Mathematics Test - 32

\(\int \frac{\sin 3 x}{\sin x} d x=\int \frac{3 \sin x-4 \sin ^{3} x}{\sin x} d x\)
\(\int 3 d x-4 \int \sin ^{2} x d x=3 x-2 \int(1-\cos 2 x) d x+c\)
\(=3 x-2 x+\sin 2 x+c=x+\sin 2 x+c\)

Multiple Angle Formulas
\(\sin (3 X)=3 \sin X-4 \sin ^{3} X\)
\(\cos (3 x)=4 \cos ^{3} x-3 \cos x\)
\(\sin (4 X)=4 \sin X \cos X-8 \sin 3 X \cos X\)
\(\cos (4 x)=8 \cos ^{4} x-8 \cos ^{2} x+1\)

Double Angle Formulas
\(\sin (2 X)=2 \sin X \cos X\)
\(\cos (2 X)=1-2 \sin ^{2} X=2 \cos ^{2} X-1\)
\(\tan (2 X)=2 \tan X /\left[1-\tan ^{2} X\right]\)

As we know that
\(p \rightarrow q \equiv(\sim p \vee q)\)
Hence, \(\sim p \rightarrow q \equiv(p \vee q)\)
Hence, \(\sim(\sim p \rightarrow q) \equiv \sim(p \vee q)\)
Hence, \(\sim(\sim p \rightarrow q) \equiv \sim p \wedge \sim q\)
Hence, \(\sim p \wedge \sim q\) is the correct answer

KEY CONCEPTS

(i) Conjunction: Any two simple statements can be connected by the word "and" to form a compound statement called the conjunction of the original statements. Conjuction is true only when both its component statement are true.

(ii) Disjunction or alternation: Any two statements can be connected by the word "or" to form a compound statement called the disjunction of the original statements. Disjunctions is faise only when both it's component statement are false
Symbolically, if p and q are two simple statements, then \(p \vee q\) denotes the disjunction of \(p\) and \(q\) and is read as "p or \(q^{\prime \prime}\)

iii) Inclusive and Exclusive OR
If both the component statements in disjunction can occur simultaneously it is called "Inclusive OR", else it's an "Exclusive OR".
(iv) Negation: The denial of a statement p is called its negation, written as \(-p\)
Negation of any statement \(p\) is formed by writing "It is not the case that \(\ldots . .\) "or" It is false that. \(\ldots\) " before \(\mathrm{p}\) or, if possible by inserting in \(\mathrm{p}\) the word "not"
Negation is called a connective although it does not combine two or more statements. In fact, it only modifies a statement.
(v) Implication or conditional statements: Any two statements connected by the connective phrase "If then" give rise to a compound statement which is known as an implication or a conditional statement.
If \(p\) and \(q\) are two statements forming the implication "f \(p\) then \(q\), then we denote this implication by \(^{n} p \Rightarrow q^{\prime \prime}\) or \(^{n} p \rightarrow q^{n}\)
In the implication \(" \mathbf{p} \Rightarrow \mathbf{q}^{\prime \prime}, p\) is the antecedent and \(q\) is the consequent.

The system of given equation are
\(K x+2 y-z=1\)
\((K-1) y-2 z=2\)
and \((K+2) z=3\)
This system of equations has a unique solution, if
\(\left|\begin{array}{ccc}K & 2 & -1 \\ 0 & K-1 & -2 \\ 0 & 0 & K+2\end{array}\right| \neq 0\)
\(\Rightarrow(K+2)(K)(K-1) \neq 0\)
\(\Rightarrow \quad K \neq-2,0,1\)
ie, \(K=-1\), is a required answer

