( x )( x) dx. Year 12 Extension 2 Term Question 1 (15 Marks) (a) Sketch the curve (x + 1)(y 2) = 1 2

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1 Yer Etension Term 7 Question (5 Mrks) Mrks () Sketch the curve ( + )(y ) (b) Write the function in prt () in the form y f(). Hence, or otherwise, sketch the curve (i) y f( ) (ii) y f () (c) Evlute (i) ( )( ) (ii) 9 (d) (i) The point P moves so tht the sum of its distnces from the points (, ) nd (, ) is 6 units. Prove tht the eqution of the locus of P is + y 9 5. (ii) Find the eccentricity of this locus. Question (5 Mrks) START A NEW PAGE Mrks () The hyperbol H is given by the eqution y 6 5 (i) Write down the equtions of the symptotes. (ii) P is n rbitrry point with coordintes (secθ, 5tnθ). Show tht P lies on H. secθ y tnθ (iii)prove tht the tngent to H t P hs eqution. 5 (iv) This tngent cuts the symptotes in L nd M. Prove tht LP PM. 7 JRAHS Y Et T Assessment - -

2 Question Continued (b) Let n be positive integer nd I n ( ln ) n. Mrks (i) Prove tht I n (ln) n ni n-. (ii) Hence, evlute ( ln ). Question (5 Mrks) START A NEW PAGE Mrks + () Find. + (b) On the ttched sheet, you re given the curve of y f(). Sketch netly on seprte digrms (i) y f() (ii) y f ( ) (c) (i) Evlute cos. (ii) Evlute 5 cos θ d θ. (d) F is the point (, ) nd d is th line. M is the foot of the perpendiculr from vrible point P to d, nd P moves so tht FP PM. (i) Derive the eqution of the locus of P. (ii) Find the cute ngle between the symptotes to the nerest degree. 7 JRAHS Y Et T Assessment - -

3 Question (5 Mrks) START A NEW PAGE () Use the substitution t tn to prove tht + sin Mrks. (b) (i) Show tht f ( ) { f ( ) + f ( ) }. (ii) Hence, or otherwise evlute. + sin (c) (i) Evlute secθ d θ. (ii) Hence, show tht sec θ θ [ + ln( + ) ] d ~ END OF ASSESSMENT~ 7 JRAHS Y Et T Assessment - -

4 Sketch netly on seprte digrms Question Prt (b) HAND IN WITH YOUR ANSWERS (i) y f() y y f() (ii) y f ( ) y y f() 7 JRAHS Y Et T Assessment - -

5 Question () ~SOLUTIONS Yer Etension Term Assessment 7~ Asymptotes & y è Œ mrk Shpe è Œ Mrk (b) (i) y f( ) is horizontl shift of y f() units to the left. è Œ Mrk (Asymptotes t y & ) (ii) y f () Asymptote t & y è ½ Mrk y intercept & intercept.5è Œ Mrk No grph below is è½ Mrk Shpe is reltively similr to the originl function no mrks wrded 7 JRAHS Y Et T Assessment - 5 -

6 Q (c) (i) ( )( ) + By Prtil Frctions [ ln( ) ln( ) ] [( ln ln) ( ln ln ) ] 8ln [ ln ] (ii) ln( + ) ln( ) 6 ln 5 ln ln ln ln 5 [( ) ( )] Correct vlues of A nd B è Œ Mrk Correct integrtion è Œ Mrk Correct nswer è Œ Mrk This is not inverse Trigonometry Question!. No mrks wrded if used incorrect method. Correct use of prtil frctions & integrtionè Œ Mrk Correct evlution (d) (i) P is (, y) then by dt ( + ) + y + ( ) + y 6 i.e y 6 8 9( + + y ) y 5 è + y 9 5 (ii) Using b ( e ) since locus is n ellipse where nd b 5, we get e. 7 JRAHS Y Et T Assessment - 6 -

7 Question () (i) Eqn. of symptotes y y ± (ii) If P is (secθ, 5tnθ) then y LHS 6 5 6sec θ 5tn θ 6 5 tn θ + tn θ RHS P lies on hyperbol H. dy 5secθ (iii) t P. tnθ eqn. of tngent t P is 5secθ y 5tnθ ( secθ ) tnθ tnθy 5 secθ (tn θ sec θ) secθ y tnθ 5 Showing LHS RHS Correct differentil Correct eqn. of tngent Correct rerrngement to get required nswer èœ Mrk 5 (iv)tngent cuts y when secθ 5 tnθ 5 nd y secθ tnθ 5 L, secθ tnθ secθ tnθ 5 Tngent cuts y t 5 M, secθ + tnθ secθ + tnθ 5 secθ tnθ. Correct coordintes of L è ½ Mrk Coordintes of M è ½Mrk If P is the midpoint of LM then LPPM. 7 JRAHS Y Et T Assessment - 7 -

8 coordinte of midpoint of LM + secθ tnθ secθ + tnθ 8secθ sec θ tn θ secθ. Correctly finding nd y coordintes of midpoint Or using distnce formul & conclusion è Mrks y coordinte of midpoint of LM secθ tnθ secθ + tnθ 5tnθ. P is the midpoint if LM. LPPM (b) (i) I n ( ln ) n ( ln ) n n [ ] ( ln ) (ln) n ni n- è Mrks (ii) I (ln) I (ln) [(ln) I ] Now I ln [ ln ] ln + ln I (ln) 6(ln) + ln 6. 7 JRAHS Y Et T Assessment - 8 -

9 Question + () ln + + tn + (b) See net pge ( ) c (i) (ii) è Mrk (c) (i) Let f() cos then f( ) cos( ) f() it s n odd function. cos Showing it s n odd function Answer (ii) 5 cos θ d θ cosθ ( cos θ ) d θ cos ( sin θ ) θ d θ cosθ cosθ sin θ + cosθ sin θ d θ sin sinθ + θ + sin 5 5 θ (d) (i) PF ( ) + y PM PF PM y 8 + y è y Correct integrtion Correct nswer ech for PF nd PM (by definition) è ½ for squring both sides & epnding è ½ for correct locus. (ii) symptotes y ± Correct equtions of symptotes line y mkes n ngle of rds with positive is nd line y mkes n ngle of rds. cute ngle is rds. Correct nswer with some eplntion 7 JRAHS Y Et T Assessment - 9 -

10 Q (b) (i) y f() y (ii) y f ( ) y 7 JRAHS Y Et T Assessment - -

11 Question () + sin [when, t nd then t ] + t dt t + + t dt dt + t + t t + t + dt t + tn + t + tn tn tn (b) (i) f ( ) f ( ) + Now f ( ) f ( u) f t + f ( ) du u) du f ( ) Since the definite integrl is independent of the vrible. Hence f ( ) f ( ) + f ( [ f ( ) + f ( ) ] ) Chnge of vribles & correct t substitution è Mrks Correct integrtion & substitution Let u then du & à u & à u. 7 JRAHS Y Et T Assessment - -

12 (ii) + sin + + sin + sin( ) + since sin sin( ) sin from prt () (c) (i) [ ] secθ d θ ln( secθ + tnθ ) ln + ln sec + tn ln( sec + tn ) ( ) Correct use of prt (i) Correct integrtion Correct nswer (ii) d sec θ d θ sec θ tn θ dθ dθ [ sec tnθ ] θ tnθ secθ tnθ d θ secθ tn θ d θ ( sec secθ ) sec θ d θ + θ d θ secθd θ + ln( + ) sec θ d θ [ + ln( + ) ] using prt (i) 7 JRAHS Y Et T Assessment - -