Mathematics Test

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Mathematics Test - 28

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Weekly Quiz Competition

Question 1
4 / -1

Let T_n denote the number of triangles which can be formed by using the vertices of a regular polygon of n sides. If T_n ₊₁ – T_n = 21, then n equals

A
5

B
7

C
6

D
4

Solution

The number of triangles that can be formed by using the vertices of a regular polygon
is \({ }^{n} C_{3},\) i.e. \(T_{n}={ }^{n} C_{3}\)
Now, \(T_{n+1}-T_{n}=21\)
\(\Rightarrow \quad n+1_{C_{3}-} n_{C_{3}}=21\)
\(\Rightarrow \quad n C_{2}+n C_{3}-n C_{3}=21 \quad\left[\because n+1_{C_{r}}={ }^{n} C_{r}-1+{ }^{n} C_{r}\right]\)
\(\Rightarrow \quad \frac{1}{2} n(n-1)=21\)
\(\Rightarrow \quad n=-6\) or 7
As \(n\) is a positive integer, \(n=7\).
Question 2
4 / -1

In an isosceles triangle ABC, AB = AC. If vertical angle A measures 20°, then what is the value of a³+ b³?

A
3a²b

B
3b²c

C
3c²a

D
abc

Solution

\(\because \angle A=20^{\circ}\)
\(\angle B=\angle C=80^{\circ}\)
Then, \(b=c\)
\(\therefore \frac{a}{\sin 20^{\circ}}=\frac{b}{\sin 80^{\circ}}=\frac{c}{\sin 80^{\circ}}\)
Or \(\frac{a}{\sin 20^{\circ}}=\frac{b}{\cos 10^{\circ}} \Rightarrow a=2 b \sin 10^{\circ}\)
\(\therefore a^{3}+b^{3}=8 b^{3} \sin ^{3} 10^{\circ}+b^{3}=b^{3}\left\{2\left(4 \sin ^{3} 10^{\circ}\right)+1\right\}=b^{3}\left\{6 \sin 10^{\circ}\right\}=3 a c^{2}\)
Here, we used \(\sin 3 A=3 \sin A-4 \sin ^{3} A\)
\(a^{3}+b^{3}=3 b^{2}\left(2 b \sin 10^{\circ}\right)=3 a b^{2}=3 a c^{2}\)
Question 3
4 / -1

The locus of the midpoint of the portion between the axes of \(x \cos \alpha+y \sin \alpha=p,\) where \(p\) is a constant, is

A
x²+ y² = 4/p²

B
x² + y² = 4p²

C
1/x²+ 1/y²= 2/p²

D
1/x² + 1/y²= 4/p²

Solution

We have \(\frac{x}{\frac{p}{\cos \alpha}}+\frac{y}{\frac{p}{\sin \alpha}}=1\)
which gives the coordinates \((\mathrm{p} / \mathrm{cos}, \alpha, 0)\) and \((0, \mathrm{p} / \mathrm{sin} \alpha),\) where the line intersects the \(\mathrm{x}\) -axis and the \(\mathrm{y}\) -axis,
respectively.
If \((h, k)\) is the mid-point, then
\[
h=p / 2 \cos \alpha, k=p / 2 \sin \alpha
\]
so \(\quad(p / 2 h)^{2}+(p / 2 k)^{2}=\cos ^{2} \alpha+\sin ^{2} \alpha=1\)
\(\Rightarrow \quad 1 / h^{2}+1 / k^{2}=4 / p^{2}\)
Locus of \((h, k)\) is \(1 / x^{2}+1 / y^{2}=4 / p^{2}\)
Question 4
4 / -1

If a circle passes through (a, b) and cuts the circle x² + y² = 4 orthogonally, then the locus of its centre is

