Show that $f\text{'}\text{'}\left(x\right)=-\frac{1}{4\sqrt{{\left(1+x\right)}^3}}$.

Use mathematical induction to prove that $f^{\left(n\right)}\left(x\right)={\left(-\frac{1}{4}\right)}^{n-1}\frac{\left(2n-3\right)!}{\left(n-2\right)!}{\left(1+x\right)}^{\frac{1}{2}-n}$ for $n\in \mathrm{\mathbb{Z}}, n\ge 2$.

Let $g\left(x\right)={\text{e}}^{mx}, m\in \mathrm{\mathbb{Q}}$.

Consider the function $h$ defined by $h\left(x\right)=f\left(x\right)\times g\left(x\right)$ for $x>-1$.

It is given that the $x^2$ term in the Maclaurin series for $h\left(x\right)$ has a coefficient of $\frac{7}{4}$.

Find the possible values of $m$.

attempt to use the chain rule            M1

$f\text{'}\left(x\right)=\frac{1}{2}{\left(1+x\right)}^{-\frac{1}{2}}$         A1

$f\text{'}\text{'}\left(x\right)=-\frac{1}{4}{\left(1+x\right)}^{-\frac{3}{2}}$         A1

$=-\frac{1}{4\sqrt{{\left(1+x\right)}^3}}$         AG

Note: Award M1A0A0 for $f\text{'}\left(x\right)=\frac{1}{\sqrt{1+x}}$ or equivalent seen

[3 marks]

let $n=2$

$f^{\text{'}\text{'}}\left(x\right)=\left(-\frac{1}{4\sqrt{{\left(1+x\right)}^3}}=\right){\left(-\frac{1}{4}\right)}^1\frac{1!}{0!}{\left(1+x\right)}^{\frac{1}{2}-2}$         R1

Note: Award R0 for not starting at $n=2$. Award subsequent marks as appropriate.

assume true for $n=k$, (so $f^{\left(k\right)}\left(x\right)={\left(-\frac{1}{4}\right)}^{k-1}\frac{\left(2k-3\right)!}{\left(k-2\right)!}{\left(1+x\right)}^{\frac{1}{2}-k}$)       M1

Note: Do not award M1 for statements such as “let $n=k$” or “$n=k$ is true”. Subsequent marks can still be awarded.

consider $n=k+1$

$\text{LHS}=f^{\left(k+1\right)}\left(x\right)=\frac{d\left(f^{\left(k\right)}\left(x\right)\right)}{dx}$            M1

$={\left(-\frac{1}{4}\right)}^{k-1}\frac{\left(2k-3\right)!}{\left(k-2\right)!}\left(\frac{1}{2}-k\right){\left(1+x\right)}^{\frac{1}{2}-k-1}$ (or equivalent)         A1

EITHER

$\text{RHS}=f^{\left(k+1\right)}\left(x\right)={\left(-\frac{1}{4}\right)}^k\frac{\left(2k-1\right)!}{\left(k-1\right)!}{\left(1+x\right)}^{\frac{1}{2}-k-1}$ (or equivalent)         A1

$={\left(-\frac{1}{4}\right)}^k\frac{\left(2k-1\right)\left(2k-2\right)\left(2k-3\right)!}{\left(k-1\right)\left(k-2\right)!}{\left(1+x\right)}^{\frac{1}{2}-k-1}$        A1

Note: Award A1 for $\frac{\left(2k-1\right)!}{\left(k-1\right)!}=\frac{\left(2k-1\right)\left(2k-2\right)\left(2k-3\right)!}{\left(k-1\right)\left(k-2\right)!}\left(=\frac{2\left(2k-1\right)\left(2k-3\right)!}{\left(k-2\right)!}\right)$

$=\left(-\frac{1}{4}\right){\left(-\frac{1}{4}\right)}^{k-1}\frac{\left(2k-1\right)\left(2k-2\right)\left(2k-3\right)!}{\left(k-1\right)\left(k-2\right)!}{\left(1+x\right)}^{\frac{1}{2}-k-1}$        A1

$\left(=\left(-\frac{1}{2}\right){\left(-\frac{1}{4}\right)}^{k-1}\frac{\left(2k-1\right)\left(2k-3\right)!}{\left(k-2\right)!}{\left(1+x\right)}^{\frac{1}{2}-k-1}\right)$

Note: Award A1 for leading coefficient of $-\frac{1}{4}$.

