1+ √x x2+1 x+√ x2+1 = √x2+1+x √x2+1 x+√ x2+1

B I (01-06 Ú)-5æ¶}

1.

x²+ 1) , t° f⁽ⁿ⁾(0) . (10 })

j. Äf⁰(x) = ¹⁺

√x x2+1

x+√ x²+1 =

√x2+1+x

√x2+1

x+√

x²+1 =√ ¹

x²+1 = (1 + x²)⁻¹²

=P∞

m=0(−¹2)(−³2)···(−^2m−12 )

m! · x^2m, x ∈ (−1, 1).

]f (x) =P∞ m=0

(−¹2)(−³2)···(−^2m−12 )

m!(2m+1) · x^2m+1

Ĥ, Jn = 2m v,^f(n)_n!⁽⁰⁾ =^f(2m)_(2m)!⁽⁰⁾ = 0, ]f^(2m)(0) = 0.

Jn = 2m+1v,^f(n)_n!⁽⁰⁾ = ^f_(2m+1)!^(2m+1)(0) =^{(−12)(−32)···(−}_m! ^{2m−12 )}(2m+1). ]f^(2m+1)(0) =

(2m)!

m! (−¹2)(−³2) · · · (−^2m−12 ) =^(2m)!_m! (−1)^m·2^1m·1·3 · · · (2m−1) = ⁽⁻¹⁾2^2mm[(2m)!](m!)^{2 2}

2.

j1. qα = tan⁻¹x,β = tan^{−1 x+1}_x−1, †tan α = x,tan β = ^x+1_x−1, /−^π₂ < α <

π

2,−^π₂ < β < ^π₂. ¢tan(α+β) = tan α+tan β 1+tan α tan β = ^x+

x+1 x−1

1−x·^x+1_x−1 = ^x_x²_−1−x^−x+x+12−x = −1, çx > 1v,^π₄ < α < ^π₂,^π₄ < β < ^π₂, ]^π₂ < α+ β < π, Ĥα + β = ^3π₄; çx < 1v,−^π2 < α < ^π₄,−^π2 < β < ^π₄,−π < α + β < ^π2, Ĥ,α + β = −^π4. j2. 5?f (x) = tan⁻¹x+ tan^{−1 x+1}_x−1, x 6= 1, Äf⁰(x) = _1+x¹2 + ⁻

2 (x−1)2

1+(^x+1

x−1)² =

1

1+x² − _(x−1)²_+(x+1)^{2 2} = _1+x¹2 −_2(1+x^{2 2}₎ = 0. ]f (x)Ê(−∞, 1)£(1, ∞)}

Ñb, qf (x) =

 c+, Jx > 1,

c₋, Jx < 1. Älimx→∞f(x) = ^π₂ +^π₄ = ^3π₄ , ] )c+=^3π₄, ¢f (0) = 0 −^π₄ = −^π₄, ])c₋= −^π₄.

3.

j . ¯x= R³

1 x·_x2¹dx

R3 1

1

x2dx =^{ln 3}₍2

3)= ³₂ln 3.

¯ y=

R3 1

1 2·(_x2¹ )²dx

R3 1

1

xdx = ¹³⁵⁴2 3 = ¹³₅₄

4.

e^2x+ e^−xdx.(10 })

1

j Ie^x= t, †e^xdx= dt, dx = ¹_tdt R e^x−1

e^2x+e^−xdx= eR t−1 t²+t⁻¹

1

tdt=R t−1 t³+1dt . w2_t^t₃⁻¹₊₁ =_(t+1)(t^t−12−t+1) =_3(t+1)⁻² +_3(t^2t−12−t+1), FJR _e^x₋₁

e^2x+e^−xdx=R ₋₂

3(t+1)dt+R _2t−1

3(t²−t+1)dt= −3²

R ₁

t+1dt+₃¹R _2t−1

t²−t+1dt=

−²3ln |t + 1| +¹3ln |t²− t + 1| + c

5.

|üM 1 , t° f (x) . (10 })

j Rt

0p1 + (f⁰(x))²dx= ¹₂(e^t+ e^−t), út}, )p1 + (f⁰(t))² = ¹₂(e^t+ e^−t), j)f⁰(x) = ±¹₂(e^x− e^−x) (t²Ax), f (x) =R f⁰(x)dx = ±¹₂(e^x+ e^−x) + C,

˛ø f (0) = ±1 + C = 1, ]C = 0C2, C = 0v,f(x) = −¹₂(e^x+ e^−x) + 2(.

¯);C = 0v,f (x) = ¹₂(e^x+ e^−x) = cosh x(¯)

6.

2−2 cos x−x¹²) . (10 })

j1. ÄÑ limx→0(_{2−2 cos x}¹ −x¹²) = limx→0x²−(2−2 cos x)

x²(2−2 cos x) = limx→0 2x−2 sin x

2x(2−2 cos x)+x²·2 sin x = limx→0 x−sin x

2x−2x cos x+x²sin x = limx→0 1−cos x

2−2 cos x+2x sin x+2x sin x+x²cos x

= limx→0 sin x

2 sin x+4 sin x+4x cos x+2x cos x−x²sin x= limx→0 sin x

6 sin x+6x cos x−x²sin x = limx→0 1

6+_{sin x}^x cos x−x² = ₁₂¹

j2. ¢ cos x = 1 −^x_2!² +^x_4!⁴ −^x6!⁶ + ... +⁽⁻¹⁾_(2n)!^nx2n + ...

