Mathematics Test

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Mathematics Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

If the line 3x + 2y = 13divides the area enclosed by the curve, 9x²+4y²−18x−16y−11=0 into two parts then the ratio of the larger area to the smaller area is

A
$=\frac{3 \pi+2}{\pi-2}$

B
$\frac{\pi+2}{\pi-2}$

C
$\frac{3 \pi+2}{3 \pi-2}$

D
None of these

Solution

Equation of given curve is $9 x^{2}+4 y^{2}-18 x-16 y-11=0$
It can be reduced to $\frac{(x-1)^{2}}{(2)^{2}}+\frac{(y-2)^{2}}{(3)^{2}}=1$ which is the equation of ellipse having it's principal axis parallel to coordinate axes
$\Rightarrow$ It's area $=\pi \mathrm{ab}=\pi(2)(3)=6 \pi$
From the figure we can see that points of intersection of $9 x^{2}+4 y^{2}-18 x-16 y-11=0$ and line $3 x+2 y=13,$ have $x=1,3$
$\Rightarrow$ Smaller area between ellipse and line
$=\int_{1}^{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right) \mathrm{dx}=\int_{1}^{3}\left[2+\frac{3}{2} \sqrt{4-(\mathrm{x}-1)^{2}}-\frac{(13-3 \mathrm{x})}{2}\right] \mathrm{dx}$
$=\left[2 \mathrm{x}+\frac{3}{2}\left\{\frac{(\mathrm{x}-1)}{2} \sqrt{4-(\mathrm{x}-1)^{2}}+\frac{4}{2} \sin ^{-1}\left(\frac{\mathrm{x}-1}{2}\right)\right\}-\frac{1}{2}\left\{13 \mathrm{x}-\frac{3 \mathrm{x}^{2}}{2}\right\}\right]_{1}^{3}$
$=\frac{3 \pi}{2}-3$

$\Rightarrow \frac{\text { Area of Larger region }}{\text { Area of smaller region }}=\frac{6 \pi-\left(\frac{3 \pi}{2}-3\right)}{\frac{3 \pi}{2}-3}$
$=\frac{3 \pi+2}{\pi-2}$
Hence, the correct option is (A)
Question 2
1 / -0

The equation of a circle C₁ is x²+y²−4x−2y−11=0. A circle C₂ of radius 1 unit rolls on the outside of the circle C₁ touching it externally. The locus of the centre of C₂ has the equation

A
x²+ y²− 4x − 2y − 20 = 0

B
x² + y²+4x − 2y − 20 = 0

C
x² + y² − 3x − 2y − 11 = 0

D
None of these

Solution

The centre of $C_{1}=(2,1)$ and the radius $=\sqrt{2^{2}+1^{2}+11}=4$
If $(\alpha, \beta)$ be the centre of $\mathrm{C}_{2}, \sqrt{(\alpha-2)^{2}+(\beta-1)^{2}}=4+1$
$\Rightarrow \alpha^{2}+\beta^{2}-4 \alpha-2 \beta-20=0$
Hence locus is $\mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{x}-2 \mathrm{y}-20=0$

Hence, the correct option is (A)
Question 3
1 / -0

If $a^{3}+b^{6}=2$, then the maximum value of the term independent of $x$ in the expansion of $\left(a x^{1 / 3}+b x^{-1 / 6}\right)^{9}$ $(a>0, b>0)$ is

