Mathematics Test - 30

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

\(2 y=\log _{e}\left(1+4 x^{2}\right)\)

\(y=\frac{1}{2} \log _{e}\left(1+4 x^{2}\right)\)

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

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

\(1+4 {x}^{2}\) will be positive for \({x} \in {R}\)

So, \(\frac{d y}{d x}=m=\frac{4 x}{1+4 x^{2}}<1\), for all \(x \in R\)

\(\Rightarrow|{m}| \leq 1\)

Let, A be any square matrix.

(adj A^T) = |A^T| (A^T) ^-1

(adj A)^T = (|A| A^-1)^T

(adj A^T) = (adj A)^T

Given, A is a square matrix.

Consider, (adj A^T) - (adj A)^T

= |A^T| (A^T)^-1 - (|A| A^-1)^T

= |A|^T (A^T)^-1 - (|A|^T(A^-1)^T

= |A|^T (A^-1)^T - (|A| ^T(A^-1)^T

= 0

= Null Matrix

Thus, If A is a square matrix, then what is adj A^T - (adj A)^T equal to null matrix.

Now, \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{3 x+\tan ^{2} x}{x}\left(\frac{0}{0}\right.\) form \()\)

\(=\lim _{x \rightarrow 0} \frac{3+2 \tan x \sec ^{2} x}{1}=3\)

Since, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\)

\(\therefore \mathrm{f}(0)=3\)

Given:

tan θ + sec θ = 4

To Find: sin θ

tan θ + sec θ = 4 .... (1)

As we know sec² θ - tan² θ = 1

⇒ (sec θ + tan θ) × (sec θ - tan θ) = 1

⇒ 4 × (sec θ - tan θ) = 1

∴ sec θ - tan θ =$\frac{1}{4}$.......... (2)

Adding equation (1) + (2), we get

⇒ 2sec θ =$4+\frac{1}{4}$

⇒ sec θ =$\frac{17}{8}$

⇒ cos θ =$\frac{8}{17}$

Now,

sin θ =$\sqrt{1-{\cos }^2\theta }$

=$\sqrt{1-{\left(\frac{8}{17}\right)}^2}=\frac{15}{17}$

The total number of caps \(=2+4+5+1=12\)

The number of ways for selecting four caps \(={ }^{12} \mathrm{C}_{4}\)

The total number of caps other than green \(=12-5=7\)

The number of ways for selecting four caps other than green \(={ }^{7} \mathrm{C}_{4}\)

The probability of selecting four caps and none of them are green P(A')\(=\frac{\text { The number of ways for selecting four caps other than green }}{\text { The number of ways for selecting four caps }}\)

\( \mathrm{P}\left(\mathrm{A}^{\prime}\right)=\frac{{ }^{7} \mathrm{C}_{4}}{{ }^{12} \mathrm{C}_{4}}\)

\(\mathrm{P}\left(\mathrm{A}^{\prime}\right)=\frac{7 !(12-4) ! 4 !}{4 !(7-4) ! 12 !}\)

\(\mathrm{P}\left(\mathrm{A}^{\prime}\right)=\frac{7}{99}\)

Number lying between \(100\) and \(1000\) formed by \(3\) digits.

And every digit have \(5\) option \((1,2,3,4,5)\) to select a number.

But repetition is not allowed.

So, number \(={ }^{5} P_{3}\)

\(=\frac{5 !}{(5-3) !}\)

\(=\frac{5 !}{2 !}\)

\({ }^{5} P_{3}\)\(=60\)

We have,

\(y=f(x)\)

Area bounded by the curve with \(x\) -axis from 0 to \(a\), means,

\(\int_{0}^{a} f(x) d x\)

\(\int_{0}^{a} f(x) d x=1+\frac{a^{2}}{2} \sin a\)

Differentiating this w.r. to \(a\), we get

\(\Rightarrow f(a)=a \sin a+\frac{a^{2}}{2} \cos a\left[\because \frac{d}{d x} (u, v)=u \frac{d}{d x} v+v \frac{d}{d x} u\right]\)

Since, A is skew-symmetric.

A^T = −A

Since, A is symmetric.

A^T = A

⇒ −A = A

⇒ 2A = 0

⇒ A = 0

Therefore, A is a null matrix.

Given,

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

Let \(P\) be any point on the plane,

\(P=(k+3,-k-2,-2 k-\lambda)\)

\(P\) lie on the plane, \(2 x-4 y+3 z=2,\) Which means \(P\) satisfy the equationn of plane.

So, \(2(k+3)-4(-k-2)+3(-2 k-\lambda)=2\)

Solving above equation, we have \(\lambda=4\)

Now, the shortest distance between the below lines:

\(\frac{(x-3)}{1}=\frac{(y-2)}{-1}=\frac{(z+4)}{-2}\) and \(\frac{(x-1)}{12}=\frac{y}{9}=\frac{z}{4}\) is:

\(\begin{aligned} \mathrm{d}^{2} &=\left|\begin{array}{ccc}x_{1}-x_{2} & y_{1}-y_{2} & z_{1}-z_{2} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2}\end{array}\right| \\ &=\left|\begin{array}{ccc}3-1 & -2-0 & -4-0 \\ 1 & -1 & -2 \\ 12 & 9 & 4\end{array}\right| \end{aligned}\)

\(=\mid2(14)+2(28)-4(21) \mid\)

\(=0\)

\(d^{2}=0\) or \({d}=0\)

Given: \({dy}=\left(4+{y}^{2}\right) {dx}\)

\(\Rightarrow \frac{{dy}}{4+{y}^{2}}={dx}\)

Integrating both sides, we get

\(\int \frac{{dy}}{2^{2}+{y}^{2}}=\int {dx}\)

\(\Rightarrow \frac{1}{2} \tan ^{-1} \frac{y}{2}=x+c\)

\(\Rightarrow \tan ^{-1} \frac{y}{2}=2 x+C\)

\(y=2 \tan (2 x+C)\)