Question Number 58816 by Tawa1 last updated on 30/Apr/19
Answered by Kunal12588 last updated on 30/Apr/19
![from looking the graph we can say that maxima of function is at x=2 for A) ((d(f(x)))/dx)=0 [for minima or maxima] −(1/3)×2(x−2)(1)=0 ⇒x=2 option A can be correct for B) ((d(f(x)))/dx)=0 −(1/3)×2(x+2)(1)=0 ⇒x=−2 option B is not correct for C) ((d(f(x)))/dx)=0 (1/3)×2(x+2)(1)=0 ⇒x=−2 option C is not correct for D) ((d(f(x)))/dx)=0 6(x−2)(1)=0 ⇒x=2 ∴ only A or D options can be correct now lets find the zeroes of A and D A. −(1/3)(x−2)^2 +5=0 ⇒(x−2)=±(√(−(5)(−3))) ⇒x=2±(√(15)) D. 3(x−2)^2 +5=0 ⇒(x−2)=±(√(−(5/3))) ⇒x=2±z ⇒x∉ R But zeroes of the equation in the graph are Real ∴ Only A is correct Ans. A. f(x)=−(1/3)(x−2)^2 +5](https://www.tinkutara.com/question/Q58820.png)
$${from}\:{looking}\:{the}\:{graph}\:{we}\:{can}\:{say}\:{that} \\ $$$${maxima}\:{of}\:{function}\:{is}\:{at}\:\:{x}=\mathrm{2} \\ $$$$\left.{for}\:{A}\right) \\ $$$$\frac{{d}\left({f}\left({x}\right)\right)}{{dx}}=\mathrm{0}\:\:\:\:\:\left[{for}\:{minima}\:{or}\:{maxima}\right] \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{2}\left({x}−\mathrm{2}\right)\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{2} \\ $$$${option}\:{A}\:{can}\:{be}\:{correct} \\ $$$$\left.{for}\:{B}\right) \\ $$$$\frac{{d}\left({f}\left({x}\right)\right)}{{dx}}=\mathrm{0} \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{2}\left({x}+\mathrm{2}\right)\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{2} \\ $$$${option}\:{B}\:{is}\:{not}\:{correct} \\ $$$$\left.{for}\:{C}\right) \\ $$$$\frac{{d}\left({f}\left({x}\right)\right)}{{dx}}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{2}\left({x}+\mathrm{2}\right)\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{2} \\ $$$${option}\:{C}\:{is}\:{not}\:{correct} \\ $$$$\left.{for}\:{D}\right) \\ $$$$\frac{{d}\left({f}\left({x}\right)\right)}{{dx}}=\mathrm{0} \\ $$$$\mathrm{6}\left({x}−\mathrm{2}\right)\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{2} \\ $$$$\therefore\:{only}\:{A}\:{or}\:{D}\:{options}\:{can}\:{be}\:{correct} \\ $$$${now}\:{lets}\:{find}\:{the}\:{zeroes}\:{of}\:{A}\:{and}\:{D} \\ $$$${A}.\:\:\:−\frac{\mathrm{1}}{\mathrm{3}}\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{2}\right)=\pm\sqrt{−\left(\mathrm{5}\right)\left(−\mathrm{3}\right)} \\ $$$$\Rightarrow{x}=\mathrm{2}\pm\sqrt{\mathrm{15}} \\ $$$${D}.\:\:\:\:\mathrm{3}\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{5}=\mathrm{0}\:\: \\ $$$$\Rightarrow\left({x}−\mathrm{2}\right)=\pm\sqrt{−\frac{\mathrm{5}}{\mathrm{3}}} \\ $$$$\Rightarrow{x}=\mathrm{2}\pm{z} \\ $$$$\Rightarrow{x}\notin\:\mathbb{R} \\ $$$${But}\:{zeroes}\:{of}\:{the}\:{equation}\:{in}\:{the}\:{graph}\:{are}\:{Real} \\ $$$$\therefore\:{Only}\:{A}\:{is}\:{correct} \\ $$$${Ans}.\:{A}.\:{f}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{3}}\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{5} \\ $$
Commented by Kunal12588 last updated on 30/Apr/19
Commented by tanmay last updated on 30/Apr/19
$${bah}\:{excellent}…{very}\:{good}.. \\ $$
Commented by Tawa1 last updated on 30/Apr/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Kunal12588 last updated on 30/Apr/19
$${It}\:{was}\:{my}\:{pleasure}.\:{It}\:{is}\:{my}\:{first}\:{try}\:{of}\: \\ $$$${such}\:{questions}. \\ $$
Answered by MJS last updated on 30/Apr/19
$$\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{hanging}\:\mathrm{parabola}\:\Rightarrow \\ $$$$\Rightarrow\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}\:\mathrm{with}\:{a}<\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{options}\:\mathrm{C},\:\mathrm{D}\:\mathrm{don}'\mathrm{t}\:\mathrm{fit} \\ $$$$\mathrm{A}\:\:{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{11}=\mathrm{0}\:\Rightarrow\:−\frac{{p}}{\mathrm{2}}=\mathrm{2}\:\Rightarrow\:\mathrm{max}\:\mathrm{at}\:\mathrm{2} \\ $$$${B}\:\:{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{11}=\mathrm{0}\:\Rightarrow\:−\frac{{p}}{\mathrm{2}}=−\mathrm{2}\:\Rightarrow\:\mathrm{max}\:\mathrm{at}\:−\mathrm{2} \\ $$$$\Rightarrow\:\mathrm{option}\:{A} \\ $$
Commented by Tawa1 last updated on 30/Apr/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$