A
42

B
68

C
148

D
84

Solution

Let $(r+1)$ th term be independent of $x,$ then
$$
\begin{aligned}
T_{r+1} &={ }^{g} C_{r}\left(a x^{\frac{1}{3}}\right)^{9-r}\left(b x^{-1 / 6}\right)^{r} \\
&={ }^{9} C_{r} a^{9-r}, b^{r} \cdot x^{3-\frac{r}{3}-\frac{r}{6}}
\end{aligned}
$$
As $T_{r}+1$ is independent of $x$
$\therefore \quad 3-\frac{r}{3}-\frac{r}{6}=0$
$\Rightarrow \quad r=6$
$\therefore \quad T_{6+1}={ }^{9} C_{6} a^{3} b^{6}={ }^{9} C_{3} a^{3} b^{6}=84 a^{3} b^{6}$
$\because \quad A M \geq G M$
$\therefore \quad \frac{a^{3}+b^{6}}{2} \geq \sqrt{a^{3} b^{6}}$
$\Rightarrow \quad \frac{2}{2} \geq \sqrt{a^{3} b^{6}}$
$\Rightarrow \quad a^{3} b^{6} \leq 1$
or $\quad 84 a^{3} b^{6} \leq 84$
or $\quad T_{7} \leq 84$
$\therefore \quad$ Maximum value $=84$
Hence, the correct option is (D)
Question 4
1 / -0

The value of $\int_{0}^{1}\left(\prod_{\mathrm{r}=1}^{\mathrm{n}}(\mathrm{x}+\mathrm{r})\right)\left(\sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{x}+\mathrm{k}}\right) \mathrm{dx}$ is

A
n ⋅ n !

B
n(n + 1) !

C
(n + 1) !

D
None of these

Solution

Let $y=\prod_{r=1}^{n}(x+r)$
$\mathbf{y}=(\mathbf{x}+\mathbf{1})(\mathbf{x}+2) \ldots \ldots(\mathbf{x}+\mathbf{n})$
$\log \mathrm{y}=\log (\mathrm{x}+1)+\log (\mathrm{x}+2)+\ldots \ldots+\log (\mathrm{x}+\mathrm{n})$
$\frac{1}{y} \frac{d y}{d x}=\frac{1}{x+1}+\frac{1}{x+2}+\ldots \ldots \ldots+\frac{1}{x+n}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=[(\mathrm{x}+1) \cdot(\mathrm{x}+2) \ldots \ldots(\mathrm{x}+\mathrm{n})]\left[\frac{1}{\mathrm{x}+1}+\frac{1}{\mathrm{x}+2}+\ldots \ldots+\frac{1}{\mathrm{x}+\mathrm{n}}\right]$
So given integration
$\mathrm{I}=\left[\prod_{\mathrm{r}=1}^{\mathrm{n}}(\mathrm{x}+\mathrm{r})\right]_{0}^{1}$
$\mathrm{I}=\prod_{\mathrm{r}=1}^{\mathrm{n}}(1+\mathrm{r})-\prod_{\mathrm{r}=1}^{\mathrm{n}}(\mathrm{r})$
$\mathrm{I}=(\mathrm{n}+1) !-\mathrm{n} !$
$\mathrm{I}=\mathrm{n} \cdot \mathrm{n} !$
Hence, the correct option is (A)
Question 5
1 / -0

The function f(x) = x³+ λx²+ 5x + sin 2x will be an invertible function if λ belongs to the set -

A
[−∞,−3]

B
[−3,3]

C
[3,∞]

D
None of these

Solution

For given function $f(x)$
$\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+2 \lambda \mathrm{x}+5+2 \cos 2 \mathrm{x}$
since $\cos 2 x \in[-1,1]$
$\Rightarrow 3 x^{2}+2 \lambda x+3 \leq f^{\prime}(x) \leq 3 x^{2}+2 \lambda x+7$
But $3 x^{2}+2 \lambda x+7 \leq 0$ is not possible for all $x \in R$ for any $\lambda$
$\therefore \mathrm{f}^{\prime}(\mathrm{x}) \geq 3 \mathrm{x}^{2}+2 \lambda \mathrm{x}+3 \geq 0$ for all $x \Rightarrow D \leq 0,$ i.e., $4 \lambda^{2}-4 \cdot 3 \cdot 3 \leq 0$
or $\lambda^{2}-9 \leq 0 \Rightarrow-3 \leq \lambda \leq 3$
$\therefore$ if $-3 \leq \lambda \leq 3, \mathrm{f}(\mathrm{x})$ is strictly m.i. and so $\mathrm{f}(\mathrm{x})$ is invertible.
Hence, the correct option is (B)
Question 6
1 / -0

If the tangent and the normal to x² - y² = 4 at a point cut off intercepts a₁ , a₂ on the x-axis respectively and b₁ , b₂ on the y-axis respectively then the value of a₁a₂ + b₁b₂ is

