Mathematics Test

Result

Mathematics Test - 27

Score
-
out of -
Rank
-
out of -

TIME Taken - -

Weekly Quiz Competition

Question 1
1 / -0

If $a \in \mathrm{R}$ and the equation, $-3(x-[x])^2+2(x-[x])+a^2=0$ (where $[x]$ denotes the greatest integer $\leq x$ ) has no integral solution, then all possible values of $a$ lie in the interval:

A
$(-2,-1)$

B
$(\infty,-2) \cup(2, \infty)$

C
$(-1,0) \cup(0,1)$

D
$(1,2)$

Solution

$a \in R$ and equation is$-3(x-[x])^2+2(x-[x])+a^2=0$
First we find the domain of the given function,So, we put$(x-[x])=t$$\Rightarrow 0 \leq t<1$
Now, $-3 t^2+2 t+a^2=0$
$t=\frac{-2 \pm \sqrt{4+12 a^2}}{-6}$$\Rightarrow 0 \leq \frac{-2 \pm \sqrt{4+12 a^2}}{-6}<1$$\Rightarrow 0 \leq \frac{1 \pm \sqrt{1+3 a^2}}{3}<1$
Taking positive sign of $0 \leq \frac{1 \pm \sqrt{1+3 a^2}}{3}<1$
$1+\sqrt{1+3 a^2}<3$$\Rightarrow \sqrt{1+3 a^2}<2$$\Rightarrow 1+3 a^2<4$$\Rightarrow 3 a^2<3$$\Rightarrow a^2<1$
$\therefore a$ belongs to $(-1,1)$
Only possibility for non-integral solution is $0Thus, \(a^2 \geq 0$, when $a=0$
So, for all integral values of $a$ the integral is belongs in the interval $(-1,0) \cup(0,1)$.
Question 2
1 / -0

The least positive integral value of $m$, if $\left(\frac{1+i}{1-i}\right)^{m}=1$ is:

A
1

B
2

C
3

D
4

Solution

We need to find the minimum value of ${m}$ which is positive as well as an integer.

Lets first find the value of $\left(\frac{1+i}{1-i}\right)$

Rationalize the denominator we get,

$=\frac{1+i}{1-i} \times \frac{1+i}{1+i}$

$=\frac{(1+i)(1+i)}{(1-i)(1+i)}$

$=\frac{(1+i)^{2}}{(1)^{2}-(i)^{2}} \quad\left(\right.$Using$(a+b)(a-b)=a^{2}-b^{2}\left.\right)$

$=\frac{1+(i)^{2}+2 \times 1 \times i}{1-i^{2}} \quad\left(\right.$Using $\left.(a+b)^{2}=a^{2}+b^{2}+2 a b\right)$

$=\frac{1+i^{2}+2 i}{1-i^{2}}$

Putting $i^{2}=-1$

$=\frac{1+(-1)+2 i}{1-(-1)} $

$=\frac{1-1+2 i}{1+1} $

$=\frac{0+2 i}{2} $

$=\frac{2 i}{2} $

$=i$

Thus, $\frac{1+i}{1-i}=i$

Given: $\left(\frac{1+i}{1-i}\right)^{m}=1$

$(i)^{m}=1$

We know that $i^{2}=-1$.

Squaring both sides.

$\left(i^{2}\right)^{2}=(-1)^{2} $

$i^{4}=1$

Thus, the minimum value of $m$ which satisfies the equation is 4.
Question 3
1 / -0

If the roots of the equation a(b - c)x² + b(c - a)x + c(a - b) = 0 are equal, then which one of the following is correct?

A
a, b and c are in AP

B
a, b and c are in GP

C
a, b and c are in HP

D
a, b and c do not follow any regular pattern

Solution

It is given that,

a(b - c)x² + b(c - a)x + c(a - b) = 0 has equal roots

That means, Discriminant (D) = 0

By comparing the given equation with the quadratic equation, ax² + bx + c = 0. We get, a’ = a(b - c), b’ = b(c - a) and c’ = c(a - b)

As, Discriminant, D = 0.

⇒ D = b² - 4ac = b² × (c - a) 2 - 4 × a (b - c) × c (a - b) = 0

⇒ D = (bc + ab - 2ac) 2 = 0

⇒ bc + ab - 2ac = 0

⇒ b × (a + c) = 2ac

⇒ b = $\frac{2 a c}{(a + c)}$ .

So, a, b and c are in HP.
Question 4
1 / -0

In how many ways can we select 6 people out of 10, of which a particular person is not included?

A
¹⁰C₃

B
⁹C₅

C
⁹C₆

D
⁹C₄

Solution

One particular person is not included we have to select 6 persons.

Out of 9 which can be done in⁹C₆ ways.
Question 5
1 / -0

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 6
1 / -0

If the sum of first n terms of a series is (n + 12), then what is its third term?

A
1

B
2

C
3

D
4

Solution

S_n denotes the sum of the first n terms of the sequence.

It is given that,

S_n = n + 12

By substituting n = 2 in the given equation, we get

⇒ S₂ = 2 + 12 = 14

Similarly by substituting n = 3 in the given equation, we get

⇒ S₃ = 3 + 12 = 15

⇒ a₃ = S₃ - S₂ = 15 - 14 = 1
Question 7
1 / -0

In an entrance test, there are multiple-choice questions, with four possible answers to each question of which one is correct, The probability that a student knows the answer to a question is $90 \%$, If the student gets the correct answer to a question, then the probability that he was guessing is:

A
$\frac{37}{40}$

B
$\frac{1}{37}$

C
$\frac{36}{37}$

D
$\frac{1}{9}$

Solution

Let,

$E_{1}$: He knows the answer.

