Mathematics Test - 16

The condition given in the question is \(x^{2}-|x+2|+x>0\)

Two cases are possible:

Case 1: When \((x+2) \geq 0\)

Therefore, \(x^{2}-x-2+x>0\)

Thus, \(x^{2}-2>0\)

So, either \(x<-\sqrt{2}\) or \(x>\sqrt{2}\).

Therefore, \(x \in[-2,-\sqrt{2}) \cup(\sqrt{2}, \infty) \quad \ldots \ldots .(1)\)

Case 2: When \((x+2)<0\) Then \(x^{2}+x+2+x>0\)

So, \(x^{2}+2 x+2>0\)

This gives \((x+1)^{2}+1>0\) and this is true for every \(x\)

Therefore, \(x \leq-2\) or \(x \in(-\infty,-2) \quad \ldots \ldots \ldots\) (2)

From equations (1) and (2) 

\(x \in(-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty)\)

Given:

\(\sqrt{3}-3 \sqrt{3 }\tan ^{2} \mathrm{A}=3 \tan \mathrm{A}-\tan ^{3} \mathrm{~A}\)

\(\Rightarrow \sqrt{3}\left(1-3 \tan ^{2} A\right)=3 \tan A-\tan ^{3} A\)

\(\Rightarrow \sqrt{3}=\frac{\left(3 \tan \mathrm{A}-\tan ^{3} \mathrm{~A}\right)}{\left(1-3 \tan ^{2} \mathrm{~A}\right)}\)

\(\Rightarrow \sqrt{3}=\tan 3 \mathrm{~A}\)

\(\because \tan 60^{\circ}=\sqrt{3}\)

\(\therefore 3 A=60^{\circ}\)

\(\Rightarrow A=\frac{60^{\circ}}{3}=20^{\circ}\)

\(\therefore 20^{\circ}\)

Here, we are asked to find the value of \(\int \tan ^{4} x d x\).

Let, \(I=\int \tan ^{4} x d x\)

Now, we can write \(\tan ^{4} x\) as \(\tan ^{2} x \cdot \tan ^{2} x\)

\(\therefore I=\int \tan ^{2} x \cdot \tan ^{2} x d x\)

Also, we know that \(\tan ^{2} x=\sec ^{2} x-1\)

\(\therefore I=\int \tan ^{2} x\left(\sec ^{2} x-1\right) d x\)

\(\Rightarrow I=\int\left(\tan ^{2} x \sec ^{2} x-\tan ^{2} x\right) d x\)

\(\Rightarrow I=\int \tan ^{2} x \sec ^{2} x d x-\int \tan ^{2} x d x\)

\(\Rightarrow I=\int \tan ^{2} x \sec ^{2} x d x-\int\left(\sec ^{2} x-1\right) d x\)

\(\Rightarrow I=\int \tan ^{2} x \sec ^{2} x d x-\int \sec ^{2} x d x+\int d x\)

\(\Rightarrow I=\int \tan ^{2} x \sec ^{2} x d x-\int \sec ^{2} x d x+x+C\)

Now, to solve further, 

Let \(\tan x=t\)

\(\therefore \sec ^{2} x d x=d t\)

Now, \( I=\int t^{2} d t-\int d t+x+C\)

\(\Rightarrow I=\frac{t^{3}}{3}-t+x+C\)

Now, we shall return back the substituted value \(\tan x=t\) in the above equation. 

\(\therefore I=\frac{\tan ^{3} x}{3}-\tan x+x+C\)

Thus, \(\int \tan ^{4} x d x=\frac{\tan ^{3} x}{3}-\tan x+x+C\).

Properties of Transpose of a Matrix:

The transpose of the transpose of a matrix is the matrix itself: \(\left(\mathrm{A}^{\prime}\right)^{\prime}=\mathrm{A}\).

The transposes of equal matrices are also equal: \(\mathrm{A}=\mathrm{B} \Rightarrow {A}^{\prime}=\mathrm{B}^{\prime}\).

