Mathematics Test

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Mathematics Test - 20

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Weekly Quiz Competition

Question 1
4 / -1

The number of ways of painting the faces of a cube with six different colours is

A
1

B
6

C
6!

D
none of these

Solution

Since, the number of faces is the same as the number of colours, and the faces are identical, therefore, the number of ways of painting them is 1.
Question 2
4 / -1

If the progressions 3, 10, 17, ....... and 63, 65, 67, ....... are such that their nth terms are equal, then n is equal to

A
13

B
15

C
19

D
18

Solution

n^th term of 1^st series =n^th term of 2^nd series

⇒ 3 + (n − 1) 7 = 63 + (n − 1) 2

⇒ (n − 1)5 = 60

⇒ n -1=12

⇒ n = 13
Question 3
4 / -1

\(\operatorname{lt}_{x \rightarrow 0}(1+2 x)^{\frac{x+3}{x}}\)is equal to

A
e³

B
e^3/2

C
e⁶

D
None of these

Solution

\(\operatorname{Lt}_{x \rightarrow 0}(1+2 x) \frac{x+3}{x}=\operatorname{Lt}_{x \rightarrow 0}(1+2 x)^{1}(1+2 x)^{3 / x}\)
\(\operatorname{lt}_{x \rightarrow 0}(1+2 x)^{3 / x}\)
\(=\operatorname{li}_{2 x \rightarrow 0}\left\{(1+2 x)^{1 / 2 x}\right\}^{6}=e^{6}\)
Question 4
4 / -1

What is the solution of the equation (y log y) dx + (x - logy) dy = 0?

A
x = C/log y + log y

B
y = C/log x + log x

C
(x + 1) log y = c

D
None of these

Solution

The given equation is \(\frac{d x}{d y}+\frac{1}{y \log y} x=\frac{1}{y},\) which is linear
\(\mid F=\int \frac{1}{y \log y} d y=e^{\log (\log y)}=\log y\)
Its solution is \(x \log y=\int \frac{1}{y} \cdot \log y d y\)
\(\Rightarrow x \log y=\frac{1}{2}(\log y)^{2}+c\)
\(\Rightarrow x=\frac{1}{2} \log y+\frac{c}{\log y}\)
Question 5
4 / -1

The value of tan (cos^-1(4/5) + tan^-1 (2/3)) is equal to

A
6/17

B
16/7

C
7/16

D
17/6

Solution

Put \(\cos ^{-1} \frac{4}{5}=A \Rightarrow \cos A=\frac{4}{5} \tan A=\frac{3}{4}\)
Put \(\tan ^{-1} \frac{2}{3}=B \tan B=\frac{2}{3}\)
Now, tan \(\left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)\)
\(=\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}=\frac{\frac{17}{12}}{\frac{6}{12}}=\frac{17}{6}\)
Question 6
4 / -1

If g [f(x)] = |sin x| and f[g(x)] = (sin√x)² then,

A
f(x) = sin² x, g(x) = √x

B
f(x) = sin x, g(x) = |x|

C
f(x) = x², g(x) = sin√x

D
f and g cannot be determined

Solution

Given \((g \circ f)(x)=|\sin x| \ldots \ldots .(1)\)
and \(f[g(x)]=(\sin \sqrt{x})^{2}\)
\((f \circ g)(x)=(\sin \sqrt{x})^{2}\)
For option\((A), f(x)=\sin ^{2} x, g(x)=\sqrt{x}\)
\((f \circ g)(x)=f[g(x)]=f(\sqrt{x})\)
\(\Rightarrow(f \circ g)(x)=f(y)=\sin ^{2} y=\sin ^{2}(\sqrt{x})\)
\(y=\sqrt{x}\)
\((g \circ f)(x)=g[f(x)]=g\left(\sin ^{2} x\right)=g(t) t-\sin ^{2} x\)
\(\Rightarrow \sqrt{t}=\sqrt{\sin ^{2} x}=|\sin x|\)
Question 7
4 / -1

If f(x) = x/1+x is defined as [0,∞) → [0, ∞), then f(x) is

∞

A
one-one and onto

B
one-one, but not onto

C
onto, but not one-one

D
neither one-one nor onto

Solution

Let \(x_{1}, x_{2} \in[0, \infty)\) such that \(f\left(x_{1}\right)=f\left(x_{2}\right)\)
\(\Rightarrow \frac{x_{1}}{1+x_{1}}=\frac{x_{2}}{1+x_{2}}\)
\(\Rightarrow 1+x_{1}=1+x_{2}\)
\(\Rightarrow x_{1}=x_{2}\)
So, \(f\) is one - one
Let \(y=\frac{x}{1+x} \Rightarrow x=\frac{y}{1-y}\)
If \(y=2\) then \(x=-2 \notin[0, \infty)\)
Hence f is not onto.
Question 8
4 / -1

If |x| > 1, then which one of the following is different from the other three?

