JEE Main 2021 (February) (25 Feb Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Differentiation: Let $f (x)$ be a polynomial of degree $6$ in $x$ , in which the coefficient of $x^{6}$ is unity and it has extrema at $x = - 1$ and $x = 1$ . If $lim_{x \to 0} \frac{f ( x )}{x ^{3}} = 1$ , then $5 \cdot f (2)$ is equal to

Q: Wait, why do we assume the lowest power is $x^3$? Could it be $x^4$ or $x^2$?

Great question! If the lowest power were $x^2$, then $\lim_{x \to 0} \frac{f(x)}{x^3}$ would involve a term like $\frac{dx^2}{x^3} = \frac{d}{x}$, which approaches infinity as $x \to 0$. If the lowest power were $x^4$, the limit would be $0$. For the limit to be exactly $1$, the term $x^3$ must be the lowest power, and its coefficient must be $1$.

Q: Why do we multiply by $5$ at the very end?

The question asks for $5 \cdot f(2)$. We do this because our coefficients $a$ and $b$ are fractions (specifically, they have a denominator of $5$ or $2$). Multiplying by $5$ is a strategic move to clear the denominators and make the final arithmetic much cleaner and less prone to error.

Enter Numerical Value:

Visualized Solution

Defining the Polynomial $f (x)$

Let the degree 6 polynomial be:
$f (x) = x^{6} + a x^{5} + b x^{4} + c x^{3} + d x^{2} + e x + g$
The coefficient of $x^{6}$ is given as $1$ .

Analyzing the Limit Condition

Given limit: $lim_{x \to 0} \frac{f ( x )}{x ^{3}} = 1$
For the limit to be finite and non-zero, the lowest power of $x$ in $f (x)$ must be $x^{3}$ .
Therefore, the coefficients of $x^{2}, x^{1},$ and $x^{0}$ must be zero.
So, $d = 0, e = 0, g = 0$ .

Finding Coefficient $c$

Substitute the remaining terms into the limit:
$lim_{x \to 0} \frac{x ^{3} ( x ^{3} + a x ^{2} + b x + c )}{x ^{3}} = 1$
Evaluating the limit at $x = 0$ gives $c = 1$ .
The polynomial is now: $f (x) = x^{6} + a x^{5} + b x^{4} + x^{3}$

Finding the Derivative $f^{'} (x)$

To find extrema, we need the derivative $f^{'} (x)$ .
Differentiating $f (x) = x^{6} + a x^{5} + b x^{4} + x^{3}$ :
$f^{'} (x) = 6 x^{5} + 5 a x^{4} + 4 b x^{3} + 3 x^{2}$

Applying Extrema at $x = 1$

At $x = 1$ , $f^{'} (1) = 0$ :
$6 (1)^{5} + 5 a (1)^{4} + 4 b (1)^{3} + 3 (1)^{2} = 0$
$6 + 5 a + 4 b + 3 = 0$
Equation 1: $5 a + 4 b = - 9$

Applying Extrema at $x = - 1$

At $x = - 1$ , $f^{'} (- 1) = 0$ :
$6 (- 1)^{5} + 5 a (- 1)^{4} + 4 b (- 1)^{3} + 3 (- 1)^{2} = 0$
$- 6 + 5 a - 4 b + 3 = 0$
Equation 2: $5 a - 4 b = 3$

Solving for $a$

Add Equation 1 and Equation 2:
$(5 a + 4 b) + (5 a - 4 b) = - 9 + 3$
$10 a = - 6$
$a = - \frac{6}{10} = - \frac{3}{5}$

Solving for $b$

Substitute $a = - \frac{3}{5}$ into Equation 2:
$5 (- \frac{3}{5}) - 4 b = 3$
$- 3 - 4 b = 3$
$- 4 b = 6 ⟹ b = - \frac{3}{2}$

The Final Polynomial $f (x)$

The complete polynomial is:
$f (x) = x^{6} - \frac{3}{5} x^{5} - \frac{3}{2} x^{4} + x^{3}$

Calculating $f (2)$

Substitute $x = 2$ into $f (x)$ :
$f (2) = 2^{6} - \frac{3}{5} (2^{5}) - \frac{3}{2} (2^{4}) + 2^{3}$
$f (2) = 64 - \frac{3 \cdot 32}{5} - \frac{3 \cdot 16}{2} + 8$
$f (2) = 64 - \frac{96}{5} - 24 + 8$

Final Result: $5 \cdot f (2)$

Multiply the entire expression by $5$ :
$5 \cdot f (2) = 5 (64 - \frac{96}{5} - 24 + 8)$
$5 \cdot f (2) = 320 - 96 - 120 + 40$
$5 \cdot f (2) = 144$

Conclusion \& Key Takeaway

Key Takeaway:
The limit $lim_{x \to 0} \frac{f ( x )}{x ^{n}} = L$ implies that the lowest power of $x$ in $f (x)$ is $x^{n}$ and its coefficient is $L$ .
Extrema at $x = x_{0}$ implies $f^{'} (x_{0}) = 0$ for a polynomial.