The transpose of the sum/difference of two matrices is equivalent to the sumvdifference of their transposes: \((\mathrm{A} \pm \mathrm{B})^{\prime}={A}^{\prime} \pm {B}^{\prime}\)

The transpose of the product of two matrices is equivalent to the product of their transposes in reversed order \((\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}\)

Using the Properties of Transpose of a Matrix:

\(3 \mathrm{~A}+4 \mathrm{~B}^{\prime}=\left[\begin{array}{ccc}7 & -10 & 17 \\ 0 & 6 & 31\end{array}\right]\)

\(\Rightarrow\left(3 \mathrm{~A}+4 \mathrm{~B}^{\prime}\right)^{\prime}=\left[\begin{array}{ccc}7 & -10 & 17 \\ 0 & 6 & 31\end{array}\right]^{\prime}\)

\(\Rightarrow 3 \mathrm{~A}^{\prime}+4\left(\mathrm{~B}^{\prime}\right)^{\prime}=\left[\begin{array}{ccc}7 & -10 & 17 \\ 0 & 6 & 31\end{array}\right]^{\prime}\)

\(\Rightarrow 3 \mathrm{~A}^{\prime}+4 \mathrm{~B}=\left[\begin{array}{cc}7 & 0 \\ -10 & 6 \\ 17 & 31\end{array}\right]\)\(\quad\)......(1)

Also,

\(2 \mathrm{~B}-3 \mathrm{~A}^{\prime}=\left[\begin{array}{rr}-1 & 18 \\ 4 & 0 \\ -5 & -7\end{array}\right]\)\(\quad\)......(2)

Adding equations (1) and (2), we get,

\(\left(3 \mathrm{~A}^{\prime}+4 \mathrm{~B}\right)+\left(2 \mathrm{~B}-3 \mathrm{~A}^{\prime}\right)=\left[\begin{array}{cc}7 & 0 \\ -10 & 6 \\ 17 & 31\end{array}\right]+\left[\begin{array}{rr}-1 & 18 \\ 4 & 0 \\ -5 & -7\end{array}\right]\)

\(\Rightarrow 6 \mathrm{~B}+0=\left[\begin{array}{cc}7-1 & 0+18 \\ -10+4 & 6+0 \\ 17-5 & 31-7\end{array}\right]\)

\(\Rightarrow 6 \mathrm{~B}=\left[\begin{array}{cc}6 & 18 \\ -6 & 6 \\ 12 & 24\end{array}\right]\)

\(\Rightarrow \mathbf{B}=\left[\begin{array}{cc}1 & 3 \\ -1 & 1 \\ 2 & 4\end{array}\right]\)

First, we will find the derivative of each term, then after we will make \(\frac{d y}{d x}\) as a subject.

Since, the given equation is \(x y+y^{2}=\tan x+y\).

Now, differentiate the above equation with respect to \(x\), we get:

\(\frac{d}{d x}(x y)+\frac{d}{d x}\left(y^{2}\right)=\frac{d}{d x}(\tan x)+\frac{d y}{d x}\)

Applying the formula of derivative of product of two functions i.e.,  \(\frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x}\) on the function \(xy\).

\(\Rightarrow x \cdot \frac{d y}{d x}+y \cdot 1+2 y \frac{d y}{d x}=\sec ^{2} x+\frac{d y}{d x}\)

Now, making \(\frac{d y}{d x}\) as subject.

\(\Rightarrow \frac{d y}{d x}(x+2 y-1)=\sec ^{2} x-y\)

\(\Rightarrow \frac{d y}{d x}=\frac{\sec ^{2} x-y}{(x+2 y-1)}\)

Given:

A coin is tossed 3 times

Sample Space are,

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

\(\therefore\) Total no of Outcomes \(=8\)

Taking two heads and one tail,

Thus, Favourable outcomes are \(=\{(\mathrm{H}, \mathrm{H}, \mathrm{T}),(\mathrm{H}, \mathrm{T}, \mathrm{H}),(\mathrm{T}, \mathrm{H}, \mathrm{H})\}=3\)

Probability of an event happening \(=\frac{\text { Number of ways it can happen }}{\text { Total number of outcomes }}\)

Hence the probability that they will fall two heads and one tail \(=\frac{3}{8}\)

I implies II means ≡ I → II

\(X^{-1}=\frac{\operatorname{Adj}(\mathrm{X})}{|X|}\)

If \(|X| \neq 0\) then \(X^{-1}\)

\(|X|=\frac{\operatorname{Adj}(X)}{X^{-1}}\)

If \(X^{-1}\) then \(|X| \neq 0\) also \(\mid\) Adj \(\left.X|=| X\right|^{n-1}\) then \(\mid\) Adj \(X \mid \neq 0\)

If \(X^{-1}\) then \(|X| \neq 0\)

I implies II and II implies I

∴ Both I and II are equivalent.

Note:

\(X^{-1}\) means \(X\) is invertible

|X| determinant of \(X\)