JEE Main 2021 (25 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Consider the function $f (x) = \frac{P ( x )}{s i n ( x - 2 )}, x \neq = 2$ and $f (x) = 7, x = 2$ . Where $P (x)$ is a polynomial such that $P^{''} (x)$ is always a constant and $P (3) = 9$ . If $f (x)$ is continuous at $x = 2$ , then $P (5)$ is equal to

Q: Wait, why do we assume $P(x)$ is a quadratic? Could it be a cubic?

Great observation! The problem explicitly states that $P''(x)$ is a constant. If $P(x)$ were a cubic, say $P(x) = Ax^3 + Bx^2 + Cx + D$, then $P''(x) = 6Ax + 2B$, which is clearly not a constant unless $A=0$. Thus, the constraint on the second derivative forces the polynomial to be at most degree 2.

Q: Why does $P(2)$ have to be 0 for the limit to exist?

Think about the behavior of the function. If $P(2)$ were some non-zero constant $k$, then the limit would look like $\frac{k}{0}$, which approaches infinity. Since the problem tells us the function is continuous at $x=2$, the limit must be a finite value (specifically 7). The only way to 'cancel out' the zero in the denominator is to have a zero in the numerator.

Q: Why did you write $P(x) = (x-2)(ax+b)$ instead of $ax^2+bx+c$?

It is all about efficiency. By identifying that $(x-2)$ is a factor, we reduce the number of unknowns from three ($a, b, c$) to two ($a, b$). In competitive exams like JEE, saving time by choosing the smartest algebraic form is just as important as knowing the calculus.

Q: Is it possible for $P(x)$ to be a constant polynomial?

If $P(x)$ were a constant, say $P(x) = k$, then $P(2)$ would be $k$. But we established that $P(2)$ must be 0 for the limit to exist. If $P(x) = 0$, then $f(x)$ would be 0 everywhere, which contradicts $f(2) = 7$. So, $P(x)$ cannot be a constant.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the function

f (x)

x = 2

. Since the denominator

sin (x - 2)

approaches zero as

x \to 2

, what must happen to the numerator

P (x)

for the limit to exist and equal a finite value like 7? This is the classic gateway to the indeterminate form.

Hint 2

Hint 2: The Arsenal. You are told

P^{''} (x)

is a constant. Recall that if the second derivative of a polynomial is constant, the polynomial must be of degree 2. Write it as

P (x) = a x^{2} + b x + c

, or even better, use the root

(x - 2)

you discovered in Hint 1 to write it as

P (x) = (x - 2) (a x + b)

Hint 3

Hint 3: The Execution. Set up the limit equation:

lim_{x \to 2} \frac{( x - 2 ) ( a x + b )}{s i n ( x - 2 )} = 7

. Use the fundamental limit

lim_{θ \to 0} \frac{s i n θ}{θ} = 1

to simplify this expression and find a linear relationship between

a

and

b

Visualized Solution

Function Definition

$f (x) = \frac{P ( x )}{s i n ( x - 2 )}$ for $x \neq = 2$
$f (2) = 7$
Given: $P^{''} (x) = constant$

Nature of $P (x)$

$P^{''} (x) = k$
Integrating twice: $P (x) = a x^{2} + b x + c$
So, $P (x)$ is a quadratic polynomial.

Continuity at $x = 2$

$f (x)$ is continuous at $x = 2$
$lim_{x \to 2} f (x) = f (2)$
$lim_{x \to 2} \frac{P ( x )}{s i n ( x - 2 )} = 7$

Indeterminate Form $\frac{0}{0}$

As $x \to 2$ , $sin (x - 2) \to 0$
For the limit to be finite, $P (2)$ must be $0$
This is the $\frac{0}{0}$ indeterminate form.

