AP CALCULUS AB JUNE 2020 MODEL SOLUTIONS - from Prof Iyer ib.mathfaculty@gmail.com

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AP Calc/AB/JUNE 2020/MS/IBI version

Including necessary theory for understanding Solution developed by Prof.Iyer Head of department -Math Curriculum Manager- IB Author – Examiner Moderator. IB World groups ib.mathfaculty@gmail.com Teaching IB , AP , A level and FRESHMEN since 2005

The solution is not copied from any source Created as a model answer. Any feedback, criticisms are most welcome © PROF.IYER IB WITH IYER © PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

AP Calc/AB/JUNE 2020/MS/IBI version

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QUESTION 1 (a) Using Mean Value theorem, we get the Average rate change over tne interval 10≤ � ≤12 v’(11) =

đ?’—(đ?&#x;?đ?&#x;?)−đ?’—(đ?&#x;?đ?&#x;Ž) đ?&#x;?đ?&#x;?−đ?&#x;?đ?&#x;Ž

=> v’(11) = => v’(11) =

đ?&#x;?đ?&#x;Žâˆ’đ?&#x;?đ?&#x;“ đ?&#x;? đ?’Ž đ?&#x;?. đ?&#x;“ đ?’”đ?&#x;?

(b) Instantaneous Acceleration of particle at 11th Second (t =11)

IB WITH IYER Š PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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(c) V is differentiable in in the given interval. Implies it is continuous in the interval đ?&#x;?đ?&#x;“−đ?&#x;”

đ?&#x;—

Gradient of the Velocity curve at v’(c)=đ?&#x;?đ?&#x;Žâˆ’đ?&#x;’ = đ?&#x;” = đ?&#x;?. đ?&#x;“ > đ?&#x;Ž for 4 < c < 10 we can understand that velocity curve is increasing , continuous and differentiable Also v (4) = 6 m/s and v (10) = 15 m/s and given that 6 < c < 10 => v (4) < v (c) < v (10) => 4 < v (c) < 15 => And, 4 < 8 < 15 is true so v (c) = 8 m/s Justified that v (c) = 8 m/s

IB WITH IYER Š PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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(d) t = time V(t) m /s

0 4 2 6

10 15

12 20

Upper Riemann Sum ( Right Riemann otherwise called as ) = Sum of Areas of rectangles = (4-0 ) (6) + (10-4)(15) + (12-10)(20) =24 + 90+ 40 =154 m/s đ?&#x;?đ?&#x;?

Therefore âˆŤđ?&#x;Ž đ?’—(đ?’•)đ?’…đ?’• = 154 m/s using Upper Riemann Sum Approximation

IB WITH IYER Š PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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(e) đ?&#x;?đ?&#x;?

âˆŤ

�′ (

đ?&#x;’

đ?’• − đ?&#x;?) đ?’…đ?’• đ?&#x;?

=> Using substitution method Put

đ?’• đ?&#x;?

đ?&#x;? đ?’…đ?’• đ?&#x;?

= du , dt = 2du

−đ?&#x;? = u

Change of limits When t= 4 , u=0 and When t= 12, u=4 đ?&#x;’

=> đ?&#x;? âˆŤđ?&#x;Ž đ?’—′ (đ?’–)đ?’…đ?’– => đ?&#x;?[đ?’—(đ?&#x;’) − đ?’—(đ?&#x;Ž)] => đ?&#x;?[đ?&#x;” − đ?&#x;?] => đ?&#x;?[đ?&#x;’] => đ?&#x;–

đ?’Ž đ?’”

IB WITH IYER Š PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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(f) đ?&#x;’đ?’™

Given h(x) = => âˆŤđ?&#x;? đ?’—(đ?&#x;‘đ?’•)đ?’…đ?’• Using Second Fundamental theorem of Calculus we get đ?’ƒ

If F(x) = âˆŤđ?’‚ đ?’‡(đ?’™)đ?’…đ?’™ then = > F’(x) = [ f(x)] Limits x=a to x =b = > F’(x) = [ f(b) – f(a)] h’(x) = => [ v(3t) ]2 4x => h’(x) = => [ v(12x). 4 - v(6).0] => h’(x) = => [ 4.v(12x)] => h’(1) = => [ 4.v(12)] => h’(1) = => [ 4.20] = 80 m Implies displacement at t=1 equals 80 m

IB WITH IYER Š PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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(g) g(t)= displacement given. => g’(t)= velocity => g ’’(t)= Acceleration Given g (4) = 7 , g’(4) = -3 and g’’(4) = -2 Velocity at t=4 equals -4 m/s Acceleration at t=4 equals -4 m/s2 If velocity has to increase at t=4 , acceleration at t=4 has to be positive If velocity has to decrease at t=4 , acceleration at t=4 has to be Negative In our problem acceleration is negative at t=4 Therefore Velocity at t=4 is decreasing

IB WITH IYER © PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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đ?&#x;?