\(\mathrm{x} \in \mathrm{A} \cup \mathrm{B} \Rightarrow \mathrm{x} \in \mathrm{A}\) or \(\mathrm{x} \in \mathrm{B},\) and
\(\mathrm{x} \notin \mathrm{A} \cup \mathrm{B} \Rightarrow \mathrm{x} \notin \mathrm{A}\) and \(\mathrm{x} \notin \mathrm{B}\)
Difference of Two sets: If A and B are two sets, then their difference \(A-B\) is defined as:
\(A-B=\{x: x \in A \text { and } x \notin B\}\)
Similarly, \(B-A=\{x: x \in B \text { and } x \notin A\}\)
By using these definitions we can find the answer.
We have
\(A \cap(B \cup C)\)
\(\mathrm{x} \in \mathrm{A} \cap(\mathrm{B} \cup \mathrm{C})\)
\(\Rightarrow x \in A\) and \x \in B \text { or } x \in C)\)
\(\Rightarrow(x \in A \text { and } x \in B)\) or \((x \in A \text { and } x \in C)\)
(\because and' distributes'or')
\(\Rightarrow x \in(A \cap B)\) or \(x \in(A \cap C)\)
\(\therefore A \cap(B \cup C) \subseteq(A \cap B) \cup(A \cap C)\)
Similarly, \((\mathrm{A} \cap \mathrm{B}) \cup(\mathrm{A} \cap \mathrm{C}) \subseteq \mathrm{A} \cap(\mathrm{B} \cup \mathrm{C})\)

Hence, \(A \cap(B \cup C)=(A \cap B) \cup(A \cap C)\)
It is one of the Distributive law.
For example:
Let \(A=\{2,3,5,7\}, B=\{2,4,6,8\}\) and \(C=\{1,3,6,8\}\) then
\((B U C)=\{2,4,6,8\} \cup\{1,3,6,8\}\)
\(=\{1,2,3,4,6,8\}\)
\(\mathrm{LHS}=\mathrm{A} \cap(\mathrm{BUC})\)
\(=\{2,3,5,7\} \cap\{1,2,3,4,6,8\}\)
\(=\{2,3\}\)
\(\mathrm{RHS}=(\mathrm{A} \cap \mathrm{B} \cup(\mathrm{A} \cap \mathrm{C})\)
\((A \cap B)=\{2,3,5,7\} \cap\{2,4,6,8\}\)
\(=\{2\}\)
\((A \cap C)=\{2,3,5,7\} \cap\{1,3,6,8\}\)
\(=\{3\}\)
\(\therefore(A \cap B) \cup(A \cap C)=\{2,3\}\)
\(\therefore \mathrm{LHS}=\mathrm{RHS}\)
i.e., \(A \cap(B \cup C)=(A \cap B) \cup(A \cap C)\)

\(\frac{f(1)-f(-1)}{2}=\frac{3^{5}-1}{2}\)
\(=\frac{243-1}{2}\)
\(=\frac{242}{2}\)
\(=121\)
Hence,
The sum of the coefficients of the odd powers of \(x\) in the expansion of \(\left(1+x+x^{2}\right)^{5}\) is

\(\int_{-1}^{1} \frac{2 x^{3}+|x|+4}{|x|^{2}+5|x|+4}=\int_{-1}^{1} \frac{2 x^{3}}{(|x|+4)(|x|+1)} d x\)
\(\left(I_{1}\right)\)
\(+\int_{-1}^{1} \frac{|x|+4}{(|x|+4)(|x|+1)} d x\)
\(\left(I_{2}\right)\)
\(I=I_{1}+I_{2}\)
\(I_{1}=\int_{-1}^{1} \frac{2 x^{3}}{(|x|+4)(|x|+1)} d x\)
\(\mathrm{I}_{1} \rightarrow\) odd function \(\mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x})\)
\(\mathrm{so}, \mathrm{I}_{1}=0\)
\(I_{2}=\int_{-1}^{1} \frac{|x|+4}{(|x|+4)(|x|+1)} d x\)
\(=\int_{-1}^{1} \frac{1}{|x|+1} d x\)
\(\frac{1}{|x|+1} \rightarrow\) even function, as \(f(-x)=f(x)\)
\(\mathrm{o}, \mathrm{I}_{2}=2 \int_{0}^{1} \frac{1}{\mathrm{x} \mid+1} \mathrm{d} \mathrm{x}\)
\(\frac{1}{|x|+1}=\frac{1}{x+1}\) for \(x>0\)

\(\mathrm{so}, \mathrm{I}_{2}=2 \int_{0}^{1} \frac{1}{\mathrm{x}+1} \mathrm{dx}\)
\(\mathrm{I}_{2}=2[\ln (\mathrm{x}+1)]_{0}^{1}\)
\(\left[\mathrm{I}_{2}=2 \ln 2\right.\)
\(\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2} \Rightarrow \mathrm{I}=0+2 \ln 2=2 \ln 2\)
\(\mathrm{Now}, \int_{0}^{\pi / 2} \ln (\sin \alpha) \mathrm{d} \alpha=\frac{-\pi}{2} \ln 2\)
\(\mathrm{so}, \frac{-4}{\pi} \int_{0}^{\pi / 2} \ln (\sin \alpha) \mathrm{d} \alpha=2 \ln 2\)