Factor of $P (x)$

$P (2) = 0 ⟹ (x - 2)$ is a factor of $P (x)$
Let $P (x) = (x - 2) (a x + b)$

Evaluating the Limit

Substitute $P (x)$ in the limit:
$lim_{x \to 2} \frac{( x - 2 ) ( a x + b )}{s i n ( x - 2 )} = 7$
$lim_{x \to 2} [\frac{x - 2}{s i n ( x - 2 )}] \cdot (a x + b) = 7$

Standard Limit Result

Using $lim_{θ \to 0} \frac{θ}{s i n θ} = 1$
$lim_{x \to 2} (a x + b) = 7$
Substitute $x = 2$ : $2 a + b = 7$

Using Given $P (3) = 9$

Given $P (3) = 9$
Substitute $x = 3$ in $P (x) = (x - 2) (a x + b)$ :
$(3 - 2) (3 a + b) = 9$
$3 a + b = 9$

Solving for $a$ and $b$

Eq 1: $2 a + b = 7$
Eq 2: $3 a + b = 9$
Subtracting Eq 1 from Eq 2:
$(3 a - 2 a) + (b - b) = 9 - 7 ⟹ a = 2$

Finding value of $b$

Substitute $a = 2$ in $2 a + b = 7$ :
$2 (2) + b = 7$
$4 + b = 7 ⟹ b = 3$

The Polynomial $P (x)$

$P (x) = (x - 2) (2 x + 3)$
Expanding: $P (x) = 2 x^{2} + 3 x - 4 x - 6$
Final form: $P (x) = 2 x^{2} - x - 6$

Final Calculation $P (5)$

Substitute $x = 5$ in $P (x) = (x - 2) (2 x + 3)$ :
$P (5) = (5 - 2) (2 (5) + 3)$
$P (5) = 3 \cdot (10 + 3)$
$P (5) = 3 \cdot 13 = 39$

Summary & Key Takeaway

Final Answer: $P (5) = 39$
Key Takeaway: Continuity at a point where denominator $\to 0$ implies numerator $\to 0$ .

00:00 / 00:00

The Sigma Insight: Continuity at a Point and in an Interval

Solution Diagram

Analyzing the Setup

Imagine you are standing on a graph, tracing the path of

f (x) = \frac{P ( x )}{s i n ( x - 2 )}

. As you approach the point

x = 2

, the denominator

sin (x - 2)

shrinks toward zero.

In the world of functions, a denominator approaching zero usually creates a vertical asymptote. However, the problem states that the function is continuous at

x = 2

and

f (2) = 7

This implies the function does not explode; it lands gracefully at the value

7

. This is the 'Spark' of our journey.

The Geometry of the Polynomial

We are told that

P^{''} (x)

is a constant. In the language of calculus, if the second derivative is constant, the function must be a quadratic polynomial of the form

P (x) = a x^{2} + b x + c

Because the limit

lim_{x \to 2} \frac{P ( x )}{s i n ( x - 2 )}

must be finite, the numerator

P (x)

must also approach zero as

x \to 2

. This implies that

(x - 2)

is a root of

P (x)

Instead of dealing with three unknown coefficients, we can elegantly write

P (x) = (x - 2) (a x + b)

. This is the 'Arsenal' we need to conquer the problem.

The Limit of Elegance

Now, let us set up our limit equation:

x \to 2 lim \frac{( x - 2 ) ( a x + b )}{sin ( x - 2 )} = 7

We know from our fundamental toolkit that

lim_{θ \to 0} \frac{θ}{s i n θ} = 1

. As

x \to 2

, the term

(x - 2)

acts as our

θ

The limit simplifies beautifully to:

x \to 2 lim (a x + b) = 7

Substituting

x = 2

, we get our first linear equation:

2 a + b = 7

Solving the System

The problem provides one more clue:

P (3) = 9

. Using our factored form

P (x) = (x - 2) (a x + b)

, we substitute

x = 3

(3 - 2) (3 a + b) = 9 ⟹ 3 a + b = 9

We now have a system of two linear equations:

2 a + b = 7

3 a + b = 9

Subtracting the first from the second, we find

a = 2

. Plugging this back into the first equation,

2 (2) + b = 7

, we find

b = 3

Final Calculation

Our polynomial is fully revealed as

P (x) = (x - 2) (2 x + 3)

. Expanding this, we get

P (x) = 2 x^{2} - x - 6

Finally, we calculate

P (5)

P (5) = (5 - 2) (2 (5) + 3) = 3 \cdot 13 = 39

The final result is 39.

