JEE MAIN - Mathematics (2020 - 7th January Evening Slot - No. 10)
Let ƒ(x) be a polynomial of degree 5 such that x = ±1 are its critical points.
If $$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4$$, then which one of the following is not true?
If $$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4$$, then which one of the following is not true?
ƒ(1) - 4ƒ(-1) = 4.
x = 1 is a point of minima and x = -1 is a point of maxima of ƒ.
x = 1 is a point of maxima and x = -1 is a point of minimum of ƒ.
ƒ is an odd function.
Explanation
let f(x) = ax5 + bx4 + cx3 + dx2 + ex + f
Given $$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4$$
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{a{x^5} + b{x^4} + c{x^3} + d{x^2} + ex + f} \over {{x^3}}}} \right)$$ = 4
As this limit exists so d = e = f = 0
$$ \therefore $$ f(x) = ax5 + bx4 + cx3
$$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{a{x^5} + b{x^4} + c{x^3}} \over {{x^3}}}} \right)$$ = 4
$$ \Rightarrow $$ 2 + c = 4
$$ \Rightarrow $$ c = 2
$$ \therefore $$ f(x) = ax5 + bx4 + 2x3
$$ \Rightarrow $$ f'(x) = 5ax4 + 4bx3 + 6x2
As x = ±1 are its critical points so f'(x) = 0 at x = ±1.
f'(1) = 5a + 4b + 6 = 0 ....(1)
and f'(-1) = 5a - 4b + 6 = 0 .....(2)
Solving (1) and (2),
a = $$ - {6 \over 5}$$ and b = 0
$$ \therefore $$ f(x) = $$ - {6 \over 5}{x^5} + 2{x^3}$$
So f'(x) = -6x4 + 6x2 = 6x2(1 + x)(1 - x)
Sign scheme for f'(x)
_7th_January_Evening_Slot_en_10_1.png)
It is clear that maxima at x = 1 and minima at x = –1.
Given $$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4$$
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{a{x^5} + b{x^4} + c{x^3} + d{x^2} + ex + f} \over {{x^3}}}} \right)$$ = 4
As this limit exists so d = e = f = 0
$$ \therefore $$ f(x) = ax5 + bx4 + cx3
$$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{a{x^5} + b{x^4} + c{x^3}} \over {{x^3}}}} \right)$$ = 4
$$ \Rightarrow $$ 2 + c = 4
$$ \Rightarrow $$ c = 2
$$ \therefore $$ f(x) = ax5 + bx4 + 2x3
$$ \Rightarrow $$ f'(x) = 5ax4 + 4bx3 + 6x2
As x = ±1 are its critical points so f'(x) = 0 at x = ±1.
f'(1) = 5a + 4b + 6 = 0 ....(1)
and f'(-1) = 5a - 4b + 6 = 0 .....(2)
Solving (1) and (2),
a = $$ - {6 \over 5}$$ and b = 0
$$ \therefore $$ f(x) = $$ - {6 \over 5}{x^5} + 2{x^3}$$
So f'(x) = -6x4 + 6x2 = 6x2(1 + x)(1 - x)
Sign scheme for f'(x)
It is clear that maxima at x = 1 and minima at x = –1.
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