Let f(x)= 7x^10+9x^8(1+x^2+2x^9)^2dx,x0,x 0limf(x)=0 and f(1)=(1)/(4).

Question

Let $f(x)=\int \frac{7{x}^{10}+9{x}^{8}}{{(1+{x}^{2}+2{x}^{9})}^{2}}dx,x>0,\underset{x	o 0}{lim}f(x)=0$ and $f(1)=\frac{1}{4}$.

If $A=[[|\begin{gathered} 0 \quad  0 \quad  1 \ \frac{1}{4} \quad  {f}^{'}(1) \quad  1 \ {\alpha }^{2} \quad  4 \quad  1 \end{gathered}]]|$ and $B=adj(adjA)$ be such that $|B|=81$, then ${\alpha }^{2}$ is equal to

Accepted Answer

$\begin{gathered} f(x)=\int \frac{7{x}^{10}+9{x}^{8}}{{(1+{x}^{2}+2{x}^{9})}^{2}}dx \ =\int \frac{7{x}^{10}+9{x}^{8}}{{x}^{18}{(\frac{1}{{x}^{9}}+\frac{1}{{x}^{7}}+2)}^{2}}=\int \frac{\frac{7}{{x}^{8}}+\frac{9}{{x}^{10}}}{{(\frac{1}{{x}^{9}}+\frac{1}{{x}^{7}}+2)}^{2}} \end{gathered}$

substitute $\frac{1}{{x}^{9}}+\frac{1}{{x}^{7}}+2=t$

$\begin{gathered} (\frac{-9}{{x}^{10}}-\frac{7}{{x}^{8}})dx=dt\Rightarrow (\frac{7}{{x}^{8}}+\frac{9}{{x}^{10}})dx=-dt \ f(t)=-\int \frac{dt}{{t}^{2}}=\frac{1}{t}+C \end{gathered}$

put value of t

$\begin{gathered} f(x)=\frac{1}{\frac{1}{{x}^{9}}+\frac{1}{{x}^{7}}+2}+C \ f(x)=\frac{{x}^{9}}{2{x}^{9}+{x}^{2}+1}+C \ \underset{x	o 0}{lim}f(x)=0 \ \underset{x	o 0}{lim}\frac{{x}^{9}}{2{x}^{9}+{x}^{2}+1}+C=0 \ 0+C=0\Rightarrow C=0 \ f(x)=\frac{{x}^{9}}{2{x}^{9}+{x}^{2}+1} \ {f}^{'}(x)=\frac{9{x}^{8}(2{x}^{9}+{x}^{2}+1)-{x}^{9}(18{x}^{8}+2x)}{{(2{x}^{9}+{x}^{2}+1)}^{2}} \ {f}^{'}(1)=\frac{9(1{)}^{8}(2(1{)}^{9}+(1{)}^{2}+1)-(1{)}^{9}(18(1{)}^{8}+2(1))}{{(2(1{)}^{9}+(1{)}^{2}+1)}^{2}} \ =\frac{9	imes 4-20}{16}=\frac{16}{16}=1 \end{gathered}$

It is given that $A=[[|\begin{gathered} 0 \quad  0 \quad  1 \ \frac{1}{4} \quad  {f}^{'}(1) \quad  1 \ {\alpha }^{2} \quad  4 \quad  1 \end{gathered}]]|$

$\begin{gathered} {f}^{'}(1)=1\Rightarrow A=[[|\begin{array}{ccc}0 \quad  0 \quad  1 \ \frac{1}{4} \quad  1 \quad  1 \ {\alpha }^{2} \quad  4 \quad  1\end{array}]]| \ |A|=1(4	imes \frac{1}{4}-{\alpha }^{2})=1-{\alpha }^{2} \end{gathered}$

It is given that $B=adj(adjA)\&|B|=81$

$|B|=|A{|}^{(n-1{)}^{2}},n=3$ as A is $3	imes 3$ matrix

$\begin{gathered} |B|=|A{|}^{(3-1{)}^{2}}=81 \ |A{|}^{4}=81={3}^{4} \ |A|=\pm 3 \end{gathered}$

$1-{\alpha }^{2}=\pm 3{|A|=1-{\alpha }^{2}}$

$1-{\alpha }^{2}=-31-{\alpha }^{2}=3$

${\alpha }^{2}=4$ or ${\alpha }^{2}=-2$ not possible

so value of ${\alpha }^{2}$ is 4

Solutions · 1