Mathematics Test

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Mathematics Test - 31

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Question 1
4 / -1

sin^-1 (sin 10 ) =

A
10

B
3π - 10

C
π + 6

D
2π - 6

Solution

$3 \pi<10<3 \pi+\frac{\pi}{2} \therefore 10 \mathrm{rad} \in Q_{3}$
$\therefore-\frac{\pi}{2}<3 \pi-10<0 \therefore 3 \pi-10 \in Q_{4}$
Also $\sin (3 \pi-10)=\sin 10$
Hence $\sin ^{-1}(\sin 10)=\sin ^{-1} \sin (3 \pi-10)=3 \pi-10$
Question 2
4 / -1

$\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)=$

A
1/√10

B
-1/√10

C
1/10

D
-1/10

Solution

Let $0 \leq x \leq \pi$
So, $\cos ^{-1} \frac{4}{5}=x \Rightarrow \cos x=\frac{4}{5} \quad \ldots .$ (i)
Now $\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)=\sin \left(\frac{x}{2}\right)$
$\ldots .$ (ii)
From (i), $\cos x=\frac{4}{5} \Rightarrow 1-2 \sin ^{2} \frac{x}{2}=\frac{4}{5}$
$\Rightarrow 2 \sin ^{2} \frac{x}{2}=1-\frac{4}{5}=\frac{1}{5} \Rightarrow \sin \frac{x}{2}=\sqrt{\frac{1}{10}}$
Question 3
4 / -1

If one root of the quadratic equation $a x^{2}+b x+c=0$ is equal to the $n^{\text {th }}$ power of the other root, then the value of $\left(a c^{n}\right)^{\frac{1}{n+1}}+\left(a^{n} c\right)^{\frac{1}{n+1}}$

A
b

B
-b

C
$b^{\frac{1}{n+1}}$

D
$-b^{\frac{1}{n+1}}$

Solution

Let $\alpha, \alpha^{n}$ be two roots,
Then $\alpha+\alpha^{n}=-\frac{b}{a^{\prime}} \alpha \alpha^{n}=\frac{c}{a}$
Eliminating $\alpha,$ we get $\left(\frac{9}{a}\right)^{\frac{1}{n+1}}+\left(\frac{c}{a}\right)^{\frac{n}{n+1}}=-\frac{b}{a}$
$\Rightarrow a \cdot a^{-\frac{1}{n+1}}, c^{\frac{1}{n+1}}+a \cdot a^{-\frac{n}{n+1}}, c^{\frac{n}{n+1}}=-b$
or $\left(a^{n} c\right)^{\frac{1}{n+1}}+\left(a c^{n}\right)^{\frac{1}{n+1}}=-b$
Question 4
4 / -1

The coefficient of x in the equation x² + px + q = 0 was taken as 17 in place of 13, its roots were found to be -2 and -15, the roots of the original equation are

A
3, 10

B
- 3, - 10

C
- 5, - 18

D
18,5

Solution

Let the equation (in written form) be $x^{2}+17 x+q=0$ Roots are -2,-15
So $q=30$ And correct equation is $x^{2}+13 x+30=0$.
Hence roots are $-3,-10 .$
Question 5
4 / -1

If the sets $A$ and $B$ are defined as
$A=\left\{(x, y): y=\frac{1}{x^{\prime}} 0 \neq x \in R\right\}$
$B=[(x, y): y=-x, x \in R],$ then

A
A ∩ B = A

B
A ∩ B = B

C
A ∩ B = ∅

D
None of these

Solution

Since $y=\frac{1}{x^{2}} y=-x$ meet when $-x=\frac{1}{x} \Rightarrow x^{2}=-1$
which does not give any real value of $x$.
Hence, $A \cap B=\emptyset$.
Question 6
4 / -1

If $x=\sin \left(2 \tan ^{-1} 2\right), y=\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)$ then

A
x = 1 - y

B
x² = 1 - y

C
x² = 1 + y

D
y² = 1 - x

Solution

$2 \tan^{- 1} 2 = π + \tan^{- 1} (\frac{2 + 2}{1 - 4})$
$= π + \tan^{- 1} \frac{4}{- 3}$
$= π - \tan \frac{- 4}{3}$
Let, $\tan \frac{- 4}{3} = θ$
$x = \sin (π - θ), y = \sin \frac{θ}{2}$
$x^{2} + y^{2} = \sin^{2} θ + \frac{1 - \cos θ}{2}$
$= \frac{16}{25} + \frac{1}{5} = \frac{21}{25}$
$y^{2} = \frac{1}{5}, 1 - x = 1 - \frac{4}{5} = \frac{1}{5}$
$∴ y^{2} = 1 - x$
Question 7
4 / -1

