Mathematics Test - 1

Sample space contains \(2^{4}=16\) elements.

The favorable events are:

\(\{ H , H , H , T \},\{ H , H , T , H \},\{ H , T , H , H \},\{ T , H , H , H \},\{ H , H , H , H \}\)

There are 5 possibilities,

\(\therefore\) The required probability is \(\frac{5}{16}\).

Given:

\(\left(\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right) \times\left(\begin{array}{cc}5 & -2 \\ -3 & 1\end{array}\right)=\left(\begin{array}{cc}1 & -1 \\ 17 & \lambda\end{array}\right)\)

\(\mathrm{A}\) is in \(2\)nd row and \(2\)nd column, so, it will get by sum of multiplication of \(2\)nd row of first matrix and \(2\)nd column of the second matrix

\(\therefore \lambda=4(-2)+1(1)\)

\(=-8+1=-7\)

 \(y=4 x^{2}, x=0\),

\(y=1\) and \(y=4\)

The curve can be drawn as:

Now, \(y=4 x^{2}\)

\( \Rightarrow x^{2}=\frac{y}{4}\)

\( \Rightarrow x=\frac{1}{2} \sqrt{y}\)

Required area \(=\int_{1}^{4} x d y\)

\(=\frac{1}{2} \int_{1}^{4} \sqrt{y} d y\)

\(=\frac{1}{2}\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4}\)

\(=\frac{1}{3}\left[y^{\frac{3}{2}}\right]_{1}^{4}\)

\(=\frac{1}{3}\left[4^{\frac{3}{2}}-1^{\frac{3}{2}}\right]\)

\(=\frac{1}{3}[8-1]\) \(=\frac{7}{3} \) sq. unit

Given: 

\(y=\frac{12 \tan x-4 \tan ^{3} x}{9-27 \tan ^{2} x}\)  

\(\Rightarrow y=\frac{4\left(3 \tan x-\tan ^{3} x\right)}{9\left(1-3 \tan ^{2} x\right)}\)

We know that, 

\(\tan 3 x=\frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x}\)

\(\Rightarrow y=\frac{4}{9} \tan 3 x\)

Differentiating both sides with respect to \(x\), we get:

\(\Rightarrow \frac{d y}{d x}=\frac{4}{9} \sec ^{2} 3 x \times 3\)

\(\Rightarrow \frac{d y}{d x}=\frac{4}{3} \sec ^{2} 3 x\)

Again, differentiating with respect to \(x\), we get:

\(\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{4}{3} \times 2 \times \sec 3 x \times \tan 3 x \times \sec 3 x \times 3\)

\(\Rightarrow \frac{d^{2} y}{d x^{2}}=8 \sec ^{2} 3 x \tan 3 x\)

Given the exprission is \(\left(2 \mathrm{x}+\frac{1}{\mathrm{x}}\right)^{8}\).

Here \(n=8(\mathrm{n}\) is even number)

\(\therefore\) Middle term \(=\left(\frac{\mathrm{n}}{2}+1\right)=\left(\frac{8}{2}+1\right)=5\) th term

General term in the expansion of \((x+y)^{n}\) is given by

\(\mathrm{T}_{(\mathrm{r}+1)}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \times \mathrm{x}^{\mathrm{n}-\mathrm{r}} \times \mathrm{y}^{\mathrm{r}}\)

\(\Rightarrow T_{5}=T_{(4+1)}={ }^{8} \mathrm{C}_{4} \times(2 \mathrm{x})^{(8-4) \times}\left(\frac{1}{\mathrm{x}}\right)^{4}\)

\(\Rightarrow \mathrm{~T}_{5}={ }^{8} \mathrm{C}_{4} \times 2^{4}\)

Given:

A pack of \(52\) cards.

As we know there are \(26\) cards in total which are black. Let A and B denote respectively the events that the first and second drawn cards are black.

