Mathematics Test

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Mathematics Test - 27

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Weekly Quiz Competition

Question 1
4 / -1

If for a real number \(x,[x]\) is the greatest integer less than or equal to \(x,\) then the value of the integral \(\int_{\pi / 2}^{3 \pi / 2}[2 \sin x] d x\)

A
-π

B
0

C
-π/2

D
π/2

Solution

We have \(I=\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}[2 \sin x] d x\)
Taking all the integral values from \(\pi / 2\) to \(3 \pi / 2\) gives the integer.
\(\int_{\pi / 2}^{3 \pi / 2}[2 \sin x] \mathrm{d} x=\int_{\pi / 2}^{5 \pi / 6}[2 \sin x] d x+\int_{5 \pi / 6}^{\pi}[2 \sin x] d x+\int_{\pi}^{7 \pi / 6}[2 \sin x] d x+\)
\(\int_{7 \pi / 6}^{3 \pi / 2}[2 \sin x] d x\)
\(=\int_{\pi / 2}^{5 \pi / 6} 1 \mathrm{dx}+\int_{5 \pi / 6}^{\pi} 0 \mathrm{dx}+\int_{\pi}^{7 \pi / 6}-1 \mathrm{dx}+\int_{7 \pi / 6}^{3 \pi / 2}-2 \mathrm{dx}\)
\(=(5 \pi / 6-\pi / 2)-(7 \pi / 6-\pi)-2(3 \pi / 2-7 \pi / 6)=-(\pi / 2)\)
Question 2
4 / -1

If \(A=\left[\begin{array}{rrr}x & 3 & 2 \\ -3 & y & -7 \\ -2 & 7 & 0\end{array}\right]\) and \(A=-A^{T},\) then \(x+y\) is equal to

A
2

B
-1

C
0

D
12

Solution

\(A=-A^{\top}\)
\(\Leftrightarrow \quad A\) is a skew-symmetric matrix.
\(\Rightarrow \quad\) Diagonal elements of A are zeroes.
\(\Rightarrow \quad x=0, y=0\)
\(x+y=0\)
Question 3
4 / -1

\(f(x)=\frac{e^{x}-e^{-x}}{2}, g(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\)

(1) f(x) has one point of inflexion.

(2) g(x) has one point of inflexion.

(3) f(x) has one point of local minima.

(4) g(x) has one point of local maxima.

Which one of the following is correct regarding the above statements?

A

(1) is true.
(2) is false.

B

(1) is false.
(2) is true.

C
(3) and (4) both are true.

D
(1) and (2) both are true.

Solution

\(\mathrm{f}(\mathrm{x})=\frac{\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}}{2} \Rightarrow \mathrm{f}^{*}(\mathrm{x})=\frac{\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}}{2}\)
\(\Rightarrow \mathrm{f}^{\prime \prime}(0)=0, \mathrm{f}^{\prime \prime}\left(0^{+}\right)>0, \mathrm{f}^{\prime \prime}\left(0^{-}\right)<0\)
\(\therefore \mathrm{x}=0\) is point of inflexion of \(f(x)\) Now, \(g(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \Rightarrow g^{\prime \prime}(x)=\frac{-8\left(e^{x}-e^{-x}\right)}{\left(e^{x}+e^{-x}\right)^{3}}\)
\(\Rightarrow g^{\prime \prime}(0)=0, g^{\prime \prime}\left(0^{+}\right)<0, g^{*}\left(0^{-}\right)>0\)
\(\therefore \mathrm{x}=0\) is point of inflexion of \(\mathrm{g}(\mathrm{x})\)
\(\therefore\) the option is correct.
Question 4
4 / -1

If \(A=\left[\begin{array}{rrr}\cos \frac{\pi}{5} & \sin \frac{\pi}{5} \\ -\sin \frac{\pi}{5} & \cos \frac{\pi}{5}\end{array}\right],\) and \(B=A+A^{2}+A^{3}+A^{4},\) which of the following statements is/are correct?

A
B is symmetric

B
B is invertible

C
A is periodic

D
All of the above

Solution

If \(P=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\)
Then, \(P^{n}=\left[\begin{array}{cc}\cos n \alpha & \sin n \alpha \\ -\sin n \alpha & \cos n \alpha\end{array}\right]\)
\(B=A+A^{2}+A^{3}+A^{4}\)
\(=\left[\begin{array}{cc}\cos \left(\frac{\pi}{5}+\frac{2 \pi}{5}+\frac{3 \pi}{5}+\frac{4 \pi}{5}\right) & \left.\sin \left(\frac{\pi}{5}+\frac{2 \pi}{5}+\frac{3 \pi}{5}+\frac{4 \pi}{5}\right)\right] \\ -\sin \left(\frac{\pi}{5}+\frac{2 \pi}{5}+\frac{3 \pi}{5}+\frac{4 \pi}{5}\right) & \cos \left(\frac{\pi}{5}+\frac{2 \pi}{5}+\frac{3 \pi}{5}+\frac{4 \pi}{5}\right)\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]\)
\(\therefore \mathrm{B}\) is symmetric and \(\mathrm{B}\) is invertible.
Also, \(A^{10}=1\)
\(\therefore \mathrm{A}\) is periodic.
So, option 4 is correct
Question 5
4 / -1

What is the circumradius of a triangle with length of sides as 6 units, 8 units and 10 units?

