IB Mathematics

Question 1

The function `f` is defined by `f(x)=e^x sinx`, where `x∈R`.

(a) Find the Maclaurin series for `f(x)` up to and including the `x^3` term.

(b) Hence, find an approximate value for `∫_0^1e^(x^2 ) sin(x^2 )dx`.

The function `g` is defined by `g(x)=e^x cosx`, where `x∈R`.

(c) (i) Show that `g(x)` satisfies the equation `g^('')(x)=2(g^' (x)-g(x))`.

(ii) Hence, deduce that `g^((4)) (x)=2(g^(''')(x)-g^('')(x))`.

(d) Using the result from part (c), find the Maclaurin series for `g(x)` up to and including the `x^4` term.

(e) Hence, or otherwise, determine the value of `lim_(x→0) (e^x cosx-1-x)/x^3`.

(a)

recognition of both known series

`e^x=1+x/(1!)+x^2/(2!)+⋯` and `sinx=x-x^3/(3!)+x^5/(5!)+⋯`

attempt to multiply the two series up to and including `x^3` term

`e^x sinx=(1+x/(1!)+x^2/(2!)+⋯)(x-x^3/(3!)+x^5/(5!)+⋯)`

`=x-x^3/(3!)+x^2+x^3/(2!)+⋯`

`e^x sinx=x+x^2+1/3 x^3+⋯`

(b) `e^(x^2 ) sin(x^2 )=x^2+x^4+1/3 x^6+⋯`

substituting their expression and attempt to integrate

`∫_0^1e^(x^2 ) sin(x^2 )dx≈∫_0^1(x^2+x^4+1/3 x^6 ) dx`

`=[x^3/3+x^5/5+x^7/21]_0^1`

`=61/105`

(c)

(i) attempt to use product rule at least once

`g^'(x)=e^x cosx-e^x sinx`

`g^('')(x)=e^x cosx-e^x sinx-e^x sinx-e^x cosx(=-2e^x sinx)`

`2(g^' (x)-g(x))=2(e^x cosx-e^x sinx-e^x cosx)=-2e^x sinx`

`g^'' (x)=2(g^' (x)-g(x))`

(ii) `g^(''')(x)=2(g^('')(x)-g^'(x))`

`g^((4)) (x)=2(g^(''')(x)-g^('')(x))`

(d) attempt to substitute `x=0` into a derivative

`g(0)=1,g^' (0)=1,g^('')(0)=0, g^(''')(0)=-2,g^((4)) (0)=-4`

attempt to substitute into Maclaurin formula

`g(x)=1+x-2/(3!) x^3-4/(4!) x^4+⋯(=1+x-1/3 x^3-1/6 x^4+⋯)`

(e)

`lim_(x→0) (e^x cosx-1-x)/x^3 =lim_(x→0) ((1+x-1/3 x^3-1/6 x^4+⋯)-1-x)/x^3`

`=lim_(x→0) (-1/3-1/6 x+⋯)`

`=-1/3`

Question 2

(a) Let `f(x)=(1-ax)^(-1/2)`, where `ax<1,a≠0`.

The `n^(th)` derivative of `f(x)` is denoted by `f^((n)) (x),n∈Z^+`.

Prove by induction that `f^((n)) (x)=(a^n (2n-1)!(1-ax)^(-(2n+1)/2))/(2^(2n-1) (n-1)!),n∈Z^+.`.

(b) By using part (a) or otherwise, show that the Maclaurin series for `f(x)=(1-ax)^(-1/2)` up to and including the `x^2` term is `1+1/2 ax+3/8 a^2 x^2`.

(c) Hence, show that `(1-2x)^(-1/2) (1-4x)^(-1/2)≈(2+6x+19x^2)/2`.

(d) Given that the series expansion for `(1-ax)^(-1/2)` is convergent for `|ax|<1`, state the restriction which must be placed on `x` for the approximation `(1-2x)^(-1/2) (1-4x)^(-1/2)≈(2+6x+19x^2)/2` to be valid.

(e) Use `x=1/10` to determine an approximate value for `sqrt3`.

Give your answer in the form `c/d`, where `c,d∈Z^+`.