A
2ax – 2by + (a²+ b² + 4) = 0

B
2ax + 2by – (a² + b² + 4) = 0

C
2ax + 2by + (a² + b² + 4) = 0

D
2ax – 2by – (a² + b² + 4) = 0

Solution

Let the equation of the circle be \(x^{2}+y^{2}-2 g x-2 f y+c=0\) As it passes through (a, b), its equation should be:
\[
a^{2}+b^{2}-2 g a-2 f b+c=0
\]
since it intersects the circle \(x^{2}+y^{2}=4\) orthogonally,
\[
28 \times 0+2 f \times 0=c-4 \Rightarrow c=4
\]
and the locus of \((\mathrm{g}, \mathrm{f}),\) the centre, is
\[
2 a x+2 b y-\left(a^{2}+b^{2}+4\right)=0
\]
Question 5
4 / -1

A
ad – bc < 0

B
ad – bc > 0

C
ab – cd > 0

D
ab – cd < 0

Solution
Question 6
4 / -1

If \(g(x)=\int_{0}^{x} \cos ^{4} t d t,\) then \(g(x+\pi)\) equals

A
g(x) + g (π)

B
g(x) – g (π)

C
g(x) g (π)

D
g(x)/g (π)

Solution

\[
\begin{array}{l}
g(x+\pi)=\int_{0}^{x+\pi} \cos ^{4} t \mathrm{d} t \\
=\int_{0}^{\pi} \cos ^{4} t \mathrm{d} t+\int_{\pi}^{x+\pi} \cos ^{4} t \mathrm{d} t \\
=g(\pi)+I_{1}
\end{array}
\]
\(\operatorname{In} I_{1},\) put \(t=\pi+u,\) so that
\[
I_{1}=\int_{0}^{x} \cos ^{4}(\pi+u) \mathrm{d} u=\int_{0}^{x} \cos ^{4} u \mathrm{d} u=g(x)
\]
\(\Rightarrow g(x+\pi)=g(x)+g(\pi)\)
Question 7
4 / -1

\(f(x)=\left\{\begin{array}{ll}x+a & x<0 \\ |x-1| & : x \geq 0\end{array}\right.\) and

\(g(x)=\left\{\begin{array}{ll}x+1 & : \text { if } x<0 \\ (x-1)^{2}+b & : x \geq 0\end{array}\right.\)

If gof is continuous, then

A
a = 2, b = 0

B
a = 0, b = 1

C
a = 1, b = 0

D
b = 1, a = 1

Solution

\[
g o f(x)=\left\{\begin{array}{ccc}
x+a+1 & : & x<-a \\
(x+a-1)^{2}+b & : & -a \leq x<0 \\
x^{2}+b & : & 0 \leq x<1 \\
(x-2)^{2}+b & : & x \geq 1
\end{array}\right.
\]
gof is continuous except possibly at \(x=-a, 0,1\).
\[
\lim _{x \rightarrow-a-} f(x)=-a+a+1=1, \text { gof }(-a)=1+b
\]
Therefore, \(b+1=1 \Rightarrow b=0\)
\[
\begin{aligned}
\lim _{x \rightarrow 0-} f(x) &=(a-1)^{2}+b \\
&=(a-1)^{2}, g o f(0)=b=0
\end{aligned}
\]
So \((a-1)^{2}=0 \Rightarrow a=1\)
Question 8
4 / -1

Let S₁, S₂, … be squares such that for each n≥ 1, the length of a side of S_n equals the length of a diagonal of S_{n + 1}. If the length of a side of S₁ is 10 cm, then the smallest value of n for which Area (S_n) < 1 is

A
7

B
8

C
9

D
10

Solution

Let \(a_{n}\) denote the length of a side of \(S_{n} .\) We are given
Length of a side of \(S_{n}=\) Length of a diagonal of \(S_{n+1}\)
\[
\Rightarrow \quad a_{n}=\sqrt{2} a_{n+1} \Rightarrow \frac{a_{n+1}}{a_{n}}=\frac{1}{\sqrt{2}}
\]
Thus, \(a_{1}, a_{2}, a_{3}, \ldots\) is a G.P. with first term 10 and common ratio \(1 / \sqrt{2}\). Therefore,
\[
a_{n}=10(1 / \sqrt{2})^{n-1}
\]
Also, Area \(\left(S_{n}\right)=a_{n}^{2}<1\)
\(\Rightarrow\left[10(1 / \sqrt{2})^{n-1}\right]^{2}<1 \Rightarrow 100<2^{n-1}\)
\(\Rightarrow\) smallest possible value of \(n\) is 8
Question 9
4 / -1

