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- Question 2 Exercise 4.3 @math-11-kpk:sol:unit04
- s & 4 n^2-11 n-225=0\end{align} This is quadratic equation with $a=4, b=-11$ and $c=-225$, then \begin{align
- Question 11 Exercise 7.3 @math-11-kpk:sol:unit07
- {2^6}+\ldots$ Adding 1 to both sides of the above equation $$ S=y+1=1+\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \f
- Question 7 and 8 Exercise 7.3 @math-11-kpk:sol:unit07
- Solution: We are taking L.H.S of the above given equation and apply the binomial theorem $$ \begin{aligned}... rac{1}{4}=2-\frac{3 x^2}{16}$. From the above two equations, we get that $a \cdot b x^2=2-\frac{3 x^2}{16}$ ... taking numerator in the L.H.S of the above given equation ====Go To==== <text align="left"><btn type="
- Question 14 Exercise 7.3 @math-11-kpk:sol:unit07
- g binomial theorem on the R.H.S of the above last equation, $$ \begin{aligned} & p x^p-q x^q \\ & =p(1+p h+\
- Question 12 Exercise 7.3 @math-11-kpk:sol:unit07
- 6}+\ldots $$ Adding 1 to both sides of the above equation, we get $S=2 y+1=1+\frac{1}{2^2}+\frac{1.3}{2 !}
- Question 10 Exercise 7.2 @math-11-kpk:sol:unit07
- *\right) x^n \cdot $$ Putting $x=1$ in the above equation, we have $(1 \div 1)^n=\left(\begin{array}{l}n \\
- Question 2 Exercise 6.3 @math-11-kpk:sol:unit06
- 0 \end{align} Let $y=n^2-3 n$ then the above last equation becomes \begin{align} & y(y+2)=840 \\ & \Rightarr
- Question 7 and 8 Exercise 6.2 @math-11-kpk:sol:unit06
- different arrangements can be formed of the word "equation" if all the vowels are to be kept together? ====S... olution==== The total number of alphabets in word equation are $8$, out of which $5$ are vowels. If all the
- Question 3 and 4 Exercise 6.2 @math-11-kpk:sol:unit06
- {r-1})$$ We are taking the right hand side of the equation \begin{align}n(^{n-1} P_{r-1})&=n \dfrac{(n-1) !}... 1} P_r+r({ }^{n-1} P_{r-1})$$ Taking R.H.S of the equation \begin{align}^{n-1} P_r+r(^{n-1} P_{r-1})&=\dfrac
- Question 5 Exercise 6.1 @math-11-kpk:sol:unit06
- ====Solution==== We are taking L.H.S of the above equation \begin{align}\dfrac{(2 n) !}{n !}&=\dfrac{1}{n !}... n-(2 n-1))]\end{align} In the L.H.S of the above equation are total $2 n$ terms \begin{align}\dfrac{(2 n) !... ====Solution==== We are taking L.H.S of the above equation \begin{align}\dfrac{(2 n+1) !}{n !}&=\dfrac{1}{n ... n+1-(2 n))]\end{align} In the L.H.S of the above equation. are total $2 n+1$ terms \begin{align}\dfrac{(2 n
- Question 3 & 4 Exercise 6.1 @math-11-kpk:sol:unit06
- on==== We are taking the L.H.S of the above given equation. \begin{align}\dfrac{1}{6 !}+\dfrac{2}{7 !}+\dfra... on==== We are taking the L.H.S of the above given equation. \begin{align}\dfrac{(n+5) !}{(n+3) !}&=\dfrac{(n
- Question 5 & 6 Review Exercise @math-11-kpk:sol:unit05
- $n$ and constants on the both sides of the above equation, we get $$A+B=0\quad \text{and}\quad A=1$$ Puttin... } Taking summation of the both sides of the above equation \begin{align} \sum_{k=1}^n T_k&=\sum_{k=1}^n(\dfr
- Question 4 Review Exercise @math-11-kpk:sol:unit05
- \cdot n$ and constant terms on the both sides of equations, we get $$A+B+C=0 \quad 15 A+6 B-3 C=0$$ $$4 A+8 B-2 C=1$$ Solving these three equations for the constants $A, B$ and $C$ we get $A=\dfr
- Question 4 Exercise 5.4 @math-11-kpk:sol:unit05
- \dfrac{A}{n+3}+\dfrac{B}{n+4}$$ Solving the above equation for $A$ and $B$, we get $A=1$ and $B=-1$, so $$u
- Question 2 & 3 Exercise 5.4 @math-11-kpk:sol:unit05
- $k$ and constants on the both sides of the above equation, we get $$3 A+3 B=0\quad \text{and}\quad 2 A-B=1$$ Solving the above two equations for $A$ and $B$ we get $$A=\dfrac{1}{3}\quad\te... }=\dfrac{A}{n}+\dfrac{B}{n-1}$$ Solving the above equation for $A$ and $B$ we get $A=1$ and $B=-1$. So, \be