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- Definitions: FSc Part 1 (Mathematics): PTB by Aurang Zaib @fsc-part1-ptb
- th values of its variables. ===Example:=== \( p \rightarrow q \leftrightarrow (\neg q \rightarrow \neg p) \) is a tautology because its truth table shows that it is alwa... h values of its variables. ===Example:=== \( (p \rightarrow q) \land (p \lor q) \) is a contingency because i... \) and \( B = \{a, b, c\} \). A function \( f: A \rightarrow B \) could be defined as \( f(1) = a, f(2) = b, f
- Definitions: FSc Part 1 (Mathematics): PTB @fsc-part1-ptb
- ble involved in it is called tautology.\\ e.g. $p\rightarrow q\leftrightarrow (\sim q \rightarrow \sim p)$ is a tautology. * **Contradiction:** A statement which is a
- MathCraft: PDF to LaTeX file: Sample-02 @mathcraft
- ment} \maketitle \vspace{-0.7 in} If $f:[a, b] \rightarrow \mathbb{R}$ is convex, then $$ f\left(\frac{a+b}{
- MathCraft: PDF to LaTeX file: Sample-01 @mathcraft
- $ is log-convex in Jensen sense. Also $\lim _{r \rightarrow 0} \phi(r)=\phi(0)$, which implies $\phi$ is cont
- Question 3 & 4 Exercise 4.3 @math-11-kpk:sol:unit04
- it becomes,\\ \begin{align} 350&=25+(n-1)(5) \\ \Rightarrow 5 n-5+25&=350 \\ \Rightarrow 5 n&=350-20=330 \\ \Rightarrow n&=66, \text { now for the sum } \\ S_n&=\dfrac{n}{2}(a_1+a_n), \text { that becomes } \\ S_{66}&=\dfrac{66}{2}(25+350) \\ \Rightarrow S_{66}&=33(375)=12375 .\end{align} =====Question
- Question 14 Exercise 4.2 @math-11-kpk:sol:unit04
- ext{ and } a_6=41.$$ Now \begin{align}& a_5=11\\ \Rightarrow &a_1+4 d=41 \\ \Rightarrow &6+4 d=41 \\ \Rightarrow &d=\dfrac{41-6}{4}\\ &=\dfrac{35}{4}.\end{align} Now \begin{align} A_1&=a+d=6+\dfrac... } a_6=32.$$ Now \begin{align} & a_6=a_1+5 d \\ \Rightarrow &17+5 d=32 \\ \Rightarrow &d=\dfrac{32-17}{5}\\ &
- Question 12 & 13 Exercise 4.2 @math-11-kpk:sol:unit04
- b^{\prime}}{2}\\ &=\dfrac{(a+b)^2+(a-b)^2}{2} \\ \Rightarrow A&=\dfrac{a^2+b^2+2 a b+a^2+b^2-2 a b}{2} \\ & =\
- Question 8 Exercise 4.2 @math-11-kpk:sol:unit04
- -b}{b} \\ \text{Let}\quad S&=\dfrac{a+b+c}{2} \\ \Rightarrow a+b+c&=2 S\\ \text{then} \Rightarrow a+b-c&=2(S-c) \text {, }\\ a+c-b&=2(S-b), \quad\text{and}\\ b+c-a&=2(S-a... b-b S+a b}{a b}&=\dfrac{b S-b c-c S+b c}{b c} \\ \Rightarrow \dfrac{(a-b) S}{a b}&=\dfrac{(b-c) S}{b c}\end{align} Dividing both sides by $S$\\ \begin{align}\Rightarrow \dfrac{a-b}{a b}&=\dfrac{b-c}{b c} \\ \Rightarrow
- Question 3 and 4 Exercise 4.1 @math-11-kpk:sol:unit04