(a)

`n=1:LHS=f^((1)) (x)=-1/2×-a(1-ax)^(-3/2) (=a/2(1-ax)^(-3/2) )`

`RHS=(a(1)!(1-ax)^(-3/2))/(2^1 (0)!)`

therefore true for `n=1`

assume (that the result is) true for `n=k`

`f^((k)) (x)=(a^k (2k-1)!(1-ax)^(-((2k+1))/2))/(2^(2k-1) (k-1)!)`

attempt to differentiate the right-hand side with respect to `x`:

`f^((k+1)) (x)=d/( dx) (f^((k)) (x))`

`(-(2k+1)×-a)/2×(a^k (2k-1)!(1-ax)^(-((2k+1))/2-1))/(2^(2k-1) (k-1)!))`

(or equivalent)

attempt to multiply top and bottom by `2k`

`=((2k+1))/2×(a^(k+1) (2k-1)!(1-ax)^(-(2k+3)/2))/(2^(2k-1) (k-1)!)×(2k)/(2k)`

`=(a^(k+1) (2k+1)!(1-ax)^(-(2k+3)/2))/(2^(2k+1) (k)!))`

hence if the result is true for `n=k` then it is true for `n=k+1` and as it is true for `n=1` it is true for all `n∈Z^+`

(b)

`f(x)=f(0)+xf^' (0)+x^2/2 f^('')(0)+⋯`

`f^('')(x)=(a^2 (3)!(1-ax)^(-5/2))/(2^3 (1)!)`

`3/4 a^2 (1-ax)^(-5/2)`

`f^('')(0)=(a^2 (3)!)/2^3`

`f(0)=1,f^' (0)=a/2,f^('')(0)=(6a^2)/8`

`f(x)=1+1/2 ax+3/8 a^2 x^2+⋯`

(c) attempt to use `a=2` or `a=4` in the expansion

`(1-2x)^(-1/2)=1+2x/2+(3×4x^2)/8 (=1+x+3/2 x^2+⋯)`

`(1-4x)^(-1/2)=1+4x/2+(3×16x^2)/8 (=1+2x+6x^2+⋯) )`

attempt to multiply their two expansions together

`(1+x+3/2 x^2+⋯)(1+2x+6x^2+⋯)=1+2x+6x^2+x+2x^2+3/2 x^2+⋯`

`=1+3x+19/2 x^2+⋯" OR " (2+4x+12x^2+2x+4x^2+3x^2+⋯)/2)`

`(1-2x)^(-1/2) (1-4x)^(-1/2)≈(2+6x+19x^2)/2`

(d) `|x|<1/4`

(e) `((2+6x+19x^2)/2)=(2+0.6+0.19)/2 (=279/200)`

attempt to substitute `x=1/10` into `(1-2x)^(1/2) (1-4x)^(1/2)`

`((1-2x)^(-1/2) (1-4x)^(-1/2)=) 10/sqrt48` (or equivalent)

`10/(4sqrt3)≈279/200` (or equivalent in terms of `sqrt3` )

`1/sqrt3≈279/500 ; sqrt3≈500/279`

Question 3

(a) Find the first two non-zero terms in the Maclaurin series of

(i) `sin(x^2 );`

(ii) `sin^2 (x^2 )`

(b) Hence, or otherwise, find the first two non-zero terms in the Maclaurin series of `4xsin(x^2 )cos(x^2 )`.

(a) (i) `sin(x^2 )=x^2-(x^2 )^3/3!+⋯`

`sin(x^2 )=x^2-x^6/3!+⋯(=x^2-x^6/6+⋯)`

(ii) attempt to square their series for `sin(x^2 )`

`(sin(x^2 ))^2=(x^2-x^6/(3!)+⋯)^2`

`=x^4-(2x^8)/(3!)+⋯(=x^4-x^8/3+⋯)`

(b) recognition that `4xsin(x^2 )cos(x^2 )=(d((sin(x^2 ))^2 ))/dx`

`=4x^3-(8x^7)/3+⋯`

Question 4

The function `f` is defined by `f(x)=ln(1+x^2 )` where `-1 < x < 1`.

(a) (i) Use the Maclaurin series for `ln(1+x)` to write down the first three non-zero terms of the Maclaurin series for `f(x)`.

(ii) Hence find the first three non-zero terms of the Maclaurin series for `x/(1+x^2 )`.

(b) Use your answer to part (a)(i) to write down an estimate for `f(0.4)`.

The seventh derivative of `f` is given by `f^((7)) (x)=(1440x(x^6-21x^4+35x^2-7))/(1+x^2 )^7 .`

(c) (i) Use the Lagrange form of the error term to find an upper bound for the absolute value of the error in calculating `f(0.4)`, using the first three non-zero terms of the Maclaurin series for `f(x)`.

(ii) With reference to the Lagrange form of the error term, explain whether your answer to part (b) is an overestimate or an underestimate for `f(0.4)`.

