5) \times(6) \Rightarrow \frac{h^{2}}{a^{2}}-\frac{k^{2}}{b^{2}}=\frac{3}{4}$
or, $4\left(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\right)=3$
$\Rightarrow k=3$
Hence, the correct option is (A)
Question 9
1 / -0

If y=7x − x³and x increases at the rate of 4 units per second, then the rate of change in slope of the curve when x = 3, is

A
72

B
75

C
64

D
50

Solution

We have : $\mathbf{y}=7 \mathrm{x}-\mathrm{x}^{3}$
$\therefore \mathrm{m},$ the slope of the curve $=\frac{\mathrm{dy}}{\mathrm{dx}}=7-3 \mathrm{x}^{2}$
Now $\mathrm{m}=7-3 \mathrm{x}^{2}$
$\therefore \frac{d m}{d t}=-6 x \frac{d x}{d t} \Rightarrow \frac{d m}{d t}=-6 x(4)$
$\Rightarrow \frac{d m}{d t}=-24 x$
$\left.\therefore \frac{d m}{d t}\right]_{x=3}=-24(3)=-72$
Hence the rate of change of slope $=72$
Hence, the correct option is (A)
Question 10
1 / -0

The equation of the common tangent to the equal parabolas y² = 4ax and x² = 4ay is

A
x + y + a = 0

B
x + y = a

C
x - y = a

D
None of these

Solution

Any tangent to $y^{2}=4 a x$ is $y=m x+\frac{a}{m}$ if touches $x^{2}=4 a y$ if $x^{2}=4 a\left(m x+\frac{a}{m}\right),$ i $e \cdot x^{2}-4 a m x-\frac{4 a^{2}}{m}=0$
has equal roots
So, $m^{3}+1=0,$ i.e. $, m=-1 .$ Hence, the common tangent is $y=-x-a$
Hence, the correct option is (A)

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Mathematics Test - 24

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Mathematics Test - 24

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SHARING IS CARING

Question 1

ℓn(2)

ℓn(3)

ℓn(4)

ℓn(8)

Solution

Question 2

7.11

8.11

6.11

5.11

Solution

Question 3

850, 155, 135, 905, 955

510, 930, 810, 630, 120

600, 600, 600, 600, 600

510, 630, 950, 100, 810

Solution

Question 4

1

-1

2

0

Solution

Question 5

√3 x + (1−√3)y = √3, x2+ y2= 4

(1 + √3) x−(1−√2) y=2, x2+ y2= 9

(2−√3)x + y = 2(√3−1), x2+y2−x3–√−y=02-3x+y=23-1, x2+y2-x3-y=0

None of these

Solution

Question 6

0.4

0.8

1.2

1.4

Solution

Question 7

The three numbers a,c,b are in AP

The three numbers a,c,b are in GP

The three numbers are a,c,b in HP

None of these

Solution

Question 8

3

5

11

23

Solution

Question 9

72

75

64

50

Solution

Question 10

x + y + a = 0

x + y = a

x - y = a

None of these

Solution

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√3 x + (1−√3)y = √3, x²+ y²= 4

(1 + √3) x−(1−√2) y=2, x²+ y²= 9