JEE Main 2019 (12 January)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: If the function $f$ given by $f (x) = x^{3} - 3 (a - 2) x^{2} + 3 a x + 7$ , for some $a \in R$ is increasing in $(0, 1]$ and decreasing in $[1, 5)$ , then a root of the equation $\frac{f ( x ) - 14}{( x - 1 ) ^{2}} = 0 (x \neq = 1)$ is :

Q: Why does the problem explicitly state $x \neq 1$?

Look at the denominator of the target equation: $(x - 1)^2$. If $x$ were $1$, we would be dividing by zero, which is undefined in mathematics. The condition $x \neq 1$ is a domain restriction to ensure the expression is well-defined, even though $x=1$ is a root of the numerator.

Q: Is there a faster way to find the other root without full polynomial division?

Yes! Since $f(x) - 14$ is a cubic polynomial with a leading coefficient of $1$, and we know it has a repeated root at $x = 1$, we can write it as $(x - 1)^2(x - k)$. By comparing the constant term or the coefficient of $x^2$, you can easily find $k$ without long division.

Select Answer:

Visualized Solution

Analyze the Function's Behavior

Given function: $f (x) = x^{3} - 3 (a - 2) x^{2} + 3 a x + 7$
Condition: $f (x)$ is increasing in $(0, 1]$ and decreasing in $[1, 5)$
This implies $x = 1$ is a point of local maximum.

The Critical Point Condition

At a local maximum, the tangent is horizontal.
Therefore, the derivative must be zero: $f^{'} (1) = 0$

Differentiate $f (x)$

Differentiating $f (x)$ with respect to $x$ :
$f^{'} (x) = \frac{d}{d x} [x^{3} - 3 (a - 2) x^{2} + 3 a x + 7]$
$f^{'} (x) = 3 x^{2} - 6 (a - 2) x + 3 a$

Substitute the Critical Point

Substitute $x = 1$ into $f^{'} (x) = 0$ :
$3 (1)^{2} - 6 (a - 2) (1) + 3 a = 0$
$3 - 6 a + 12 + 3 a = 0$

Solve for Constant $a$

Combine the terms:
$15 - 3 a = 0$
$\Rightarrow 3 a = 15$
$\Rightarrow a = 5$

Reconstruct the Function $f (x)$

Substitute $a = 5$ back into $f (x)$ :
$f (x) = x^{3} - 3 (5 - 2) x^{2} + 3 (5) x + 7$
$f (x) = x^{3} - 9 x^{2} + 15 x + 7$

Evaluate $f (1)$

The target equation is: $\frac{f ( x ) - 14}{( x - 1 ) ^{2}} = 0$
Let's evaluate $f (1)$ :
$f (1) = 1^{3} - 9 (1)^{2} + 15 (1) + 7 = 14$

The Repeated Root Theorem

Since $f (1) = 14$ and $f^{'} (1) = 0$ , the line $y = 14$ is tangent to $f (x)$ at $x = 1$ .
This means $x = 1$ is a repeated root of $f (x) - 14 = 0$ .
Therefore, $(x - 1)^{2}$ is a factor of $f (x) - 14$ .

Form the Numerator Polynomial

Let's form the numerator:
$f (x) - 14 = (x^{3} - 9 x^{2} + 15 x + 7) - 14$
$f (x) - 14 = x^{3} - 9 x^{2} + 15 x - 7$

Factorize the Cubic Expression

Factorize the cubic expression using the known factor $(x - 1)^{2} = x^{2} - 2 x + 1$ :
By polynomial division:
$x^{3} - 9 x^{2} + 15 x - 7 = (x - 1)^{2} (x - 7)$

Solve the Target Equation

Substitute the factorized form into the target equation:
$\frac{( x - 1 ) ^{2} ( x - 7 )}{( x - 1 ) ^{2}} = 0$
Since $x \neq = 1$ , we can cancel $(x - 1)^{2}$ .

Final Conclusion

We are left with:
$x - 7 = 0$
$\Rightarrow x = 7$
This is the required root.

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The Sigma Insight: Monotonicity

Solution Diagram

Analyzing the Setup

The function is given by

f (x) = x^{3} - 3 (a - 2) x^{2} + 3 a x + 7

. We are told that

f (x)

is increasing on

(0, 1]

and decreasing on

[1, 5)

This implies that the function reaches a local maximum at

x = 1

. In calculus, a local maximum at a point where the function is differentiable requires the derivative to be zero.

The Master Equation

We calculate the derivative of

f (x)

using the power rule:

f^{'} (x) = 3 x^{2} - 6 (a - 2) x + 3 a

Since the summit occurs at

x = 1

, we set

f^{'} (1) = 0

3 (1)^{2} - 6 (a - 2) (1) + 3 a = 0

Expanding this equation, we get:

3 - 6 a + 12 + 3 a = 0

15 - 3 a = 0

\Rightarrow a = 5

Determining the Function Identity

Substituting

a = 5

back into the original function, we obtain:

f (x) = x^{3} - 3 (5 - 2) x^{2} + 3 (5) x + 7

f (x) = x^{3} - 9 x^{2} + 15 x + 7

Next, we evaluate the function at the peak

x = 1

f (1) = 1 - 9 + 15 + 7 = 14

Solving the Final Expression

We are asked to find the root of the equation:

\frac{f ( x ) - 14}{( x - 1 ) ^{2}} = 0

Since

f (1) = 14

, the value

x = 1

is a root of

f (x) - 14 = 0

. Because

x = 1

is a local maximum, it must be a double root, meaning

(x - 1)^{2}

is a factor of

f (x) - 14

We can express the cubic as:

f (x) - 14 = (x - 1)^{2} (x - k)

By comparing the constant term of

f (x) - 14

, which is

7 - 14 = - 7

, we have:

(- 1)^{2} \times (- k) = - 7

- k = - 7 \Rightarrow k = 7

The equation simplifies to:

\frac{( x - 1 ) ^{2} ( x - 7 )}{( x - 1 ) ^{2}} = 0

For $x eq 1$, we cancel the common terms to find the root:

x - 7 = 0

\Rightarrow x = 7

Common Traps & FAQs

Wait, why do we assume

f^{'} (1) = 0

? Could it be a sharp corner?