A
4 units

B
5 units

C
3 units

D
None of the above

Solution

As the given triangle is a right triangle, the circumradius is half of the hypotenuse which is 5 units

Hence, answer option 2 is correct.
Question 6
4 / -1

Let f(x) = sin x + cos x. Then,

A
x = 17π/4 is a point of minima

B
x = 13π/4 is a point of maxima

C
x = 21π/4 is a point of minima

D
x = 29π/4 is a point of maxima

Solution

\[
\begin{aligned}
f^{\prime}(x)=& \cos x-\sin x=0 \Rightarrow x=n \pi+\pi / 4 \\
f^{\prime \prime}(x)=&-\sin x-\cos x=-(\sin x+\cos x) \\
& f^{\prime \prime}\left(n \pi+\frac{\pi}{4}\right)=(-1)^{n+1} \\
&\left(\sin \frac{\pi}{4}+\cos \frac{\pi}{4}\right)=(-1)^{n+1} \sqrt{2}
\end{aligned}
\]
If \(n\) is even then \(f(x)\) has maxima and if \(n\) is odd then
Question 7
4 / -1

A
A.P.

B
G.P.

C
H.P.

D
None of these

Solution
Question 8
4 / -1

The number of subsets of the set A = {a₁, a₂, …, a_n} which contain even number of elements is

A
2^{(n – 1)}

B
2ⁿ – 1

C
2^{n – 2}

D
2ⁿ

Solution

For each of the first \((n-1)\) elements a \(1, a_{2}, \ldots, a_{n-1},\) we have two choices: either a \(_{i}(1\)

\(\leq i \leq n-1\) ) lies in the subset or a a doesn't lie in the subset. For the last element, we

have just one choice. If even number of elements have already been selected, we do not include an in the subset, otherwise (when odd number of elements have been

selected), we include it in the subset. Thus, the number of subsets of \(A=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}\) which contain even number of

elements is equal to \(2^{\mathrm{n}}-1\)
Question 9
4 / -1

A line 'm' is drawn as reflection of the line \({ }^{\prime} \ell^{\prime}\) on the \(\mathrm{x}\) -axis as shown in the figure given below. What will be the slope of line 'm'? (It is given that the reflection is about the x-axis.)

A
2/3

B
-2/3

C
3/2

D
-3/2

Solution

The reflected line 'm' passes through (0, - 2) and (3, 0). So, slope \(=\frac{0-(-2)}{3-0}=\frac{2}{3}\)
Question 10
4 / -1

If \(\lim _{x \rightarrow 0}\left(1+x \log \left(1+b^{2}\right)\right)^{1 / x}=2 b \sin ^{2} \theta, b / 0\) and \(\theta \in[-\pi, \pi],\) then the value of \(\theta\) is

A
\(\pm \frac{\pi}{4}\)

B
\(\pm \frac{\pi}{3}\)

C
\(\pm \frac{\pi}{6}\)

D
\(\pm \frac{\pi}{2}\)

Solution

\(\lim _{x \rightarrow 0}\left(1+x \log \left(1+b^{2}\right)\right)^{1 / x}\)
\(=\lim _{x \rightarrow 0}\left(1+x \log \left(1+b^{2}\right)^{\frac{1}{x \log \left(1+b^{2}\right)} \times \log \left(1+b^{2}\right)}\right.\)
\(=e^{\log \left(1+b^{2}\right)}=1+b^{2}\)
Thus,
\(1+b^{2}=2 b \sin ^{2} \theta\)
\(\Rightarrow \quad \sin ^{2} \theta=\frac{1+b^{2}}{2 b} \geq 1\)
\(\Rightarrow \quad \sin ^{2} \theta=1 \Rightarrow \sin \theta=\pm 1\) or \(\quad \theta=\pm \pi / 2\)

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Mathematics Test - 27

Result

Mathematics Test - 27

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SHARING IS CARING

Question 1

-π

0

-π/2

π/2

Solution

Question 2

2

-1

0

12

Solution

Question 3

(1) is true. (2) is false.

(1) is false. (2) is true.

(3) and (4) both are true.

(1) and (2) both are true.

Solution

Question 4

B is symmetric

B is invertible

A is periodic

All of the above

Solution

Question 5

4 units

5 units

3 units

None of the above

Solution

Question 6

x = 17π/4 is a point of minima

x = 13π/4 is a point of maxima

x = 21π/4 is a point of minima

x = 29π/4 is a point of maxima

Solution

Question 7

A.P.

G.P.

H.P.

None of these

Solution

Question 8

2(n – 1)

2n – 1

2n – 2

2n

Solution

Question 9

2/3

-2/3

3/2

-3/2

Solution

Question 10

\(\pm \frac{\pi}{4}\)

\(\pm \frac{\pi}{3}\)

\(\pm \frac{\pi}{6}\)

\(\pm \frac{\pi}{2}\)

Solution

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(1) is true.
(2) is false.

(1) is false.
(2) is true.

2^{(n – 1)}

2ⁿ – 1

2^{n – 2}

2ⁿ