Mathematics Test

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Mathematics Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

The function $f: R \rightarrow[-1,1]$ defined by $f(x)=\frac{|x|}{1+|x|}, x \in R$ is

A
Invertible

B
Injective but not surjective.

C
Surjective but not injective

D
Neither injective nor surjective.

Solution

$f(x)=\frac{|x|}{1+|x|}=\left\{\begin{array}{ll}\frac{x}{1+x}, & x \geq 0 \\ \frac{-x}{1-x} & , x<0\end{array}\right.$

Graph of f(x)

From graph we can see that, for two value of ‘x’, f(x) has same value so f(x) is not injective.

Also, range of f(x) is [0, 1)

So it is not surjective also.
Hence, the correct option is (D)
Question 2
1 / -0

Combine terms: 12a + 26b -4b – 16a.

A
4a + 22b

B
-28a + 30b

C
-4a + 22b

D
28a + 30b

Solution

12a + 26b -4b – 16a.

= 12a – 16a + 26b – 4b.

= -4a + 22b

Hence the correct option is C.
Question 3
1 / -0

Find the differential equation of the family of ellipse such that its centre is on the origin

A
$-\frac{y}{x} \cdot \frac{d y}{d x}+y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0$

B
$\frac{1}{x} \cdot \frac{d y}{d x}+y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0$

C
$\frac{d y}{d x}-x y \frac{d^{2} y}{d x^{2}}-\left(\frac{d y}{d x}\right)^{2}=0$

D
$\frac{y}{x} \cdot \frac{d y}{d x}+y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0$

Solution

Let the centre be at (0,0) Thus, the equation of the ellipse becomes $\frac{(x)^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Differentiating w.r.t $x,$ we get $\Rightarrow \frac{2(x)}{a^{2}}+\frac{2 y}{b^{2}} \cdot \frac{d y}{d x}=0$
Thus,we get,$\frac{d y}{d x}=-\left(\frac{x}{y}\right) \cdot \frac{b^{2}}{a^{2}}$
Differntiating the equation 1 w. r.t $x$ $\Rightarrow \frac{2}{a^{2}}+\frac{2 y}{b^{2}} \cdot \frac{d^{2} y}{d x^{2}}+\frac{2}{b^{2}} \cdot\left(\frac{d y}{d x}\right)^{2}=0$
$\Rightarrow \frac{b^{2}}{a^{2}}+y \cdot \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0$
$\Rightarrow-\frac{y}{x} \cdot \frac{d y}{d x}+y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0$
Question 4
1 / -0

A
$\frac{x}{\sqrt{1+x^{2}}}$

B
x

C
$x \sqrt{1 + x^{2}}$

D
$\sqrt{1 + x^{2}}$

Solution
Question 5
1 / -0

If Area occupied by the curves with equation given below is A then Value of A/2 is-

Equation $1:(x-5)^{2}+y^{2}=25$

Equation $2: y^{2}=9 x$

A
$\frac{25}{2} \cos ^{-1}\left(\frac{4}{5}\right)-8-\frac{25 \pi}{4}$

B
$\frac{25}{2} \sin ^{-1}\left(\frac{3}{5}\right)-8-\frac{25 \pi}{4}$

C
$\frac{25}{2} \sin ^{-1}\left(\frac{4}{5}\right)-8-\frac{25 \pi}{2}$

D
$\frac{25 \pi}{4}-8-\frac{25}{2} \sin ^{-1}\left(\frac{4}{5}\right)$

Solution
Question 6
1 / -0

A
³⁴C₃

B
³³C₃

C
³⁵C₃

D
³⁶C₄

Solution

$(x-5)^{2}+y^{2}=25$ and $y^{2}=9 x$
$(x-5)^{2}+9 x=25$
$\Rightarrow x^{2}-10 x+9 x=0$
$\Rightarrow x^{2}-x=0$
$x=\{0,1\}$
So area in one quadrant $\int_{0}^{1} \operatorname{circle}(y) d x-\operatorname{Parabola}(y) d x$
$\Rightarrow \int_{0}^{1} \sqrt{25-(x-5)^{2}} d x-\sqrt{9 x} d x$
$\Rightarrow\left|\left(\frac{x-5}{2}\right) \sqrt{25-(x-5)^{2}}+\frac{25}{2} \sin ^{-1}\left(\frac{x-5}{5}\right)-\frac{3.2}{3} x^{\frac{3}{2}}\right|_{0}^{1}$
$\Rightarrow\left(-6-\frac{25}{2} \sin ^{-1}\left(\frac{4}{5}\right)-2\right)-\left(0-\frac{2}{2} \cdot \frac{\pi}{2}\right)$
$\Rightarrow \frac{2 \pi}{4}-8-\frac{25}{2} \sin ^{-1}\left(\frac{4}{5}\right)$
Question 7
1 / -0

If $y=\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right),$ then $\frac{d y}{d x}=$

