Mathematics Test

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Mathematics Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

The function $f(x)=\int_{0}^{\pi} \log \left(t+\sqrt{1+t^{2}}\right) d t$ is:

A
an even function

B
an odd function

C
a periodic function

D
None of these

Solution

$f(x)=\int_{0}^{x} \log \left(t+\sqrt{1+t^{2}}\right) d t$
$f(-x)=\int_{0}^{-x} \log \left(t+\sqrt{1+t^{2}}\right) d t, p u t \quad t=-z$
$\left.f(-x)=\int_{0}^{x} \log \left(-z+\sqrt{1+z^{2}}\right)-d z\right)$
$f(-x)=-\int_{0}^{x} \log \frac{\left(\sqrt{1+z^{2}}-z\right)\left(\sqrt{1+z^{2}}+z\right)}{\left(\sqrt{1+z^{2}}+z\right)} d z$
$f(-x)=-\int_{0}^{x} \log \left(\frac{1}{\sqrt{1+z^{2}}+z}\right) d z$
$f(-x)=\int_{0}^{x} \log \left(z+\sqrt{1+z^{2}}\right) d z f(-x)=f(x)$
$\mathrm{f}(\mathrm{x})$ is an even function.
NOTE: Integration of every even function is odd function and integration of every odd function is even function, provided the integrands are integrable and the limit is 0 to x. Similarly differentiation of every odd function is even function and that of every even function is odd function, provided the they are differentiable. Here $\mathrm{h}(\mathrm{t})=10 \mathrm{g}\left(\mathrm{t}+\sqrt{\mathrm{t}^{2}+1}\right),$ which is an odd
function, therefore $\int_{0}^{\mathrm{x}} \mathrm{h}(\mathrm{t}) \mathrm{dt}$ is even function.
Hence, the correct option is A.
Question 2
1 / -0

The area bounded by curves y = f(x), the x-axis and the ordinates x = 1 and x = b is (b - 1) sin (3b + 4). Then f(x) is-

A
(x-1) cos (3x + 4)

B
sin (3x + 4)

C
sin (3x + 4) + 3 (x – 1) cos (3x + 4)

D
sin (3x + 4) + (x – 1) cos (3x + 4)

Solution

1. Area bounded by curve y = f(x), x =1 and x = b is

$\int_{1}^{b} f(x) d x=(b-1) \sin (3 b+4)$

Now differentiating both sides with respect to b we get $f B .=\sin (3 b+4)+3(b-1) \cos (3 b+4)$
$f(x)=\sin (3 x+4)+3(x-1) \cos (3 x+4)$
Hence, the correct option is C.
Question 3
1 / -0

A tangent having slope of $-\frac{4}{3}$ to the ellipse $\frac{x^{2}}{18}+\frac{y^{2}}{32}=1$ intersects the major and minor axes in points A and B respectively. If $C$ is
the center of the ellipse then the area of the triangle $\mathrm{ABC}$ is:

A
12 sq. units

B
24 sq. units

C
36 sq. units

D
48 sq. units

Solution

Equation of the tangent to $\frac{x^{2}}{18}+\frac{y^{2}}{32}=1$ whose slope $m=-\frac{4}{3}$ is:
$y=\left(-\frac{4}{3}\right) x \pm \sqrt{18 \times\left(-\frac{4}{3}\right)^{2}+32}$
$y=-\frac{4 x}{3} \pm \sqrt{64} y=\left(-\frac{4}{3}\right) x \pm 8$
From symmetry of ellipse it is obvious that area of the triangle $\mathrm{ABC}$ will be same with respect either tangent. Let us consider $y=\left(-\frac{4}{3}\right) x+8=\frac{x}{6}+\frac{y}{8}=1$
$\therefore A \equiv(6,0)$ and $B=(0,8)$
Hence area $\triangle$ of the triangle $\mathrm{ABC}$ is:
$\Delta=\frac{1}{2} \times \mathrm{OB} \times \mathrm{OA}=\frac{1}{2} \times 8 \times 6=24 \mathrm{sq} .$ units
Hence, the correct option is B.
Question 4
1 / -0

