Mathematics Test

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Mathematics Test - 22

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Weekly Quiz Competition

Question 1
4 / -1

If f(x) is a function whose domain is symmetric about the origin, then f(x) + f(–x) is

A
One-one

B
Even

C
Odd

D
Both even and odd

Solution

(a, b)

g(x) = f(x) + f(–x)

g(–x) = f(–x) + f(x) = g(x)

therefore g(x) is even
Question 2
4 / -1

If $f(9)=9, f^{\prime}(9)=4,$ then $\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}$ equals

A
2

B
4

C
6

D
0

Solution

Given that $f(9)=9, f^{\prime}(9)=4$
$\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}$
On rationalisation we get -
$=\lim _{x \rightarrow 9}\left(\frac{\sqrt{f(x)}-3}{\sqrt{x}-3}\right)\left(\frac{\sqrt{x}+3}{\sqrt{x}+3}\right)\left(\frac{\sqrt{f(x)}+3}{\sqrt{f(x)}+3}\right)$
$=\lim _{x \rightarrow 9} \frac{f(x)-f(9)}{x-9} \cdot \frac{6}{6}$
$=\lim _{x \rightarrow 9} \frac{f(x)-f(9)}{x-9}$
$=f^{\prime}(9)$
$=4$
Question 3
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 4
4 / -1

If $f(x)=\sqrt{x}, g(x)=e^{x-1},$ and $\int f o g(x) d x=A f o g(x)+B t a n^{-1}(f o g(x))+C,$ then

$A+B$ is equal to

A
1

B
2

C
3

D
0

Solution

$\operatorname{fog}(x)=\sqrt{e^{x}-1}$
$\therefore I=\int \sqrt{e^{x}-1} d x=\int \frac{2 t^{2}}{t^{2}+1} d t\{$ where $t=\sqrt{e^{x}-1}\}$
$=2 t-2 \tan ^{-1} t+C$
$=2 \sqrt{e^{x}-1}-2 \tan ^{-1}(\sqrt{e^{x}-1})+C$
$=2 \operatorname{fog}(x)-2 \tan ^{-1}(\operatorname{fog}(x))+C$
$\therefore A+B=2+(-2)=0$
Question 5
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 6
4 / -1

In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then number of families which buy A only is

A
3100

B
3300

C
2900

D
1400

Solution

n(A) = 40% of 10,000 = 4,000

n(B) = 20% of 10,000 = 2,000

n(C) = 10% of 10,000 = 1,000

n(A ∩ B) = 5% of 10,000 = 500

n(B ∩ C) = 3% of 10,000 = 300

n(C ∩ A) = 4% of 10,000 = 400

n(A ∩ B ∩ C) = 2% of 10,000 = 200

We want to find the number of families which buy only A = n(A) - [n(A ∩ B) + n(A ∩ C) - n(A ∩ B ∩ C)]

=4000 - [500 + 400 - 200] = 4000 - 700 = 3300
Question 7
4 / -1

$\lim _{x \rightarrow-\infty}\left\{\frac{x^{4} \sin \left(\frac{1}{x}\right)+x^{2}}{1+|x|^{3}}\right\}$ is equal to

A
2

B
1

C
-1

D
does not exist

Solution

$\lim _{x \rightarrow-\infty}\left[\frac{x^{4} \sin \left(\frac{1}{x}\right)+x^{2}}{1+|x|^{3}}\right]$
$\lim _{x \rightarrow-\infty} \frac{x^{3}}{1+|x|^{3}}\left[x \sin \left(\frac{1}{x}\right)+\frac{1}{x}\right]$
As $x \rightarrow-\infty$
$\frac{x^{3}}{1+|x|^{3}}=-1$
$x \sin \left(\frac{1}{x}\right)+\frac{1}{x}=1$
$\Rightarrow \lim _{x \rightarrow-\infty} \frac{x^{3}}{1+|x|^{3}}\left(x \sin \frac{1}{x}+\frac{1}{x}\right)=-1$
Question 8
4 / -1

The distance moved by the particle in time $t$ is given by $x=t^{3}-12 t^{2}+6 t+8 .$ At

the instant when its acceleration is zero, then the velocity is

A
42

B
-42

C
48

D
-48

Solution

$x=t^{3}-12 t^{2}+6 t+8$
$\Rightarrow \frac{d x}{d t}=3 t^{2}-24 t+6 a n d \frac{d^{2} x}{d t^{2}}=6 t-24$
Now, Acceleration =0 $\Rightarrow \frac{d^{2} x}{d t^{2}}=0 \Rightarrow 6 t-24=0 \Rightarrow t=4$
$\mathrm{Att}=4,$ we have Velocity $=\left(\frac{d x}{d t}\right)_{r-4}=3 \times 4^{2}-24 \times 4+6=-42$
Hence (b) is the correct answer.
Question 9
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 10
4 / -1

If $\sin x+\operatorname{cosec} x=2,$ then $\sin ^{n} x+\operatorname{cosec}^{n} x$ is equal to

A
2

B
1

C
2^n-1

D
2^n-2

Solution

$\sin x+\operatorname{cosec} x=2 \Rightarrow(\sin x-1)^{2}=0 \Rightarrow \sin x=1$
$\therefore \sin ^{n} x+\operatorname{cosec}^{n} x=1+1=2$

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 22

Result

Mathematics Test - 22

Score

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Rank

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SHARING IS CARING

Question 1

One-one

Even

Odd

Both even and odd

Solution

Question 2

2

4

6

0

Solution

Question 3

A ∩ B = A

A ∩ B = B

A ∩ B = ∅

None of these

Solution

Question 4

1

2

3

0

Solution

Question 5

x = 1 - y

x2 = 1 - y

x2 = 1 + y

y2 = 1 - x

Solution

Question 6

3100

3300

2900

1400

Solution

Question 7

2

1

-1

does not exist

Solution

Question 8

42

-42

48

-48

Solution

Question 9

16/15

14/15

12/15

11/15

Solution

Question 10

2

1

2n-1

2n-2

Solution

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x² = 1 - y

x² = 1 + y

y² = 1 - x

2^n-1

2^n-2