(a) (i) substitution of `x^2` in `ln(1+x)=x-x^2/2+x^3/3-…`

`x^2-x^4/2+x^6/3`

(ii) `d/dx (ln(1+x^2 ))=(2x)/(1+x^2 )`

attempt to differentiate their answer in part (a)

`(2x)/(1+x^2 )=2x-(4x^3)/2+(6x^5)/3`

`x/(1+x^2 )=x-x^3+x^5`

(b) `f(0.4)≈0.149`

(c) (i) attempt to find the maximum of `|f^((7)) (c)|` for `c∈[0,0.4]`

maximum of `|f^((7)) (c)|` occurs at `c=0.199`

`|f^((7)) (c)|<1232.97…` (for all `c∈]0,0.4[` )

use of `x=0.4`

substitution of `n=6` and `a=0` and their value of `x` and their value of `f^((7)) (c)` into Lagrange error term

`|R_6 (0.4)|<(1232.97(0.4)^7)/(7!)`

upper bound `=0.000401`

(ii) `f^((7)) (c)<0` (for all `c∈]0,0.4[`)

The answer in (b) is an overestimate

Question 5

Let `f(x)=sqrt(1+x)` for `x > -1`.

(a) Show that `f^('')(x)=-1/(4sqrt((1+x)^3 )).`

(b) Use mathematical induction to prove that `f^((n)) (x)=(-1/4)^(n-1) ((2n-3)!)/((n-2)!)(1+x)^(1/2-n)`for `n∈Z,n ≥ 2`.

Let `g(x)=e^(mx),m∈Q.`.

Consider the function `h` defined by `h(x)=f(x)×g(x)` for `x > -1`.

It is given that the `x^2` term in the Maclaurin series for `h(x)` has a coefficient of `7/4`.

(c) Find the possible values of `m`.

(a) attempt to use the chain rule

`f^' (x)=1/2(1+x)^(-1/2)`

`f^('')(x)=-1/4(1+x)^(-3/2)`

`=-1/(4sqrt((1+x)^3 ))`

(b) let `n=2`

`f^('')(x)=(-1/(4sqrt((1+x)^3 ))=) (-1/4)^1 (1!)/(0!)(1+x)^(1/2-2)`

assume true for `n=k`, (so `f^((k)) (x)=(-1/4)^(k-1) ((2k-3)!)/((k-2)!)(1+x)^(1/2-k)` )

consider `n=k+1`

`LHS=f^((k+1)) (x)=(d(f^((k)) (x)))/dx`

`=(-1/4)^(k-1) ((2k-3)!)/((k-2)!) (1/2-k)(1+x)^(1/2-k-1) " (or equivalent) "`

`RHS=f^((k+1)) (x)=(-1/4)^k ((2k-1)!)/((k-1)!)(1+x)^(1/2-k-1) " (or equivalent) "`

`=(-1/4)^k ((2k-1)(2k-2)(2k-3)!)/((k-1)(k-2)!)(1+x)^(1/2-k-1)`

`=(-1/4) (-1/4)^(k-1) ((2k-1)(2k-2)(2k-3)!)/((k-1)(k-2)!)(1+x)^(1/2-k-1)`

`(=(-1/2) (-1/4)^(k-1) ((2k-1)(2k-3)!)/((k-2)!)(1+x)^(1/2-k-1) )`

`=(1/2-k) (-1/4)^(k-1) ((2k-3)!)/((k-2)!)(1+x)^(1/2-k-1)`

since true for `n=2`, and true for `n=k+1` if true for `n=k`, the statement is true for all `n ∈ Z,n ≥ 2` by mathematical induction

(c) `h(x)=sqrt(1+x) e^(mx)`

using product rule to find `h^' (x)`

`h^' (x)=sqrt(1+x) me^(mx)+1/(2sqrt(1+x)) e^(mx)`

`h^('')(x)=m(sqrt(1+x) me^(mx)+1/(2sqrt(1+x)) e^(mx) )+1/(2sqrt(1+x)) me^(mx)-1/(4sqrt((1+x)^3 )) e^(mx)`

substituting `x=0` into `h^('')(x)`

`h^('')(0)=m^2+1/2 m+1/2 m-1/4 (=m^2+m-1/4)`

`h(x)=h(0)+xh^' (0)+x^2/(2!) h^('')(0)+⋯`

equating `x^2` coefficient to `7/4`

`(h^('')(0))/(2!)=7/4 (⇒h^('')(0)=7/2)`

`4m^2+4m-15=0`

`(2m+5)(2m-3)=0`

`m=-5/2 " or " m=3/2`

Question 6

The function `f` is defined by `f(x)=arcsin(2x)`, where `-1/2 ≤ x ≤ 1/2`.

