Mathematics Test

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Mathematics Test - 6

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Weekly Quiz Competition

Question 1
3 / -1

The derivative of sin^-1 (2x² - 1) w.r.t sin^-1 x is

A
$\sqrt{2}$

B
2

C
$2 \sqrt{2}$

D
1

Solution

Let , y = sin^-1 (2x² - 1) and z = sin^-1 x
Differentiate both the function w.r.t to x
$\Rightarrow \frac{d y}{d x}=\frac{d \sin ^{-1}\left(2 x^{2}-1\right)}{d x}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{1-\left(2 \mathrm{x}^{2}-1\right)^{2}}} \times(4 \mathrm{x}-0)$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{\sqrt{1-\mathrm{x}^{2}}}$ .....(i)
Similarly,
$\frac{\mathrm{d} z}{\mathrm{dx}}=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}$ .....(ii)
We know that, $\frac{\mathrm{dy}}{\mathrm{d} z}=\frac{\frac{\mathrm{dy}}{\mathrm{d} x}}{\frac{\mathrm{d} \mathrm{z}}{\mathrm{dx}}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dz}}=\frac{\frac{2}{\sqrt{1-\mathrm{x}^{2}}}}{\frac{1}{\sqrt{1-\mathrm{x}^{2}}}}=2$
Question 2
3 / -1

Consider the following statements in respect of the differential equation
$\frac{d^{2} y}{d x^{2}} + \cos (\frac{d y}{d x}) = 0$
1. The degree of the differential equation is not defined.
2. The order of the differential equation is 2
Which of the above statements is correct?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

$\frac{d^{2} y}{d x^{2}} + \cos (\frac{d y}{d x}) = 0$
1. As the given differential equation is not a polynomial equation in derivatives, the degree of this equation is not defined
So, this statement is true
2. Here Highest derivative: $\frac{d^{2} y}{d x^{2}}$
∴ Order = 2
Question 3
3 / -1

In the following trigonometry expression, find the value of (MN).
$\frac{(1+\cos x)}{(1-\sin x)}(\sin x+\cos x-1)^{2}=M \sin ^{N} x$

A
$3$

B
$6$

C
$4$

D
$2$

Solution

Given,
$\frac{(1+\cos x)}{(1-\sin x)}(\sin x+\cos x-1)^{2}=M \sin ^{N} x$
$\Rightarrow \frac{(1+\cos x)}{(1-\sin x)}(\sin x+\cos x-1)^{2}=M \sin ^{N} x$
$\Rightarrow \frac{(1+\cos x)}{(1-\sin x)} \times 2(1-\sin x)(1-\cos x)=M \sin ^{N} x$
$\Rightarrow 2(1+\cos x)(1-\cos x)=\mathrm{M} \sin ^{N} x$
$\Rightarrow 2\left(1-\cos ^{2} x\right)=M \sin ^{N} x$
$\Rightarrow 2 \sin ^{2} x=M \sin ^{N} x$
$\text { Comparing both sides, we have- }$
$\Rightarrow M N=(2 \times 2)=4$
Question 4
3 / -1

Find the equation of line through (-4,1,3) and parallel to the plane $x+y+z=3$ while the line intersects another line whose equation is $x+y-z=0=x+2 y-3 z+5$:

A
$\frac{x+4}{-3}=\frac{y-1}{-2}=\frac{z-3}{1}$

B
$\frac{x+4}{1}=\frac{y-1}{2}=\frac{z-3}{1}$

C
$\frac{x+4}{-3}=\frac{y-1}{2}=\frac{z-3}{1}$

D
$\frac{x+4}{-1}=\frac{y-1}{2}=\frac{z-3}{-3}$

Solution

Family of planes containing the line of intersection of planes is, ${A}+\lambda {B}=0$
Where $A=(x+y-z)$ and $B=(x+2 y-3 z+5)$
That is,
$(x+y-z)+\lambda(x+2 y-3 z+5)=0$
Above line passing through the point $(-4,1,3),$ so satisfy the equation.
On substituting the above equation, we get,
$\lambda=-1$
$\Rightarrow$ Equation of plane is $y-2 z+5=0$
Required line is lie in this plane and is parallel to $x+y+z=5$.
Now.
Direction of required line, $=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 1 & -2\end{array}\right|=-3 \hat{i}+2 \hat{j}+\hat{k}$
Required line is, $\frac{x+4}{-3}=\frac{y-1}{2}=\frac{z-3}{1}$
Question 5
3 / -1

Evaluate: $\lim _{\mathrm{x} \rightarrow \infty} \mathrm{x}^{\frac{2}{\mathrm{x}}}=?$

A
2

B
∞

C
1

D
0

Solution

Let$\lim _{\mathrm{x} \rightarrow \infty} \mathrm{x}^{\frac{2}{\mathrm{x}}}$
By taking log on both sides we get.
$\Rightarrow \log y=\lim _{x \rightarrow \infty} \log \left(x^{\frac{2}{\mathrm{x}}}\right)$
$\Rightarrow \log y=\lim _{x \rightarrow \infty} \frac{2 \log x}{x}=\frac{\infty}{\infty}$
So, by using L'Hospital's rule i.e. Differentiating numerator and denominator w.r.t. x.
$\Rightarrow \log y=\lim _{x \rightarrow \infty} \frac{\frac{2}{\mathrm{x}}}{1}=0$
⇒ log y = 0
⇒ y = e$^{\circ}$ = 1
Question 6
3 / -1

A line with direction cosines proportional to $(2,1,2)$ meets each of the line ${x}={y}+{a}={z}$ and ${x}+{a}=2 {y}=2 {z}$. The co-ordinates of each of the points of intersection are given by:

A

$(3 a, 3 a, 3 a),(a, a, a)$

B

$(3 a, 2 a, 3 a),(a, a, a)$

C

$(3 a, 2 a, 3 a),(a, a, 2 a)$

D

$(2 {a}, 3 {a}, 3 {a}),(2 {a}, {a}, {a})$

Solution

Let the equation of line ${AB}$ is $\frac{{x}-0}{1}=\frac{{y}+{a}}{1}=\frac{{z}-0}{1}={k}$ (let)
Therefore coordinate of ${E}$ is $({k}, {k}-{a}, {k})$
Also the equation of other line ${CD}$ is $\frac{{x}+{a}}{2}=\frac{{y}-0}{1}=\frac{{z}-0}{1}=\lambda$ (let)
Therefore coordinate of ${F}$ is $(2 \lambda-{a}, \lambda, \lambda)$.
Direction ratio of ${EF}$ are $({k}-2 \lambda+{a}),({k}-\lambda-{a}),({k}-\lambda)$
$\therefore \frac{{k}-2 \lambda+{a}}{2}=\frac{{k}-\lambda-{a}}{1}=\frac{{k}-\lambda}{2}$
On solving first and second fraction, we get,
$\frac{k-2 \lambda+a}{2}=\frac{k-\lambda-a}{1}$
$k-2 \lambda+a=2 k-2 \lambda-2 a$
${k}=3 {a}$
On solving second and third fraction, we get,
$\frac{{k}-\lambda-{a}}{1}=\frac{{k}-\lambda}{2}$
$2 {k}-2 \lambda-2 {a}={k}-\lambda$
$k-\lambda=2 a$
$\lambda={k}-2 {a}=3 {a}-2 {a}$
$\lambda=a$
Therefore coordinate of ${E}=(3 {a}, 2 {a}, 3 {a})$ and ${F}=({a}, {a}, {a})$
Question 7
3 / -1