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The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

We are tasked with reconstructing a degree six polynomial

f (x)

given specific constraints on its limit at zero and its extrema. A general degree six polynomial is defined as:

f (x) = x^{6} + a x^{5} + b x^{4} + c x^{3} + d x^{2} + e x + g

The Limit Detective

Our first clue is the limit:

lim_{x \to 0} \frac{f ( x )}{x ^{3}} = 1

. When we divide the polynomial by

x^{3}

, we obtain:

\frac{f ( x )}{x ^{3}} = x^{3} + a x^{2} + b x + c + \frac{d}{x} + \frac{e}{x ^{2}} + \frac{g}{x ^{3}}

For this limit to be a finite, non-zero value as

x \to 0

, the coefficients of the terms with negative powers of

x

must be zero. Thus,

d = 0

e = 0

, and

g = 0

Furthermore, the constant term

c

must equal the limit value. Therefore,

c = 1

. Our polynomial simplifies to:

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

The Derivative Dance

To find the constants

a

and

b

, we utilize the information that the polynomial has extrema at

x = 1

and

x = - 1

. This implies that the derivative

f^{'} (x)

must vanish at these points.

The derivative is calculated as:

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

The System of Equations

We apply the conditions

f^{'} (1) = 0

and

f^{'} (- 1) = 0

to generate a system of linear equations. For

x = 1

6 (1)^{5} + 5 a (1)^{4} + 4 b (1)^{3} + 3 (1)^{2} = 0 \Rightarrow 5 a + 4 b = - 9

For

x = - 1

6 (- 1)^{5} + 5 a (- 1)^{4} + 4 b (- 1)^{3} + 3 (- 1)^{2} = 0 \Rightarrow - 6 + 5 a - 4 b + 3 = 0 \Rightarrow 5 a - 4 b = 3

Solving this system by adding the two equations:

(5 a + 4 b) + (5 a - 4 b) = - 9 + 3 \Rightarrow 10 a = - 6 \Rightarrow a = - \frac{3}{5}

Substituting

a = - \frac{3}{5}

into

5 a - 4 b = 3

5 (- \frac{3}{5}) - 4 b = 3 \Rightarrow - 3 - 4 b = 3 \Rightarrow - 4 b = 6 \Rightarrow b = - \frac{3}{2}

Final Calculation

The reconstructed polynomial is

f (x) = x^{6} - \frac{3}{5} x^{5} - \frac{3}{2} x^{4} + x^{3}

. We now evaluate

5 \cdot f (2)

f (2) = 2^{6} - \frac{3}{5} (2^{5}) - \frac{3}{2} (2^{4}) + 2^{3} = 64 - \frac{96}{5} - 24 + 8

Multiplying the entire expression by

5

5 \cdot f (2) = 5 (64) - 5 (\frac{96}{5}) - 5 (24) + 5 (8)

5 \cdot f (2) = 320 - 96 - 120 + 40 = 144

The final result is 144.

Common Traps & FAQs

Wait, why do we assume the lowest power is

x^{3}

? Could it be

x^{4}

x^{2}

Great question! If the lowest power were

x^{2}

, then

lim_{x \to 0} \frac{f ( x )}{x ^{3}}

would involve a term like

\frac{d x ^{2}}{x ^{3}} = \frac{d}{x}

, which approaches infinity as

x \to 0

. If the lowest power were

x^{4}

, the limit would be

0

. For the limit to be exactly

1

, the term

x^{3}

must be the lowest power, and its coefficient must be

1

Why can't we just use

f (1) = f (- 1)

for the extrema?

That is a common trap! An extremum (a local maximum or minimum) means the slope of the tangent line is zero, which is defined by the derivative

f^{'} (x) = 0

. It does not necessarily mean the function values at those points are equal. Always use the derivative for extrema!

Why do we multiply by

5

at the very end?