Common Traps & FAQs

Wait, why do we assume

P (x)

is a quadratic? Could it be a cubic?

Great observation! The problem explicitly states that

P^{''} (x)

is a constant. If

P (x)

were a cubic, say

P (x) = A x^{3} + B x^{2} + C x + D

, then

P^{''} (x) = 6 A x + 2 B

, which is clearly not a constant unless

A = 0

. Thus, the constraint on the second derivative forces the polynomial to be at most degree 2.

Why does

P (2)

have to be 0 for the limit to exist?

Think about the behavior of the function. If

P (2)

were some non-zero constant

k

, then the limit would look like

\frac{k}{0}

, which approaches infinity. Since the problem tells us the function is continuous at

x = 2

, the limit must be a finite value (specifically 7). The only way to 'cancel out' the zero in the denominator is to have a zero in the numerator.

Can I use L'Hôpital's Rule here instead of the standard limit?

Absolutely! L'Hôpital's Rule is a powerful tool. Since we have a

\frac{0}{0}

form, you can differentiate the numerator and denominator:

lim_{x \to 2} \frac{P ^{'} ( x )}{c o s ( x - 2 )} = 7

. Since

cos (0) = 1

, this simplifies to

P^{'} (2) = 7

. This gives you another elegant way to solve for your coefficients!

Why did you write

P (x) = (x - 2) (a x + b)

instead of

a x^{2} + b x + c

It is all about efficiency. By identifying that

(x - 2)

is a factor, we reduce the number of unknowns from three (

a, b, c

) to two (

a, b

). In competitive exams like JEE, saving time by choosing the smartest algebraic form is just as important as knowing the calculus.

Is it possible for

P (x)

to be a constant polynomial?

P (x)

were a constant, say

P (x) = k

, then

P (2)

would be

k

. But we established that

P (2)

must be 0 for the limit to exist. If

P (x) = 0

, then

f (x)

would be 0 everywhere, which contradicts

f (2) = 7

. So,

P (x)

cannot be a constant.

Enter Numerical Value:

Visualized Solution

Function Definition

Nature of $P (x)$

Continuity at $x = 2$

Indeterminate Form $\frac{0}{0}$

Factor of $P (x)$

Evaluating the Limit

Standard Limit Result

Using Given $P (3) = 9$

Solving for $a$ and $b$

Finding value of $b$

The Polynomial $P (x)$

Final Calculation $P (5)$

Summary & Key Takeaway

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Geometry of the Polynomial

The Limit of Elegance

Solving the System

Final Calculation

Similar Questions

If the function $f (x) = ⎩ ⎨ ⎧ \frac{2}{x} {sin (k_{1} + 1) x + sin (k_{2} - 1) x}, 4, \frac{2}{x} lo g_{e} (\frac{2 + k _{1} x}{2 + k _{2} x}), x < 0 x = 0 x > 0$ is continuous at $x = 0$ , then $k_{1}^{2} + k_{2}^{2}$ is equal to

Let $f (x) = {\frac{( x ^{3} + x ^{2} - 16 x + 20 )}{( x - 2 ) ^{2}} k if x \neq = 2 if x = 2$ . If $f (x)$ is continuous for all $x$ , then $k = .........$

Let $f (x) = {\frac{a x ^{2} + 2 a x + 3}{4 x ^{2} + 4 x - 3}, b, x \neq = - \frac{3}{2}, \frac{1}{2} x = - \frac{3}{2}, \frac{1}{2}$ be continuous at $x = - \frac{3}{2}$ . If $f \circ f (x) = \frac{7}{5}$ , then $x$ is equal to:

If the function $f (x) = ⎩ ⎨ ⎧ \frac{1}{x} lo g_{e} (\frac{1 + \frac{x}{a}}{1 - \frac{x}{b}}), k, \frac{c o s ^{2} x - s i n ^{2} x - 1}{x ^{2} + 1 - 1}, x < 0 x = 0 x > 0$ is continuous at $x = 0$ , then $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ is equal to :

