JEE Main 2021 (01 Sep Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f (x) = x^{6} + 2 x^{4} + x^{3} + 2 x + 3, x \in R$ . Then the natural number $n$ for which $lim_{x \to 1} \frac{x ^{n} f ( 1 ) - f ( x )}{x - 1} = 44$ is.

Q: Why can't I just plug in $x=1$ immediately?

If you try to plug in $x=1$ directly, you get $\frac{1^n f(1) - f(1)}{1 - 1} = \frac{f(1) - f(1)}{0} = \frac{0}{0}$. This is an indeterminate form. It doesn't mean the limit doesn't exist; it means the function is hiding its true value behind a veil of division by zero. We must use calculus to lift that veil.

Q: Wait, why do we treat $f(1)$ as a constant?

This is a common point of confusion! $f(x)$ is a function, but $f(1)$ is just the value of that function at a specific point. Once you calculate $f(1) = 9$, it is just a number. When differentiating with respect to $x$, constants behave like any other number—their derivative is zero, or they act as coefficients. Don't let the notation intimidate you!

Enter Numerical Value:

Visualized Solution

Analyze the Function $f (x)$

Given: $f (x) = x^{6} + 2 x^{4} + x^{3} + 2 x + 3$
Goal: Find $n \in N$ for $lim_{x \to 1} \frac{x ^{n} f ( 1 ) - f ( x )}{x - 1} = 44$

Calculate $f (1)$

Substitute $x = 1$ into $f (x)$
$f (1) = 1^{6} + 2 (1)^{4} + 1^{3} + 2 (1) + 3$
$f (1) = 1 + 2 + 1 + 2 + 3 = 9$

Check the Limit Form

Evaluate the limit as $x \to 1$
Numerator: $1^{n} \cdot f (1) - f (1) = 9 - 9 = 0$
Denominator: $1 - 1 = 0$
The limit is in the $\frac{0}{0}$ indeterminate form.

Apply L'Hopital's Rule

Since the form is $\frac{0}{0}$ , apply L'Hopital's Rule.
Differentiate the numerator and denominator separately with respect to $x$ .
$lim_{x \to 1} \frac{\frac{d}{d x} [ x ^{n} f ( 1 ) - f ( x )]}{\frac{d}{d x} [ x - 1 ]} = 44$

Differentiate the Numerator

$\frac{d}{d x} [x^{n} f (1) - f (x)]$
$f (1)$ is a constant.
Derivative is $n x^{n - 1} f (1) - f^{'} (x)$

Differentiate the Denominator

$\frac{d}{d x} [x - 1]$
Derivative of $x$ is $1$ .
Derivative of $- 1$ is $0$ .
Result is $1$ .

Evaluate the New Limit

The limit becomes: $lim_{x \to 1} \frac{n x ^{n - 1} f ( 1 ) - f ^{'} ( x )}{1} = 44$
Substitute $x = 1$ :
$n (1)^{n - 1} f (1) - f^{'} (1) = 44$
$n f (1) - f^{'} (1) = 44$

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

$f (x) = x^{6} + 2 x^{4} + x^{3} + 2 x + 3$
Apply power rule: $\frac{d}{d x} (x^{k}) = k x^{k - 1}$
$f^{'} (x) = 6 x^{5} + 8 x^{3} + 3 x^{2} + 2$

Evaluate $f^{'} (1)$

Substitute $x = 1$ into $f^{'} (x)$ :
$f^{'} (1) = 6 (1)^{5} + 8 (1)^{3} + 3 (1)^{2} + 2$
$f^{'} (1) = 6 + 8 + 3 + 2 = 19$

Substitute Values into the Equation

Recall the equation: $n f (1) - f^{'} (1) = 44$
Substitute $f (1) = 9$ and $f^{'} (1) = 19$ :
$n (9) - 19 = 44$
$9 n - 19 = 44$

Solve for $n$

$9 n = 44 + 19$
$9 n = 63$
$n = \frac{63}{9}$
$n = 7$

Final Conclusion

Final Answer: $n = 7$
Key Takeaway: L'Hopital's Rule simplifies limits of the form $\frac{0}{0}$ by converting them into relations involving derivatives.

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The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Solution Diagram

Analyzing the Setup

The given limit expression is:

x \to 1 lim \frac{x ^{n} f ( 1 ) - f ( x )}{x - 1} = 44

This expression involves a function

f (x) = x^{6} + 2 x^{4} + x^{3} + 2 x + 3

. Our goal is to determine the value of the integer

n

The Indeterminate Trap

The first step is to test the limit by substituting

x = 1

. The numerator becomes

1^{n} f (1) - f (1) = 0

, and the denominator becomes

1 - 1 = 0

We have arrived at the

\frac{0}{0}

indeterminate form. This confirms that the expression is well-defined and allows us to apply L'Hopital's Rule to resolve the limit.