In a triangle \(\mathrm{ABC}, \mathrm{AD}\) is the altitude from A. Given \(b>c, \angle C=23^{\circ}\) and \(A D=\frac{a b c}{b^{2}-c^{2}}\). What is the measure of \(\angle B ?\)

A
53^o

B
113^o

C
87^o

D
67^o

Solution

\(\angle C=23^{\circ}\)
\(\because A D=\frac{a b c}{b^{2}-c^{2}}=\frac{k \sin A b k \sin C}{k^{2}\left(\sin ^{2} B-\sin ^{2} C\right)},\) where \(k=2 R\)
\(A D=\frac{b \sin A \sin C}{\sin (B+C) \sin (B-C)}=\frac{b \sin A \sin C}{\sin \left(180^{\circ}-A\right) \sin (B-C)}\)
\(A D=\frac{b \sin C}{\sin (B-C)}=b \sin C(i n \Delta A D C)\)
\(\sin (B-C)=1\)
\(B-C=90^{\circ}\)
\(B=90^{\circ}+23^{\circ}=113^{\circ}\)
Question 10
4 / -1

\(\operatorname{li}_{x \rightarrow \infty}\left(\frac{x+5}{x+1}\right)^{x+4}\) is equal to

A
e⁴

B
e⁵

C
e³

D
None of these

Solution

\(\lim _{x \rightarrow \infty}\left(\frac{x+5}{x+1}\right)^{x+4}\)
\(=\operatorname{Lif}_{x \rightarrow \infty}\left\{\left.\left(1+\frac{4}{x+1}\right)^{\frac{x+1}{4}}\right|^{\frac{4(x+4)}{x+1}}=e^{4}\right.\)
\(\left(\operatorname{Lit}_{x \rightarrow \infty}\left(1+\frac{4}{x+1}\right)^{\frac{x+1}{4}}=\operatorname{es} x \rightarrow \infty\right.\)
\(\Rightarrow \frac{4}{x+1} \rightarrow 0\) and \(\left.\operatorname{Lim}_{x \rightarrow \infty} \frac{(x+4) 4}{x+1}=\operatorname{Lt}_{x \rightarrow \infty} \frac{4\left(1+\frac{4}{x}\right)}{1+\frac{1}{x}}=4\right)\)

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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

5

7

6

4

Solution

Question 2

3a2b

3b2c

3c2a

abc

Solution

Question 3

x2 + y2 = 4/p2

x2 + y2 = 4p2

1/x2 + 1/y2 = 2/p2

1/x2 + 1/y2 = 4/p2

Solution

Question 4

2ax – 2by + (a2 + b2 + 4) = 0

2ax + 2by – (a2 + b2 + 4) = 0

2ax + 2by + (a2 + b2 + 4) = 0

2ax – 2by – (a2 + b2 + 4) = 0

Solution

Question 5

ad – bc < 0

ad – bc > 0

ab – cd > 0

ab – cd < 0

Solution

Question 6

g(x) + g (π)

g(x) – g (π)

g(x) g (π)

g(x)/g (π)

Solution

Question 7

a = 2, b = 0

a = 0, b = 1

a = 1, b = 0

b = 1, a = 1

Solution

Question 8

7

8

9

10

Solution

Question 9

53o

113o

87o

67o

Solution

Question 10

e4

e5

e3

None of these

Solution

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3a²b

3b²c

3c²a

x²+ y² = 4/p²

x² + y² = 4p²

1/x²+ 1/y²= 2/p²

1/x² + 1/y²= 4/p²

2ax – 2by + (a²+ b² + 4) = 0

2ax + 2by – (a² + b² + 4) = 0

2ax + 2by + (a² + b² + 4) = 0

2ax – 2by – (a² + b² + 4) = 0

53^o

113^o

87^o

67^o

e⁴

e⁵

e³