MHT CET · Maths · Mathematical Reasoning
The value of \((1+\Delta)(1-\nabla)\) is
- A 0
- B \(-1\)
- C 1
- D None of these
Answer & Solution
Correct Answer
(C) 1
Step-by-step Solution
Detailed explanation
We have, \((1+\Delta)(1-\nabla) f(x)\)
\(
\begin{array}{l}
=(1+\Delta)\{(1-\nabla) f(x)\} \\
=(1+\Delta)\{f(x)-\nabla f(x)\} \\
=(1+\Delta)[f(x)-\{f(x)-f(x-h)\}] \\
=(1+\Delta) f(x-h)=E f(x-h) \\
[\because(E=1+\Delta)] \\
=f(x)=1 \cdot f(x)
\end{array}
\)
Thus, \((1+\Delta)(1-\nabla) f(x)=1 \cdot f(x)\), for any function \(f(x)\).
\(
\therefore (1+\Delta)(1+\nabla)=1
\)
\(
\begin{array}{l}
=(1+\Delta)\{(1-\nabla) f(x)\} \\
=(1+\Delta)\{f(x)-\nabla f(x)\} \\
=(1+\Delta)[f(x)-\{f(x)-f(x-h)\}] \\
=(1+\Delta) f(x-h)=E f(x-h) \\
[\because(E=1+\Delta)] \\
=f(x)=1 \cdot f(x)
\end{array}
\)
Thus, \((1+\Delta)(1-\nabla) f(x)=1 \cdot f(x)\), for any function \(f(x)\).
\(
\therefore (1+\Delta)(1+\nabla)=1
\)
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