Let  $$f_n : [0, 1] \to \mathbb{R}$$​  be given by:

​$$f_n(t) = (n+2)(n+1)t^n(1-t), \quad \forall t \in [0, 1].$$​​

Which of the following is true?

Correct option is B

The given function is:

​$$f_n(t) = (n+2)(n+1)t^n(1-t).$$​​

The pointwise limit is:

​$$f(t) = \lim_{n \to \infty} f_n(t) = 0, \quad \forall t \in [0, 1].$$​​

Uniform Convergence:

To check uniform convergence, consider:

​$$|f_n(t) - f(t)| = f_n(t) = (n+2)(n+1)t^n(1-t).$$​​

Let:
​$$g(t) = (n+2)(n+1)t^n(1-t).$$​​

The derivative is:

​$$g'(t) = (n+2)(n+1)t^{n-1}\left[-t + n(1-t)\right] = (n+2)(n+1)t^{n-1}\left[n - (n+1)t\right].$$​​

Setting  g'(t) = 0 gives the critical point:

​$$t = \frac{n}{n+1}.$$​​

At $$t = \frac{n}{n+1}, \text{the local maximum of } \ (g(t)\ is:$$​​

​$$g\left(\frac{n}{n+1}\right) = (n+2)(n+1)\left(\frac{n}{n+1}\right)^n \left(1 - \frac{n}{n+1}\right).$$​​

Simplify:

​$$g\left(\frac{n}{n+1}\right) = (n+2)(n+1) \cdot \frac{n^n}{(n+1)^n} \cdot \frac{1}{n+1}.$$​​

This gives:

​$$M_n = (n+2) \cdot \frac{1}{\left(1 + \frac{1}{n}\right)^n}.$$​​

As $$n \to \infty$$​:

​$$M_n \to \infty.$$​​


Hence, the convergence is not uniform; it is only pointwise.

​$$\implies \textbf{Option B is correct}$$​​