The slope of the tangent at $(x, y)$ to a curve passing through $\left[1,\left(\frac{\pi}{4}\right)\right]$ is given by $\left(\frac{y}{x}\right)-\cos ^{2}\left(\frac{y}{x}\right)$, then the equation of the curve is:

A
$y=\tan ^{-1}\left[\log \left(\frac{e}{x}\right)\right]$

B
$y=x \cdot \tan ^{-1}\left[\log \left(\frac{e}{x}\right)\right]$

C
$y=x \tan ^{-1}\left(\frac{x}{e}\right)$

D
None of these

Solution

Given:
$\left(\frac{d y}{d x}\right)=\left(\frac{y}{x}\right)-\cos ^{2}\left(\frac{y}{x}\right)$
$\text { Let }\left(\frac{y}{x}\right)=t $
$\Rightarrow\left(\frac{d y}{d x}\right)=x\left(\frac{d t}{d x}\right)+t $
$\Rightarrow x\left(\frac{d t}{d x}\right)+t=t-\cos ^{2} t $
$\Rightarrow x\left(\frac{d t}{d x}\right)+\cos ^{2} t=0 $
$\Rightarrow\left[\frac{(x-d t)}{\left(d x-\cos ^{2} t\right)}\right]+1=0 $
$\Rightarrow [\frac{(d t)}{(\cos ^{2} t)}]+(\frac{d x}{x})$
On integrating,
$\Rightarrow \int \sec ^{2} t d t+\int\left(\frac{d x}{x}\right)=0 $
$\Rightarrow \tan \left(\frac{y}{x}\right)+\log x=c $
$\Rightarrow \tan \left(\frac{y}{x}\right)+\log x=c$
when $x=1, y=\left(\frac{\pi}{4}\right)$ hence
$\Rightarrow \tan \left(\frac{\pi}{4}\right)=c \text { i.e. } c=1 $
$\Rightarrow \tan \left(\frac{y}{x}\right)+\log x=\log e $
$\Rightarrow \tan \left(\frac{y}{x}\right)+\log \left(\frac{e}{x}\right)$
$\therefore$ The equation of the curve is $y=x \tan ^{-1}\left[\log \left(\frac{e}{x}\right)\right]$
Question 8
3 / -1

The value of $[\vec{a} \vec{b}+\vec{c} \vec{a}+\vec{b}+\vec{c}]$ is:

A
$0$

B
$[\vec{a} \vec{b} \vec{c}]$

C
$2[\vec{a} \vec{b} \vec{c}]$

D
$3[\vec{a} \vec{b} \vec{c}]$

Solution

If $\vec{a}, \vec{b}, \vec{c}$ are any non-zero vector then
$[\vec{a} \vec{b} \vec{c}]=\vec{a} \cdot(\vec{b} \times \vec{c})$
Given,
$[\vec{a} \vec{b}+\vec{c} \vec{a}+\vec{b}+\vec{c}]$
$=\vec{a} \cdot[(\vec{b}+\vec{c}) \times(\vec{a}+\vec{b}+\vec{c})] $
$=\vec{a} \cdot[\vec{b} \times \vec{a}+\vec{b} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{c} \times \vec{b}+\vec{c} \times \vec{c}] $
$=\vec{a} \cdot[\vec{b} \times \vec{a}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{c} \times \vec{b}] $
$=[\vec{a} \vec{b} \vec{a}]+[\vec{a} \vec{b} \vec{c}]+[\vec{a} \vec{c}]+[\vec{a} \vec{c}] \quad(\text { Since }[\vec{a} \vec{b} \vec{a}]=0) $
$=[\vec{a} \vec{b} \vec{c}]-[\vec{a} \vec{b} \vec{c}] $
$=0$
Question 9
3 / -1

What should be the value of k such that the function $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ccc}\frac{k \sin (\pi-x)}{\pi-x} & \text { if } & x \neq \pi \\ 1 & \text { if } & x=\pi\end{array}\right.$ is continuous at $x=\pi$.

