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The-tangent-to-the-curve-y-x-3-bx-where-b-is-constant-at-x-1-passes-through-the-points-A-1-6-and-2-15-Find-the-value-of-b-




Question Number 56738 by pieroo last updated on 22/Mar/19
The tangent to the curve y=x^3 +bx, where  b is constant, at x=1 passes through the  points A(−1,6) and (2,−15). Find the  value of b.
$$\mathrm{The}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{y}=\mathrm{x}^{\mathrm{3}} +\mathrm{bx},\:\mathrm{where} \\ $$$$\mathrm{b}\:\mathrm{is}\:\mathrm{constant},\:\mathrm{at}\:\mathrm{x}=\mathrm{1}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the} \\ $$$$\mathrm{points}\:\mathrm{A}\left(−\mathrm{1},\mathrm{6}\right)\:\mathrm{and}\:\left(\mathrm{2},−\mathrm{15}\right).\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{b}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Mar/19
(dy/dx)=3x^2 +b→m=3+b  eqn y=(3+b)x+c  6=(3+b)×−1+c  −15=(3+b)×2+c  21=(3+b)(−1−2)  3+b=−7  b=−10
$$\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} +{b}\rightarrow{m}=\mathrm{3}+{b} \\ $$$${eqn}\:{y}=\left(\mathrm{3}+{b}\right){x}+{c} \\ $$$$\mathrm{6}=\left(\mathrm{3}+{b}\right)×−\mathrm{1}+{c} \\ $$$$−\mathrm{15}=\left(\mathrm{3}+{b}\right)×\mathrm{2}+{c} \\ $$$$\mathrm{21}=\left(\mathrm{3}+{b}\right)\left(−\mathrm{1}−\mathrm{2}\right) \\ $$$$\mathrm{3}+{b}=−\mathrm{7} \\ $$$${b}=−\mathrm{10} \\ $$
Commented by pieroo last updated on 22/Mar/19
Thanks boss
$$\mathrm{Thanks}\:\mathrm{boss} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 22/Mar/19
most welcome
$${most}\:{welcome} \\ $$

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