Speed = | v(t)| = √(đ?’—(đ?’•)) ‌‌‌‌..(1) Speed is decreasing when

đ?’… đ?’…đ?’•

{|đ?’—(đ?’•)|} < 0

đ?’… đ?&#x;? đ?’—(đ?’•). đ?’—′ (đ?’•) {|đ?’—(đ?’•)|} = đ?’…đ?’• đ?&#x;? đ?&#x;?√(đ?’—(đ?’•)) At t= 4 đ?’… đ?&#x;? đ?’—(đ?&#x;’). đ?’‚′ (đ?&#x;’) {|đ?’—(đ?&#x;’)|} = đ?’…đ?’• đ?&#x;? √ ( ) đ?&#x;? (đ?’— đ?&#x;’ ) đ?’… đ?&#x;? (−đ?&#x;‘). (−đ?&#x;?) {|đ?’—(đ?&#x;’)|} = đ?’…đ?’• đ?&#x;?√(−đ?&#x;‘)đ?&#x;? đ?’… {|đ?’—(đ?&#x;’)|} đ?’…đ?’•

(đ?&#x;?đ?&#x;?)

= đ?&#x;?(đ?&#x;‘)=2 =+ ve

Therefore speed is increasing otherwise to say theoretically when v(t) and a(t) have same signs, then Speed is increasing . For unlike signs, speed is decreasing.

IB WITH IYER Š PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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If the velocity and Acceleration are in same sign, Speed is increasing If the velocity and Acceleration are in Opposite sign, Speed is decreasing

0< t< a , Velocity is decreasing , Acceleration < 0 a< t< c, Velocity is increasing, Acceleration > 0 c< t< e , Velocity is decreasing, Acceleration < 0 e< t< f , Velocity is increasing, Acceleration > 0

if we observe the graph, the orange graph is the reflection of velocity graph v(t) = | v(t)| = Speed Notice Speed is increasing when Velocity v(t)<0 and a(t) <0 when 0< t < a U d < t < e You can see in these intervals, speed ( orange ) graph is increasing above the x-axis

IB WITH IYER Š PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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(h) Given y = f(x) Equation of tangent is given by y -y1= m (x-x1) where m is the slope of the tangent at the point of contact (x1, y1) on the curve y=f(x) Given f(2)= 1 => (2,1) is the point of contact Given f ‘(x) = 3x2 -y3 => f’(2) = 3(2)2-(1)3 = 11 => Slope of tangent , m =11 Equation of tangent, y -1 = 11 (x -2) => y = 11x -22+1 => y = 11x -21 Also it can be written as f(x)=11x-21

IB WITH IYER © PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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Given f(2)= 1 => (2,1) is the point of contact Given f ‘(x) = 3x2 -y3 => f’(2) = 3(2)2-(1)3 = 11 => Slope of tangent , m =11 Equation of tangent, y -1 = 11 (x -2) => y = 11x -22+1 => y = 11x -21 Also it can be written as f(x)=11x-21

IB WITH IYER © PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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QUESTION 2

Graph of h(x) for

-5≤đ?’• ≤đ?&#x;”

IB WITH IYER Š PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

Not to scale

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(a) đ?’™

Given p(x) = âˆŤđ?&#x;? đ?’‰(đ?’•)đ?’…đ?’• for - 5 ≤ đ?’• ≤ đ?&#x;” Using Fundamental theorem of calculus we get p’(x) = h(x)‌.(1) Given p’(-1) = h(-1) = 0 from graph Therefore p’(x) = 0 at x =-1 Implies, p(x) has stationary point at x=-1 To review it is minima or maxima , observe the sign change Using First derivative test for x < -1 h (x) < 0 and x> -1 , h(x) > 0 ie for x < -1 p’ (x) < 0 and x> -1 , p’(x) > 0 from (1) Using First derivative test, At =-1 , p has a relative minima

IB WITH IYER Š PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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(b) đ?’™

Given p(x) = âˆŤđ?&#x;? đ?’‰(đ?’•)đ?’…đ?’• for - 5 ≤ đ?’• ≤ đ?&#x;” By FTOC We get p’(x) = h(x) Let the absolute minimum be at x= k Implies p’(k) =0 => h(k)=0 Possible values of k from the graph are h(-1)= 0 Therefore k =-1 But to prove absolute minimum at k =-1 for the function p(x) we need to analyze the given interval ensuring the minimum value of p(x) is only at k =-1

IB WITH IYER Š PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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X

P(x)

-5

0

Remarks −đ?&#x;“

âˆŤ đ?’‰(đ?’•)đ?’…đ?’• =-

đ?&#x;? đ?&#x;?

− âˆŤ đ?’‰(đ?’•)đ?’…đ?’• −đ?&#x;“

0 ( Area sum ) -1

-4 At x=-1 p(x) is the least value among the points between - = 5 and 6

−đ?&#x;?

