Mathematics Test

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Mathematics Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

The value of $s i n^{2} 5^{\circ} + s i n^{2} 10^{\circ} + s i n^{2} 15^{\circ}$ + ..............+

$s i n^{2} 85^{\circ} + s i n^{2} 90^{\circ}$ is equal to

A
7

B
8

C
9

D
9 1/2

Solution

Given expression is
\[
\sin ^{2} 5^{\circ}+\sin ^{2} 10^{\circ}+\sin ^{2} 15^{\circ}+\ldots \ldots \ldots \ldots . .+\sin ^{2} 85^{\circ}+\sin ^{2} 90^{\circ}
\]
We know that $\sin 90^{\circ}=1$ or $\sin ^{2} 90^{\circ}=1$
Similarly, $\sin 45^{\circ}=\frac{1}{\sqrt{2}}$ or $\sin ^{2} 45^{\circ}=\frac{1}{2}$ and the angles are in A.P. of 18 terms We also know that
\[
\sin ^{2} 85^{\circ}=\left[\sin \left(90^{\circ}-5^{\circ}\right)\right]^{2}=\cos ^{2} 5^{\circ}
\]
Therefore from the complementary rule, we find
\[
\sin ^{2} 5^{\circ}+\sin ^{2} 85^{\circ}=\sin ^{2} 5^{\circ}+\cos ^{2} 5^{\circ}=1
\]
Therefore,
\[
\begin{array}{l}
\sin ^{2} 5^{\circ}+\sin ^{2} 10^{\circ}+\sin ^{2} 15^{\circ}+\ldots \ldots \ldots \ldots . .+\sin ^{2} 85^{\circ}+\sin ^{2} 90^{\circ} \\
\qquad=(1+1+1+1+1+1+1+1)+1+\frac{1}{2}=9 \frac{1}{2}
\end{array}
\]
Question 2
1 / -0

If $\int \frac{\left(x^{2}+1\right)\left(x^{2}+4\right)}{\left(x^{2}+3\right)\left(x^{2}-5\right)} d x=\int\left\{1+\frac{f(x)}{\left(x^{2}+3\right)\left(x^{2}-5\right)}\right\} d x$
$x+A \tan ^{-1}\left(\frac{x}{A^{\prime}}\right)+B \log \left(\frac{x-l}{x+m}\right)+K \quad$ then which of the following is correct

A
$A=\frac{1}{4 \sqrt{3}}, B=\frac{27}{8 \sqrt{5}}, K \in R$

B
$f(x)=7 x^{2}+19, A^{\prime}=\sqrt{3}, K \in R$

C
$l=m=\sqrt{5}, L=1, K \in R$

D
All of these

Solution

$=\int 1+\frac{7 x^{2}+19}{\left(x^{2}+3\right)\left(x^{2}-5\right)} d x=\int\left(1+\frac{1}{4\left(x^{2}+3\right)}+\frac{27}{4} \frac{1}{x^{2}-5}\right) d x$
$\therefore \int \frac{\left(x^{2}+1\right)\left(x^{2}+4\right)}{\left(x^{2}+3\right)\left(x^{2}-5\right)} d x=x+\frac{1}{4 \sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+\frac{27}{8 \sqrt{5}} \log \left(\frac{x-\sqrt{5}}{x+\sqrt{5}}\right)+K$
$\therefore \int \frac{\left(x^{2}+1\right)\left(x^{2}+4\right)}{\left(x^{2}+3\right)\left(x^{2}-5\right)} d x=L x+A \tan ^{-1}\left(\frac{x}{A^{\prime}}\right)+B \log \left(\frac{x-l}{x+m}\right)+K$
$\therefore L=1, A=\frac{1}{4 \sqrt{3}}, A^{\prime}=\sqrt{3}, B=\frac{27}{8 \sqrt{5}}, K \in R$
$l=m=\sqrt{5} ; \quad f(x)=7 x^{2}+19$
Question 3
1 / -0

If $\log _{\frac{1}{\sqrt{2}}} \sin x>0, x \epsilon[0,4 \pi]$ then the number of values of $x$ which are integral
multiples of $\frac{\pi}{4}$ is

A
4

B
12

C
3

D
None of these

Solution

Solution $0<\frac{1}{\sqrt{2}}<1$

\[
\log _{\frac{1}{\sqrt{2}}} \sin x>0, x \epsilon[0,4 \pi] \Rightarrow 0<\sin x<1
\]
$\therefore$ Integral multiple of $\frac{\pi}{4}$ will be $\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{9 \pi}{4}, \frac{11 \pi}{4}$
Number of required values $=4$
Question 4
1 / -0

The values of x and y for which the numbers 3+ ix²y andx²+y+ 4i are conjugate complex are

A
(-2,-1) or (2,-1)

B
(1,-2) or (-2,1)

C
(1,2) or (-1,-2)

D
None of these

Solution

According to condition, $3-i x^{2} y=x^{2}+y+4 i$
\[
\begin{array}{l}
\Rightarrow x^{2}+y=3 \text { And } x^{2} y=-4 \Rightarrow x=\pm 2, y=-1 \\
\Rightarrow(x, y)=(2,-1) \text { or }(-2,-1)
\end{array}
\]
Question 5
1 / -0

