Let f : R [] R be a function defined by f(x) = (x - 3)n1 (x - 5)n2, n1, n2 [] N.

Question

Let f : R $$\to$$ R be a function defined by f(x) = (x - 3)n1 (x - 5)n2, n1, n2 $$\in$$ N. Then, which of the following is NOT true?

Accepted Answer

Given,

$f(x) = {( - 3)^{{n_1}}}{(x - 5)^{{n_2}}}$

Differentiating both side with respect to x, we get

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

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

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

$$ = {(x - 3)^{{n_1} - 1}}{(x - 5)^{{n_2} - 1}}[{n_1}(x - 5) + {n_2}(x - 3)]$$
Option A :
When n1 = 3 and n2 = 4 then

$$f'(x) = {(x - 3)^{3 - 1}}{(x - 5)^{4 - 1}}\left[ {3(x - 5) + 4(x - 3)} \right]$$

$= {(x - 3)^2}{(x - 5)^3}(7x - 27)$

Critical point is at x = 3, 5 and $${{27} \over 7}$$

When value of f'(x) is less than $${{27} \over 5}$$, f'(x) is positive means slope of f(x) graph is positive.

And when value of f'(x) is greater than $${{27} \over 7}$$ but less than 5 then f'(x) is negative means slope of f(x) is negative.

So at $${{27} \over 7}$$ between 3 and 5, f(x) attain local maxima.

$$\therefore$$ Option A is correct.
Option B :
When n1 = 4 and n2 = 3 then

$$f'(x) = {(x - 3)^{4 - 1}}{(x - 5)^{3 - 1}}\left[ {4(x - 5) + 3(x - 3)} \right]$$

$= {(x - 3)^3}{(x - 5)^2}(7x - 29)$

Critical point is at x = 3, 5 and $${{29} \over 7}$$

When f'(x) is less than $${{29} \over 7}$$ but greater than 3 then f'(x) is negative means slope of graph of f(x) is negative.

And when f'(x) is greater than $${{29} \over 7}$$ but less than 5 then f'(x) is positive means slope of graph of f(x) is positive.

So at $${{29} \over 7}$$ between 3 and 5, f(x) attain local minima.

$$\therefore$$ Option (B) is correct.
Option C :
When n1 = 3 and n2 = 5 then

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

$= {(x - 3)^2}{(x - 5)^4}(8x - 30)$

Critical point is at x = 3, 5 and $${{30} \over 8}$$

When f'(x) is less than $${{30} \over 8}$$ but greater than 3 then f'(x) is negative means slope of graph of f(x) is negative.

And when f'(x) is greater than $${{30} \over 8}$$ but less than 5 then f'(x) is positive means slope of graph of f(x) is positive.

So, at $${{30} \over 8}$$ between 3 and 5, f(x) attain local minima.

$$\therefore$$ Option C is not true.

Similarly you can check option D also.

Solutions · 1