Mathematics Test

Result

Mathematics Test - 26

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The equation of the ellipse whose centre is at origin \((0,0),\) foci (±1,0) and eccentricity \(\frac{1}{2}\), is

A

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

B

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

C

\(\frac{x^{2}}{3}+\frac{y^{2}}{4}=12\)

D

\(x^{2}+y^{2}=12\)

Solution

Here foci \(=(\pm 1,0)\) and eccentricity, \(e=\frac{1}{2}\)
\(\therefore\) ae. \(=1\) and \(e=\frac{1}{2} \Rightarrow a=2\)
\(\therefore b^{2}=a^{2}\left(1-e^{2}\right) \Rightarrow b^{2}=4\left(1-\frac{1}{4}\right)=4 \times \frac{3}{4}=3\)
So, the equation of required ellipse is \(\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)
Question 2
1 / -0

If in the triangle ABC, we have cotA + cotB = 2 cotC then,

A

\(a^{2}+b^{2}=2 c^{2}\)

B

\(a^{2}+b^{2}+c^{2}=2\)

C

\(a^{2}+c^{2}=-2 b^{2}\)

D

\(a^{2}+\frac{b^{2}}{2}+\frac{c^{2}}{2}=1\)

Solution

\(\cot A+\cot B=2 \cot C\)
\(\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}=\frac{2 \cos C}{\sin C}\)
We know that, \(\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}, \cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}\) and \(\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}\)

\(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=m\)
\(\Rightarrow \sin A=\frac{a}{m}, \sin B=\frac{b}{m}, \sin C=\frac{c}{m}\)
By substitution
\(2 c^{2}=2\left(a^{2}+b^{2}-c^{2}\right)\)
\(\Rightarrow 2 c^{2}=a^{2}+b^{2}\)
Question 3
1 / -0

The coefficient of \(x^{53}\) in the following expansion \(\sum_{m=0}^{100} 100 C_{m}(x-3)^{100-m} .2^{m}\) is

A

\(^{100} C_{47}\)

B

\(^{100} C_{53}\)

C

\(-^{100} C_{53}\)

D

\(-^{100} C_{100}\)

Solution

Given sigma expression can be rewritten as
\((x-3+2)^{100}=(x-1)^{100}=(1-x)^{100}\)
So, \(x^{53}\) will occur in \(T_{54}\)
\(\therefore T_{54}=^{100} C_{53}(-x)^{53}\)
Hence coefficient of \(x^{53}\) is \(-^{100} C_{53}\)
Question 4
1 / -0

From a group of 7 men & 4 women a committee of 6 persons is formed. What will be the probability that the committee will consist of exactly two women?

A
5/11

B
3/11

C
4/11

D
2/11

Solution

\(7 \mathrm{M}, 4 \mathrm{W},\) member in committee \(=6\)
The favourable case may be \(\rightarrow(2 \mathrm{W}, 4 \mathrm{M})\)
\(\therefore\) Required Probability \(=\frac{^{4} C_{2} \times^{7} C_{4}}{^{12} C_{6}}\)
\(=\frac{6 \times 35}{462}=\frac{5}{11}\)
Question 5
1 / -0

The value of x for which \(\left[\begin{array}{lll}1 & 1 & x\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 1 & 0\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=0\) is

A
3

B
-2

C
3

D
–3

Solution

\(\left[\begin{array}{lll}1 & 1 & x\end{array}\right]\left[\begin{array}{l}3 \\ 3 \\ 3\end{array}\right]=0\)
\(3+3+3 x=0\)
\(3 x=-6\)
\(x=-2\)
Question 6
1 / -0

If \(\mathrm{x}=\mathrm{k}(\theta+\sin \theta)\) and \(\mathrm{y}=\mathrm{k}(1+\cos \theta),\) then what is the derivative of \(\mathrm{y}\) w.r.t \(\mathrm{x}\) at \(\theta=\pi / 2 ?\)

A
–1

B
0

C
1

D
2

Solution

\(\because x=k(\theta+\sin \theta)\)
\(\mathbf{y}=\mathrm{k}(1+\cos \theta)\)
\(\Rightarrow \frac{d x}{d \theta}=k(1+\cos \theta)\) and \(\frac{d y}{d \theta}=-k \sin \theta\)
\(\therefore \frac{d y}{d x}=\frac{-k \sin \theta}{k(1+\cos \theta)}\)
\(=\frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2}=\tan \frac{\theta}{2}\)
\(\Rightarrow\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{2}}=-\tan \frac{\pi}{4}=-1\)
Question 7
1 / -0

What is the area of a triangle whose vertices are at (3, –1, 2), (1, –1, –3) and (4, –3, 1) ?

A

\(\frac{\sqrt{165}}{2}\) sq units

B

\(\frac{\sqrt{135}}{2}\) sq units

C

\(4 \mathrm{sq}\) units

D

2 sq units

Solution

Let the vertices of the \(\Delta A B C\) be \(A(3,-1,2), B(1,-1,-3)\) and \(C(4,-3,1)\)
\(\therefore\) area of \(\Delta A B C\)
\[
\begin{array}{l}
=\frac{1}{2}\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{k} \\
1-3 & -1+1 & -3-2 \\
4-3 & -3+1 & 1-2
\end{array}\right| \\
=\frac{1}{2}\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{k} \\
-2 & 0 & -5 \\
1 & -2 & -1
\end{array}\right|
\end{array}
\]

\(=\frac{1}{2}|\hat{\imath}(-10)-\hat{\jmath}(7)+\hat{k}(4)|\)
\(=\frac{1}{2} \sqrt{100+49+16}\)
\(=\frac{1}{2} \sqrt{165} \mathrm{sq}\) units
Question 8
1 / -0

If P ,q and r are in AP as well as in GP, then which one of the following is correct?

