Mathematics Test

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Mathematics Test - 8

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Weekly Quiz Competition

Question 1
4 / -1

Let f(x) = [x], then f′(1) = ?

A
1

B
0

C
Does not exist

D
None of these

Solution

\(R f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{[1+h]-[1]}{h}=\lim _{h \rightarrow 0} \frac{(1-1)}{h}=0\)
\(L f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}=\lim _{h \rightarrow 0} \frac{[1-h]-[1]}{-h}\)
\(=\lim _{h \rightarrow 0} \frac{0-1}{-h}=\infty\)
\(R f^{\prime}(1)^{1} L f^{\prime}(1)\)
So. \(f^{\prime}(1)\) does not exist.
Question 2
4 / -1

If \(\theta\) be an acute angle and \(7 \sin ^{2} \theta+3 \cos ^{2} \theta=4\), then the value of \(\tan \theta\) is:

A
\(\sqrt{3}\)

B
\(\frac{1}{ \sqrt{3}}\)

C
\(1\)

D
\(0\)

Solution

Given, \(7 \sin ^{2} \theta+3 \cos ^{2} \theta=4\)

\(\Rightarrow 7 \frac{\sin ^{2} \theta}{\cos ^{2} \theta}+3=\frac{4}{\cos ^{2} \theta}=4 \sec ^{2} \theta\)

\(\Rightarrow 7 \tan ^{2} \theta+3=4\left(1+\tan ^{2} \theta\right)\)

\(\Rightarrow 7 \tan ^{2} \theta-4 \tan ^{2} \theta=4-3\)

\(\Rightarrow 3 \tan ^{2} \theta=1\)

\(\Rightarrow \tan ^{2} \theta=\frac{1}{3}\)

\(\Rightarrow \tan \theta=\frac{1}{ \sqrt{3}}\)
Question 3
4 / -1

Period of cot 3 x - cos (4x + 3) is

A
π/3

B
π/4

C
π

D
π/2

Solution

Period of cot 3x is π/3 and period of cos (4x + 3) is π/2

⇒ L.C.M. is π, hence the answer.
Question 4
4 / -1

\(\ln \Delta A B C, \frac{b-c \cos A}{c-b \cos A}=\)

A
sin B/sin C

B
cos C/cos B

C
cos B/cos C

D
None of these

Solution

\(\frac{b-c \cos A}{c-b \cos A}=\frac{b-\frac{b^{2}+c^{2}-a^{2}}{2 b}}{c-\frac{b^{2}+c^{2}-a^{2}}{2 c}}=\left(\frac{b^{2}+a^{2}-c^{2}}{c^{2}+a^{2}-b^{2}}\right) \frac{c}{b}\)
\(=\frac{b^{2}+a^{2}-c^{2}}{2 a b} \cdot \frac{2 a c}{c^{2}+a^{2}-b^{2}}=\frac{\cos C}{\cos B}\)
Question 5
4 / -1

The area of the figure bounded by the curves y = | x - 1 | and y = 3 - | x | is

A
2

B
3

C
4

D
None of these

Solution

Required area \(=\int_{-1}^{0}\{(3+x)-(-x+1)\}\)
\(\left.+\int_{0}^{1}\{(3-x)\}-(-x+1)\right] d x+\int\{(3-x)-(x-1)\} d x=\)
\(\int_{-1}^{2}(2+2 x) d x+\int_{0}^{1} 2 d x+\int_{0}^{2}(4-2 x) d x\)
\(=\left[2 x+x^{2}\right]_{-1}^{0}+[2 x]_{0}^{1}+\left[4 x-x^{2}\right]_{1}^{2}\)
\(-0-(-2+1)+(2-0)+(8-4)-(4-1)=1+2+4-3=4\)
Question 6
4 / -1

The area bounded by the curve y = 2x - x²and the straight line y = - x is given by

A
9/2

B
43/6

C
35/6

D
None of these

Solution

Required area \(=\int_{0}^{\left(y_{1}-y_{2}\right)} d x\)
\(=\int_{0}^{3}\left(2 x-x^{2}\right)-(-x) d x=\int_{0}^{3}\left(3 x-x^{2}\right) d x\)
\(=\left[\frac{3 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{3}=\frac{27}{2}-9=\frac{9}{2}\)
Question 7
4 / -1

The area under the curve y = sin 2x + cos 2x between x = 0 and x = π/4 is

A
2 sq. units

B
1 sq. unit

C
3 sq. units

D
4 sq. units

Solution

Required area
\(=\int_{0}^{\pi / 4}(\sin 2 x+\cos 2 x) d x\)
\(=\left[\frac{-\cos 2 x}{2}+\frac{\sin 2 x}{2}\right]_{0}^{\pi / 4}\)
\(=\frac{1}{2}\left[-\cos \frac{\pi}{2}+\sin \frac{\pi}{2}+\cos 0-\sin 0\right]\)
\(=1\) sq. unit
Question 8
4 / -1

\(\int_{0}^{\pi} \sqrt{1-\cos x} d x=\)

A
2

B
1

C
√2

D
2√2

Solution

\(1=\int_{0}^{\pi} \sqrt{2 \sin ^{2}(x / 2)} d x=\sqrt{2} \int_{0}^{\pi} \sin (x / 2) d x=\sqrt{2} \cdot\left[\frac{-\cos (x / 2)}{(1 / 2)}\right]_{0}^{\pi}\)
\(=2 \sqrt{2}\left[-\cos \frac{x}{2}+\cos 0\right]=2 \sqrt{2}\)
Question 9
4 / -1

If \(y=\tan ^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right),\) then \(\frac{d y}{d x}=?\)

A
0

B
1 +π /4

C
1

D
3/2

Solution

\(y=\tan ^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)\)
\(=\tan ^{-1}\left(\frac{\tan x+1}{1-\tan x}\right)=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+x\right)\right]=\frac{\pi}{4}+x\)
\(\therefore \frac{d y}{d x}=1\)
Question 10
4 / -1

The angles of a triangle are (x + 5)°, (2x – 3)° and (3x + 4)°. The value of x is:

A
28

B
29

C
31

D
25

Solution

Sum of angles of a triangle = 180°

∴ x + 5 + 2x – 3 + 3x + 4 = 180°

⇒ 6x + 6 = 180°

⇒ 6x = 174°

⇒ x = 29

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Mathematics Test - 8

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Mathematics Test - 8

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SHARING IS CARING

Question 1

1

0

Does not exist

None of these

Solution

Question 2

\(\sqrt{3}\)

\(\frac{1}{ \sqrt{3}}\)

\(1\)

\(0\)

Solution

Question 3

π/3

π/4

π

π/2

Solution

Question 4

sin B/sin C

cos C/cos B

cos B/cos C

None of these

Solution

Question 5

2

3

4

None of these

Solution

Question 6

9/2

43/6

35/6

None of these

Solution

Question 7

2 sq. units

1 sq. unit

3 sq. units

4 sq. units

Solution

Question 8

2

1

√2

2√2

Solution

Question 9

0

1 +π /4

1

3/2

Solution

Question 10

28

29

31

25

Solution

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