JEE Main 2020 (7 January Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Differentiation: Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Q: Why does the limit $\lim_{x \to 0} (2 + \frac{f(x)}{x^3}) = 4$ imply that the coefficients of $x^2$, $x$, and the constant term are zero?

If $f(x)$ had a term like $dx^2$, $ex$, or $f$, then the expression $\frac{f(x)}{x^3}$ would contain terms like $\frac{d}{x}$, $\frac{e}{x^2}$, or $\frac{f}{x^3}$. As $x \to 0$, these terms would approach infinity, making the limit undefined. For the limit to be a finite value, these lower-degree terms must be zero.

Q: Why are $x = 1$ and $x = -1$ called critical points?

In calculus, a critical point of a function $f(x)$ is a point where the derivative $f'(x)$ is either zero or undefined. Since $f(x)$ is a polynomial, it is differentiable everywhere, so the critical points must be the roots of the equation $f'(x) = 0$.

Select Answer:

Visualized Solution

Analyzing the Limit at $x \to 0$

Given: $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$
Subtract 2 from both sides: $lim_{x \to 0} \frac{f ( x )}{x ^{3}} = 2$

Deducing the Polynomial Structure

For the limit $lim_{x \to 0} \frac{f ( x )}{x ^{3}}$ to be finite, $f (x)$ cannot have terms of degree less than 3.
Let $f (x) = a x^{5} + b x^{4} + c x^{3} + d x^{2} + e x + f$ .
Therefore, $d = 0$ , $e = 0$ , and $f = 0$ .

Finding the Coefficient $c$

The polynomial simplifies to: $f (x) = a x^{5} + b x^{4} + c x^{3}$ .
Substitute into the limit: $lim_{x \to 0} \frac{a x ^{5} + b x ^{4} + c x ^{3}}{x ^{3}} = 2$ .
Simplifying: $lim_{x \to 0} (a x^{2} + b x + c) = 2$ .
Thus, $c = 2$ .

The Simplified Polynomial

Current form of the polynomial: $f (x) = a x^{5} + b x^{4} + 2 x^{3}$ .
We need to find $a$ and $b$ using the critical points.

Utilizing Critical Points

Critical points occur where the derivative is zero.
Given critical points at $x = 1$ and $x = - 1$ .
Therefore, $f^{'} (1) = 0$ and $f^{'} (- 1) = 0$ .

Differentiating the Function

Differentiate $f (x) = a x^{5} + b x^{4} + 2 x^{3}$ .
$f^{'} (x) = 5 a x^{4} + 4 b x^{3} + 6 x^{2}$ .

Setting Up the Equations

Substitute $x = 1$ : $f^{'} (1) = 5 a (1)^{4} + 4 b (1)^{3} + 6 (1)^{2} = 0 ⟹ 5 a + 4 b + 6 = 0$ .
Substitute $x = - 1$ : $f^{'} (- 1) = 5 a (- 1)^{4} + 4 b (- 1)^{3} + 6 (- 1)^{2} = 0 ⟹ 5 a - 4 b + 6 = 0$ .

Solving for $a$ and $b$

Add the two equations: $(5 a + 4 b + 6) + (5 a - 4 b + 6) = 0$ .
$10 a + 12 = 0 ⟹ a = - \frac{6}{5}$ .
Subtract the equations: $(5 a + 4 b + 6) - (5 a - 4 b + 6) = 0$ .
$8 b = 0 ⟹ b = 0$ .

The Final Polynomial and its Parity

Substitute $a$ and $b$ back: $f (x) = - \frac{6}{5} x^{5} + 2 x^{3}$ .
Notice that all powers of $x$ are odd (5 and 3).
Therefore, $f (- x) = - f (x)$ , which means $f (x)$ is an odd function.
This makes Option 3 True.

Evaluating $f (1) - 4 f (- 1)$

Calculate $f (1) = - \frac{6}{5} (1)^{5} + 2 (1)^{3} = \frac{4}{5}$ .
Calculate $f (- 1) = - \frac{6}{5} (- 1)^{5} + 2 (- 1)^{3} = - \frac{4}{5}$ .
Evaluate: $f (1) - 4 f (- 1) = \frac{4}{5} - 4 (- \frac{4}{5}) = \frac{4}{5} + \frac{16}{5} = \frac{20}{5} = 4$ .
This makes Option 1 True.