The question asks for

5 \cdot f (2)

. We do this because our coefficients

a

and

b

are fractions (specifically, they have a denominator of

5

2

). Multiplying by

5

is a strategic move to clear the denominators and make the final arithmetic much cleaner and less prone to error.

Animated Solution for Mathematics - Differentiation: Let $f (x)$ be a polynomial of degree $6$ in $x$ , in which the coefficient of $x^{6}$ is unity and it has extrema at $x = - 1$ and $x = 1$ . If $lim_{x \to 0} \frac{f ( x )}{x ^{3}} = 1$ , then $5 \cdot f (2)$ is equal to

Enter Numerical Value:

Visualized Solution

Defining the Polynomial $f (x)$

Analyzing the Limit Condition

Finding Coefficient $c$

Finding the Derivative $f^{'} (x)$

Applying Extrema at $x = 1$

Applying Extrema at $x = - 1$

Solving for $a$

Solving for $b$

The Final Polynomial $f (x)$

Calculating $f (2)$

Final Result: $5 \cdot f (2)$

Conclusion \& Key Takeaway

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Limit Detective

The Derivative Dance

The System of Equations

Final Calculation

Similar Questions

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 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 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 :-

If $p (x)$ be a polynomial of degree 3 satisfying $p (- 1) = 10, p (1) = - 6$ and $p (x)$ has maxima at $x = - 1$ and $p^{'} (x)$ has minima at $x = 1$ . Find the distance between the local maxima and local minima of the curve.

$f (x)$ is cubic polynomial with $f (2) = 18$ and $f (1) = - 1$ . Also $f (x)$ has local maxima at $x = - 1$ and $f^{'} (x)$ has local minima at $x = 0$ , then

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 $p (x)$ be a real polynomial of least degree which has a local maximum at $x = 1$ and a local minimum at $x = 3$ . If $p (1) = 6$ and $p (3) = 2$ , then $p^{'} (0)$ is

Let $a > 0$ . If the function $f (x) = 6 x^{3} - 45 a x^{2} + 108 a^{2} x + 1$ attains its local maximum and minimum values at the points $x_{1}$ and $x_{2}$ respectively such that $x_{1} x_{2} = 54$ , then $a + x_{1} + x_{2}$ is equal to :-

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?

Enter Numerical Value:

Visualized Solution

Defining the Polynomial .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Analyzing the Limit Condition

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

Finding the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Applying Extrema at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1

Applying Extrema at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−1

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

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

The Final Polynomial .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(2)

Final Result: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5⋅f(2)

Conclusion \& Key Takeaway

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Limit Detective

The Derivative Dance

The System of Equations

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let p(x) be a polynomial of degree 4 having extremum at x=1,2 and limx→0​(1+x2p(x)​)=2. Then the value of p(2) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a polynomial of degree four having extreme values at x=1 and x=2. If limx→0​[1+x2f(x)​]=3, then f(2) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a polynomial function of degree four having extreme values at x=4 and x=5. If limx→0​x2f(x)​=5, then f(2) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If limx→0​(x2f(x)​+1)=3 then f(-1) is equal to :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) is cubic polynomial with f(2)=18 and f(1)=−1. Also f(x) has local maxima at x=−1 and f′(x) has local minima at x=0, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a polynomial of degree 5 such that x=±1 are its critical points. If limx→0​(2+x3f(x)​)=4, then which one of the following is not true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let p(x) be a real polynomial of least degree which has a local maximum at x=1 and a local minimum at x=3. If p(1)=6 and p(3)=2, then p′(0) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a polynomial of degree 5 such that x=±1 are its critical points. If limx→0​(2+x3f(x)​)=4, then which one of the following is not true?

Defining the Polynomial $f (x)$

Finding Coefficient $c$

Finding the Derivative $f^{'} (x)$

Applying Extrema at $x = 1$

Applying Extrema at $x = - 1$

Solving for $a$

Solving for $b$

The Final Polynomial $f (x)$

Calculating $f (2)$

Final Result: $5 \cdot f (2)$

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 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 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 :-

$f (x)$ is cubic polynomial with $f (2) = 18$ and $f (1) = - 1$ . Also $f (x)$ has local maxima at $x = - 1$ and $f^{'} (x)$ has local minima at $x = 0$ , then

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 $p (x)$ be a real polynomial of least degree which has a local maximum at $x = 1$ and a local minimum at $x = 3$ . If $p (1) = 6$ and $p (3) = 2$ , then $p^{'} (0)$ is

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?