(a) By finding a suitable number of derivatives of `f`, find the first two non-zero terms in the Maclaurin series for `f`.

(b) Hence or otherwise, find `lim_(x→0) (arcsin(2x)-2x)/((2x)^3 )`.

(a) `f(x)=arcsin(2x)`

`f^' (x)=2/sqrt(1-4x^2 )`

`f^('')(x)=(8x)/(1-4x^2 )^(3/2)`

`f^(''')(x)=(8(1-4x^2 )^(3/2)-8x(3/2(-8x)(1-4x^2 )^(1/2) ))/(1-4x^2 )^3`

`=(8(1-4x^2 )^(3/2)+96x^2 (1-4x^2 )^(1/2))/(1-4x^2 )^3`

substitute `x=0` into `f` or any of its derivatives

`f(0)=0,f^' (0)=2` and `and f^('')(0)=0`

`f^(''')(0)=8`

the Maclaurin series is

`f(x)=2x+(8x^3)/6+⋯(=2x+(4x^3)/3+⋯)`

(b)

`lim_(x→0) (arcsin(2x)-2x)/((2x)^3 )=lim_(x→0) (2x+(4x^3)/3+⋯-2x)/(8x^3 )`

`=lim_(x→0) ( 4/3+⋯" terms with " x)/8`

`=1/6`

Question 7

(a) Prove by mathematical induction that `d^n/dx^n (x^2 e^x )=[x^2+2nx+n(n-1)] e^x` for `n∈Z^+`.

(b) Hence or otherwise, determine the Maclaurin series of `f(x)=x^2 e^x` in ascending powers of `x`, up to and including the term in `x^4`.

(c) Hence or otherwise, determine the value of `lim_(x→0) [(x^2 e^x-x^2 )^3/x^9 ].`.

(a) For `n=1`

LHS: `d/dx (x^2 e^x )=x^2 e^x+2xe^x (=e^x (x^2+2x))`

RHS: `(x^2+2(1)x+1(1-1)) e^x (=e^x (x^2+2x))`

so true for `n=1`

now assume true for `n=k`; i.e. `d^k/( dx^k ) (x^2 e^x )=[x^2+2kx+k(k-1)] e^x`

attempt to differentiate the RHS

`d^(k+1)/( dx^(k+1) ) (x^2 e^x )=d/( dx) ([x^2+2kx+k(k-1)] e^x )`

`=(2x+2k)e^x+(x^2+2kx+k(k-1)) e^x )`

`=[x^2+2(k+1)x+k(k+1)] e^x`

so true for `n=k` implies true for `n=k+1`

therefore `n=1` true and `n=k` true `⇒n=k+1` true

therefore, true for all `n∈Z^+`

(b ' `x^2×` Maclaurin series of `e^x` '

`x^2 (1+x+x^2/(2!)+⋯) ⇒f(x)≈x^2+x^3+1/2 x^4`

(c) attempt to substitute `x^2 e^x≈x^2+x^3+1/2 x^4` into `(x^2 e^x-x^2 )^3/x^9`

`(x^2 e^x-x^2 )^3/x^9 ≈(x^2+x^3+1/2 x^4 (+⋯)-x^2 )^3/x^9`

`=(x^3+1/2 x^4+⋯)^3/x^9 =(x^9 (+higherorderterns))/x^9`

`=1` (+ higher order terms) So `lim_(x→0) [(x^2 e^x-x^2 )^3/x^9 ]=1`

Question 8

The function `f` is defined by `f(x)=(arcsinx)^2,-1 ≤ x ≤ 1.`.

(a) Show that `f^' (0)=0`.

The function `f` satisfies the equation `(1-x^2 ) f^('')(x)-xf^' (x)-2=0`.

(b) By differentiating the above equation twice, show that `(1-x^2 ) f^((4)) (x)-5xf^((3)) (x)-4f^('')(x)=0`

where `f^((3)) (x)` and `f^((4)) (x)` denote the 3rd and 4th derivative of `f(x)` respectively.

(c) Hence show that the Maclaurin series for `f(x)` up to and including the term in `x^4` is `x^2+1/3 x^4`.

(d) Use this series approximation for `f(x)` with `x=1/2` to find an approximate value for `π^2`.

