The value of (1+Δ) (1- nabla) is

Question

The value of (1+Δ) (1- nabla) is (A)  0  (B)  -1  (C)  1  (D) None of these

Tardigrade Admin · Accepted Answer

We have, (1+Δ)(1- nabla) f(x) =(1+Δ) (1- nabla) f(x) = (1+Δ) f(x)- nabla f(x) = (1+Δ)[f(x)- f(x)-f(x-h) ] = (1+Δ) f(x-h)=E f(x-h) [∵(E=1+Δ)] =f(x)=1 ⋅ f(x) Thus, (1+Δ)(1- nabla) f(x)=1 ⋅ f(x), for any function f(x) ∴ (1+Δ)(1+ nabla) =1

Q. The value of $(1 + Δ) (1 - \nabla)$ is

$0$

$- 1$

$1$

None of these

Q. The value of (1+Δ)(1−∇) is

0

−1

1

None of these

We have, (1+Δ)(1−∇)f(x) =(1+Δ){(1−∇)f(x)} =(1+Δ){f(x)−∇f(x)} =(1+Δ)[f(x)−{f(x)−f(x−h)}] =(1+Δ)f(x−h)=Ef(x−h) [∵(E=1+Δ)] =f(x)=1⋅f(x) Thus, (1+Δ)(1−∇)f(x)=1⋅f(x), for any function f(x) ∴(1+Δ)(1+∇)=1

Q. The value of $(1 + Δ) (1 - \nabla)$ is

$0$

$- 1$

$1$