The Surgical Precision of L'Hopital

First, we calculate the constant value

f (1)

f (1) = 1^{6} + 2 (1)^{4} + 1^{3} + 2 (1) + 3 = 9

Applying L'Hopital's Rule, we differentiate the numerator and denominator with respect to

x

x \to 1 lim \frac{\frac{d}{d x} [ x ^{n} f ( 1 ) - f ( x )]}{\frac{d}{d x} [ x - 1 ]} = 44

This simplifies to the following expression:

x \to 1 lim [n x^{n - 1} f (1) - f^{'} (x)] = 44

The Derivative Dance

Next, we find the derivative of the polynomial

f (x) = x^{6} + 2 x^{4} + x^{3} + 2 x + 3

f^{'} (x) = 6 x^{5} + 8 x^{3} + 3 x^{2} + 2

Evaluating this derivative at

x = 1

f^{'} (1) = 6 (1)^{5} + 8 (1)^{3} + 3 (1)^{2} + 2 = 19

Final Calculation

Now, we substitute the known values

f (1) = 9

and

f^{'} (1) = 19

into our simplified limit equation:

n (1)^{n - 1} (9) - 19 = 44

9 n - 19 = 44

Solving for

n

9 n = 63

$n = 7$

Common Traps & FAQs

Why can't I just plug in

x = 1

immediately?

If you try to plug in

x = 1

directly, you get

\frac{1 ^{n} f ( 1 ) - f ( 1 )}{1 - 1} = \frac{f ( 1 ) - f ( 1 )}{0} = \frac{0}{0}

. This is an indeterminate form. It doesn't mean the limit doesn't exist; it means the function is hiding its true value behind a veil of division by zero. We must use calculus to lift that veil.

Wait, why do we treat

f (1)

as a constant?

This is a common point of confusion!

f (x)

is a function, but

f (1)

is just the value of that function at a specific point. Once you calculate

f (1) = 9

, it is just a number. When differentiating with respect to

x

, constants behave like any other number—their derivative is zero, or they act as coefficients. Don't let the notation intimidate you!

Is there a way to solve this without L'Hopital's Rule?

Absolutely! You could use the definition of the derivative. Recall that

f^{'} (1) = lim_{x \to 1} \frac{f ( x ) - f ( 1 )}{x - 1}

. You could manipulate the original expression to force this definition to appear. However, L'Hopital's Rule is the 'surgical' tool designed exactly for this scenario. It is faster, cleaner, and less prone to algebraic errors.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f (x) = x^{6} + 2 x^{4} + x^{3} + 2 x + 3, x \in R$ . Then the natural number $n$ for which $lim_{x \to 1} \frac{x ^{n} f ( 1 ) - f ( x )}{x - 1} = 44$ is.

Enter Numerical Value:

Visualized Solution

Analyze the Function $f (x)$

Calculate $f (1)$

Check the Limit Form

Apply L'Hopital's Rule

Differentiate the Numerator

Differentiate the Denominator

Evaluate the New Limit

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

Evaluate $f^{'} (1)$

Substitute Values into the Equation

Solve for $n$

Final Conclusion

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Indeterminate Trap

The Surgical Precision of L'Hopital

The Derivative Dance

Final Calculation

Similar Questions

If $lim_{x \to 1} \frac{x + x ^{2} + x ^{3} + ... + x ^{n} - n}{x - 1} = 820$ , $(n \in N)$ then the value of $n$ is equal to

Let $f : R \to R$ be such that $f (1) = 3$ and $f^{'} (1) = 6$ . Then $lim_{x \to 0} (\frac{f ( 1 + x )}{f ( 1 )})^{1/ x}$ equals

$lim_{x \to \infty} (\frac{x + 6}{x + 1})^{x + 4} = .........$

Let $f (x) = {x - 1, 2 x, x is even, x is odd, x \in N$ . If for some $a \in N, f (f (f (a))) = 21$ , then $lim_{x \to a^{-}} {\frac{∣ x ∣ ^{3}}{a} - [\frac{x}{a}]}$ , where $[t]$ denotes the greatest integer less than or equal to $t$ , is equal to :

The integer $n$ for which $lim_{x \to 0} \frac{( c o s x - 1 ) ( c o s x - e ^{x} )}{x ^{n}}$ is a finite non-zero number is

