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The-curve-y-ax-b-2x-1-has-the-stationary-point-at-2-7-Find-the-value-of-a-and-b-




Question Number 52124 by 786786AM last updated on 03/Jan/19
The curve y = ax + (b/(2x− 1)) has the stationary point at  (2, 7) . Find the value of a and b .
$$\mathrm{The}\:\mathrm{curve}\:\mathrm{y}\:=\:\mathrm{ax}\:+\:\frac{\mathrm{b}}{\mathrm{2x}−\:\mathrm{1}}\:\mathrm{has}\:\mathrm{the}\:\mathrm{stationary}\:\mathrm{point}\:\mathrm{at}\:\:\left(\mathrm{2},\:\mathrm{7}\right)\:.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jan/19
2xy−y=2ax^2 −ax+b  partial derivative w.r.t x  2y−0=4ax−a  14=8a−a  a=7  y=ax+(b/(2x−1))  7=7×2+(b/(2×2−1))  (b/3)=−7  b=−21    pls check is it correct...
$$\mathrm{2}{xy}−{y}=\mathrm{2}{ax}^{\mathrm{2}} −{ax}+{b} \\ $$$${partial}\:{derivative}\:{w}.{r}.{t}\:{x} \\ $$$$\mathrm{2}{y}−\mathrm{0}=\mathrm{4}{ax}−{a} \\ $$$$\mathrm{14}=\mathrm{8}{a}−{a} \\ $$$${a}=\mathrm{7} \\ $$$${y}={ax}+\frac{{b}}{\mathrm{2}{x}−\mathrm{1}} \\ $$$$\mathrm{7}=\mathrm{7}×\mathrm{2}+\frac{{b}}{\mathrm{2}×\mathrm{2}−\mathrm{1}} \\ $$$$\frac{{b}}{\mathrm{3}}=−\mathrm{7} \\ $$$${b}=−\mathrm{21} \\ $$$$ \\ $$$${pls}\:{check}\:{is}\:{it}\:{correct}… \\ $$

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