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Question Number 33651 by rahul 19 last updated on 21/Apr/18
![Let function f(x) be defined as f(x)= x^2 +bx+c , where b,c∈R . And f(1) − 2f(5) +f(9) =32. Find no. of ordered pairs (b,c) such that ∣f(x)∣≤8 ∀ x∈ [1,9] ?](Q33651.png)
$${Let}\:{function}\:{f}\left({x}\right)\:{be}\:{defined}\:{as}\: \\ $$$${f}\left({x}\right)=\:{x}^{\mathrm{2}} +{bx}+{c}\:,\:{where}\:{b},{c}\in{R}\:. \\ $$$${And}\:{f}\left(\mathrm{1}\right)\:−\:\mathrm{2}{f}\left(\mathrm{5}\right)\:+{f}\left(\mathrm{9}\right)\:=\mathrm{32}. \\ $$$${Find}\:{no}.\:{of}\:{ordered}\:{pairs}\:\left({b},{c}\right) \\ $$$${such}\:{that}\:\mid{f}\left({x}\right)\mid\leqslant\mathrm{8}\:\forall\:{x}\in\:\left[\mathrm{1},\mathrm{9}\right]\:? \\ $$
Answered by MJS last updated on 21/Apr/18
![something about polynomial functions of 2^(nd) degree there are 2 different points of view usually we are focused on zeros x^2 +px+q=0 ⇔ ⇔ (x−x_1 )(x−x_2 ) ⇔ ⇔ p=−(x_1 +x_2 ) ∧ q=x_1 x_2 but we should think about steepness, orientation and shifts in both x− and y−directions y=ax^2 +bx+c a>0 ⇔ “hanging” parabola a<0 ⇔ “standing parabola” ∣a∣ determines steepness y′=2ax+b ∣2a∣<1 ⇔ “flat” or “wide” parabola ∣2a∣>1 ⇔ “steep” or “narrow” parabola b and c have got something to do with the x− and y−shifts let′s look at the basic function y=x^2 shift up/down y−v=x^2 ⇔ y=x^2 +v shift left/right y=(x+u)^2 +v ⇔ y=x^2 +2ux+u^2 +v b=2u ∧ c=u^2 +v ⇔ u=(b/2) ∧ v=c−(b^2 /4) so y=x^2 +bx+c is a shifted basic parabola. we are looking for a part of this function which fits into a window of width 8 and height 16 (because the width of the given interval [1;9] is 8 and abs(f(x))≤8 ⇔ ⇔ −8≤f(x)≤8 ⇔ 0≤f(x)+8≤16 we can look at y=x^2 and afterwards do the shifting 1. try on left arm (symm. to right arm) f(x): y=x^2 x≤0 f(x)±16=f(x−8) x^2 ±16=x^2 −16x+64 16x=64±16 [x=3 ∨ x=5 ⇒ no solution on left arm f(±3)=9; f(±5)=25 but f(0)=0 ⇒ our window would be 8×25 instead of 8×16] 2. try center f(−4)=16 f(0)=0 f(4)=16 is the only solution but we need x∈[1;9] instead of x∈[−4;4] ⇒ ⇒ shift 5 to the right y=(x−5)^2 and we need y∈[−8;8] instead of y∈[0;16] ⇒ ⇒ shift 8 down y=(x−5)^2 −8 y=x^2 −10x+17 there′s only one pair (b;c)=(−10;17) [f(1)−2f(5)+f(9)=32 is true for any pair (b;c), so we didn′t need it: f(1)=1+b+c f(5)=25+5b+c f(9)=81+9b+c (1+b+c)−2(25+5b+c)+(81+9b+c)=32]](Q33653.png)
$$\mathrm{something}\:\mathrm{about}\:\mathrm{polynomial}\:\mathrm{functions}\:\mathrm{of} \\ $$$$\mathrm{2}^{\mathrm{nd}} \:\mathrm{degree} \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{2}\:\mathrm{different}\:\mathrm{points}\:\mathrm{of}\:\mathrm{view} \\ $$$$\mathrm{usually}\:\mathrm{we}\:\mathrm{are}\:\mathrm{focused}\:\mathrm{on}\:\mathrm{zeros} \\ $$$${x}^{\mathrm{2}} +{px}+{q}=\mathrm{0}\:\Leftrightarrow \\ $$$$\Leftrightarrow\:\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)\:\Leftrightarrow \\ $$$$\Leftrightarrow\:{p}=−\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)\:\wedge\:{q}={x}_{\mathrm{1}} {x}_{\mathrm{2}} \\ $$$$\mathrm{but}\:\mathrm{we}\:\mathrm{should}\:\mathrm{think}\:\mathrm{about}\:\mathrm{steepness},\:\mathrm{orientation} \\ $$$$\mathrm{and}\:\mathrm{shifts}\:\mathrm{in}\:\mathrm{both}\:{x}−\:\mathrm{and}\:{y}−\mathrm{directions} \\ $$$${y}={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${a}>\mathrm{0}\:\Leftrightarrow\:``\mathrm{hanging}''\:\mathrm{parabola} \\ $$$${a}<\mathrm{0}\:\Leftrightarrow\:``\mathrm{standing}\:\mathrm{parabola}'' \\ $$$$\mid{a}\mid\:\mathrm{determines}\:\mathrm{steepness} \\ $$$$\:\:\:\:\:\:\:\:\:\:{y}'=\mathrm{2}{ax}+{b} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mid\mathrm{2}{a}\mid<\mathrm{1}\:\Leftrightarrow\:``\mathrm{flat}''\:\mathrm{or}\:``\mathrm{wide}''\:\mathrm{parabola} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mid\mathrm{2}{a}\mid>\mathrm{1}\:\Leftrightarrow\:``\mathrm{steep}''\:\mathrm{or}\:``\mathrm{narrow}''\:\mathrm{parabola} \\ $$$${b}\:\mathrm{and}\:{c}\:\mathrm{have}\:\mathrm{got}\:\mathrm{something}\:\mathrm{to}\:\mathrm{do}\:\mathrm{with}\:\mathrm{the} \\ $$$${x}−\:\mathrm{and}\:{y}−\mathrm{shifts} \\ $$$$\mathrm{let}'\mathrm{s}\:\mathrm{look}\:\mathrm{at}\:\mathrm{the}\:\mathrm{basic}\:\mathrm{function} \\ $$$${y}={x}^{\mathrm{2}} \\ $$$$\mathrm{shift}\:\mathrm{up}/\mathrm{down} \\ $$$${y}−{v}={x}^{\mathrm{2}} \:\Leftrightarrow\:{y}={x}^{\mathrm{2}} +{v} \\ $$$$\mathrm{shift}\:\mathrm{left}/\mathrm{right} \\ $$$${y}=\left({x}+{u}\right)^{\mathrm{2}} +{v}\:\Leftrightarrow\:{y}={x}^{\mathrm{2}} +\mathrm{2}{ux}+{u}^{\mathrm{2}} +{v} \\ $$$${b}=\mathrm{2}{u}\:\wedge\:{c}={u}^{\mathrm{2}} +{v}\:\Leftrightarrow\:{u}=\frac{{b}}{\mathrm{2}}\:\wedge\:{v}={c}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{so}\:{y}={x}^{\mathrm{2}} +{bx}+{c}\:\mathrm{is}\:\mathrm{a}\:\mathrm{shifted}\:\mathrm{basic}\:\mathrm{parabola}. \\ $$$$\mathrm{we}\:\mathrm{are}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{a}\:\mathrm{part}\:\mathrm{of}\:\mathrm{this}\:\mathrm{function} \\ $$$$\mathrm{which}\:\mathrm{fits}\:\mathrm{into}\:\mathrm{a}\:\mathrm{window}\:\mathrm{of}\:\mathrm{width}\:\mathrm{8}\:\mathrm{and} \\ $$$$\mathrm{height}\:\mathrm{16}\:\left(\mathrm{because}\:\mathrm{the}\:\mathrm{width}\:\mathrm{of}\:\mathrm{the}\:\mathrm{given}\right. \\ $$$$\mathrm{interval}\:\left[\mathrm{1};\mathrm{9}\right]\:\mathrm{is}\:\mathrm{8}\:\mathrm{and}\:\mathrm{abs}\left({f}\left({x}\right)\right)\leqslant\mathrm{8}\:\Leftrightarrow\: \\ $$$$\Leftrightarrow\:−\mathrm{8}\leqslant{f}\left({x}\right)\leqslant\mathrm{8}\:\Leftrightarrow\:\mathrm{0}\leqslant{f}\left({x}\right)+\mathrm{8}\leqslant\mathrm{16} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{look}\:\mathrm{at}\:{y}={x}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{afterwards}\:\mathrm{do}\:\mathrm{the} \\ $$$$\mathrm{shifting} \\ $$$$\mathrm{1}.