A
$\frac{1}{3}$

B
2

C
$\frac{1}{2}$

D
$\frac{3}{2}$

Solution

$y=\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)$
$=\tan ^{-1}\left\{\frac{\sin (2 \alpha)}{1+\cos (2 \alpha)}\right\}$
where $x=2 a$
$=\tan ^{-1}\left(\frac{2 \sin \alpha \cos \alpha}{2 \cos ^{2} \alpha}\right)$
$=\tan ^{-1}(\tan \alpha)$
$\Rightarrow y=\alpha=\frac{1}{2} x$
$\Rightarrow y=\frac{x}{2}$
$\frac{d y}{d x}=\frac{1}{2}$
Question 8
1 / -0

The number of values of k for which the system of equations (k + 1) x + 8y = 4k and
kx + (k + 3)y = 3k - 1 has infinitely many solutions is

A
0

B
1

C
2

D
infinite

Solution

For infinitely many solutions, the two equations become identical $\Rightarrow \frac{k+1}{k}=\frac{8}{k+3}=\frac{4 k}{3 k-1}$
from first 2
$\frac{k+1}{k}=\frac{8}{k+3}$
$\Rightarrow k=1,3$
From last 2
$\frac{8}{k+3}=\frac{4 k}{3 k-1}$
$\Rightarrow k=1,2$
so value of $k$ which satisfy both is 1
so $k=1$ so number of values of $\mathrm{k}$ is one.
Question 9
1 / -0

The eccentricity of an ellipse with its centre at the origin is 1/2. If one of the directrix is x = 4, then the equation of the ellipse is

A
4x² + 3y² = 1

B
3x² + 4y² = 12

C
4x² + 3y² = 12

D
3x² + 4y² = 1

Solution

$e=\frac{1}{2}$
directrix, $x=\frac{a}{e}=4$
$\therefore a=4 \times \frac{1}{2}=2$
$b=2 \sqrt{1-\frac{1}{4}}=\sqrt{3}$
Equation of ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$
$\Rightarrow 3 x^{2}+4 y^{2}=12$
Question 10
1 / -0

The value of $\int_{0}^{1} \mathrm{x}(1-\mathrm{x})^{99} \mathrm{dx}$ is

A
1/10010

B
1/10110

C
100/10110

D
1/10100

Solution

$\begin{aligned} I &=\int_{0}^{1} x(1-x)^{99} d x \\ &=\int_{0}^{1}(1-x)\{1-(1-x)\}^{99} d x \\ &=\int_{0}^{1}(1-x) x^{99} d x \\ &=\int_{0}^{1}\left(x^{99}-x^{100}\right) d x \\ &=\left[\frac{x^{100}}{100}-\frac{x^{101}}{101}\right]_{0}^{1} \\=& \frac{1}{100}-\frac{1}{101}=\frac{1}{10100} \end{aligned}$

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Mathematics Test - 15

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Mathematics Test - 15

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SHARING IS CARING

Question 1

Invertible

Injective but not surjective.

Surjective but not injective

Neither injective nor surjective.

Solution

Question 2

4a + 22b

-28a + 30b

-4a + 22b

28a + 30b

Solution

Question 3

\(-\frac{y}{x} \cdot \frac{d y}{d x}+y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0\)

\(\frac{1}{x} \cdot \frac{d y}{d x}+y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0\)

\(\frac{d y}{d x}-x y \frac{d^{2} y}{d x^{2}}-\left(\frac{d y}{d x}\right)^{2}=0\)

\(\frac{y}{x} \cdot \frac{d y}{d x}+y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0\)

Solution

Question 4

\(\frac{x}{\sqrt{1+x^{2}}}\)

x

x1+x2

1+x2

Solution

Question 5

\(\frac{25}{2} \cos ^{-1}\left(\frac{4}{5}\right)-8-\frac{25 \pi}{4}\)

\(\frac{25}{2} \sin ^{-1}\left(\frac{3}{5}\right)-8-\frac{25 \pi}{4}\)

\(\frac{25}{2} \sin ^{-1}\left(\frac{4}{5}\right)-8-\frac{25 \pi}{2}\)

\(\frac{25 \pi}{4}-8-\frac{25}{2} \sin ^{-1}\left(\frac{4}{5}\right)\)

Solution

Question 6

34C3

33C3

35C3

36C4

Solution

Question 7

13

2

12

32

Solution

Question 8

0

1

2

infinite

Solution

Question 9

4x2 + 3y2 = 1

3x2 + 4y2 = 12

4x2 + 3y2 = 12

3x2 + 4y2 = 1

Solution

Question 10

1/10010

1/10110

100/10110

1/10100

Solution

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$x \sqrt{1 + x^{2}}$

$\sqrt{1 + x^{2}}$

³⁴C₃

³³C₃

³⁵C₃

³⁶C₄

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{3}{2}$

4x² + 3y² = 1

3x² + 4y² = 12

4x² + 3y² = 12

3x² + 4y² = 1