A
1

B
-1

C
0

D
4

Solution

Consider a general point $P(\theta) \equiv(2 \sec \theta, 2 \tan \theta)$ On the given hyperbola
The tangent at $(2 \sec \phi, 2 \tan \phi)$ is $\mathrm{x} \sec \phi-\mathrm{y} \tan \phi=2$
$\therefore \mathrm{a}_{1}=2 \cos \phi, \mathrm{b}_{1}=-2 \cot \phi$
The normal at $(2 \sec \phi, 2 \tan \phi)$ is $\mathrm{x} \cos \phi+\mathrm{y} \cot \phi=4$
$\therefore \mathrm{a}_{2}=4 \sec \phi, \mathrm{b}_{2}=4 \tan \phi$
$\therefore \mathrm{a}_{1} \mathrm{a}_{2}+\mathrm{b}_{1} \mathrm{b}_{2}=8 \cos \phi \sec \phi+(-2 \cot \phi)(4 \tan \phi)=8-8=0$
Hence, the correct option is (C)
Question 7
1 / -0

A line makes angles of 45° and 60° with the positive directions of x and y axes respectively. An angle, which the line can make with the positive direction of z-axis is:

A
60°

B
40°

C
80°

D
100°

Solution

Let $\alpha, \beta, \gamma$ be the angles, which the line makes with the positive directions of $x$ -axis, $y$ -axis and z-axis respectively.
$\therefore \quad \cos \alpha=\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \cos \beta=\cos 60^{\circ}=\frac{1}{2}$
since $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$
$\therefore \quad \frac{1}{2}+\frac{1}{4}+\cos ^{2} \gamma=1$
$\Rightarrow \quad \cos ^{2} \gamma=\frac{1}{4}$
$\Rightarrow \quad \cos \gamma=\pm \frac{1}{2}$
$\Rightarrow \quad \gamma=60^{\circ}$ or $120^{\circ}$
Hence the line makes an angle of $60^{\circ}$ with the positive direction of z-axis.
Hence, the correct option is (A)
Question 8
1 / -0

The co-ordinates of the point which divides the line segment joining the points (5,4, 2) and (−1,−2, 4) in the ratio 2 : 3 externally is

A
(17, 16,−2)

B
(15, 12,−3)

C
(14, 10,−2)

D
(11, 13,−4)

Solution

Let $\mathrm{A}(5,4,2)$ and $\mathrm{B}(-1,-2,4)$ be the given points.
Let $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ be the point, which divides the line segment $[\mathrm{AB}]$ in the ratio -2: 3

$\therefore$ The co-ordinates of P are:
$\left(\frac{(-2)(-1)+3(5)}{-2+3}, \frac{(-2)(-2)+3(4)}{-2+3}, \frac{(-2)(4)+3(2)}{-2+3}\right)$
$=\left(\frac{2+15}{1}, \frac{4+12}{1}, \frac{-8+6}{1}\right)=(17,16,-2)$
Hence, the correct option is (A)
Question 9
1 / -0

The third vertex of the triangle whose centroid is (7,−2, 5) and whose other two vertices are (2, 6, −4) and (4,−2, 3) is

A
(15,−10, 16)

B
(10,−12, 16)

C
(13,−14, 11

D
(6,−11, 10)

Solution

Let $\mathrm{A}(2,6,-4)$ and $\mathrm{B}(4,-2,3)$ be the two given vertices of the triangle. Let $\mathrm{C}(\alpha, \beta, \gamma)$ be the third vertex
since $\mathrm{G}(7,-2,5)$ is the centroid of $\Delta \mathrm{ABC}$,
$\therefore 7=\frac{2+4+\alpha}{3},-2=\frac{6-2+\beta}{3}$
$5=\frac{-4+3+\gamma}{3}$
$\Rightarrow 6+\alpha=21,4+\beta=-6,-1+\gamma=15$
$\Rightarrow \alpha=15, \beta=-10, \gamma=16$
Hence $\mathrm{C}(15,-10,16)$ is the third vertex.
Hence, the correct option is (A)
Question 10
1 / -0