A
$\pi$

B
$1$

C
$-1$

D
$0$

Solution

Given,
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cl}\frac{k \sin (\pi-x)}{\pi-x} & \text { if } x \neq \pi \\ 1 & \text { if } x=\pi\end{array}\right.$ is continuous at $x=\pi .$
As we know that, if a function $\mathrm{f}$ is continuous at point say a then $\lim _{x \rightarrow a} f(x)=l=f(a)$
$\Rightarrow \lim _{x \rightarrow \pi} \frac{\mathrm{ksin}(\pi-x)}{\pi-x}=\mathrm{f}(\pi)=1$
$\lim _{x \rightarrow \pi} \frac{\mathrm{k} \sin (\pi-x)}{\pi-x}=\frac{0}{0},$ we can use L'Hospitals rule.
$\Rightarrow \lim _{x \rightarrow \pi} \frac{-\mathrm{k} \cos (\pi-x)}{-1}=1$
$\Rightarrow \frac{\mathrm{k} \cos (\pi-\pi)}{1}=1$
$\Rightarrow k \cos (0)=1$
$\Rightarrow k=1$
Question 10
3 / -1

Find the real and imaginary part of the complex number $z=\frac{1-i}{1+i}$.

A
$1,1$

B
$-1,1$

C
$0,1$

D
$0,-1$

Solution

Given,
Complex number $z=\frac{1-i}{1+i}$
Multiplying by $(1-i)$ in the numerator and denominator, we get $z=\frac{1-i}{1+i} \times \frac{1-i}{1-i}$
$\Rightarrow z=\frac{(1-i)(1-i)}{1-i^{2}}$
$\Rightarrow z=\frac{1+i^{2}-2 i}{1+1} \quad\left[\because i^{2}=-1\right]$
$\Rightarrow z=\frac{1-1-2 i}{2}$
$\Rightarrow z=\frac{0-2 i}{2}$
$\Rightarrow z=0-1 i$
$\therefore \operatorname{Re}(\mathrm{z})=0$ and $\operatorname{lm}(\mathrm{z})=-1$

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 6

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Mathematics Test - 6

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SHARING IS CARING

Question 1

\(\sqrt{2}\)

2

\(2 \sqrt{2}\)

1

Solution

Question 2

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 3

\(3\)

\(6\)

\(4\)

\(2\)

Solution

Question 4

\(\frac{x+4}{-3}=\frac{y-1}{-2}=\frac{z-3}{1}\)

\(\frac{x+4}{1}=\frac{y-1}{2}=\frac{z-3}{1}\)

\(\frac{x+4}{-3}=\frac{y-1}{2}=\frac{z-3}{1}\)

\(\frac{x+4}{-1}=\frac{y-1}{2}=\frac{z-3}{-3}\)

Solution

Question 5

2

∞

1

0

Solution

Question 6

\((3 a, 3 a, 3 a),(a, a, a)\)

\((3 a, 2 a, 3 a),(a, a, a)\)

\((3 a, 2 a, 3 a),(a, a, 2 a)\)

\((2 {a}, 3 {a}, 3 {a}),(2 {a}, {a}, {a})\)

Solution

Question 7

\(y=\tan ^{-1}\left[\log \left(\frac{e}{x}\right)\right]\)

\(y=x \cdot \tan ^{-1}\left[\log \left(\frac{e}{x}\right)\right]\)

\(y=x \tan ^{-1}\left(\frac{x}{e}\right)\)

None of these

Solution

Question 8

\(0\)

\([\vec{a} \vec{b} \vec{c}]\)

\(2[\vec{a} \vec{b} \vec{c}]\)

\(3[\vec{a} \vec{b} \vec{c}]\)

Solution

Question 9

\(\pi\)

\(1\)

\(-1\)

\(0\)

Solution

Question 10

\(1,1\)

\(-1,1\)

\(0,1\)

\(0,-1\)

Solution

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