Mathematics Test

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Mathematics Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

The interval on which f(x)=1−x√2 is continuous is:

A
(0,∞)

B
(1,∞)

C
[−1,1]

D
(−∞,−1)

Solution

The function will be continuous on an interval where it is completely defined.
Since, we know, a negative quantity cannot go inside the square root sign,
Hence, 1−x²>=0
Or,
x²<=1
Hence, the function is continuous for all x lying between [−1,1]
Hence, the correct option is (C)
Question 2
1 / -0

The value of \(\lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}\) is equal to

A
1/5

B
186

C
1/4

D
1/2

Solution

\(\lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}\)
\(=\lim _{x \rightarrow 0} \frac{2 \sin \left(\frac{x+\sin x}{2}\right) \sin \left(\frac{x-\sin x}{2}\right)}{x^{4}}\)
As \(x \rightarrow 0 \Rightarrow \sin x \rightarrow 0\)
\(=\lim _{x \rightarrow 0}\left(\frac{x+\sin x}{2}\right)\left(\frac{x-\sin x}{2}\right) \frac{2 \sin \left(\frac{x+\sin x}{2}\right) \sin \left(\frac{x-\sin x}{2}\right)}{x^{4}\left(\frac{x+\sin x}{2}\right)\left(\frac{x-\sin x}{2}\right)}\)
\(=\lim _{x \rightarrow 0} \frac{x^{2}-\sin ^{2} x}{2 x^{4}}\)
It is of the form \(\frac{0}{0},\) so applying L-Hospitals' rule
\(=\lim _{x \rightarrow 0} \frac{2 x-2 \sin x \cos x}{8 x^{3}}\)
\(=\lim _{x \rightarrow 0} \frac{2 x-\sin 2 x}{8 x^{3}}\)
Again applying L-Hospital's rule \(=\lim _{x \rightarrow 0} \frac{2-2 \cos 2 x}{24 x^{2}}\)
Again applying L-Hospital's rule \(=\lim _{x \rightarrow 0} \frac{4 \sin 2 x}{48 x}\)
Again applying L-Hospital's rule \(=\lim _{x \rightarrow 0} \frac{8 \cos 2 x}{48}\)
\(=\frac{1}{6}\)
Hence, the correct option is (B)
Question 3
1 / -0

If \(A=\tan ^{-1}\left(\frac{x \sqrt{3}}{2 K-x}\right)\) and \(B=\tan ^{-1}\left(\frac{2 x-K}{K \sqrt{3}}\right),\) then the value of \(A-B\) is

A
0^o

B
45^o

C
60^o

D
30^o

Solution

In the above question, if we carefully observe, we can see some symmetry in A and
B. Hence we substitute \(k=x\) Therefore \(A=\tan ^{-1}(\sqrt{3})=60^{\circ}\)
\(B=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=30^{\circ}\)
Hence \(A-B\) \(=60^{\circ}-30^{\circ}\)
\(=30^{\circ}\)
Hence, the correct option is (D)
Question 4
1 / -0

The value of \(\lim _{\alpha \rightarrow \beta}\left[\frac{\sin ^{2} \alpha-\sin ^{2} \beta}{\alpha^{2}-\beta^{2}}\right]\) is:

A
0

B
1

C
sinβ/β

D
sin2β/2β

Solution

\(\lim _{\alpha \rightarrow \beta}\left[\frac{\sin ^{2} \alpha-\sin ^{2} \beta}{\alpha^{2}-\beta^{2}}\right]=\lim _{\alpha \rightarrow \beta}\left[\frac{\sin (\alpha-\beta) \sin (\alpha+\beta)}{(\alpha-\beta)(\alpha+\beta)}\right]\)
\(=\lim _{\alpha \rightarrow \beta}\left[\frac{\sin (\alpha-\beta)}{(\alpha-\beta)}\right] \times \lim _{\alpha \rightarrow \beta}\left[\frac{\sin (\alpha+\beta)}{(\alpha+\beta)}\right]\)
\(=\lim _{\alpha-\beta \rightarrow 0}\left[\frac{\sin (\alpha-\beta)}{(\alpha-\beta)}\right] \cdot\left(\frac{\sin 2 \beta}{2 \beta}\right) \ldots \ldots[\because \alpha \rightarrow \beta \Longrightarrow \alpha-\beta \rightarrow 0]\)
\(=1 . \frac{\sin 2 \beta}{2 \beta} \ldots .\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]\)
\(=\frac{\sin 2 \beta}{2 \beta}\)
Hence, the correct option is (D)
Question 5
1 / -0

\(\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^{3}}=\)