Analyzing the Derivative for Extrema

We need to classify the critical points as maxima or minima.
$f^{'} (x) = - 6 x^{4} + 6 x^{2} = 6 x^{2} (1 - x^{2})$ .
Factorizing further: $f^{'} (x) = 6 x^{2} (1 - x) (1 + x)$ .

Concluding Maxima and Minima

Use the First Derivative Test to check sign changes of $f^{'} (x)$ .
At $x = 1$ : As $x$ increases through 1, $(1 - x)$ changes from positive to negative. $f^{'} (x)$ changes from $+$ to $-$ . Thus, $x = 1$ is a Local Maxima.
At $x = - 1$ : As $x$ increases through -1, $(1 + x)$ changes from negative to positive. $f^{'} (x)$ changes from $-$ to $+$ . Thus, $x = - 1$ is a Local Minima.
Option 4 claims $x = 1$ is minima and $x = - 1$ is maxima, which is False.

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The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

We are tasked with determining a polynomial

f (x)

of degree 5. We are given the limit condition:

x \to 0 lim (2 + \frac{f ( x )}{x ^{3}}) = 4

By subtracting 2 from both sides, we obtain:

x \to 0 lim \frac{f ( x )}{x ^{3}} = 2

The Limit as a Filter

f (x)

contained terms with powers of

x

less than 3 (i.e.,

x^{2}, x^{1}, x^{0}

), the expression

\frac{f ( x )}{x ^{3}}

would result in terms with

x

in the denominator. As

x \to 0

, these terms would diverge to infinity.

Since the limit is a finite value of 2, the coefficients of

x^{2}, x,

and the constant term must be zero. Thus, the polynomial simplifies to:

f (x) = a x^{5} + b x^{4} + c x^{3}

Dividing by

x^{3}

yields

a x^{2} + b x + c

. Taking the limit as

x \to 0

, we find

c = 2

. Our polynomial is now:

f (x) = a x^{5} + b x^{4} + 2 x^{3}

The Geometry of Slopes

We are given that

f (x)

has critical points at

x = 1

and

x = - 1

. This implies that the derivative

f^{'} (x)

must vanish at these points.

First, we compute the derivative:

f^{'} (x) = 5 a x^{4} + 4 b x^{3} + 6 x^{2}

Setting

f^{'} (1) = 0

and

f^{'} (- 1) = 0

provides the following system of equations:

5 a + 4 b + 6 = 0

5 a - 4 b + 6 = 0

Adding these equations yields

10 a + 12 = 0

, which gives

a = - \frac{6}{5}

. Subtracting them yields

8 b = 0

, so

b = 0

The Final Verdict

The reconstructed polynomial is:

f (x) = - \frac{6}{5} x^{5} + 2 x^{3}

Since the polynomial contains only odd powers of

x

, it is an odd function. We can verify the value of

f (1) - 4 f (- 1)

f (1) = - \frac{6}{5} + 2 = \frac{4}{5}

f (- 1) = \frac{6}{5} - 2 = - \frac{4}{5}

f (1) - 4 f (- 1) = \frac{4}{5} - 4 (- \frac{4}{5}) = \frac{20}{5} = 4

Finally, analyzing the derivative

f^{'} (x) = - 6 x^{4} + 6 x^{2} = 6 x^{2} (1 - x) (1 + x)

, we observe sign changes around the critical points. The function reaches a local maximum at $x = 1$ and a local minimum at $x = - 1$ .

Common Traps & FAQs

Why does the limit

lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4

imply that the coefficients of

x^{2}

x

, and the constant term are zero?

f (x)

had a term like

d x^{2}

e x

, or

f

, then the expression

\frac{f ( x )}{x ^{3}}

would contain terms like

\frac{d}{x}

\frac{e}{x ^{2}}

, or

\frac{f}{x ^{3}}

. As

x \to 0

, these terms would approach infinity, making the limit undefined. For the limit to be a finite value, these lower-degree terms must be zero.

Why are

x = 1

and

x = - 1

called critical points?

In calculus, a critical point of a function

f (x)

is a point where the derivative

f^{'} (x)

is either zero or undefined. Since

f (x)

is a polynomial, it is differentiable everywhere, so the critical points must be the roots of the equation

f^{'} (x) = 0

How do I know if

x = 1

is a maximum or a minimum without using the second derivative test?

The first derivative test is often more intuitive. By factoring

f^{'} (x)

, you can determine the sign of the slope on either side of the critical point. If the slope changes from positive to negative as you pass through the point, it is a local maximum. If it changes from negative to positive, it is a local minimum.

