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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. Then, which of the following is NOT true?

Option C: For n 1 = 3, n 2 = 5, there exists \alpha \in (3, 5) where f attains local maxima. Explanation: 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 n 1 = 3 and n 2 = 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 n 1 = 4 and n 2 = 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 n 1 = 3 and n 2 = 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.

JEE Main 2022MathematicsApplication Of DerivativesTangent And NormalmediumMCQ

JEE Main 2022 — Application Of Derivatives Question with Solution

From: JEE Main 2022 (Online) 29th June Evening Shift

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?

Choose an option

Show full solutionCorrect option: C

Correct answer

CFor n₁ = 3, n₂ = 5, there exists

α

\in

(3, 5) where f attains local maxima.

Step-by-step explanation

Given,

$f (x) = (- 3)^{n_{1}} (x - 5)^{n_{2}}$

Differentiating both side with respect to x, we get

$f^{'} (x) = (x - 5)^{n_{2}} (n_{1} (x - 3)^{n_{1} - 1}) + (x - 3)^{n_{1}} (n_{2} (x - 5)^{n_{2} - 1})$

$= (x - 3)^{n_{1}} (x - 5)^{n_{2}} [\frac{n _{1}}{x - 3} + \frac{n _{2}}{x - 5}]$

$= (x - 3)^{n_{1}} (x - 5)^{n_{2}} [\frac{n _{1} ( x - 5 ) + n _{2} ( x - 3 )}{( x - 3 ) ( x - 5 )}]$

$= (x - 3)^{n_{1} - 1} (x - 5)^{n_{2} - 1} [n_{1} (x - 5) + n_{2} (x - 3)]$

Option A :

When n₁ = 3 and n₂ = 4 then

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

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

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

JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Application of Derivatives Question 91 English Explanation 1

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

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

JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Application of Derivatives Question 91 English Explanation 2

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

$∴$ Option A is correct.

Option B :

When n₁ = 4 and n₂ = 3 then

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

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

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

JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Application of Derivatives Question 91 English Explanation 3

When f'(x) is less than $\frac{29}{7}$ 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 $\frac{29}{7}$ but less than 5 then f'(x) is positive means slope of graph of f(x) is positive.

JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Application of Derivatives Question 91 English Explanation 4

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

$∴$ Option (B) is correct.

Option C :

When n₁ = 3 and n₂ = 5 then

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

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

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

JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Application of Derivatives Question 91 English Explanation 5

When f'(x) is less than $\frac{30}{8}$ 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 $\frac{30}{8}$ but less than 5 then f'(x) is positive means slope of graph of f(x) is positive.

JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Application of Derivatives Question 91 English Explanation 6

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

$∴$ Option C is not true.

Similarly you can check option D also.

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About this question

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