If the function $f (x) = \frac{s i n 3 x + α s i n x - β c o s 3 x}{x ^{3}}, x \in R$ , is continuous at $x = 0$ , then $f (0)$ is equal to :

Function is continuous at $x = 0$ , find $a + 2 b$ . $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 2 ) x + s i n x}{x} b \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}; x < 0; x = 0; x > 0$

If $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 2 ) x + s i n x}{x}, b, \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}, x < 0 x = 0 x > 0$ is continuous at $x = 0$ then $a + 2 b$ is equal to:

If the function $f$ defined on $(- \frac{1}{3}, \frac{1}{3})$ by $f (x) = {\frac{1}{x} lo g_{e} (\frac{1 + 3 x}{1 - 2 x}), k, when x \neq = 0 when x = 0$ is continuous, then $k$ is equal to . . . .

If the function $f (x) = {\frac{7 2 ^{x} - 9 ^{x} - 8 ^{x} + 1}{2 - 1 + c o s x}, a lo g_{e} 2 lo g_{e} 3, x \neq = 0 x = 0$ is continuous at $x = 0$ , then the value of $a^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Function Definition

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

Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2

Indeterminate Form .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 00​

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

Evaluating the Limit

Standard Limit Result

Using Given .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(3)=9

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

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

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

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

Summary & Key Takeaway

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Geometry of the Polynomial

The Limit of Elegance

Solving the System

Final Calculation

Similar Questions

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Nature of $P (x)$

Continuity at $x = 2$

Indeterminate Form $\frac{0}{0}$

Factor of $P (x)$

Using Given $P (3) = 9$

Solving for $a$ and $b$

Finding value of $b$

The Polynomial $P (x)$

Final Calculation $P (5)$

If the function $f (x) = ⎩ ⎨ ⎧ \frac{2}{x} {sin (k_{1} + 1) x + sin (k_{2} - 1) x}, 4, \frac{2}{x} lo g_{e} (\frac{2 + k _{1} x}{2 + k _{2} x}), x < 0 x = 0 x > 0$ is continuous at $x = 0$ , then $k_{1}^{2} + k_{2}^{2}$ is equal to

Let $f (x) = {\frac{( x ^{3} + x ^{2} - 16 x + 20 )}{( x - 2 ) ^{2}} k if x \neq = 2 if x = 2$ . If $f (x)$ is continuous for all $x$ , then $k = .........$

Let $f (x) = {\frac{a x ^{2} + 2 a x + 3}{4 x ^{2} + 4 x - 3}, b, x \neq = - \frac{3}{2}, \frac{1}{2} x = - \frac{3}{2}, \frac{1}{2}$ be continuous at $x = - \frac{3}{2}$ . If $f \circ f (x) = \frac{7}{5}$ , then $x$ is equal to:

If the function $f (x) = \frac{s i n 3 x + α s i n x - β c o s 3 x}{x ^{3}}, x \in R$ , is continuous at $x = 0$ , then $f (0)$ is equal to :

Function is continuous at $x = 0$ , find $a + 2 b$ . $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 2 ) x + s i n x}{x} b \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}; x < 0; x = 0; x > 0$

If $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 2 ) x + s i n x}{x}, b, \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}, x < 0 x = 0 x > 0$ is continuous at $x = 0$ then $a + 2 b$ is equal to:

If the function $f$ defined on $(- \frac{1}{3}, \frac{1}{3})$ by $f (x) = {\frac{1}{x} lo g_{e} (\frac{1 + 3 x}{1 - 2 x}), k, when x \neq = 0 when x = 0$ is continuous, then $k$ is equal to . . . .

If the function $f (x) = {\frac{7 2 ^{x} - 9 ^{x} - 8 ^{x} + 1}{2 - 1 + c o s x}, a lo g_{e} 2 lo g_{e} 3, x \neq = 0 x = 0$ is continuous at $x = 0$ , then the value of $a^{2}$ is equal to