If the sum of the n terms of G.P. is S product is P and sum of their inverse is R, than P² is equal to

A
R/S

B
S/R

C
(R/S)ⁿ

D
(S/R)ⁿ

Solution
Question 6
1 / -0

The odds in favour of A solving a problem are 3 to 4 and the odds against B solving the same problem are 5 to 7. If they both try the problem, the probability that the problem is solved is:

A
41/84

B
16/21

C
5/21

D
1/4
Question 7
1 / -0

The value $\operatorname{lt} \sum_{n \rightarrow \infty} \sum_{r=0}^{r-n} \frac{^{n} C_{r}}{(r+3)}$ is

A
e

B
1

C
e - 2

D
None of these

Solution

$\because \int_{0}^{1} x^{r+2} d x=\frac{1}{r+3}$
$\therefore \operatorname{lt}_{n \rightarrow \infty} \sum_{r=0}^{r=n} C_{r} \frac{1}{n^{r}} \int_{0}^{1} x^{r+2} d x$
$\left.=\int_{0}^{1} x^{2}\left\{\left(\begin{array}{c}l t \sum_{n \rightarrow \infty}^{r-n} n \\ r=0\end{array}\right) r\left(\frac{x}{n}\right)^{r}\right)\right\} d x$
$=\int_{0}^{1} x^{2}\left(\begin{array}{c}\left.l t\left(1+\frac{x}{n}\right)^{n}\right) d x=\int_{0}^{1} x^{2} e^{x} d x \\ =\left[e^{x}\left(x^{2}-2 x+2\right)\right]_{0}^{1}=e-2\end{array}\right.$
Question 8
1 / -0

If sinθ, cosθ and tanθ are in G.P. then cot⁶θ − cot²θ is

A
1

B
1/2

C
2

D
3

Solution

Given that $\sin \theta, \cos \theta$ and $\tan \theta$ are in G.P
So, we have
\[
\begin{array}{l}
\cos ^{2} \theta=\tan \theta \sin \theta \\
\Rightarrow \cos ^{3} \theta=\sin ^{2} \theta \ldots .(i)
\end{array}
\]
Now, $\cot ^{6} \theta-\cot ^{2} \theta$
\[
\begin{array}{l}
=\cot ^{2} \theta\left(\cot ^{4} \theta-1\right) \\
=\frac{\cos ^{2} \theta}{\sin ^{2} \theta}\left(\frac{\cos ^{4} \theta}{\sin ^{4} \theta}-1\right) \\
=\frac{\cos ^{2} \theta}{\cos ^{3} \theta}\left(\frac{\cos ^{4} \theta}{\cos ^{6} \theta}-1\right) \ldots \ldots \text { From }(i) \\
=\frac{1}{\cos \theta} \frac{\sin ^{2} \theta}{\cos ^{2} \theta} \\
=\frac{\cos ^{3} \theta}{\cos ^{3} \theta} \\
=1
\end{array}
\]
Question 9
1 / -0

The value of $4+5\left(-\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)^{334}+3\left(-\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)^{335}$ is

A
1−i√3

B
−1+i√3

C
4√3i

D
−i√3

Solution

\[
\begin{array}{l}
4+5\left(-\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)^{334}+3\left(-\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)^{335} \\
=4+5 \omega^{334}+3 \omega^{670}\left(\omega^{2}=-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right) \\
=4+5 \omega+3 \omega=4+9 \omega \\
=4+8\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)=4 \sqrt{3} i
\end{array}
\]
Hence choice (c) is correct.
Question 10
1 / -0

The degree and order respectively of the differential equation of all the parabolas whose axis is x-axis, are

A
2, 1

B
1, 2

C
2, 2

D
1, 1

Solution

Equation of all parabolas whose axis is x-axis are $y^{2}=4 a(x+c)$ Differentiating w.r.t $x$ we get $2 y y^{\prime}=4 a$ Again differentiating we get $2\left(y y^{\prime}+y^{\prime} y^{\prime}\right)=0$ $\Rightarrow 2 y y^{\prime \prime}+2\left(y^{\prime}\right)^{2}=0$
$\therefore$ degree is the natural power on highest order differential coefficient $=1$ Order is the highest order differential coefficient $=2$

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 20

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Mathematics Test - 20

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SHARING IS CARING

Question 1

7

8

9

9 1/2

Solution

Question 2

\(A=\frac{1}{4 \sqrt{3}}, B=\frac{27}{8 \sqrt{5}}, K \in R\)

\(f(x)=7 x^{2}+19, A^{\prime}=\sqrt{3}, K \in R\)

\(l=m=\sqrt{5}, L=1, K \in R\)

All of these

Solution

Question 3

4

12

3

None of these

Solution

Question 4

(-2,-1) or (2,-1)

(1,-2) or (-2,1)

(1,2) or (-1,-2)

None of these

Solution

Question 5

R/S

S/R

(R/S)n

(S/R)n

Solution

Question 6

41/84

16/21

5/21

1/4

Question 7

e

1

e - 2

None of these

Solution

Question 8

1

1/2

2

3

Solution

Question 9

1−i√3

−1+i√3

4√3i

−i√3

Solution

Question 10

2, 1

1, 2

2, 2

1, 1

Solution

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(R/S)ⁿ

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