A
√ x² - 1

B
sec(cosec^-1x)

C
cosec(sec^-1x)

D
|x | / √x² - 1

Solution

sec (cosec^-1x) = cosec (sec^-1x) = |x|/√x² -1.

So options (2) = (3) = (4); Hence option (1) is different from the other three.
Question 9
4 / -1

The solution of the boundary value problem 2y dy/dx + y² cot x = 2 cos x, y(π/2) = 1 is

A
y² = sin x

B
y = sin² x

C
y² = cos x + 1

D
None of these

Solution

Given equation \(2 y \frac{d y}{d x}+y^{2} \cot x=2 \cos x\)
Putting \(y^{2}=v\) and \(2 y \frac{d y}{d x}=\frac{d v}{d x}\). we get \(\frac{d v}{d x}+(\cot x) v=2 \cos x,\) which is linear
\(\mid F_{0}=\theta^{\int \cot x d x}-e^{\log \sin x}=\sin x\)
Its solution is \(v \sin x=\int 2 \sin x \cos x d x\)
\(\Rightarrow v \sin x=-\frac{1}{2} \cos 2 x+c\)
\(\Rightarrow 2 y^{2} \sin x+\cos 2 x=2 c\)
Putting \(x=\frac{\pi}{2}\) and \(y=1\) in (i), we get \(c=\frac{1}{2}\)
\(\Rightarrow 2 y^{2} \sin x+\cos 2 x=1\)
\(\Rightarrow y^{2}=\sin x\)
Question 10
4 / -1

The equation of the plane through the line of intersection of planes x + y + z + 3 = 0 and 2x − y + 3z + 1 =0 and parallel to the line x/1 = y/2 =z/3 is

A
x − 5y + 3z = 7

B
x − 5y + 3z = − 7

C
x + 5y + 3z = 7

D
none of these

Solution

Any plane through the given line is 2x − y + 3z + 1 + λ (x + y + z + 3) = 0 If this plane is parallel to the line x/1 = y/2 = z/3, then the normal to the plane is also perpendicular to the above line.

∴(l₁l₂ + m₁ m₂ + n₁ n₂ = 0

⇒ (2 + λ) 1 + (λ −1) 2 + (3 + λ) 3 = 0

⇒ λ= -3/2

and the required plane is

2x − y + 3z + 1 + (-3/2) (x + y + z + 3) = 0

⇒ x − 5y + 3z − 7 = 0

or

x − 5y + 3z = 7

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 20

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Mathematics Test - 20

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SHARING IS CARING

Question 1

1

6

6!

none of these

Solution

Question 2

13

15

19

18

Solution

Question 3

e3

e3/2

e6

None of these

Solution

Question 4

x = C/log y + log y

y = C/log x + log x

(x + 1) log y = c

None of these

Solution

Question 5

6/17

16/7

7/16

17/6

Solution

Question 6

f(x) = sin2 x, g(x) = √x

f(x) = sin x, g(x) = |x|

f(x) = x2, g(x) = sin√x

f and g cannot be determined

Solution

Question 7

one-one and onto

one-one, but not onto

onto, but not one-one

neither one-one nor onto

Solution

Question 8

√ x2 - 1

sec(cosec-1x)

cosec(sec-1x)

|x | / √x2 - 1

Solution

Question 9

y2 = sin x

y = sin2 x

y2 = cos x + 1

None of these

Solution

Question 10

x − 5y + 3z = 7

x − 5y + 3z = − 7

x + 5y + 3z = 7

none of these

Solution

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e³

e^3/2

e⁶

f(x) = sin² x, g(x) = √x

f(x) = x², g(x) = sin√x

√ x² - 1

sec(cosec^-1x)

cosec(sec^-1x)

|x | / √x² - 1

y² = sin x

y = sin² x

y² = cos x + 1