$E_{2}$: He does not know the answer.

$X:$ He gets the correct answer.

Therefore, $P\left(E_{1}\right)=90 \%=\frac{9}{10}$

$P\left(E_{2}\right)=1-\frac{9}{10}=\frac{1}{10}$

$P\left(X \mid E_{1}\right)=1$

$P\left(X \mid E_{2}\right)=\frac{1}{4}$

As we know that according to Bayes theorem:

$\mathrm{P}\left(\mathrm{E}_{\mathrm{i}} \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{\mathrm{i}}\right) \times \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{\mathrm{i}}\right)}{\sum_{\mathrm{i=1}}^{\mathrm{n}} \mathrm{P}\left(\mathrm{E}_{\mathrm{i}}\right) \times \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{\mathrm{i}}\right)}, \mathrm{i}=1,2, \ldots, \mathrm{n}$

$\therefore \mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{X}\right)=\frac{\frac{1}{10}. \frac{1}{4}}{\left[\frac{9}{10}.1+\frac{1}{10}.\frac{1}{4}\right]}=\frac{1}{37}$
Question 8
1 / -0

The order of the differential equation:

$\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}+\mathrm{y} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}-\sin ^{-1} \frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{x}^{2}}{\mathrm{y}}=0$

A
3

B
2

C
1

D
Not defined

Solution

Given the differential equation is:

$\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}+\mathrm{y} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}-\sin ^{-1} \frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{x}^{2}}{\mathrm{y}}=0$

Highest derivate is $\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}$.

Here, The given equation is not a polynomial equation.

∴ The order of the differential equation is not defined.
Question 9
1 / -0

Rolle's theorem is not applicable for the function $\mathrm{f}(\mathrm{x})=|\mathrm{x}|$ in the interval $[-1,1]$ because:

A
f' (1) does not exist

B
f' (-1) does not exist

C
f (x) is discontinuous at x = 0

D
f' (0) does not exist

Solution

Rolle's theorem is not applicable for the function $\mathrm{f}(\mathrm{x})=|\mathrm{x}|$ in $[-1,1]$

$\because \mathrm{f}^{\prime}(0)$ does not exist.
Question 10
1 / -0

The general solution y(x) of the differential equation

$\frac{{dy}}{{dx}}=\frac{{x} \sqrt{1-{y}^{2}}}{{y} \sqrt{1-{x}^{2}}}$

A
y² - 1 = C ln (x² - 1)

B
y² - 1 = C(x² - 1)

C
y² + 1 = C(x² + 1)

D
ln (y² - 1) = C(x² - 1)

Solution

The given differential equation is

$\frac{{dy}}{{dx}}=\frac{{x} \sqrt{1-{y}^{2}}}{{y} \sqrt{1-{x}^{2}}}$

$\frac{{ydy}}{\sqrt{1-{y}^{2}}}=\frac{{xdx}}{\sqrt{1-{x}^{2}}}$

Integrating both the sides

$\int \frac{{ydy}}{\sqrt{1-{y}^{2}}}=\int \frac{{xdx}}{\sqrt{1-{x}^{2}}}$

$-\ln \sqrt{1-{y}^{2}}=-\ln \sqrt{1-{x}^{2}}+{c}$

$\ln \sqrt{1-{y}^{2}}-\ln \sqrt{1-{x}^{2}}+{c}=0$

$\ln \frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}=c \quad[\because \log m-\log n=\log (\frac{m}{n})]$

$\sqrt{\frac{1-y^{2}}{1-x^{2}}}=e^{c}=a$

$\frac{1-y^{2}}{1-x^{2}}=a^{2}=C$

$y^{2}-1=C\left(x^{2}-1\right)$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Mathematics Test - 27

Result

Mathematics Test - 27

Score

-

Rank

-

Question 1

\((-2,-1)\)

\((\infty,-2) \cup(2, \infty)\)

\((-1,0) \cup(0,1)\)

\((1,2)\)

Solution

Question 2

1

2

3

4

Solution

Question 3

a, b and c are in AP

a, b and c are in GP

a, b and c are in HP

a, b and c do not follow any regular pattern

Solution

Question 4

10C3

9C5

9C6

9C4

Solution

Question 5

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

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

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

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

Solution

Question 6

1

2

3

4

Solution

Question 7

\(\frac{37}{40}\)

\(\frac{1}{37}\)

\(\frac{36}{37}\)

\(\frac{1}{9}\)

Solution

Question 8

3

2

1

Not defined

Solution

Question 9

f' (1) does not exist

f' (-1) does not exist

f (x) is discontinuous at x = 0

f' (0) does not exist

Solution

Question 10

y2 - 1 = C ln (x2 - 1)

y2 - 1 = C(x2 - 1)

y2 + 1 = C(x2 + 1)

ln (y2 - 1) = C(x2 - 1)

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

¹⁰C₃

⁹C₅

⁹C₆

⁹C₄

y² - 1 = C ln (x² - 1)

y² - 1 = C(x² - 1)

y² + 1 = C(x² + 1)

ln (y² - 1) = C(x² - 1)