Mathematics Test

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Mathematics Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

\(\text { Find } \theta \in\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right] \text { satisfying } 2 \cos ^{2} \theta+\sin \theta \leq 2\)

A

\(\left[\frac{\pi}{2}, \frac{3 x}{2}\right] \cap\left[\pi, \frac{3 \pi}{2}\right]\)

B

\(\left[\frac{\pi}{2}, \frac{5 \pi}{6}\right] \cup\left[\pi, \frac{3 \pi}{2}\right]\)

C
\(\left[\frac{\pi}{2}, \frac{4 \pi}{3}\right] \cap\left[\pi, \frac{4 \pi}{2}\right]\)

D

\(\left[\frac{\pi}{2}, \frac{2 \pi}{3}\right] \cap\left[\pi, \frac{2 \pi}{2}\right]\)

Solution

\(2 \cos ^{2} \theta+\sin \theta \leq 2\)
\(\cos ^{2} \theta+\sin ^{2} \theta=1\)
\(=2\left(1-\sin ^{2} \theta\right)+\sin \theta \leq 2\)
\(=2-2 \sin ^{2} \theta+\sin \theta \leq 2\)
\(=2 \sin ^{2} \theta-\sin \theta \geq 0\)
\(=2 t^{2}-t \geq 0\)
let \(\sin \theta=t\)
\(t(2 t-1) \geq 0\)
\((t)(2 t-1) \geq 0\)
\(t \in(-\infty, 0]\) and \([1 / 2,-\infty)\)
\(t=\sin \theta\) and \(\sin \theta \epsilon[-1,-1]\)
so
\(\sin \theta \in[-1,0]\) and \([1 / 2,1]\)

Hence , the correct option is B
Question 2
1 / -0

The locus of the midpoints of the chords of the circle \(x^{2} y^{2}=16\) which are tangents to the hyperbola \(9 x^{2}-16 y^{2}=144\) is

A
\(\left(x^{2} y^{2}\right)^{2}=16 x^{2}-9 y^{2}\)

B
\(\left(x^{2}-y^{2}\right)^{2}=16 x^{2}-9 y^{2}\)

C
\(\left(x^{2} y^{2}\right)^{2}=16^{2} 9 y^{2}\)

D
\(\left(x^{2}-y^{2}\right)^{2}=16 x^{2} 9 y^{2}\)

Solution

Let \((\mathrm{h}, \mathrm{k})\) be the middle point of a chord of the circle \(x^{2}+y^{2}=16\)
Then its equation is \(h x+k y-16=h^{2}+k^{2}-16 i . e .\)
\(h x+k y=h^{2}+k^{2} \ldots\) (i)
Let (i) touch the hyperbola
\(9 x^{2}-16 y^{2}=144\) i.e., ... (ii)
At the point (ii) say, then (i) is identical with \(\frac{x \alpha}{16}-\frac{y \beta}{9}=1\)
(iii)
Thus \(\frac{\alpha}{16 h}=-\frac{\beta}{9 k}=\frac{1}{h^{2}+k^{2}}\)
\(\alpha=\frac{16 h}{h^{2}+k^{2}}\) and \(\beta=-\frac{9 k}{h^{2}+k^{2}}\)
since (i) lies on the hyperbola (ii),
\(\frac{1}{16}\left(\frac{16 h}{h^{2}+k^{2}}\right)^{2}-\frac{1}{9}\left(\frac{9}{h^{2}+k^{2}}\right)^{2}=1\)
\(16 h^{2}-9 k^{2}=\left(h^{2}+k^{2}\right)^{2}\)
Hence the required locus of \((\mathrm{h}, \mathrm{k})\) is \(\left(x^{2}+y^{2}\right)^{2}=16 x^{2}-9 y^{2}\)
Hence, the correct option is (A)
Question 3
1 / -0

In an acute - angled triangle ABC, points D, E and F are the feet of the perpendiculars from A, B and C onto BC, AC and AB, respectively. H is the orthocentre of ∆ABC . If sin A = 3/5 and BC = 39, then the length of AH is:

A
50

B
42

C
55

D
52

Solution
Question 4
1 / -0

lf Z=x+iy is a complex number then |x|+|y|≤ ?

