Mathematics Test

Result

Mathematics Test - 22

Score
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Rank
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Weekly Quiz Competition

Question 1
1 / -0

Let $f(x)$ be a continuous and differentiable function satisfying $f(x+y)=f x f(y)$ for all $x, y \in R$. If $f(x)$ can be expressed as $f(x)=1+x p(x)+x^{2} q(x)$ where $\lim x \rightarrow 0 p x=a$ and $\lim x \rightarrow 0 q x=b,$ then $b$ is

A

B

C

D

Solution
Question 2
1 / -0

A
5/3

B
7/3

C
2/3

D
4/3

Solution
Question 3
1 / -0

The plane $x+2 y-z=4$ cuts the sphere $x^{2}+y^{2}+z^{2}-x+z-2=0$ in a circle of radius

A
$2$

B
$\sqrt{2}$

C
$3$

D
$1$

Solution
Question 4
1 / -0

Forces acting on a particle have magnitude 5,3 and 1 unit and act in the direction of the vectors $6 \hat{i}+2 \hat{j}+3 \hat{k}, 3 \hat{i}-2 \hat{j}+6 \hat{k}, 3 \hat{i}-2 \hat{j}+6 \hat{k}$ and $2 \hat{i}-3 \hat{j}-6 \hat{k}$ respectively. They remain constant while theparticle is displaced from the points $A(2,-1,-3)$ to $B(5,-1,1)$. The work done is

A
11 unit

B
33 unit

C
12 unit

D
30 unit

Solution
Question 5
1 / -0

The value of $\sin \left[2 \cos ^{-1} \frac{\sqrt{5}}{3}\right]$ is

A
$\frac{\sqrt{5}}{3}$

B
$\frac{2 \sqrt{5}}{3}$

C
$\frac{4 \sqrt{5}}{9}$

D
$\frac{2 \sqrt{5}}{9}$

Solution

$\sin \left[2 \cos ^{-1} \frac{\sqrt{5}}{3}\right]=\sin \left[\cos ^{-1}\left\{2 \cdot\left(\frac{\sqrt{5}}{3}\right)^{2}-1\right\}\right]$
$\left[\because 2 \cos ^{-1} x=\cos ^{-1}\left(2 x^{2}-1\right)\right]$
$\frac{4 \sqrt{5}}{9}$
Question 6
1 / -0

A
$0$

B
$8$

C
$- 8$

D
None of these

Solution
Question 7
1 / -0

The equation of the sphere concentric with the sphere $2 x^{2}+2 y^{2}+2 z^{2}-6 x+2 y-4 z=1$ and double its radius is

A

B

C

D

Solution
Question 8
1 / -0

A
0

B

C
1

D

Solution
Question 9
1 / -0

Find the first derivative of $\frac{2}{x+1}-\frac{x^{2}}{3 x-1}$

A

B

C

D

Solution
Question 10
1 / -0

The argument of the complex number$\frac{13-5 i}{4-9 i}$ is

A
$π / 3$

B
$π / 4$

C
$π / 5$

D
$π / 6$

Solution

Let $z=\frac{13-5 i}{4-9 i} \times \frac{4+9 i}{4+9 i}=\frac{97+97 i}{97}=1+i$
$\therefore \quad \arg (z)=\tan ^{-1}\left(\frac{1}{1}\right)=\frac{\pi}{4}$