Animated Solution for Mathematics - Differentiation: Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Select Answer:

Visualized Solution

Analyzing the Limit at $x \to 0$

Deducing the Polynomial Structure

Finding the Coefficient $c$

The Simplified Polynomial

Utilizing Critical Points

Differentiating the Function

Setting Up the Equations

Solving for $a$ and $b$

The Final Polynomial and its Parity

Evaluating $f (1) - 4 f (- 1)$

Analyzing the Derivative for Extrema

Concluding Maxima and Minima

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Limit as a Filter

The Geometry of Slopes

The Final Verdict

Similar Questions

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If $lim_{x \to 0} (\frac{f ( x )}{x ^{2}} + 1) = 3$ then f(-1) is equal to :-

Let $f (x)$ be a polynomial of degree four having extreme values at $x = 1$ and $x = 2$ . If $lim_{x \to 0} [1 + \frac{f ( x )}{x ^{2}}] = 3$ , then $f (2)$ is equal to

Let $f : R \to R$ be a polynomial function of degree four having extreme values at $x = 4$ and $x = 5$ . If $lim_{x \to 0} \frac{f ( x )}{x ^{2}} = 5$ , then $f (2)$ is equal to :

Let $p (x)$ be a polynomial of degree 4 having extremum at $x = 1, 2$ and $lim_{x \to 0} (1 + \frac{p ( x )}{x ^{2}}) = 2$ . Then the value of $p (2)$ is

Let $f (x)$ be a polynomial of degree 3 such that $f (- 1) = 10, f (1) = - 6, f (x)$ has a critical point at $x = - 1$ and $f^{'} (x)$ has a critical point at $x = 1$ . Then the local minima at $x = ______$

Let $f (x)$ be a polynomial of degree $3$ such that $f (- 1) = 10$ , $f (1) = - 6$ , $f (x)$ has a critical point at $x = - 1$ and $f^{'} (x)$ has a critical point at $x = 1$ . Then $f (x)$ has a local minima at $x =$

Let $P (x) = a_{0} + a_{1} x^{2} + a_{2} x^{4} + \dots + a_{n} x^{2 n}$ be a polynomial in a real variable $x$ with $0 < a_{0} < a_{1} < a_{2} < \dots < a_{n}$ . The function $P (x)$ has

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?

The number of critical points of the function $f (x) = (x - 2)^{2/3} (2 x + 1)$ is

Select Answer:

Visualized Solution

Analyzing the Limit at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x→0

Deducing the Polynomial Structure

Finding the Coefficient .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

The Simplified Polynomial

Utilizing Critical Points

Differentiating the Function

Setting Up the Equations

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

The Final Polynomial and its Parity

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1)−4f(−1)

Analyzing the Derivative for Extrema

Concluding Maxima and Minima

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Limit as a Filter

The Geometry of Slopes

The Final Verdict

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of critical points of the function f(x)=(x−2)2/3(2x+1) is

Analyzing the Limit at $x \to 0$

Finding the Coefficient $c$

Solving for $a$ and $b$

Evaluating $f (1) - 4 f (- 1)$

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If $lim_{x \to 0} (\frac{f ( x )}{x ^{2}} + 1) = 3$ then f(-1) is equal to :-

Let $f (x)$ be a polynomial of degree four having extreme values at $x = 1$ and $x = 2$ . If $lim_{x \to 0} [1 + \frac{f ( x )}{x ^{2}}] = 3$ , then $f (2)$ is equal to

Let $f : R \to R$ be a polynomial function of degree four having extreme values at $x = 4$ and $x = 5$ . If $lim_{x \to 0} \frac{f ( x )}{x ^{2}} = 5$ , then $f (2)$ is equal to :

Let $p (x)$ be a polynomial of degree 4 having extremum at $x = 1, 2$ and $lim_{x \to 0} (1 + \frac{p ( x )}{x ^{2}}) = 2$ . Then the value of $p (2)$ is

Let $P (x) = a_{0} + a_{1} x^{2} + a_{2} x^{4} + \dots + a_{n} x^{2 n}$ be a polynomial in a real variable $x$ with $0 < a_{0} < a_{1} < a_{2} < \dots < a_{n}$ . The function $P (x)$ has

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?

The number of critical points of the function $f (x) = (x - 2)^{2/3} (2 x + 1)$ is