Mathematics Test

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Mathematics Test - 25

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Weekly Quiz Competition

Question 1
4 / -1

The two curves \(x^{3}-3 x y^{2}+2=0\) and \(3 x^{2} y-y^{3}-2=0\)

A
Cut at right angles

B
Touch each other

C
Cut at an angle π/3

D
Cut at an angle π/4

Solution

\(x^{3}-3 x y^{2}+2=0 \ldots(1)\)
\(3 x^{2} y-y^{3}-2=0 \ldots(2)\)
On differentiating equations ( 1 ) and (2) w.r.t \(x\), we obtain \(\left(\frac{d y}{d x}\right)_{c 1}=\frac{x^{2}-y^{2}}{2 x y}\) and \(\left(\frac{d y}{d x}\right)_{c 2}=\frac{-2 x y}{x^{2}-y^{2}}\)
since \(m_{1} \cdot m_{2}=-1\). Therefore the two curves cut at right angles. Hence (a) is the correct answer.
Question 2
4 / -1

\(\lim _{x \rightarrow 0} \frac{\sin x-x+\frac{x^{3}}{6}}{x^{5}}\) is equal to

A
1/120

B
1/160

C
1/2

D
0

Solution

\(\lim _{x \rightarrow 0} \frac{\sin x-x+\frac{x^{3}}{6}}{x^{5}} \ldots \ldots\)
We know expansion of Sinx. \(\sin x=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\ldots \ldots . .\)
Substituting (2) in (1) we get
\(\lim _{x \rightarrow 0} \frac{x^{5}}{5 ! \cdot x^{5}}\)
\(=\frac{1}{5 !}\)
\(=\frac{1}{120}\)
Question 3
4 / -1

A relation from P to Q is

A
A universal set of P × Q

B
P × Q

C
An equivalent set of P × Q

D
A subset of P × Q

Solution

A relation from P to Q is a subset of P × Q.

Hence option D is correct.
Question 4
4 / -1

\(3 \cos ^{-1} x-\pi x-\frac{\pi}{2}=0\) has :

A
One solution

B
Infinite solutions

C
No solution

D
None of these

Solution

\(3 \cos ^{-1} x-\pi x+\frac{\pi}{2}\) Clearly graphs of \(y=3 \cos ^{-1} x\) and \(y=\pi x+\frac{\pi}{2}\) in the domain of \(\cos ^{-1} x\) i.e., in [-1,1] intersect only once, therefore there is only one solution of the given equation.
Question 5
4 / -1

Let \(f(x)=\frac{2 \sin ^{2} x-1}{\cos x}+\frac{\cos x(2 \sin x+1)}{1+\sin x}\) then \(\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x\) equals
(where \(c\) is the constant of integeration)

A
e^x tanx + c

B
e^x cotx + c

C
e^xcosec² x + c

D
None of these

Solution

\(\frac{\cos x(1+2 \sin x)}{1+\sin x}-\frac{\cos ^{2} x-\sin ^{2} x}{\cos x}\)
\(=\frac{\cos ^{2} x(1+2 \sin x)-(1+\sin x)\left(\cos ^{2} x-\sin ^{2} x\right)}{\cos x(1+\sin x)}=\frac{\sin x(1-\sin x)+\sin ^{2} x}{\cos x}=\tan x\)
\(=\frac{\sin x \cos ^{2} x+\sin ^{2} x(1+\sin x)}{\cos x(1+\sin x)}=\frac{x}{\sin x}\)

=e^x tanx + c
Question 6
4 / -1

If the sum of two of the roots of x³+ px² + qx + r = 0is zero, then pq =

A
- r

B
r

C
2 r

D
- 2 r

Solution

Given that, \(a+\beta=0\)
\(\alpha+\beta+y=-p \Rightarrow y=-p\)
Substituting \(\mathrm{Y}=-\mathrm{p}\) in the given equation
\(\Rightarrow-p^{3}+p^{3}-p q+r=0 \Rightarrow p q=r\)
Question 7
4 / -1

\(\int \frac{\cos ^{4} x d x}{\sin ^{3} x\left(\sin ^{5} x+\cos ^{5} x\right)^{\frac{3}{5}}}=-\frac{1}{2}\left(1+\cot ^{A} x\right)^{B}+C\) then \(\mathbf{A B}=\)

A
1

B
2

C
1/2

D
None of these

Solution

\(I=\int \frac{\cos x^{4} d x}{\sin ^{3} x\left(\sin ^{5} x+\cos ^{5} x\right)^{\frac{3}{5}}}\)
\(=\int \frac{\operatorname{cosec}^{2} x \cot ^{4} x d x}{\left(1+\cot ^{5} x\right)^{3 / 5}}\)
\(1+\cot ^{5} x=t\)
\(5 \cot ^{4} x\left(-\operatorname{cosec}^{2} x\right) d x=d t\)
\(=\frac{-1}{5} \int \frac{d t}{t^{3 / 5}}\)
\(=\frac{-t^{2 / 5}}{2}\)
\(=\frac{-1}{2}\left(1+\cot ^{5} x\right)^{2 / 5}+C\)
Question 8
4 / -1

\(\int \frac{(1+\sqrt{\tan x})\left(1+\tan ^{2} x\right)}{2 \tan x} d x\) equal \(\operatorname{to}\)

A
\(\log \tan ^{2} x+\sqrt{\tan x}+c\)

B
\(\log \tan ^{2} x+\frac{1}{2 \sqrt{\tan x}}+c\)

C
\(\log |\tan x|+2 \sqrt{\tan x}+c\)

D
\(\log |\tan x|+\sqrt{\tan x}+c\)

Solution

\(\int \frac{1}{2 \sin x \cos x} d x+\frac{1}{2} \int \frac{\sqrt{\tan x}}{\sin x \cos x} d x\)
\(=\frac{1}{2} \int \frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x} d x+\frac{1}{2} \int \frac{\sec ^{2} x}{\sqrt{\tan x}} d x=\log \left(\tan ^{2} x\right)+\sqrt{\tan x}+c\)
Question 9
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 10
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 \(-3,-10 .\)

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Mathematics Test - 25

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Mathematics Test - 25

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SHARING IS CARING

Question 1

Cut at right angles

Touch each other

Cut at an angle π/3

Cut at an angle π/4

Solution

Question 2

1/120

1/160

1/2

0

Solution

Question 3

A universal set of P × Q

P × Q

An equivalent set of P × Q

A subset of P × Q

Solution

Question 4

One solution

Infinite solutions

No solution

None of these

Solution

Question 5

ex tanx + c

ex cotx + c

ex cosec2 x + c

None of these

Solution

Question 6

- r

r

2 r

- 2 r

Solution

Question 7

1

2

1/2

None of these

Solution

Question 8

\(\log \tan ^{2} x+\sqrt{\tan x}+c\)

\(\log \tan ^{2} x+\frac{1}{2 \sqrt{\tan x}}+c\)

\(\log |\tan x|+2 \sqrt{\tan x}+c\)

\(\log |\tan x|+\sqrt{\tan x}+c\)

Solution

Question 9

10

3π - 10

π + 6

2π - 6

Solution

Question 10

3, 10

- 3, - 10

- 5, - 18

18,5

Solution

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e^x tanx + c

e^x cotx + c

e^xcosec² x + c