The operations are as follows:
Addition: if \(A\) and \(B\) are matrices of the same size \(m \times n,\) then \(A+B\) their sum, is a matrix of size \(m \times n\)
Multiplication by scalars: if \(A\) is a matrix of size \(m \times n\) and \(c\) is a scalar, then \(c A\) is a matrix of size \(m \times n\)
Matrix multiplication: if \(A\) is a matrix of size \(m \times n\) and \(B\) is a matrix of size \(n \times p,\) then the product \(A B\) is a matrix of size \(m \times p\)
Vectors: a vector of length \(n\) can be treated as a matrix of size \(n \times\) 1, and the operations of vector addition, multiplication by scalars, and multiplying a matrix by a vector agree with the corresponding matrix operations.
Transpose: if \(A\) is a matrix of size \(m \times n\), then its transpose \(A^{T}\) is a matrix of size \(n \times m\).
Identity matrix: \(I_{n}\) is the \(n \times n\) identity matrix; its diagonal elements are equal to 1 and its offdiagonal elements are equal to 0 .
Zero matrix: we denote by 0 the matrix of all zeroes (of relevant size).
Inverse: if \(A\) is a square matrix, then its inverse \(A^{-1}\) is a matrix of the same size. Not every square matrix has an inverse! (The matrices that have inverses are called invertible.

In general, \(A B \neq B A\), even if \(A\) and \(B\) are both square. If \(A B=B A\) then we say that \(A\) and \(B\) commute.
For a general matrix \(A,\) we cannot say that \(A B=A C\) yields \(B=C\) (However, if we know that \(A\) is invertible, then we can multiply both sides of the equation \(\left.A B=A C \text { to the left by } A^{-1} \text {and get } B=C .\right)\)
The equation \(A B=0\) does not necessarily yield \(A=0\) or \(B=0 .\) For example, take
\[
A=\left[\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right], B=\left[\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right]
\]

\(\Rightarrow 2 \sec 2 \alpha=\frac{1+\tan ^{2} \beta}{\tan \beta}=\frac{\sec ^{2} \beta}{\tan \beta}=\frac{2}{2 \cos \beta \sin \beta}=2 \operatorname{cosec} 2 \beta\)
\(\therefore \sec 2 \alpha=\sec \left(\frac{\pi}{2}-2 \beta\right)\)
\(\Rightarrow 2 \alpha=2 n \pi \pm\left(\frac{\pi}{2}-2 \beta\right)\)
Taking +ve sign, we have
\(2(\alpha+\beta)=2 n \pi+\frac{\pi}{2}\)
\(\Rightarrow \alpha+\beta=n \pi+\frac{\pi}{4}, n \in I\)
For, \(n=0, \alpha+\beta=\frac{\pi}{4}\)

KEY CONCEPTS
Trigonometric Equations Solution for cos Function General solution of the equation \(\cos \theta=\cos \alpha:\) If \(\cos \theta=\cos \alpha \Rightarrow \cos \theta-\cos \alpha=0 \Rightarrow\)
\(-2 \sin \left(\frac{\theta+\alpha}{2}\right) \cdot \sin \left(\frac{\theta-\alpha}{2}\right)=0 \Rightarrow \sin \left(\frac{\theta+\alpha}{2}\right)=0\) or \(\sin \left(\frac{\theta-\alpha}{2}\right)=0, \Rightarrow \frac{\theta+\alpha}{2}=n \pi ; n \in I\) or
\(\left(\frac{\theta-\alpha}{2}\right)=n \pi ; n \in I\)

\(\Rightarrow \theta=2 n \pi-\alpha ; n \in I\) or \(\theta=2 n \pi+\alpha ; n \in I .\) for the general solution of \(\cos \theta=\cos \alpha,\) combine these two result which
gives \(\theta=2 n \pi \pm \alpha ; n \in I\)
Note:
The equation \(\sec \theta=\sec \alpha\) is equivalent to \(\cos \theta=\cos \alpha,\) so the general solution of these two equations are same.