1

2−2 cos x−x¹² =^x2−2(1−cos x) 2(1−cos x)x² =

x2

2−(^x2!²−^x4!⁴+^x_6!⁶−...⁽⁻¹⁾(2n)!^n+1x2n+...) x²(^x2_2!−^x44!+^x6_6!−...−(−1)n+1 x2n

(2n)! +...) = ^x

4(₂₄¹−^x6!²+...+^(−1)n+1_(2n)!^x2n−4−...) x⁴(¹₂−^x24!+^x4_6!−...+(−1)n+1 x2n−2

(2n)! +...)

⇒ ₁₂¹ as x → 0

7.

x , ·úwÇ$ .(5 }) (ii) tzp π^e< e^π .(5 })

j

(i) f (x) =^{ln x}_x , x > 0, f⁰(x) = ^{1−ln x}_x2 .

On x > e ln x > 1, f⁰(x) < 0, f (x) decreasing on (0, ∞).

On 0 < x < e, ln x < 1, f⁰(x) > 0, f (x) increasing on (0, e).

Hence f (e) =^{ln e}_e = ¹_e is the absolutly maximum value of f . f⁰⁰(x) = ^(−x1)x2−(1−ln x)2x

x⁴ =^{2(ln x−}_x3 ³²⁾.

On x > e³², ln x >³₂, f⁰⁰(x) > 0, f concave upward.

2

On 0 < x < e²³, ln x < ³₂, f⁰⁰(x) < 0, f concave downward.

Hence (e³², f(e³²)) = (e³², ³

2e³²) is the point of inflection.

limx→∞ln x

x = limx→∞

1 x

1 = 0, y = 0 is a horizontal asymptote.

limx→0⁺ ln x

x = −∞, x = 0 is a vertical asymptote (ii) f (e) > f (π), ^{ln e}_e > ^{ln π}_π , π ln e > e ln π, e^π> π^e.

8.

0 sin⁶xdx.(10 }) jR^π₂

0 sin⁶xdx= −R^π₂

0 sin⁵xdcos x = − sin⁵xcos x|0^π2+R^π₂

0 cos x·5 sin⁴xcos xdx = 5R^π₂

0 sin⁴xdx− 5R^π₂

0 sin⁶xdx, FJ 6R^π₂

0 sin⁶xdx= 5R^π₂

0 sin⁴xdx,R^π₂

0 sin⁶xdx= ⁵₆R^π₂

0 sin⁴xdx. °Ü, R^π₂

0 sin⁴xdx= ³₄R^π₂

0 sin²xdx= ³₄R^π₂

0 1−cos 2x

2 dx= ³₄ ·¹₂(x −¹₂sin 2x)|0^π2 =

3

8·^π2 . FJ R^π₂

0 sin⁶xdx=⁵₆׳8×^π2 = ₃₂⁵π.

9.

P_∞

n=2 1

n(ln n)^p . (10 }) j

(i)R _dx

x(ln x)^p =R (ln x)^−pdln x =

 ln ln x , if p = 1;

(ln x)^1−p ,otherwise . HenceR_∞

2 dx x(ln x)^p = lima→∞R^a

2 dx x(ln x)^p =







lima→∞(ln ln a − ln ln 2) = ∞ , if p = 1;

lima→∞((ln a)^1−p− (ln 2)^1−p) = ∞ , if p < 1;

lima→∞((ln a)^1−p− (ln 2)^1−p) = −(ln 2)^1−p , if p > 1;

ThereforeR_∞

2 dx

x(ln x)^p converges iff p > 1.

(ii)P∞

n=1 1

n(ln n)^p converges iff p > 1, by part(ii) and the integral test.

10.

n=1n(−x)ⁿ⁺¹ = x²− 2x³+ 3x⁴+ ... + n(−x)ⁿ⁺¹+ ...

íY¹š .(5 }) (ii) I f (x) =P∞

n=1n(−x)ⁿ⁺¹ , úF |x| < r. t° f(x) D f(0.5), f⁰(0.5)s 65M . (5 })

(i) By the ratio test, limx→∞|^(n+1)(−x)_|n(−x)ⁿ⁺¹ⁿ⁺²_| | = |x|, so, if |x| < 1 then it’s abs.

conv ; if |x| > 1, then it is div. so the radius of conv. r = 1 (ii) Let f (x) =P∞

n=1n(−x)ⁿ⁺¹ = x²P∞

n=1n(−x)ⁿ⁻¹

= x²(1 − 2x + 3x²+ ... + n(−x)ⁿ⁻¹+ ...) . LetP_∞

n=1(−x)ⁿ= 1 + (−x) + (−x)²+ ... + (−x)ⁿ+ ... , 3

so _1+x¹ =P∞

n=0(−x)ⁿ for |x| < 1, (_1+x¹ )⁰=P∞

n=1n(−x)ⁿ⁻¹· (−1) for |x| < 1 , i.e. (−1)(_(1+x)^{1 2}) =P∞

n=1n(−x)ⁿ⁻¹· (−1) for |x| < 1, and (_(1+x)¹ 2) =P_∞

n=1n(−x)ⁿ⁻¹ for |x| < 1 . Then f (x) = _(1+x)^x2 2, f⁰(x) = (−2)(1+x)^{−1 2} +_(1+x)⁻² 3 . Hencef (0.5) = ₍3¹⁴

2)² =¹₄×⁴9 = ¹₉, f⁰(0.5) = ₂₇⁸ .

4