Let $$f : R \to R$$ be a function defined by $$f(x) = (x-3)^{n_1}(x-5)^{n_2}$$, $$n_1, n_2 \in N$$. The, which of the following is NOT true?

Given $$f(x) = (x - 3)^{n_1}(x - 5)^{n_2}$$ where $$n_1, n_2 \in \mathbb{N}$$. We need to determine which statement about local maxima in $$(3, 5)$$ is NOT true.

Finding the critical point: Differentiating:

$$f'(x) = (x-3)^{n_1-1}(x-5)^{n_2-1}\left[n_1(x-5) + n_2(x-3)\right]$$

Setting the linear factor to zero: $$n_1(x-5) + n_2(x-3) = 0$$ gives $$\alpha = \dfrac{5n_1 + 3n_2}{n_1 + n_2}$$, which lies in $$(3, 5)$$ for all positive $$n_1, n_2$$.

Nature of the critical point: Since $$f(3) = f(5) = 0$$, if $$f(x) > 0$$ in $$(3, 5)$$, then $$f$$ has a local maximum; if $$f(x) < 0$$ in $$(3, 5)$$, then $$f$$ has a local minimum.

In $$(3, 5)$$: $$(x-3)^{n_1} > 0$$ always (since $$x - 3 > 0$$). The sign of $$(x-5)^{n_2}$$ depends on the parity of $$n_2$$ (since $$x - 5 < 0$$).

• $$n_2$$ even $$\Rightarrow$$ $$(x-5)^{n_2} > 0$$ $$\Rightarrow$$ $$f(x) > 0$$ $$\Rightarrow$$ local maximum.
• $$n_2$$ odd $$\Rightarrow$$ $$(x-5)^{n_2} < 0$$ $$\Rightarrow$$ $$f(x) < 0$$ $$\Rightarrow$$ local minimum.

Checking Option C ($$n_1 = 3, n_2 = 5$$): Since $$n_2 = 5$$ is odd, $$f(x) < 0$$ throughout $$(3, 5)$$ while $$f(3) = f(5) = 0$$. The function dips below zero and returns, giving a local minimum at $$\alpha$$, not a local maximum.

We can verify using the first derivative test at $$\alpha = \dfrac{5(3)+3(5)}{3+5} = \dfrac{30}{8} = 3.75$$:

$$f'(x) = (x-3)^2(x-5)^4(8x - 30)$$

For $$x < 3.75$$: $$(x-3)^2 > 0$$, $$(x-5)^4 > 0$$, $$(8x-30) < 0$$ $$\Rightarrow$$ $$f'(x) < 0$$.

For $$x > 3.75$$: $$(8x-30) > 0$$ $$\Rightarrow$$ $$f'(x) > 0$$.

Since $$f'$$ changes from negative to positive, $$\alpha = 3.75$$ is a point of local minimum, confirming that Option C's claim of local maxima is false.

The correct answer is Option C.