A
√2|z|

B
2|Z|

C
√2/Z

D
|Z|

Solution

\(z=x+i y=r \cos \theta+i r \sin \theta\) where \(|z|=r\)
Now, \(|x|+|y|=|r \cos \theta|+|r \sin \theta|\)
\(=|r|\{\cos \theta|+| \sin \theta \mid\}\)
But \(-\sqrt{2}<\cos s \theta+\sin \theta<\sqrt{2}\)
\(\Rightarrow|x|+|y| \leqslant \sqrt{2}|r|\)
\(\leqslant \sqrt{2}|z|\)
Hence, the correct option is (A)
Question 5
1 / -0

Let \(\omega\) be a complex number such that \(2 \omega+1=z\) where \(z=\sqrt{-3}\), if \(\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & -\omega^{2}-1 & \omega^{2} \\ 1 & \omega^{2} & \omega^{7}\end{array}\right|=3 k,\) then \(k\) ie equal to.

A
−z

B
z

C
−1

D
1

Solution

\(2 w+1=z\)
\(2 w+1=\sqrt{3} i\)
\(w=\frac{-1+\sqrt{3} i}{2}\)
\(\Rightarrow w^{2}=\frac{-1-\sqrt{3} i}{2}=-w-1\)
Now, \(\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & -\omega^{2}-1 & \omega^{2} \\ 1 & \omega^{2} & \omega^{7}\end{array}\right|=3 k\)...Given
\(\Rightarrow\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & w & w^{2} \\ 1 & w^{2} & w\end{array}\right|=3 k\)
\(1\left(w^{2}-w^{4}\right)-1\left(w-w^{2}\right)+1\left(w^{2}-w\right)=3 k\)
\(3 w^{2}-3 w=3 k\)
\(3\left(w^{2}-w\right)=3 k\)
\(k=w^{2}-w\)
\(k=-1-2 w\)
\(=-1-(-1+\sqrt{3} i)\)
\(=-\sqrt{3} i\)
\(=-z\)
Hence, the correct option is (A)
Question 6
1 / -0

Let ω be a cube root of unity not equal to 1. Then the maximum possible value of |a+bω+cω²| where a, b, c ϵ {+1,-1} is:

A
0

B
2

C
√3

D
1+√3

Solution

\(\omega^{3}=1\)
\(\therefore \omega^{3}-1=0\) or \((\omega-1)\left(\omega^{2}+\omega+1\right)=0\)
since \(\omega \neq 1, \omega=\frac{-1+i \sqrt{3}}{2}, \omega^{2}=\frac{-1-i \sqrt{3}}{2}\)
\(\left|a+b \omega+\alpha \omega^{2}\right|=\left|a-\frac{(b+c)}{2}+i\left(\frac{\sqrt{3}}{2}\right)(b-c)\right|\)
Square of the modulus is given by \(\left(\frac{2 a-b-c}{2}\right)^{2}+\left(\frac{\sqrt{3}(b-c)}{2}\right)^{2}\)
This can be maximised by substituting \(a=1, b=1, c=-1\) or \(a=-1, b=1, c=1\)
Both the substitutions yield the same result, of 4 \(\Rightarrow\) maximum modulus \(=2\)
Hence, the correct option is (B)
Question 7
1 / -0

Number of solutions of the equation |z|²+7z=0 is:

A
1

B
2

C
4

D
6

Solution

Let \(z=x+i y\) Then, \(|z|^{2}+7 z=x^{2}+y^{2}+7(x+i y)=0\)
\(\Rightarrow x^{2}+y^{2}+7 x=0\) \& \(7 y=0\)
\(\Rightarrow x^{2}+7 x=0\) and \(y=0\)
\(\Rightarrow x=-7\) or \(x=0\) and \(y=0\)
So, solutions are \(z=7\) or \(z=0\) Hence, two solutions.
Hence, the correct option is (B)
Question 8
1 / -0

If ω is a cube root of unity but not equal to 1, then minimum value of ∣a+b+cω²∣ (where a,b,ca,b,c are integers but not all equal) is:

A
0

B
√3/2

C
1

D
2

Solution

Let \(y=\left|a+b w+c w^{2}\right|\)
For \(y\) to be minimum \(y^{2}\) must be minimum \(y^{2}=\left|a+b w+c w^{2}\right|^{2}\)
\(=\left(a+b w+c w^{2}\right)\left(a+b w+c w^{2}\right)\)
\(=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]\)
since, \(a, b, c\) are not equal at a time. So, minimum value of \(y^{2}\) occurs when any two are same and third is differ by \(1 .\) Thus minimum of \(y=1\) (as \(a, b, c\) are integers).
Hence, the correct option is (C)
Question 9
1 / -0