$=\left(\frac{1}{2}-k\right){\left(-\frac{1}{4}\right)}^{k-1}\frac{\left(2k-3\right)!}{\left(k-2\right)!}{\left(1+x\right)}^{\frac{1}{2}-k-1}$        A1

OR

Note: The following A marks can be awarded in any order.

$={\left(-\frac{1}{4}\right)}^{k-1}\frac{\left(2k-3\right)!}{\left(k-2\right)!}\left(\frac{1-2k}{2}\right){\left(1+x\right)}^{\frac{1}{2}-k-1}$

$=\left(-\frac{1}{2}\right){\left(-\frac{1}{4}\right)}^{k-1}\frac{\left(2k-1\right)\left(2k-3\right)!}{\left(k-2\right)!}{\left(1+x\right)}^{\frac{1}{2}-k-1}$        A1

Note: Award A1 for isolating $(2k-1)$ correctly.

$=\left(-\frac{1}{2}\right){\left(-\frac{1}{4}\right)}^{k-1}\frac{\left(2k-1\right)!}{\left(2k-2\right)\left(k-2\right)!}{\left(1+x\right)}^{\frac{1}{2}-k-1}$        A1

Note: Award A1 for multiplying top and bottom by $(k-1)$ or $2(k-1)$.

$=\left(-\frac{1}{4}\right){\left(-\frac{1}{4}\right)}^{k-1}\frac{\left(2k-1\right)!}{\left(k-1\right)\left(k-2\right)!}{\left(1+x\right)}^{\frac{1}{2}-k-1}$        A1

Note: Award A1 for leading coefficient of $-\frac{1}{4}$.

$={\left(-\frac{1}{4}\right)}^k\frac{\left(2k-1\right)!}{\left(k-1\right)!}{\left(1+x\right)}^{\frac{1}{2}-k-1}$        A1

$={\left(-\frac{1}{4}\right)}^{\left(k+1\right)-1}\frac{\left(2\left(k+1\right)-3\right)!}{\left(\left(k+1\right)-2\right)!}{\left(1+x\right)}^{\frac{1}{2}-\left(k+1\right)}=\text{RHS}$

THEN

since true for $n=2$, and true for $n=k+1$ if true for $n=k$, the statement is true for all, $n\in \mathrm{\mathbb{Z}}, n\ge 2$  by mathematical induction           R1

Note: To obtain the final R1, at least four of the previous marks must have been awarded.

[9 marks]

METHOD 1

$h\left(x\right)=\sqrt{1+x }{\text{e}}^{mx}$

using product rule to find $h\text{'}\left(x\right)$        (M1)

$h\text{'}\left(x\right)=\sqrt{1+x }m{\text{e}}^{mx}\mathrm{+}\frac{1}{2\sqrt{1+\mathrm{x}}}{\text{e}}^{mx}$         A1

$h\text{'}\text{'}\left(x\right)=m\left(\sqrt{1+x }m{\text{e}}^{mx}\mathrm{+}\frac{1}{2\sqrt{1+\mathrm{x}}}{\text{e}}^{mx}\right)+\frac{1}{2\sqrt{1+\mathrm{x}}}m{\text{e}}^{mx}-\frac{1}{4\sqrt{{\left(1+x\right)}^3}}{\text{e}}^{mx}$         A1

substituting $x=0$ into $h\text{'}\text{'}\left(x\right)$       M1

$h\text{'}\text{'}\left(0\right)=m^2+\frac{1}{2}m+\frac{1}{2}m-\frac{1}{4}\left(=m^2+m-\frac{1}{4}\right)$         A1

$h\left(x\right)=h\left(0\right)+xh\text{'}\left(0\right)+\frac{x^2}{2!}h\text{'}\text{'}\left(0\right)+\dots$

equating $x^2$ coefficient to $\frac{7}{4}$       M1

$\frac{h\text{'}\text{'}\left(0\right)}{2!}=\frac{7}{4} \left(\Rightarrow h\text{'}\text{'}\left(0\right)=\frac{7}{2}\right)$