That is a brilliant question! In JEE problems, unless specified otherwise, polynomials are smooth, continuous, and differentiable everywhere. Since

f (x)

is a cubic polynomial, it is infinitely differentiable. Therefore, at any local extremum, the derivative must vanish. There are no sharp corners here!

Why does the problem explicitly state

x \neq = 1

Look at the denominator of the target equation:

(x - 1)^{2}

. If

x

were

1

, we would be dividing by zero, which is undefined in mathematics. The condition

x \neq = 1

is a domain restriction to ensure the expression is well-defined, even though

x = 1

is a root of the numerator.

Is there a faster way to find the other root without full polynomial division?

Yes! Since

f (x) - 14

is a cubic polynomial with a leading coefficient of

1

, and we know it has a repeated root at

x = 1

, we can write it as

(x - 1)^{2} (x - k)

. By comparing the constant term or the coefficient of

x^{2}

, you can easily find

k

without long division.

Animated Solution for Mathematics - Differentiation: If the function $f$ given by $f (x) = x^{3} - 3 (a - 2) x^{2} + 3 a x + 7$ , for some $a \in R$ is increasing in $(0, 1]$ and decreasing in $[1, 5)$ , then a root of the equation $\frac{f ( x ) - 14}{( x - 1 ) ^{2}} = 0 (x \neq = 1)$ is :

Select Answer:

Visualized Solution

Analyze the Function's Behavior

The Critical Point Condition

Differentiate $f (x)$

Substitute the Critical Point

Solve for Constant $a$

Reconstruct the Function $f (x)$

Evaluate $f (1)$

The Repeated Root Theorem

Form the Numerator Polynomial

Factorize the Cubic Expression

Solve the Target Equation

Final Conclusion

The Sigma Insight: Monotonicity

Analyzing the Setup

The Master Equation

Determining the Function Identity

Solving the Final Expression

Similar Questions

The function, $f (x) = (3 x - 7) x^{2/3}$ , $x \in R$ , is increasing for all $x$ lying in :

Let $f (x) = {x e^{a x}, x + a x^{2} - x^{3}, x \leq 0 x > 0$ . Where $a$ is a positive constant. Find the interval in which $f^{'} (x)$ is increasing.

The set of all $a \in R$ for which the equation $x ∣ x - 1∣ + ∣ x + 2∣ + a = 0$ has exactly one real root, is

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Let $g (x) = 3 f (\frac{x}{3}) + f (3 - x)$ and $f^{''} (x) > 0$ for all $x \in (0, 3)$ . If $g$ is decreasing in $(0, α)$ and increasing in $(α, 3)$ , then $8 α$ is

If the maximum value of $a$ , for which the function $f_{a} (x) = tan^{- 1} 2 x - 3 a x + 7$ is non-decreasing in $(- \frac{π}{6}, \frac{π}{6})$ , is $\overset{a}{ˉ}$ , then $f_{\overset{a}{ˉ}} (\frac{π}{8})$ is equal to

Let the function $f (x) = \frac{x}{3} + \frac{3}{x} + 3, x \neq = 0$ be strictly increasing in $(- \infty, α_{1}) \cup (α_{2}, \infty)$ and strictly decreasing in $(α_{3}, α_{4}) \cup (α_{4}, α_{5})$ . Then $\sum_{i = 1}^{5} α_{i}^{2}$ is equal to :-

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Let $f$ be a real valued function, defined on $R - {- 1, 1}$ and given by $f (x) = 3 lo g_{e} \frac{x - 1}{x + 1} - \frac{2}{x - 1}$ . Then in which of the following intervals, function $f (x)$ is increasing?

Select Answer:

Visualized Solution

Analyze the Function's Behavior

The Critical Point Condition

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

Substitute the Critical Point

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

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

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

The Repeated Root Theorem

Form the Numerator Polynomial

Factorize the Cubic Expression

Solve the Target Equation

Final Conclusion

The Sigma Insight: Monotonicity

Analyzing the Setup

The Master Equation

Determining the Function Identity

Solving the Final Expression

Similar Questions

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Solve for Constant $a$

Reconstruct the Function $f (x)$

Evaluate $f (1)$

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The set of all $a \in R$ for which the equation $x ∣ x - 1∣ + ∣ x + 2∣ + a = 0$ has exactly one real root, is

Let $f : R \to R$ be defined as $f (x) = {- \frac{4}{3} x^{3} + 2 x^{2} + 3 x 3 x e^{x}, x > 0, x \leq 0$ . Then $f$ is increasing function in the interval

Let $g (x) = 3 f (\frac{x}{3}) + f (3 - x)$ and $f^{''} (x) > 0$ for all $x \in (0, 3)$ . If $g$ is decreasing in $(0, α)$ and increasing in $(α, 3)$ , then $8 α$ is

If the maximum value of $a$ , for which the function $f_{a} (x) = tan^{- 1} 2 x - 3 a x + 7$ is non-decreasing in $(- \frac{π}{6}, \frac{π}{6})$ , is $\overset{a}{ˉ}$ , then $f_{\overset{a}{ˉ}} (\frac{π}{8})$ is equal to