âˆŤ đ?’‰(đ?’•)đ?’…đ?’• đ?&#x;? đ?&#x;?

âˆŤ đ?’‰(đ?’•)đ?’…đ?’• −đ?&#x;?

-4 (Area) 1

0

−đ?&#x;?

âˆŤ đ?’‰(đ?’•)đ?’…đ?’• đ?&#x;?

Limits are same

6

22.5

đ?&#x;”

âˆŤ đ?’‰(đ?’•)đ?’…đ?’• đ?&#x;?

=0.5(5) (9) ( trapezium area)

=22.5 Therefore at x=-1, p(x) has a absolute minimum

IB WITH IYER Š PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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(c) đ?’™

Given p(x) = âˆŤđ?&#x;? đ?’‰(đ?’•)đ?’…đ?’• for - 5 ≤ đ?’• ≤ đ?&#x;” Using FTOC P’(x) = h(x)

A function f(x) is a decreasing function when f’(x) < 0 A function f(x) is an increasing function when f’(x) > 0 Here p is decreasing function when p’(x) < 0 ie. h(x)< 0 when -5 < x< -1 ‌‌. (1)

A function f(x) is a Concave up when f’’(x) > 0 A function f(x) is a Concave down when f’’(x) < 0 Here p is concave up function when p’’(x) > 0 ie. h’(x)> 0 => Slope of the h curve is positive h’(x) > 0 when -5 < x < 6 ( everywhere)‌.(2) From (1) and (2), we get p (x) is decreasing and concave up for -5 < x< 1 ( intersection of 2 intervals )

IB WITH IYER Š PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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(d) đ?’™

Given p(x) = âˆŤđ?&#x;? đ?’‰(đ?’•)đ?’…đ?’• for - 5 ≤ đ?’• ≤ đ?&#x;” Using FTOC p’(x) = h(x) n=> p’’(x) = h’(x) => p’’(1)= h’(1) For h’(1) not to exist, we not to prove h(x) is not differentiable at x=1 or h’(x) is not continuous at x=1 Using the theory of continuity A function f(x) is continuous, if the left hand limit at a point= Right hand limit at same point = f(point) Using the piece-wise function Using analytical geometry fundamentals, we shall construct the straight-line functions for h(x) in the intervals đ?&#x;?đ?’™ + đ?&#x;? , đ?’™â‰¤đ?&#x;? đ?’‘′(đ?’™) = đ?’‰(đ?’™) = {đ?’™ đ?&#x;” − + đ?&#x;“, đ?’™>đ?&#x;? đ?&#x;“ đ?&#x;“ đ?&#x;?, đ?’™â‰¤đ?&#x;? đ?’‘′′(đ?’™) = đ?’‰â€˛(đ?’™) = { đ?&#x;? , đ?’™>đ?&#x;? đ?&#x;“ đ??Ľđ??˘đ??Śâˆ’ đ?’‰â€˛ (đ?’™) = đ?’‰(đ?&#x;?) = đ?&#x;? ≠đ??Ľđ??˘đ??Ś+ đ?’‰â€˛(đ?’™) đ?’™âˆ’→đ?&#x;?

đ?’™â†’đ?&#x;?

As LHL = At point ≠RHL at x=1 , h’(1) does not exist .ie p’’(1) does not exist IB WITH IYER Š PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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(e) đ??Ľđ??˘đ??Ś đ?’™â†’đ?&#x;?

đ?’‘(đ?’™) đ?’™+đ?&#x;’

đ?’‘(đ?&#x;?)

= đ?&#x;?+đ?&#x;’ = đ?&#x;?

=

âˆŤđ?&#x;? đ?’‰(đ?’•)đ?’…đ?’•

=0

đ?&#x;“

[ From the definition of p(x) ]

IB WITH IYER Š PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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(f) đ?’™đ?&#x;? âˆŤđ?&#x;? đ?’‰(đ?’•)đ?’…đ?’•

Given k(x)= p(x )= ‌.(1) but by FTOC P’(X2)= h(x2). 2x‌.(2) 2

Differentiating (1) with respect to x we get => k ’(x)=p’(x2).2x => k ’(x)=h(x2) 2x .2x (from (2) => k’ (√đ?&#x;?) = h(2) 4 .2 => k’ (√đ?&#x;?) =

đ?&#x;?đ?&#x;? .đ?&#x;– đ?&#x;“

( From similar triangle)

=> k’ (√đ?&#x;?) = đ?’„đ?’?đ?’?đ?’”đ?’•đ?’‚đ?’?đ?’• => k(x) must be a linear function but k(x)= p(x2) is a region and not a line Therefore k ’ (√đ?&#x;?) does not exist Proved

IB WITH IYER Š PROF.IYER TEACHING IB./ A LEVEL / AP and FRESHMEN since 2005

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