The eccentricity of the hyperbola whose latus rectum is half of its transverse axis, is

A
$\sqrt{\frac{5}{2}}$

B
$\sqrt{\frac{7}{4}}$

C
$\sqrt{\frac{3}{2}}$

D
$\sqrt{\frac{5}{4}}$

Solution

Length of latus rectum of hyperbola = Length of the transverse axis $\frac{2 b^{2}}{a}=\frac{1}{2}(2 a)$
$\frac{2 b^{2}}{a}=a$
$2 b^{2}=a^{2}$
$2 a^{2}\left(e^{2}-1\right)=a^{2} \cdot\left(\because b^{2}=a^{2}\left(e^{2}-1\right)\right.$
$2 e^{2}-1=1$
$e=\sqrt{\frac{3}{2}}$
Hence, the correct option is C.
Question 5
1 / -0

If $\sin ^{-1} \frac{x}{5}+\operatorname{cosec}^{-1} \frac{5}{4}=\frac{\pi}{2}$ then a value of $x$ is:

A
1

B
3

C
4

D
6

Solution

$\sin ^{-1} \frac{x}{5}+\operatorname{cosec}^{-1} \frac{5}{4}=\frac{\pi}{2}$
$\sin ^{-1} \frac{x}{5}=\frac{\pi}{2}-\operatorname{cosec}^{-1} \frac{5}{4}$
$\sin ^{-1} \frac{x}{5}=\sec ^{-1} \frac{5}{4}$
$\sin ^{-1} \frac{x}{5}=\cos ^{-1} \frac{4}{5}$
$\sin ^{-1} \frac{x}{5}=\sin ^{-1} \sqrt{1-\left(\frac{4}{5}\right)^{2}}$
$\frac{x}{5}=\frac{3}{5}$
$x=3$
Hence, the correct option is B.
Question 6
1 / -0

If $\sin 5 x+\sin 3 x+\sin x=0,$ then the value of $x$ other than zero, lying between $0 \leq x \leq \pi / 2$ is...

A
π/6

B
π/12

C
π/3

D
π/4

Solution

1. sin 5x + sin x + sin 3x = 0
2 sin 3x . cos2x + sin3x = 0
sin 3x (2 cos 2x + 1) = 0, 0 ≤ x ≤π/2
(i) $\sin 3 x=0 \quad 3 x=n \pi \quad x=\frac{n \pi}{3}$
(ii) $\cos 2 x=\frac{-1}{2}=\cos \frac{2 \pi}{3} \quad 2 x=2 k \pi \pm \frac{2 \pi}{3}$
$x=k \pi \pm \frac{\pi}{3}$
From both $x=n / 3$ (other than 0 ).
Hence, the correct option is C.
Question 7
1 / -0

Sum of the series
1 + 3 + 6 + 10 + 15 + ……………………….n terms is,

A
$\frac{1}{6} n(n-1)(n+2)$

B
$\frac{1}{6} n(n+1)(n-2)$

C
$\frac{1}{6} n(n+1)(n+2)$

D
$\frac{1}{6} n(n-1)(n-2)$

Solution

$S=1+3+6+10+15+\ldots+t_{n-1}+t_{n}$
$S=1+3+6+10+\ldots \ldots \ldots .+t_{n-1}+t_{n}$
$0=1+[2+3+4+\ldots(n-1)$ terms $]-t_{n}$
$t_{n}=1+[2+3+4+\ldots+n]=\frac{1}{2} n(n+1)$
$\therefore S_{n}=\sum_{n=1}^{n} t_{n}=\frac{1}{6} n(n+1)(n+2)$
Hence, the correct option is C.
Question 8
1 / -0

In an experiment with 15 observations on X, the following results where available:$\sum X^{2}=2830, \Sigma \mathrm{x}=170$. One observation that was 20 was found to be wrong and was replaced by the correct value 30. The corrected variance is

