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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\text{y}=\log_\text{e}\Big(\frac{\text{x}}{\text{a}+\text{bx}}\Big)^\text{x}$ then $\text{x}^3\text{y}_2=$
  • ✓
    $(\text{xy}_1-\text{y})^2$
  • B
    $(1+\text{y})^2$
  • C
    $\Big(\frac{\text{y}-\text{xy}_1}{\text{y}_1}\Big)^2$
  • D
    None of these
✓

Answer

Correct option: A.
$(\text{xy}_1-\text{y})^2$
$\text{y}=\log_\text{e}\Big(\frac{\text{x}}{\text{a}+\text{b}}\Big)^\text{x}$
$\text{y}=\text{x}(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))$
$\frac{\text{dy}}{\text{dx}}=(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))+\text{x}\Big(\frac{1}{\text{x}}-\frac{\text{b}}{\text{a}+\text{bx}}\Big)$
$=(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))+1-\frac{\text{bx}}{\text{a}+\text{bx}}$
$(\text{a}+\text{bx})\frac{\text{dy}}{\text{dx}}=(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))(\text{a}+\text{bx})+\text{a}$
Again differentiating $\text{w.r.t.  x},$ we get
$(\text{a}+\text{bx})\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{b}\frac{\text{dy}}{\text{dx}}=\text{b}(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))$
$+(\text{a}+\text{bx})\Big(\frac{1}{\text{x}}\frac{\text{b}}{\text{a}+\text{bx}}\Big)$
$\Rightarrow(\text{a}+\text{bx})\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{b}\frac{\text{dy}}{\text{dx}}=\text{b}(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))+\frac{\text{a}}{\text{x}}$
$\Rightarrow(\text{a}+\text{bx})\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{b}\frac{\text{dy}}{\text{dx}}=\frac{\text{by}}{\text{x}}+\frac{\text{a}}{\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{a}+\text{by}}{\text{x}(\text{a}+\text{bx})}-\frac{\text{b}}{(\text{a}+\text{bx})}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} =\frac{\text{a}+\text{by}}{\text{x}(\text{a}+\text{bx})}-\frac{\text{b}}{(\text{a}+\text{bx})}\Big(\frac{\text{y}}{\text{x}}+\frac{\text{a}}{\text{a}+\text{bx}}\Big)$
$=\frac{(\text{a}+\text{by})(\text{a}+\text{bx})-\text{b}(\text{ay}+\text{bxy}+\text{ax})}{\text{x}(\text{a}+\text{b})^2}$
$=\frac{\text{a}^2+\text{abx}+\text{aby}+\text{b}^2\text{xy}-\text{bay}-\text{b}^2\text{xy}-\text{abx}}{\text{x}(\text{a}+\text{bx})^2}$
$=\frac{\text{a}^2}{\text{x}(\text{a}+\text{bx})^2}$
$\text{x}^3\text{y}_2=\frac{\text{x}^2\text{a}^2}{(\text{a}+\text{bx})^2}$
$(\text{xy}_1-\text{y})=\frac{\text{ax}}{\text{a}+\text{bx}}$
$\text{x}^3\text{y}_2=\frac{\text{x}^2\text{a}^2}{(\text{a}+\text{bx})^2}=(\text{xy}_1-\text{y})^2$

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