(a) `f^' (x)=(2arcsin(x))/sqrt(1-x^2 )`

`f^' (0)=0`

(b) differentiating gives `(1-x^2 ) f^((3)) (x)-2xf^('')(x)-f^' (x)-xf^('')(x)(=0)`

differentiating again gives `(1-x^2 ) f^((4)) (x)-2xf^((3)) (x)-3f^('')(x)-3xf^((3)) (x)-f^('')(x)(=0)`

`(1-x^2 ) f^((4)) (x)-5xf^((3)) (x)-4f^('')(x)=0`

(c) attempting to find one of `f^('')(0),f^((3)) (0)` or `f^((4)) (0)` by substituting `x=0` into relevant differential equation(s)

`(f(0)=0,f^' (0)=0)`

`f^('')(0)=2 and f^((4)) (0)-4f^('')(0)=0⇒f^((4)) (0)=8`

`f^((3)) (0)=0 and so 2/(2!) x^2+8/(4!)x^4`

so the Maclaurin series for `f(x)` up to and including the term in `x^4` is `x^2+1/3 x^4`

(d) substituting `x=1/2` into `x^2+1/3 x^4`

the series approximation gives a value of `13/48`

so `π^2=13/48×36 =9.75(=39/4)`

Question 9

Consider the function `f(x)=sin(parcsinx),-1 < x < 1 and p∈R`

(a) Show that `f^' (0)=p`

The function `f` and its derivatives satisfy

`(1-x^2 ) f^((n+2)) (x)-(2n+1)xf^((n+1)) (x)+(p^2-n^2 ) f^((n)) (x)=0,n∈N`

where `f^((n)) (x)` denotes the `n`th derivative of `f(x)` and `f^((0)) (x)` is `f(x)`.

(b) Show that `f^((n+2)) (0)=(n^2-p^2 ) f^((n)) (0)`.

(c) For `p∈R∖{±1,±3}`, show that the Maclaurin series for `f(x)`, up to and including the `x^5` term, is

`px+(p(1-p^2 ))/(3!) x^3+(p(9-p^2 )(1-p^2 ))/(5!) x^5`

(d) Hence or otherwise, find `lim_(x→0) (sin(parcsinx))/x`.

(e) If `p` is an odd integer, prove that the Maclaurin series for `f(x)` is a polynomial of degree `p`.

(a) `f^' (x)=(pcos(parcsinx))/sqrt(1-x^2 )`

`f^' (0)=p`

(b) `f^((n+2)) (0)+(p^2-n^2 ) f^((n)) (0)=0" (or equivalent)`

`" f^((n+2)) (0)=(n^2-p^2 ) f^((n)) (0)`

(c) considering `f` and its derivatives at `x=0`

`f(0)=0` and `f^' (0)=p` from (a)

`f^('')(0)=0,f^((4)) (0)=0`

`f^((3)) (0)=(1-p^2 ) f^((1)) (0)=(1-p^2 )p,`

`f^((5)) (0)=(9-p^2 ) f^((3)) (0)=(9-p^2 )(1-p^2 )p`

`px+(p(1-p^2 ))/(3!) x^3+(p(9-p^2 )(1-p^2 ))/(5!) x^5`

(d)

`lim_(x→0) (sin(parcsinx))/x=lim_(x→0) (px+p(1-p^2 )/(3!) x^3+⋯)/x`

`=p`

(e) the coefficients of all even powers of `x` are zero

the coefficient of `x^p` for ( `p` odd) is non-zero (or equivalent eg, the coefficients of all odd powers of `x` up to `p` are non-zero)

`f^((p+2)) (0)=(p^2-p^2 ) f^((p)) (0)=0` and so the coefficient of `x^(p+2)` is zero

the coefficients of all odd powers of `x` greater than `p+2` are zero (or equivalent)

so the Maclaurin series for `f(x)` is a polynomial of degree `p`

Question 10

Let the Maclaurin series for `tanx` be

`tanx=a_1 x+a_3 x^3+a_5 x^5+⋯`

where `a_1,a_3` and `a_5` are constants.

(a) Find series for `sec^2 x`, in terms of `a_1,a_3` and `a_5`, up to and including the `x^4` term

(i) by differentiating the above series for `tanx`;

(ii) by using the relationship `sec^2 x=1+tan^2 x`.

(b) Hence, by comparing your two series, determine the values of `a_1,a_3` and `a_5`.

(a)

(i) `(sec^2 x=) a_1+3a_3 x^2+5a_5 x^4+⋯`

(ii) `sec^2 x=1+(a_1 x+a_3 x^3+a_5 x^5+⋯)^2`

`=1+a_1^2 x^2+2a_1 a_3 x^4+⋯`

(b) equating constant terms: a_1=1

equating `x^2` terms: `3a_3=a_1^2=1⇒a_3=1/3`

equating `x^4` terms: `5a_5=2a_1 a_3=2/3⇒a_5=2/15`