32sp10-fake-ex1-solutions

# 32sp10-fake-ex1-solutions - A GACCEEXA h e×eÔ ÖÓb ÐeÑ...

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Unformatted text preview: A GACCEEXA# h e×eÔ ÖÓb ÐeÑ ×Û e Öe Ñ Ó ×Ø ÐÝ Øak eÒÓÓ fÓ Ðda ØheÜaÑ ×ÖeeØ iÒgØh eÑ a Øe Ö ia ÐØha ØÛ iÐÐ b eÓÚ e ÖedÓÒÓÙ ÖÖ×ØÑ id Øe ÖÑ eÜaÑ 665779885 C e ÖØa iÒ ÐÝ Øh e Öea Öea feÛ ØÓÓÑ aÒÝ Ô ÖÓb ÐeÑ ×h e ÖeØÓÓ ÖÖe×ÔÓÒd ØÓahÓÙ ÖeÜaÑ bÙ ØØh eÝg iÚ eÝÓÙ Øh e id eaÓ fØh e×Ó ÖØ×Ó fÔ ÖÓb ÐeÑ × Û h ihÑ igh Øb eÓÒ Øh eeÜaÑ F iÒd : a lim x →∞ parenleftBig 1 + r x parenrightBig x Ó ÐÙ Ø iÓÒ : iÒe ln parenleftBig lim x →∞ parenleftBig 1 + r x parenrightBig x parenrightBig = lim x →∞ ln parenleftBigparenleftBig 1 + r x parenrightBig x parenrightBig = lim x →∞ x parenleftBig ln parenleftBig 1 + r x parenrightBigparenrightBig = lim x →∞ ln ( 1 + r x ) 1 x = r lim x →∞ ln ( 1 + r x ) r x . ÓÛ ×e Ø h = r x ;aÒdÛ eÓÒ Ø iÒÙ eØh eabÓÚ eÛ iØh Øh eÓb ×e ÖÚa Ø iÓÒ Øha Ø ln(1) = 0 : = r lim h → ln (1 + h )- ln(1) h = r (ln( x )) ′ vextendsingle vextendsingle x =1 = r · 1 . b lim h → 4 h- 1 h Ó ÐÙ Ø iÓÒ : h i×i×b e×ØeÜÔ Ða iÒ ed iÒ Øe ÖÑ ×Ó fØh ed e Ö iÚa Ø iÚ eÓ feÜÔÓÒ eÒ Ø ia Ð× lim h → 4 h- 1 h = f ′ (0) , Û h e Öe f ( x ) = 4 x B Ù Ø× iÒe 4 x = e x ln(4) bÝ Øh eÖÙ Ðe×Ó fÐÓga Ö iØhÑ × lim h → 4 h- 1 h = f ′ (0) = d dx vextendsingle vextendsingle vextendsingle vextendsingle x =0 e x ln(4) = ln(4) ( e x ) ′ vextendsingle vextendsingle x =0 = ln(4) · 1 . d dx ( e − 2 x + 3 ln | x | ) vextendsingle vextendsingle vextendsingle vextendsingle x = − 1 / 2 Ó ÐÙ Ø iÓÒ : d dx ( e − 2 x + 3 ln | x | ) vextendsingle vextendsingle vextendsingle vextendsingle x = − 1 / 2 = parenleftbigg- 2 e − 2 x + 3 x parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle x = − 1 / 2 =- 2 e- 6 . A GACCEEXA# d d dx (arctan( x )) 2 vextendsingle vextendsingle vextendsingle vextendsingle x = − 1 Ó ÐÙ Ø iÓÒ : d dx (arctan( x )) 2 vextendsingle vextendsingle vextendsingle vextendsingle x = − 1 = 2(arctan( x )) 1 1 + x 2 vextendsingle vextendsingle vextendsingle vextendsingle x = − 1 = 1 . e sin(tan − 1 ( 1 2 )) Ó ÐÙ Ø iÓÒ : sin(tan − 1 ( 1 2 )) = sin( θ ) = 1 √ 5 , Û h e Öe tan( θ ) = 1 / 2 ×Ó sin( θ ) = 1 √ 5 . f tan(sin − 1 (1 / 3)) Ó ÐÙ Ø iÓÒ : tan(sin − 1 (1 / 3)) = tan( θ ) = 1 2 √ 2 , Û h e Öe sin( θ ) = 1 / 3 ×Ó tan( θ ) = 1 2 √ 2 g sinh(ln(2)) Ó ÐÙ Ø iÓÒ : sinh(ln(2)) = 1 2 ( e ln(2)- e − ln(2) ) = 1 2 (2- 1 2 ) = 3 4 . F iÒd Øh e iÒd ia Øed iÒ Øeg Öa Ð× :hÓÛ Øh e×ØeÔ × iÒÚÓ ÐÚ ed a integraldisplay x 2 sin x dx Ó ÐÙ Ø iÓÒ : a ÖØ× :ak e u = x 2 aÒd dv = sin( x ) dx h eÒ du = 2 xdx aÒd v =- cos( x ) dx integraldisplay x 2 sin x dx =- x 2 cos( x ) + integraldisplay 2 x cos( x ) dx, aÒdÒÓÛ dÓÔa ÖØ×aga iÒÛ iØh u = 2 x aÒd dv = cos( x ) dx ×Ó Øha Ø du = 2 dx aÒd v = sin( x ) h i×Û iÐÐÛ h iØØ ÐeaÛ aÝa...
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32sp10-fake-ex1-solutions - A GACCEEXA h e×eÔ ÖÓb ÐeÑ...

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