$4m^2+4m-15=0$         A1

$\left(2m+5\right)\left(2m-3\right)=0$

$m=-\frac{5}{2}$  or  $m=\frac{3}{2}$         A1

METHOD 2

EITHER

attempt to find $f\left(0\right), f\text{'}\left(0\right), f\text{'}\text{'}\left(0\right)$        (M1)

$f\left(x\right)={\left(1+x\right)}^{\frac{1}{2}} f\left(0\right)=1$

$f\text{'}\left(x\right)=\frac{1}{2}{\left(1+x\right)}^{-\frac{1}{2}} f\text{'}\left(0\right)=\frac{1}{2}$

$f\text{'}\text{'}\left(x\right)=-\frac{1}{4}{\left(1+x\right)}^{-\frac{3}{2}} f\text{'}\text{'}\left(0\right)=-\frac{1}{4}$

$f\left(x\right)=1+\frac{1}{2}x-\frac{1}{8}{x}^2+\dots$         A1

OR

attempt to apply binomial theorem for rational exponents        (M1)

$f\left(x\right)={\left(1+x\right)}^{\frac{1}{2}}=1+\frac{1}{2}x+\frac{\left({\displaystyle \frac{1}{2}}\right)\left(-{\displaystyle \frac{1}{2}}\right)}{2!}{x}^2\dots$

$f\left(x\right)=1+\frac{1}{2}x-\frac{1}{8}{x}^2+\dots$         A1

THEN

$g\left(x\right)=1+mx+\frac{m^2}{2}{x}^2+\dots$        (A1)

$h\left(x\right)=\left(1+\frac{1}{2}x-\frac{1}{8}{x}^2+\dots \right)\left(1+mx+\frac{m^2}{2}{x}^2+\dots \right)$        (M1)

coefficient of $x^2$ is $\frac{m^2}{2}+\frac{m}{2}-\frac{1}{8}$         A1

attempt to set equal to $\frac{7}{4}$ and solve             M1

$\frac{m^2}{2}+\frac{m}{2}-\frac{1}{8}=\frac{7}{4}$

$4m^2+4m-15=0$          A1

$\left(2m+5\right)\left(2m-3\right)=0$

$m=-\frac{5}{2}$  or  $m=\frac{3}{2}$         A1

METHOD 3

$g\text{'}\left(x\right)=m{\text{e}}^{mx}$ and $g\text{'}\text{'}\left(x\right)=m^2{\text{e}}^{mx}$        (A1)

$h\left(x\right)=h\left(0\right)+xh\text{'}\left(0\right)+\frac{x^2}{2!}h\text{'}\text{'}\left(0\right)+\dots$

equating $x^2$ coefficient to $\frac{7}{4}$       M1

$\frac{h\text{'}\text{'}\left(0\right)}{2!}=\frac{7}{4} \left(\Rightarrow h\text{'}\text{'}\left(0\right)=\frac{7}{2}\right)$

using product rule to find $h\text{'}\left(x\right)$ and $h\text{'}\text{'}\left(x\right)$        (M1)

$h\text{'}\left(x\right)=f\left(x\right)g\text{'}\left(x\right)+f\text{'}\left(x\right)g\left(x\right)$

$h\text{'}\text{'}\left(x\right)=f\left(x\right)g\text{'}\text{'}\left(x\right)+2f\text{'}\left(x\right)g\text{'}\left(x\right)+f\text{'}\text{'}\left(x\right)g\left(x\right)$         A1

substituting $x=0$ into $h\text{'}\text{'}\left(x\right)$       M1

$h\text{'}\text{'}\left(0\right)=f\left(0\right)g\text{'}\text{'}\left(0\right)+2g\text{'}\left(0\right)f\text{'}\left(0\right)+g\left(0\right)f\text{'}\text{'}\left(0\right)$

$=1\times m^2+2m\times \frac{1}{2}+1\times \left(-\frac{1}{4}\right) \left(=m^2+m-\frac{1}{4}\right)$         A1

$4m^2+4m-15=0$          A1

$\left(2m+5\right)\left(2m-3\right)=0$

$m=-\frac{5}{2}$  or  $m=\frac{3}{2}$         A1

[8 marks]