A
1

B
1/2

C
1/4

D
-1/3

Solution

\(=\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^{3}}\)
\(=\lim _{x \rightarrow 0} \frac{\sin x(1-\cos x)}{x^{3} \cos x}\)
\(=\lim _{x \rightarrow 0} \frac{\sin x\left(2 \sin ^{2} \frac{x}{2}\right)}{x^{3} \cos x}\)
\(=\frac{2}{4}=\frac{1}{2}\)
Hence, the correct option is (B)
Question 6
1 / -0

\(\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}}=\)

A
2

B
-2

C
1/2

D
-1/2

Solution

\(\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}}\)
\(=\lim _{x \rightarrow 0} \frac{x \frac{2 \tan x}{1-\tan ^{2} x}-2 x \tan x}{\left(1-\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\right)^{2}}\)
\(=\lim _{x \rightarrow 0} \frac{2 x \tan ^{3} x}{4 \tan ^{4} x}\)
\(=\lim _{x \rightarrow 0} \frac{2 x}{4 \tan x}\)
\(=\frac{1}{2}\)
Hence, the correct option is (B)
Question 7
1 / -0

\(\lim _{x \rightarrow 0} \frac{1-\cos 2 x+\tan ^{2} x}{x \sin x}\)

A
0

B
1

C
(π/180)²

D
3

Solution

\(\operatorname{lt} \frac{1-\cos ^{2} x+\tan ^{2} x}{x \sin x}\)
\(=\lim _{x \rightarrow 0} \frac{2 \sin ^{2} x+\tan ^{2} x}{x \sin x}\)
\(=\lim _{x \rightarrow 0} \frac{\sin ^{2} x\left(2+\frac{1}{\cos ^{2} x}\right)}{x \sin x}\)
\(=\operatorname{lt} \frac{\sin x}{x \rightarrow 0} \cdot \operatorname{lt}_{x \rightarrow 0}\left(2+\sec ^{2} x\right)\)
\(=1\left(2+\sec ^{2} 0\right)=2+1=3\)
Hence, the correct option is (D)
Question 8
1 / -0

\(\lim _{x \rightarrow 0} \frac{\sqrt{\cos x}-\sqrt[3]{\cos x}}{\sin ^{2} x}=\)

A
1/3

B
-1/3

C
1/12

D
-1/12

Solution

\(\lim _{x \rightarrow 0} \frac{\sqrt{\cos x}-\sqrt[3]{\cos x}}{\sin ^{2} x}\)
It is of the form \(\frac{0}{0},\) so using L-Hospital's rule, \(=\lim _{x \rightarrow 0} \frac{\frac{-\sin x}{2 \sqrt{\cos x}}-\frac{-\sin x \cos \frac{-2}{3} x}{3}}{2 \sin x \cos x}\)
\(=-\frac{1}{12}\)
Hence, the correct option is (D)
Question 9
1 / -0

If the function defined by \(f(x)=\frac{\sin 3(x-p)}{\sin 2(x-p)}\) for \(x \neq p\) is continuous at \(x=p\) then \(f(p)=\)

A
3/2

B
2/3

C
6

D
1/6

Solution

since, it is given the function is continuous at \(x=p,\) we calculate \(\lim _{x \rightarrow p} \frac{\sin 3(x-p)}{\sin 2(x-p)}\)
\(=\lim _{x \rightarrow p} \frac{\sin 3(x-p)}{3(x-p)} \times \frac{2(x-p)}{\sin 2(x-p)} \times \frac{3}{2}\)
Using the property of limits, \(f(p)=1 \times 1 \times \frac{3}{2}\)
Hence, the correct option is (A)
Question 10
1 / -0

If \(f(x)=x^{2}-3 x+5\), then \(\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}\)

A
0

B
1

C
1/2

D
4

Solution

The above given expression, \(\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}\) is equivalent to \(f^{\prime}(x)\) at \(x=2\)
\(f(x)=x^{2}-3 x+5\)
\(\Rightarrow f^{\prime}(x)=2 x-3\)
\(\Rightarrow f^{\prime}(2)=2(2)-3\)
\(=1\)
Hence, the correct option is (B)

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Mathematics Test - 31

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Mathematics Test - 31

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SHARING IS CARING

Question 1

(0,∞)

(1,∞)

[−1,1]

(−∞,−1)

Solution

Question 2

1/5

186

1/4

1/2

Solution

Question 3

0o

45o

60o

30o

Solution

Question 4

0

1

sinβ/β

sin2β/2β

Solution

Question 5

1

1/2

1/4

-1/3

Solution

Question 6

2

-2

1/2

-1/2

Solution

Question 7

0

1

(π/180)2

3

Solution

Question 8

1/3

-1/3

1/12

-1/12

Solution

Question 9

3/2

2/3

6

1/6

Solution

Question 10

0

1

1/2

4

Solution

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My Perfomance

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0^o

45^o

60^o

30^o

(π/180)²