The value of $\sin \left[2 \tan ^{-1}\left(\frac{1}{3}\right)\right]+\cos \left[\tan ^{-1}(2 \sqrt{2})\right]=$

A
16/15

B
14/15

C
12/15

D
11/15

Solution

$\sin \left[2 \tan ^{-1}\left(\frac{1}{3}\right)\right]+\cos \left[\tan ^{-1}(2 \sqrt{2})\right]$
$=\sin \left[\tan ^{-1} \frac{\frac{2}{3}}{1-\frac{1}{9}}\right]+\cos \left[\tan ^{-1}(2 \sqrt{2})\right]$
$=\sin \left[\tan ^{-1} \frac{3}{4}\right]+\cos \left[\tan ^{-1} 2 \sqrt{2}\right]$
$=\tan ^{-1} 2 \sqrt{2}=\cos ^{-1} \frac{1}{3}$
Also $\tan ^{-1} \frac{3}{4}=\sin ^{-1} \frac{3}{5}$
$=\frac{3}{5}+\frac{1}{3}=\frac{14}{15}$
Question 8
4 / -1

$\sin ^{-1}|\cos x|-\cos ^{-1}|\sin x|=a$ has at least one solution if $\mathbf{a} \in$

A
$\left[0, \frac{\pi}{2}\right]$

B
$\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]$

C
(0,π)

D
0

Solution

$\sin ^{-1}|\cos x|-\cos ^{-1}|\sin x|=\frac{\pi}{2}-\cos ^{-1}|\cos x|-\frac{\pi}{2}+\sin ^{-1}|\sin x|=a$
$\Rightarrow \sin ^{-1}|\sin x|-\cos ^{-1}|\cos x|=a$
$\Rightarrow a=0 $
Question 9
4 / -1

$4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}$ is equal to

A
π

B
π/2

C
π/3

D
π/4

Solution

$\quad$ Since $2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}$
$\therefore 4 \tan ^{-1} \frac{1}{5}=2\left[2 \tan ^{-1} \frac{1}{5}\right]=2 \tan ^{-1} \frac{\frac{2}{5}}{1-\frac{1}{25}}$
$=2 \tan ^{-1} \frac{10}{24}=\tan ^{-1} \frac{\frac{20}{24}}{1-\frac{100}{576}}=\tan ^{-1} \frac{120}{119}$
So, $4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=\tan ^{-1} \frac{120}{119}-\tan ^{-1} \frac{1}{239}$
$=\tan ^{-1} \frac{\frac{120}{119} \frac{1}{239}}{1+\frac{120}{119} \frac{1}{239}}=\tan ^{-1} \frac{(120 \times 239)-119}{(119 \times 239)+120}$
$\Rightarrow \tan ^{-1} 1=\frac{\pi}{4}$
Question 10
4 / -1

Let $n(U)=700, n(A)=200, n(B)=300$ and $n(A \cap B)=100$,
Then $n\left(A^{c} \cap B^{c}\right)=$

A
400

B
600

C
300

D
200

Solution

$n\left(A^{c} \cap B^{c}\right)=n(U)-n(A \cup B)$
$=n(U)-[n(A)+n(B)-n(A \cap B)]$
$=700-[200+300-100]=300$

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Mathematics Test - 31

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Mathematics Test - 31

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Question 1

10

3π - 10

π + 6

2π - 6

Solution

Question 2

1/√10

-1/√10

1/10

-1/10

Solution

Question 3

b

-b

\(b^{\frac{1}{n+1}}\)

\(-b^{\frac{1}{n+1}}\)

Solution

Question 4

3, 10

- 3, - 10

- 5, - 18

18,5

Solution

Question 5

A ∩ B = A

A ∩ B = B

A ∩ B = ∅

None of these

Solution

Question 6

x = 1 - y

x2 = 1 - y

x2 = 1 + y

y2 = 1 - x

Solution

Question 7

16/15

14/15

12/15

11/15

Solution

Question 8

\(\left[0, \frac{\pi}{2}\right]\)

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

(0,π)

0

Solution

Question 9

π

π/2

π/3

π/4

Solution

Question 10

400

600

300

200

Solution

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x² = 1 - y

x² = 1 + y

y² = 1 - x