Mathematics Test

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Mathematics Test - 20

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Question 1/10
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

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A
7

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B
8

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C
9

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D
9 1/2

Solutions

Given expression is
$\sin^{2} 5^{\circ} + \sin^{2} 10^{\circ} + \sin^{2} 15^{\circ} + \dots \dots \dots \dots . . + \sin^{2} 85^{\circ} + \sin^{2} 90^{\circ}$
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
$\sin^{2} 85^{\circ} = {[\sin (90^{\circ} - 5^{\circ})]}^{2} = \cos^{2} 5^{\circ}$
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,
$\begin{array}{l} \sin^{2} 5^{\circ} + \sin^{2} 10^{\circ} + \sin^{2} 15^{\circ} + \dots \dots \dots \dots . . + \sin^{2} 85^{\circ} + \sin^{2} 90^{\circ} \\ = (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) + 1 + \frac{1}{2} = 9 \frac{1}{2} \end{array}$
Question 2/10
1 / -0

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

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

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

Correct

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

Correct

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D
All of these

Solutions

$= \int 1 + \frac{7 x^{2} + 19}{(x^{2} + 3) (x^{2} - 5)} d x = \int (1 + \frac{1}{4 (x^{2} + 3)} + \frac{27}{4} \frac{1}{x^{2} - 5}) d x$
$∴ \int \frac{(x^{2} + 1) (x^{2} + 4)}{(x^{2} + 3) (x^{2} - 5)} d x = x + \frac{1}{4 \sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}}) + \frac{27}{8 \sqrt{5}} \log (\frac{x - \sqrt{5}}{x + \sqrt{5}}) + K$
$∴ \int \frac{(x^{2} + 1) (x^{2} + 4)}{(x^{2} + 3) (x^{2} - 5)} d x = L x + A \tan^{- 1} (\frac{x}{A^{'}}) + B \log (\frac{x - l}{x + m}) + K$
$∴ L = 1, A = \frac{1}{4 \sqrt{3}}, A^{'} = \sqrt{3}, B = \frac{27}{8 \sqrt{5}}, K \in R$
$l = m = \sqrt{5}; f (x) = 7 x^{2} + 19$
Question 3/10
1 / -0

If $\log_{\frac{1}{\sqrt{2}}} \sin x > 0, x ϵ [0, 4 π]$ then the number of values of $x$ which are integral
multiples of $\frac{π}{4}$ is

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A
4

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Wrong

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B
12

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C
3

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D
None of these

Solutions

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

$\log_{\frac{1}{\sqrt{2}}} \sin x > 0, x ϵ [0, 4 π] \Rightarrow 0 < \sin x < 1$
$∴$ Integral multiple of $\frac{π}{4}$ will be $\frac{π}{4}, \frac{3 π}{4}, \frac{9 π}{4}, \frac{11 π}{4}$
Number of required values $= 4$
Question 4/10
1 / -0

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

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

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

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

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D
None of these

Solutions

According to condition, $3 - i x^{2} y = x^{2} + y + 4 i$
$\begin{array}{l} \Rightarrow x^{2} + y = 3 And x^{2} y = - 4 \Rightarrow x = \pm 2, y = - 1 \\ \Rightarrow (x, y) = (2, - 1) or (- 2, - 1) \end{array}$
Question 5/10
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

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A
R/S

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B
S/R

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C
(R/S)ⁿ

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D
(S/R)ⁿ

Solutions
Question 6/10
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:

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A
41/84

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B
16/21

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C
5/21

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D
1/4
Question 7/10
1 / -0

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

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A
e

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B
1

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C
e - 2

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D
None of these

Solutions

$∵ \int_{0}^{1} x^{r + 2} d x = \frac{1}{r + 3}$
$∴ {lt}_{n \to \infty} \sum_{r = 0}^{r = n} C_{r} \frac{1}{n^{r}} \int_{0}^{1} x^{r + 2} d x$
$= \int_{0}^{1} x^{2} {(\begin{matrix} l t \sum_{n \to \infty}^{r - n} n \\ r = 0 \end{matrix}) r {(\frac{x}{n})}^{r})} d x$
$= \int_{0}^{1} x^{2} (\begin{matrix} l t {(1 + \frac{x}{n})}^{n}) d x = \int_{0}^{1} x^{2} e^{x} d x \\ = {[e^{x} (x^{2} - 2 x + 2)]}_{0}^{1} = e - 2 \end{matrix}$
Question 8/10
1 / -0

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

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A
1

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B
1/2

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C
2

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D
3

Solutions

Given that $\sin θ, \cos θ$ and $\tan θ$ are in G.P
So, we have
$\begin{array}{l} \cos^{2} θ = \tan θ \sin θ \\ \Rightarrow \cos^{3} θ = \sin^{2} θ \dots . (i) \end{array}$
Now, $\cot^{6} θ - \cot^{2} θ$
$\begin{array}{l} = \cot^{2} θ (\cot^{4} θ - 1) \\ = \frac{\cos^{2} θ}{\sin^{2} θ} (\frac{\cos^{4} θ}{\sin^{4} θ} - 1) \\ = \frac{\cos^{2} θ}{\cos^{3} θ} (\frac{\cos^{4} θ}{\cos^{6} θ} - 1) \dots \dots From (i) \\ = \frac{1}{\cos θ} \frac{\sin^{2} θ}{\cos^{2} θ} \\ = \frac{\cos^{3} θ}{\cos^{3} θ} \\ = 1 \end{array}$
Question 9/10
1 / -0

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

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A
1−i√3

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B
−1+i√3

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C
4√3i

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D
−i√3

Solutions

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

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

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A
2, 1

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B
1, 2

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C
2, 2

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D
1, 1

Solutions

Equation of all parabolas whose axis is x-axis are $y^{2} = 4 a (x + c)$ Differentiating w.r.t $x$ we get $2 y y^{'} = 4 a$ Again differentiating we get $2 (y y^{'} + y^{'} y^{'}) = 0$ $\Rightarrow 2 y y^{''} + 2 {(y^{'})}^{2} = 0$
$∴$ degree is the natural power on highest order differential coefficient $= 1$ Order is the highest order differential coefficient $= 2$

Question 1/10

7

8

9

9 1/2

Question 2/10

A=143,B=2785,K∈R

f(x)=7x2+19,A′=3,K∈R

l=m=5,L=1,K∈R

All of these

Question 3/10

4

12

3

None of these

Question 4/10

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

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

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

None of these

Question 5/10

R/S

S/R

(R/S)n

(S/R)n

Question 6/10

41/84

16/21

5/21

1/4

Question 7/10

e

1

e - 2

None of these

Question 8/10

1

1/2

2

3

Question 9/10

1−i√3

−1+i√3

4√3i

−i√3

Question 10/10

2, 1

1, 2

2, 2

1, 1

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

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

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

(R/S)ⁿ

(S/R)ⁿ