\:\mathrm{try}\:\mathrm{on}\:\mathrm{left}\:\mathrm{arm}\:\left(\mathrm{symm}.\:\mathrm{to}\:\mathrm{right}\:\mathrm{arm}\right) \\ $$$${f}\left({x}\right):\:{y}={x}^{\mathrm{2}} \\ $$$${x}\leqslant\mathrm{0} \\ $$$${f}\left({x}\right)\pm\mathrm{16}={f}\left({x}−\mathrm{8}\right) \\ $$$${x}^{\mathrm{2}} \pm\mathrm{16}={x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{64} \\ $$$$\mathrm{16}{x}=\mathrm{64}\pm\mathrm{16} \\ $$$$\:\:\:\:\:\left[{x}=\mathrm{3}\:\vee\:{x}=\mathrm{5}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution}\:\mathrm{on}\:\mathrm{left}\:\mathrm{arm}\right. \\ $$$$\:\:\:\:\:{f}\left(\pm\mathrm{3}\right)=\mathrm{9};\:{f}\left(\pm\mathrm{5}\right)=\mathrm{25}\:\mathrm{but}\:{f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow\:\mathrm{our} \\ $$$$\left.\:\:\:\:\:\mathrm{window}\:\mathrm{would}\:\mathrm{be}\:\mathrm{8}×\mathrm{25}\:\mathrm{instead}\:\mathrm{of}\:\mathrm{8}×\mathrm{16}\right] \\ $$$$\mathrm{2}.\:\mathrm{try}\:\mathrm{center} \\ $$$${f}\left(−\mathrm{4}\right)=\mathrm{16} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left(\mathrm{4}\right)=\mathrm{16} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution} \\ $$$$ \\ $$$$\mathrm{but}\:\mathrm{we}\:\mathrm{need}\:{x}\in\left[\mathrm{1};\mathrm{9}\right]\:\mathrm{instead}\:\mathrm{of}\:{x}\in\left[−\mathrm{4};\mathrm{4}\right]\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{shift}\:\mathrm{5}\:\mathrm{to}\:\mathrm{the}\:\mathrm{right} \\ $$$${y}=\left({x}−\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{need}\:{y}\in\left[−\mathrm{8};\mathrm{8}\right]\:\mathrm{instead}\:\mathrm{of}\:{y}\in\left[\mathrm{0};\mathrm{16}\right]\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{shift}\:\mathrm{8}\:\mathrm{down} \\ $$$${y}=\left({x}−\mathrm{5}\right)^{\mathrm{2}} −\mathrm{8} \\ $$$$ \\ $$$${y}={x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{17} \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{only}\:\mathrm{one}\:\mathrm{pair}\:\left({b};{c}\right)=\left(−\mathrm{10};\mathrm{17}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\left[{f}\left(\mathrm{1}\right)−\mathrm{2}{f}\left(\mathrm{5}\right)+{f}\left(\mathrm{9}\right)=\mathrm{32}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:\mathrm{any}\:\mathrm{pair}\right. \\ $$$$\:\:\:\:\:\left({b};{c}\right),\:\mathrm{so}\:\mathrm{we}\:\mathrm{didn}'\mathrm{t}\:\mathrm{need}\:\mathrm{it}: \\ $$$$\:\:\:\:\:{f}\left(\mathrm{1}\right)=\mathrm{1}+{b}+{c} \\ $$$$\:\:\:\:\:{f}\left(\mathrm{5}\right)=\mathrm{25}+\mathrm{5}{b}+{c} \\ $$$$\:\:\:\:\:{f}\left(\mathrm{9}\right)=\mathrm{81}+\mathrm{9}{b}+{c} \\ $$$$\left.\:\:\:\:\:\left(\mathrm{1}+{b}+{c}\right)−\mathrm{2}\left(\mathrm{25}+\mathrm{5}{b}+{c}\right)+\left(\mathrm{81}+\mathrm{9}{b}+{c}\right)=\mathrm{32}\right] \\ $$
Commented by rahul 19 last updated on 22/Apr/18
$${thank}\:{u}\:{so}\:{much}\:{sir}! \\ $$