Let G : [0, 1] × [0, 1] → R be defined as:

$$G(t, x) =\begin{cases} t(1 - x), & \text{if } t \leq x \leq 1, \\ x(1 - t), & \text{if } x \leq t \leq 1.\end{cases}$$ 

For a continuous function f on [0, 1],

define:$$I[f] = \int_0^1 \int_0^1 G(t, x) f(x) \ dt \ dx.$$​​

Which of the following is true?

Correct option is A

​​$$\text{The given integral is:} \\[10pt]I[f] = \int_0^1 \int_0^1 G(t, x) f(x) \ dt \ dx.\\[10pt]\text{Splitting into two regions:} \\[10pt]I[f] = \int_0^1 \int_0^x t(1 - x) f(x) \ dt \ dx + \int_0^1 \int_0^t x(1 - t) f(x) \ dx \ dt.\\[10pt]\text{Computing the inner integrals:} \\[10pt]\begin{array}{l}\bullet \quad \textbf{First integral:} \quad \int_0^x t(1 - x) dt = (1 - x) \frac{x^2}{2} = \frac{x^2(1 - x)}{2}.\\[10pt]\bullet \quad \textbf{Second integral:} \quad \int_0^t x(1 - t) dx = (1 - t) \frac{t^2}{2} = \frac{t^2(1 - t)}{2}.\end{array}\\[10pt]\text{Since both expressions are the same, the integral simplifies to:} \\[10pt]I[f] = \int_0^1 x^2(1 - x) f(x) dx.$$

Checking the Given Options :

Option A. I[f] > 0 if f is not identically zero.-

The function  $$x^2$$​(1 − x) is non-negative for all x ∈ [0, 1] and positive forx ∈ (0, 1).

 If f(x) is not identically zero and continuous, then the integralcannot be negative or zero. 

Thus, I[f] > 0,confirming Option A is correct.

Option B. There exists a nonzero f such that I[f] = 0.-

This requires x$$^2$$​(1 − x)f(x) to integrate to zero,

which is impossible sincex$$^2$$​(1 − x) > 0 for all x $$\neq$$​ 0, 1.- So this option is false.

Option C. There exists f such that I[f] < 0.-

Since x2(1 − x) ≥ 0 and strictly positive for x ∈ (0, 1), the integral cannot benegative.

- Thus, Option C is false.

Option D. I[sin(πx)] = 1.

- The integral to check is:$$I[\sin(\pi x)] = \int_0^1 x^2(1 - x) \sin(\pi x) dx.$$​​

This does not necessarily simplify to 1, making Option D false.

Final Answer :

Correct option: Option A. I[f] > 0 if f is not identically zero.