Let $a$ be an integer such that $lim_{x \to 7} \frac{18 - [ 1 - x ]}{[ x ] - 3 a}$ exists, where $[t]$ is greatest integer $\leq t$ . Then $a$ is equal to :

Let $f : R \to (0, \infty)$ be strictly increasing function such that $lim_{x \to \infty} \frac{f ( 7 x )}{f ( x )} = 1$ . Then, the value of $lim_{x \to \infty} [\frac{f ( 5 x )}{f ( x )} - 1]$ is equal to

Let $f : R - {0} \to R$ be a function such that $f (x) - 6 f (\frac{1}{x}) = \frac{35}{3 x} - \frac{5}{2} .$ If the $lim_{x \to 0} (\frac{1}{α x} + f (x)) = β;$ $α, β \in R,$ then $α + 2 β$ is equal to

If $lim_{x \to 1} \frac{( 5 x + 1 ) ^{1/3} - ( x + 5 ) ^{1/3}}{( 2 x + 3 ) ^{1/2} - ( x + 4 ) ^{1/2}} = \frac{m 5}{n ( 2 n ) ^{2/3}}$ , where $g cd (m, n) = 1$ , then $8 m + 12 n$ is equal to ______

The largest value of non-negative integer $a$ for which $lim_{x \to 1} {\frac{- a x + s i n ( x - 1 ) + a}{x + s i n ( x - 1 ) - 1}}^{\frac{1 - x}{1 - x}} = \frac{1}{4}$ is .........

Enter Numerical Value:

Visualized Solution

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

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

Check the Limit Form

Apply L'Hopital's Rule

Differentiate the Numerator

Differentiate the Denominator

Evaluate the New Limit

Find the Derivative .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)

Substitute Values into the Equation

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

Final Conclusion

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Indeterminate Trap

The Surgical Precision of L'Hopital

The Derivative Dance

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R−{0}→R be a function such that f(x)−6f(x1​)=3x35​−25​. If the limx→0​(αx1​+f(x))=β; α,β∈R, then α+2β is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limx→1​(2x+3)1/2−(x+4)1/2(5x+1)1/3−(x+5)1/3​=n(2n)2/3m5​​, where gcd(m,n)=1, then 8m+12n is equal to ______

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The largest value of non-negative integer a for which limx→1​{x+sin(x−1)−1−ax+sin(x−1)+a​}1−x​1−x​=41​ is .........

Analyze the Function $f (x)$

Calculate $f (1)$

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

Evaluate $f^{'} (1)$

Solve for $n$

If $lim_{x \to 1} \frac{x + x ^{2} + x ^{3} + ... + x ^{n} - n}{x - 1} = 820$ , $(n \in N)$ then the value of $n$ is equal to

Let $f : R \to R$ be such that $f (1) = 3$ and $f^{'} (1) = 6$ . Then $lim_{x \to 0} (\frac{f ( 1 + x )}{f ( 1 )})^{1/ x}$ equals

$lim_{x \to \infty} (\frac{x + 6}{x + 1})^{x + 4} = .........$

The integer $n$ for which $lim_{x \to 0} \frac{( c o s x - 1 ) ( c o s x - e ^{x} )}{x ^{n}}$ is a finite non-zero number is

Let $a$ be an integer such that $lim_{x \to 7} \frac{18 - [ 1 - x ]}{[ x ] - 3 a}$ exists, where $[t]$ is greatest integer $\leq t$ . Then $a$ is equal to :

Let $f : R \to (0, \infty)$ be strictly increasing function such that $lim_{x \to \infty} \frac{f ( 7 x )}{f ( x )} = 1$ . Then, the value of $lim_{x \to \infty} [\frac{f ( 5 x )}{f ( x )} - 1]$ is equal to

Let $f : R - {0} \to R$ be a function such that $f (x) - 6 f (\frac{1}{x}) = \frac{35}{3 x} - \frac{5}{2} .$ If the $lim_{x \to 0} (\frac{1}{α x} + f (x)) = β;$ $α, β \in R,$ then $α + 2 β$ is equal to

If $lim_{x \to 1} \frac{( 5 x + 1 ) ^{1/3} - ( x + 5 ) ^{1/3}}{( 2 x + 3 ) ^{1/2} - ( x + 4 ) ^{1/2}} = \frac{m 5}{n ( 2 n ) ^{2/3}}$ , where $g cd (m, n) = 1$ , then $8 m + 12 n$ is equal to ______

The largest value of non-negative integer $a$ for which $lim_{x \to 1} {\frac{- a x + s i n ( x - 1 ) + a}{x + s i n ( x - 1 ) - 1}}^{\frac{1 - x}{1 - x}} = \frac{1}{4}$ is .........