Now, \(P(A)=P(\) black card in first draw \()=\frac{26}{52}=\frac{1}{2}\)

Because the second card is drawn without replacement so, now the total number of black cards will be \(25\) and the total cards will be \(51\) that is the conditional probability of B given that A has already occurred.

Now, \(P\left(\frac{B}{A}\right)=\mathrm{P}(\) black card in second \(\mathrm{draw})=\frac{25}{51}\)

Thus the probability that both the cards are black.

\(=P(A \cap B)=\frac{1}{2} \times \frac{25}{51}=\frac{25}{102}\)

Therefore, the probability that both the cards are black is \(\frac{25}{102}\).

Vector equation of a plane whose perpendicular distance from the origin is \(d\) and \(\hat{n}\) is the unit normal vector to the plane through origin is given by \(\vec{r} . \hat{n}=d\)

Now,

Let \(\vec{n}=\hat{i}+2 \hat{j}+2 \hat{k}\) is the normal to the required plane from the origin and it is at a distance of \(\frac{1}{3}\) units from the origin.

\(\Rightarrow|\vec{n}|=|\hat{i}+2 \hat{j}+2 \hat{k}|=\sqrt{1^2+2^2+2^2}=3 \text {. }\)

So, the unit normal vector \(\hat{n}=\frac{1}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\)

As we know that, the equation of a plane whose perpendicular distance from the origin is \(d\) and \(\hat{n}\) is the unit normal vector to the plane through origin is given by \(\vec{r} . \hat{n}=d\)

Here, \(d =\frac{1 }{ 3}, \hat{n}=\frac{1}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\) and let \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)

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

\(\Rightarrow x +2 y +2 z =1 \)

So, the equation of the required plane is \(x+2 y+2 z=1\)

A = {1, 2, 4}, B = {2, 4, 5}, C={2, 5}

⇒ A − C = {1, 4}

⇒ B − C = {4}

⇒ (A − C) × (B − C) = {1, 4} × {4}

= {1, 4}, {4, 4}

Let OAB be the equilateral triangle inscribed in parabola \(y^2=4 a x\)

Let AB intersect the x-axis at point C. Let OC = k

From the equation of the given parabola, we have \(y^2=4 a k \Rightarrow y=\) \(\pm 2 \sqrt{a k}\)

\(\therefore\) The respective coordinates of points A and B are \((k, 2 \sqrt{a k})\), and \((k,-2 \sqrt{a k})\)

\(A B=C A+C B=2 \sqrt{a k}+2 \sqrt{a k}=4 \sqrt{a k}\)

Let AB intersect the x-axis at point C.

Let OC = k

From the equation of the given parabola, we have \(y^2=4 a k \Rightarrow y=\) \(\pm 2 \sqrt{a k}\)

\(\therefore\) The respective coordinates of points A and B are

\((k, 2 \sqrt{a k})\), and \((k,-2 \sqrt{a k})\)

AB = CA + CB = \(2 \sqrt{a k}+2 \sqrt{a k}=4 \sqrt{a k}\)

since OAB is an equilateral triangle, \(O A^2=A B^2\)

\(\therefore k^2+(2 \sqrt{a k})^2=(4 \sqrt{a k})^2\)

\(\Rightarrow k^2+4 a k=16 a k\)

\(\Rightarrow k^2=12 a k\)

\(\Rightarrow k=12 a\)

\(\therefore A B=4 \sqrt{a k}=4 \sqrt{a \times 12 a}=4 \sqrt{12 a^2}=8 \sqrt{3} a\)

Let the six numbers be a, b, c, d, e, f

So, Mean \(=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}}{6}=47\)

\(\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}=282\)     .....(i)

Let, the excluded number be a,

So, mean of remaining five numbers \(=\frac{\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}}{5}=41\)

\(\Rightarrow b+c+d+e+f=205\)     .....(ii)

∴ a + 205 = 282              (from (i) and (ii))

⇒ a = 77