- -a_n.$$ For $n=1$ \begin{align}a_{1+1}&=5-a_1\\ \Rightarrow a_2&=5-3=2\end{align} For $n=2$ \begin{align}a_{2+1}&=5-a_2\\ \Rightarrow a_3&=5-2=3\end{align} For $n=3$ \begin{align}a_{3+1}&=5-a_3\\ \Rightarrow a_4&=5-3=2\end{align} For $n=4$ \begin{align}a_{4+1}&=5-a_4\\ \Rightarrow a_5&=5-2=3\end{align} Hence the first five terms
- Question 5 & 6 Review Exercise 7 @math-11-kpk:sol:unit07
- t of $x$ is possible only if $x^{4 r-20}=x^0$ $$ \Rightarrow 4 r-20=0 \Rightarrow r=5 \text {. } $$ Putting in $T_{r+1}$ we get $$ T_{5+1}=\frac{10 !}{5 ! 5 !} \cdot 2^0 \cdot x^0 $$ $$ \Rightarrow \quad T_6=252 . $$ Hence $T_6$ is constant term
- Question 7 & 8 Review Exercise 7 @math-11-kpk:sol:unit07
- gin{aligned} & =4.7^k+3\left[7^k-3^k\right] \\ & \Rightarrow 7^{k+1}-3^{k+1}=4.7^k+3.4 Q \end{aligned} $$ by induction hypothesis $$ \begin{aligned} & \Rightarrow 7^{k=1}-3^{k+1}=4\left[7^k+3 Q\right] \\ & \Rightarrow 7^{k-1}-3^{k+1} \text { is divisible by } 4 . \end{ali... 2 \\ & =1+(k+1) x+k x^2 \\ & \geq 1+(k+1) x \\ & \Rightarrow(1+x)^{k+1} \geq[1+(k+1) x] . \end{aligned} $$ He
- Question 3 & 4 Review Exercise 7 @math-11-kpk:sol:unit07
- t\left(2^4\right) \cdot(-4)^3 \cdot x^4 y^3 \\ & \Rightarrow T_4=-35840 x^4 y^3 . \end{aligned} $$ Q4 $2^7 x ... only if $$ x^{4 \prime \prime} y^{\prime}=x y^3 \Rightarrow r=3 \text {. } $$ Thus the term is: $$ \begin{al... 3 !} 2^3 a^4 x^{4-3} y^3 . \\ & =2^7 x y^3 \\ & \Rightarrow 4 \cdot 2^3 \cdot a^4=2^7 \\ & \Rightarrow a^4=\frac{2^7}{2^3 \cdot 2^2}=2^2 \\ & \Rightarrow a^2=\sqrt{2^2}=
- Question 11 Exercise 7.3 @math-11-kpk:sol:unit07
- !} \cdot \frac{1}{2^4} \cdot 16=\frac{3}{2} \\ & \Rightarrow 6 n=2 n-2 \text { or } n=-\frac{1}{2} . \\ & \tex... Eq.(1), we get } \\ & -\frac{1}{2} x=\frac{1}{4} \Rightarrow x=-\frac{1}{2} \text {. Thus } \\ & y+1=\left(1-\... e both sides $$ \begin{aligned} & (y+1)^2=2 \\ & \Rightarrow y^2+2 y+1-2=0 \\ & \Rightarrow y^2+2 y-1=0 . \end{aligned} $$ Which is the desired result. ====Go To====
- Question 10 Exercise 7.3 @math-11-kpk:sol:unit07
- n-1}{2 n}=\frac{3}{32} \cdot 16=\frac{3}{2} \\ & \Rightarrow 3 n=n-1 \Rightarrow n=-\frac{1}{2} . \end{aligned} $$ Putting $n=-\frac{1}{2}$ in Eq.(1), we get $$ \begin{aligned} & -\frac{1}{2} x=-\frac{1}{4} \\ & \Rightarrow x=\frac{1}{2} . \text { Thus } \\ & \left(1+\frac... {1 \cdot 3}{2 !} \cdot \frac{1}{2^4}+\ldots \\ & \Rightarrow\left(\frac{3}{2}\right)^{\frac{1}{2}}=1-\frac{1}{
- Question 7 and 8 Exercise 7.3 @math-11-kpk:sol:unit07
- e get that $a \cdot b x^2=2-\frac{3 x^2}{16}$ $$ \Rightarrow a=2 \text { and } b=-\frac{3}{16} . $$ Q8 If $x$