Q100 / 150 · MHT CET 2011 · Mathematics

The value of $(1+\Delta)(1-\nabla)$ is

  • A
    0
  • B
    $-1$
  • Correct answer
    C
    1
  • D
    None of these

Solution

We have, $(1+\Delta)(1-\nabla) f(x)$ $$ \begin{array}{l} =(1+\Delta)\{(1-\nabla) f(x)\} \\ =(1+\Delta)\{f(x)-\nabla f(x)\} \\ =(1+\Delta)[f(x)-\{f(x)-f(x-h)\}] \\ =(1+\Delta) f(x-h)=E f(x-h) \\ \quad[\because(E=1+\Delta)] \\ =f(x)=1 \cdot f(x) \end{array} $$ Thus, $(1+\Delta)(1-\nabla) f(x)=1 \cdot f(x)$, for any function $f(x)$. $$ \therefore \quad(1+\Delta)(1+\nabla)=1 $$

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

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