The vertices of a triangle are A(x₁, x₁ tan α), B (x₂, x₂ tan β) and C(x₃, x₃tan γ) . If the circumcentre of ΔABC coincides with the origin and H (a,b) be its orthocentre, then a/b is equal to: (Here α, β and γ are acute angles)

A

$\frac{\cos \alpha+\tan \beta+\cos \gamma}{\cos \alpha \cos \beta \cos \gamma}$

B

$\frac{\sin \alpha+\sin \beta+\sin \gamma}{\sin \alpha \sin \beta \sin \gamma}$

C
$\frac{\tan \alpha+\tan \beta+\tan \gamma}{\tan \alpha \tan \beta \tan \gamma}$

D
$\frac{\cos \alpha+\cos \beta+\cos \gamma}{\sin \alpha+\sin \beta+\sin \gamma}$

Solution

Let $R$ be the radius of the circumcircle and $O$ be the origin, then, $A O=\sqrt{\left(x_{1}^{2}+x_{1}^{2} \tan ^{2} \alpha\right)}$
$\Rightarrow \quad R = x _{1} \sec \alpha$
$\Rightarrow \quad x_{1}=R \cos \alpha$
Similarly, $x_{2}=R \cos \beta$ and $x_{3}=R \cos \gamma$
So, the coordinates of vertices are $A ( R \cos \alpha, R \sin \alpha), B ( R \cos \beta, R \sin \beta), C ( R \cos \gamma, R \sin \gamma)$
Hence, the coordinates of centroid $G$ are
$\left(\frac{\Sigma R \cos \alpha}{3}, \frac{\Sigma R \sin \alpha}{3}\right)$
since, the orthocentre $H ( a , b ),$ circumcentre $O (0,0)$ and the centroid G are collinear therefore, slope of $OH =$ slope of $OG$ $\Rightarrow \frac{b}{a}=\frac{R(\sin a+\sin \beta+\sin \gamma)}{ R (\cos \alpha+\cos \beta+\cos \gamma)}$
$\therefore \quad \frac{ a }{ b }=\frac{\cos \alpha+\cos \beta+\cos \gamma}{\sin \alpha+\sin \beta+\sin \gamma}$
Hence, the correct option is (D)

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Mathematics Test - 25

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Mathematics Test - 25

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Question 1

\(=\frac{3 \pi+2}{\pi-2}\)

\(\frac{\pi+2}{\pi-2}\)

\(\frac{3 \pi+2}{3 \pi-2}\)

None of these

Solution

Question 2

x2+ y2− 4x − 2y − 20 = 0

x2 + y2+4x − 2y − 20 = 0

x2 + y2 − 3x − 2y − 11 = 0

None of these

Solution

Question 3

42

68

148

84

Solution

Question 4

n ⋅ n !

n(n + 1) !

(n + 1) !

None of these

Solution

Question 5

[−∞,−3]

[−3,3]

[3,∞]

None of these

Solution

Question 6

1

-1

0

4

Solution

Question 7

60°

40°

80°

100°

Solution

Question 8

(17, 16,−2)

(15, 12,−3)

(14, 10,−2)

(11, 13,−4)

Solution

Question 9

(15,−10, 16)

(10,−12, 16)

(13,−14, 11

(6,−11, 10)

Solution

Question 10

\(\frac{\cos \alpha+\tan \beta+\cos \gamma}{\cos \alpha \cos \beta \cos \gamma}\)

\(\frac{\sin \alpha+\sin \beta+\sin \gamma}{\sin \alpha \sin \beta \sin \gamma}\)

\(\frac{\tan \alpha+\tan \beta+\tan \gamma}{\tan \alpha \tan \beta \tan \gamma}\)

\(\frac{\cos \alpha+\cos \beta+\cos \gamma}{\sin \alpha+\sin \beta+\sin \gamma}\)

Solution

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x²+ y²− 4x − 2y − 20 = 0

x² + y²+4x − 2y − 20 = 0

x² + y² − 3x − 2y − 11 = 0