Q86 / 150 · MHT CET 2008 · Mathematics

The value of $\Delta \log f(x)+\Delta^{2}\left(3^{x}\right)$ is

  • Correct answer
    A
    $\log \left[1+\frac{\Delta f(x)}{f(x)}\right]+4 \cdot 3^{x}$
  • B
    $\log \left[1+\frac{\Delta f(x)}{f(x)}\right]+3^{x}$
  • C
    $\log \left[\frac{\Delta f(x)}{1+f(x)}\right]+4 \cdot 3^{x}$
  • D
    $\log \left[\frac{\Delta f(x)}{1+f(x)}\right]+3^{x}$

Solution

$\begin{aligned} \Delta & \log f(x)+\Delta^{2}\left(3^{x}\right) \\ &=\log f(x+h)-\log f(x)+(E-1)^{2} 3^{x} \\ &=\log \left[\frac{f(x+h)}{f(x)}\right]+\left(E^{2}-2 E+1\right) 3^{x} \\ &=\log \left[\frac{E f(x)]}{f(x)}\right]+E^{2}\left(3^{x}\right)-2 E\left(3^{x}\right)+3^{x} \\ &=\log \left[\frac{(1+\Delta) f(x)}{f(x)}\right]+3^{x+2}-2 \cdot 3^{x+1}+3^{x} \\ &=\log \left[1+\frac{\Delta f(x)}{f(x)}\right]+3^{x}(9-6+1) \\ &=\log \left[1+\frac{\Delta f(x)}{f(x)}\right]+4 \cdot 3^{x} \end{aligned}$

Hence, the correct answer is option: 'A'.

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