A
8.33

B
78.00

C
18866

D
177.33

Solution

Let $\sum \mathrm{X}^{\prime 2}$ and $\sum \mathrm{X}$ denotes correct value of sum of squares and sum of the given data.
$\therefore \Sigma \mathrm{X}=\sum \mathrm{X}-20+30=170-20+30=180$
$\sum X^{2}=\sum X^{2}-20^{2}+30^{2}=2830-400+900=3330$
$\therefore$ Variance $=\frac{1}{\mathrm{n}} \sum \mathrm{X}^{/ 2}-\left(\frac{1}{\mathrm{n}} \Sigma \mathrm{X}^{\prime}\right)^{2}$
$=\frac{1}{15} 3330-\left(\frac{1}{15} 180\right)^{2}=222-144=78$
Hence, the correct option is B.
Question 9
1 / -0

There are two balls in an urn whose colours are not known (each ball can be either white or black). A white ball is put into the urn. A ball is drawn from the urn. The probability that it is white is,

A
1/4

B
1/3

C
2/3

D
1/6

Solution

Let $\mathrm{E}_{\mathrm{i}}(0 \leq \mathrm{i} \leq 2)$ denote the event that urn contains 'i' white and $^{\prime}(2-\mathrm{i})^{\prime}$ black balls.
Let A denote the event that a white ball is drawn from the urn.
We have $\mathrm{P}\left(\mathrm{E}_{\mathrm{i}}\right)=1 / 3$ for $\mathrm{i}=0,1,2$
$P\left(A \mid E_{1}\right)=1 / 3, P\left(A \mid E_{2}\right)=2 / 3, P\left(A \mid E_{3}\right)=1$
By the total probability rule, $P(A)=P\left(E_{1}\right) P\left(A \mid E_{1}\right)+P\left(E_{2}\right) P\left(A \mid E_{2}\right)+P\left(E_{3}\right) P\left(A \mid E_{3}\right)$
$=\frac{1}{3}\left[\frac{1}{3}+\frac{2}{3}+1\right]=\frac{2}{3}$
Hence, the correct option is C.
Question 10
1 / -0

The number of ways in which a mixed double game can be arranged from amongst 9 married couples if no husband and wife play in the same game is,

A
756

B
1512

C
3024

D
None of these

Solution

We can choose two men out of 9 in 9C2 ways. Since no husband and wife are to play in the same game, two women out of the remaining 7 can be chosen in 72 two women out of the remaining 7 can be chosen in 7C2 ways. If M1, M2, W1 & W2 are chosen, then a team may consist of M1 and W1 or M1 and W2. Thus, the number of ways of arranging the game is
$\left(9 \mathrm{C}_{2}\right)\left(\mathrm{T}_{2}\right)(2)=\frac{9 \times 8}{2} \times \frac{7 \times 6}{2} \times 2=1512$
Hence, the correct option is B.

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Mathematics Test - 19

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Mathematics Test - 19

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SHARING IS CARING

Question 1

an even function

an odd function

a periodic function

None of these

Solution

Question 2

(x-1) cos (3x + 4)

sin (3x + 4)

sin (3x + 4) + 3 (x – 1) cos (3x + 4)

sin (3x + 4) + (x – 1) cos (3x + 4)

Solution

Question 3

12 sq. units

24 sq. units

36 sq. units

48 sq. units

Solution

Question 4

52

74

32

54

Solution

Question 5

1

3

4

6

Solution

Question 6

π/6

π/12

π/3

π/4

Solution

Question 7

\(\frac{1}{6} n(n-1)(n+2)\)

\(\frac{1}{6} n(n+1)(n-2)\)

\(\frac{1}{6} n(n+1)(n+2)\)

\(\frac{1}{6} n(n-1)(n-2)\)

Solution

Question 8

8.33

78.00

18866

177.33

Solution

Question 9

1/4

1/3

2/3

1/6

Solution

Question 10

756

1512

3024

None of these

Solution

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$\sqrt{\frac{5}{2}}$

$\sqrt{\frac{7}{4}}$

$\sqrt{\frac{3}{2}}$

$\sqrt{\frac{5}{4}}$