A
p=q≠r

B
p≠q≠r

C
p≠q=r

D
p= q = r

Solution

Given that \(p, q\) and \(r\) are in AP. \(\therefore 2 q=p+r\)
As well as are in GP.
\(\therefore q^{2}=p r\)
From equation
(1) \(\&(2),\) we get \(\mathrm{p}+\mathrm{r}=2 \sqrt{p r}\)
\(\Rightarrow(\sqrt{p})^{2}-2 \sqrt{p} \cdot \sqrt{r}+(\sqrt{r})^{2}=0\)
\(\Rightarrow(\sqrt{p}-\sqrt{r})^{2}=0\)
\(\Rightarrow \sqrt{p}=\sqrt{r} \Rightarrow p=r\)
From eq(2), we get \(q^{2}=r \cdot r=r^{2} \Rightarrow q=r\)
Now from equations
(3) \(\&(4)\) we get \(p=q=r\)
Question 9
1 / -0

Equation of circle which passes through the points (1, –2) and (3, –4) and touch the x axis is

A

\(x^{2}+y^{2}+6 x+2 y+9=0\)

B

\(x^{2}+y^{2}+10 x+20 y+25=0\)

C

\(x^{2}+y^{2}+6 x+4 y+9=0\)

D

None of the above

Solution

since the circle touches x axis
\(\therefore(x-h)^{2}+(y-k)^{2}=k^{2}\)
Also, it passes through given points
\((1-h)^{2}+(-2-k)^{2}=k^{2}\)_______(1)
\(\&(3-h)^{2}+(-4-k)^{2}=k^{2}\) _______(2)
On subtracting eq (3) from eq \((2),\) we get \(h=k+5\)
On solving these equations, we get \(k=-10,-2 \& h=-5,3\)
On putting the values of \((\mathrm{h}, \mathrm{k})=(-5,-10)\) or (3,-2)
In eq(1), we get \(x^{2}+y^{2}+10 x+20 y+25=0\)
Question 10
1 / -0

In the mean value Theorem \(\frac{f(b)-f(a)}{b-a}=f^{\prime}(c), i f a=0, b=\frac{1}{2}\) and \(f(x)=x(x-1)(x-2),\) the value of \(C\) is

A

\(1-\frac{\sqrt{15}}{6}\)

B

\(1+\sqrt{15}\)

C

\(\frac{1-\sqrt{21}}{6}\)

D

\(1+\sqrt{21}\)

Solution

From mean value theorem \(f(c)=\frac{f(b)-f(a)}{b-a}\)
Given \(a=0 \Rightarrow f(a)=0\)
And \(b=\frac{1}{2} \Rightarrow f(b)=\frac{3}{8}\)
Now, \(f^{\prime}(x)=(x-1)(x-2)+x(x-2)+x(x-1)\)
\(\therefore f^{\prime}(c)=(c-1)(c-2)+c(c-2)+c(c-1)\)
\(=c^{2}-3 c+2+c^{2}-2 c+c^{2}-c\)
\(\Rightarrow f^{\prime}(c)=3 c^{2}-6 c+2\)
By definition of mean value therorem

\[
\begin{array}{l}
f(c)=\frac{f(b)-f(a)}{b-a} \\
\Rightarrow 3 c^{2}-6 c+2=\frac{\left(\frac{a}{b}\right)-0}{\left(\frac{1}{2}\right)-0}=\frac{3}{4} \\
\Rightarrow 3 c^{2}-6 c+\frac{5}{4}=0
\end{array}
\]
since this is a quadratic equation in \(c\)
\[
\therefore c=\frac{6 \pm \sqrt{36-15}}{2 \times 3}
\]
\[
=\frac{6 \pm \sqrt{21}}{6}=1 \pm \frac{\sqrt{21}}{6}
\]

since, \(C\) lies between \(\left[0, \frac{1}{2}\right]\)
\(\therefore c=1-\frac{\sqrt{21}}{6}\)
\(\left[\text { neglecting } c=1+\frac{\sqrt{21}}{6}\right]\)

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Mathematics Test - 26

Result

Mathematics Test - 26

Score

-

Rank

-

SHARING IS CARING

Question 1

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

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

\(\frac{x^{2}}{3}+\frac{y^{2}}{4}=12\)

\(x^{2}+y^{2}=12\)

Solution

Question 2

\(a^{2}+b^{2}=2 c^{2}\)

\(a^{2}+b^{2}+c^{2}=2\)

\(a^{2}+c^{2}=-2 b^{2}\)

\(a^{2}+\frac{b^{2}}{2}+\frac{c^{2}}{2}=1\)

Solution

Question 3

\(^{100} C_{47}\)

\(^{100} C_{53}\)

\(-^{100} C_{53}\)

\(-^{100} C_{100}\)

Solution

Question 4

5/11

3/11

4/11

2/11

Solution

Question 5

3

-2

3

–3

Solution

Question 6

–1

0

1

2

Solution

Question 7

\(\frac{\sqrt{165}}{2}\) sq units

\(\frac{\sqrt{135}}{2}\) sq units

\(4 \mathrm{sq}\) units

2 sq units

Solution

Question 8

p=q≠r

p≠q≠r

p≠q=r

p= q = r

Solution

Question 9

\(x^{2}+y^{2}+6 x+2 y+9=0\)

\(x^{2}+y^{2}+10 x+20 y+25=0\)

\(x^{2}+y^{2}+6 x+4 y+9=0\)

None of the above

Solution

Question 10

\(1-\frac{\sqrt{15}}{6}\)

\(1+\sqrt{15}\)

\(\frac{1-\sqrt{21}}{6}\)

\(1+\sqrt{21}\)

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.