If \(\left|z_{1}\right|=1,\left|z_{2}\right|=2,\left|z_{3}\right|=3\) and \(\left|9 z_{1} z_{2}+4 z_{1} z_{3}+z_{2} z_{3}\right|=12,\) then the value of \(\left|z_{1}+z_{2}+z_{3}\right|\)
is

A
56

B
60

C
2

D
64

Solution

Given \(\left|z_{1}\right|=1,\left|z_{2}\right|=2,\left|z_{3}\right|=3\) and \(\left|9 z_{1} z_{2}+4 z_{1} z_{3}+z_{2} z_{3}\right|=12\)
\(\left|9 z_{1} z_{2}+4 z_{1} z_{3}+z_{2} z_{3}\right|=12\)
\(\Rightarrow\left|z_{1}\right|\left|z_{2}\right|\left|z_{3}\right|\left|\frac{9}{z_{3}}+\frac{4}{z_{2}}+\frac{1}{z_{1}}\right|=12\)
\(\Rightarrow 6\left|\frac{9 \bar{z}_{3}}{\left|z_{3}\right|^{2}}+\frac{4 \bar{z}_{2}}{\left|z_{2}\right|^{2}}+\frac{\bar{z}_{1}}{\left|z_{1}\right|^{2}}\right|=12\)
\(\Rightarrow 6\left|z_{1}+z_{2}+z_{3}\right|=12\)
\(\Rightarrow 6\left|z_{1}+z_{2}+z_{3}\right|=12\)
\(\therefore\left|z_{1}+z_{2}+z_{3}\right|=2\)
Hence, the correct option is (C)
Question 10
1 / -0

Let \(z=x+i y\) be a complex number where \(x\) and \(y\) are integers. Then the area of the
rectangle whose vertices are the roots of the equation \(\bar{z} z^{3}+z \bar{z}^{3}=350\) is:

A
48

B
32

C
40

D
80

Solution

\(\bar{z} z^{3}+z \bar{z}^{3}=350\)
\(z \bar{z}\left(\bar{z}^{2}+z^{2}\right)=350\)
Put \(z=x+i y\)
\(\left(x^{2}+y^{2}\right)\left(x^{2}-y^{2}\right)=175\)
since \(\mathrm{x}\) and \(\mathrm{y}\) are integer so only possibility is \(x^{2}+y^{2}=25\)
and
\(x^{2}-y^{2}=7\)
so solving these equation we get
\(x=\pm 4, y=\pm 3\)
Thus required area \(=8 \times 6=48\) sq. unit.
Hence, the correct option is (A)

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 28

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Mathematics Test - 28

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SHARING IS CARING

Question 1

\(\left[\frac{\pi}{2}, \frac{3 x}{2}\right] \cap\left[\pi, \frac{3 \pi}{2}\right]\)

\(\left[\frac{\pi}{2}, \frac{5 \pi}{6}\right] \cup\left[\pi, \frac{3 \pi}{2}\right]\)

\(\left[\frac{\pi}{2}, \frac{4 \pi}{3}\right] \cap\left[\pi, \frac{4 \pi}{2}\right]\)

\(\left[\frac{\pi}{2}, \frac{2 \pi}{3}\right] \cap\left[\pi, \frac{2 \pi}{2}\right]\)

Solution

Question 2

\(\left(x^{2} y^{2}\right)^{2}=16 x^{2}-9 y^{2}\)

\(\left(x^{2}-y^{2}\right)^{2}=16 x^{2}-9 y^{2}\)

\(\left(x^{2} y^{2}\right)^{2}=16^{2} 9 y^{2}\)

\(\left(x^{2}-y^{2}\right)^{2}=16 x^{2} 9 y^{2}\)

Solution

Question 3

50

42

55

52

Solution

Question 4

√2|z|

2|Z|

√2/Z

|Z|

Solution

Question 5

−z

z

−1

1

Solution

Question 6

0

2

√3

1+√3

Solution

Question 7

1

2

4

6

Solution

Question 8

0

√3/2

1

2

Solution

Question 9

56

60

2

64

Solution

Question 10

48

32

40

80

Solution

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