JEE Main 2019 (12 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : R \to R$ be a continuously differentiable function such that $f (2) = 6$ and $f^{'} (2) = 1/48$ . If $\int_{6}^{f (x)} 4 t^{3} d t = (x - 2) g (x)$ , then $lim_{x \to 2} g (x)$ is equal to :

Q: Wait, why do we reject the negative root here? Or do we even have a root?

Great question! In this problem, we are not solving for $x$ or $f(x)$ as an equation to find roots. We are evaluating a limit. The expression $f(x)^4 - 6^4$ becomes $0$ at $x=2$ because $f(2)=6$. We are not looking for roots of the function, but rather analyzing the behavior of the ratio as $x$ approaches $2$. The indeterminate form $0/0$ is exactly what we expect in limit problems involving derivatives.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Do not be intimidated by the integral

\int_{6}^{f (x)} 4 t^{3} d t

. Think of it as a function of

x

. Evaluate the definite integral first using the Fundamental Theorem of Calculus to simplify the expression into a form involving

f (x)

and constants.

Hint 2

Hint 2: The Arsenal. Once you have the equation

f (x)^{4} - 6^{4} = (x - 2) g (x)

, isolate

g (x)

. You will see that finding

lim_{x \to 2} g (x)

leads to a

0/0

indeterminate form. Remember that

f (2) = 6

is given, which is crucial for evaluating the limit.

Hint 3

Hint 3: The Execution. Since you have a

0/0

form, apply L'Hopital's Rule. Differentiate the numerator

f (x)^{4} - 6^{4}

with respect to

x

using the chain rule, and differentiate the denominator

x - 2

with respect to

x

. Then, substitute

x = 2

and the given values

f (2) = 6

and

f^{'} (2) = 1/48

Visualized Solution

Analyze the Integral Equation

Given function $f : R \to R$ is continuously differentiable.
Given values: $f (2) = 6$ and $f^{'} (2) = \frac{1}{48}$ .
Integral equation: $\int_{6}^{f (x)} 4 t^{3} d t = (x - 2) g (x)$ .

Apply the Power Rule for Integration

Use the power rule: $\int t^{n} d t = \frac{t ^{n + 1}}{n + 1}$ .
Evaluate the indefinite integral: $\int 4 t^{3} d t = 4 \cdot \frac{t ^{4}}{4} = t^{4}$ .

Evaluate the Definite Integral

Apply limits from $6$ to $f (x)$ : $[t^{4}]_{6}^{f (x)}$ .
Substitute limits: $f (x)^{4} - 6^{4}$ .

Equate to the Right-Hand Side

Equate the results: $f (x)^{4} - 6^{4} = (x - 2) g (x)$ .

Isolate $g (x)$

Solve for $g (x)$ : $g (x) = \frac{f ( x ) ^{4} - 6 ^{4}}{x - 2}$ .

Set Up the Limit

We need to find $lim_{x \to 2} g (x)$ .
$lim_{x \to 2} g (x) = lim_{x \to 2} \frac{f ( x ) ^{4} - 6 ^{4}}{x - 2}$ .

Identify the Indeterminate Form

Check the form at $x = 2$ .
Numerator $\to f (2)^{4} - 6^{4} = 6^{4} - 6^{4} = 0$ .
Denominator $\to 2 - 2 = 0$ .
This is a $\frac{0}{0}$ indeterminate form.

Apply L'Hopital's Rule

Apply L'Hopital's Rule: $lim_{x \to 2} \frac{\frac{d}{d x} ( f ( x ) ^{4} - 6 ^{4} )}{\frac{d}{d x} ( x - 2 )}$ .
Differentiate numerator using Chain Rule: $4 f (x)^{3} \cdot f^{'} (x)$ .
Differentiate denominator: $1$ .

Evaluate the Limit

Substitute $x = 2$ into the derivative expression.
$lim_{x \to 2} g (x) = \frac{4 f ( 2 ) ^{3} \cdot f ^{'} ( 2 )}{1}$ .

Substitute the Given Values

Plug in $f (2) = 6$ and $f^{'} (2) = \frac{1}{48}$ .
Expression becomes: $4 (6)^{3} \cdot (\frac{1}{48})$ .

Final Calculation

Calculate $6^{3} = 216$ .
Expression: $\frac{4 \cdot 216}{48}$ .
Simplify: $\frac{864}{48} = 18$ .

Conclusion

Final Answer: $lim_{x \to 2} g (x) = 18$

00:00 / 00:00

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

Analyzing the Setup

We are given a continuously differentiable function

f (x)

with the specific values

f (2) = 6

and

f^{'} (2) = \frac{1}{48}

The relationship between

f (x)

and

g (x)

is defined by the integral equation:

\int_{6}^{f (x)} 4 t^{3} d t = (x - 2) g (x)

Our objective is to determine the value of the limit

lim_{x \to 2} g (x)

Unmasking the Integral

First, we evaluate the left-hand side of the equation. Using the power rule for integration, the antiderivative of

4 t^{3}

t^{4}

Applying the Fundamental Theorem of Calculus, we evaluate the integral from

6

f (x)

\int_{6}^{f (x)} 4 t^{3} d t = [t^{4}]_{6}^{f (x)} = f (x)^{4} - 6^{4}

This simplifies our original equation to:

f (x)^{4} - 6^{4} = (x - 2) g (x)

The Algebraic Bridge

To isolate

g (x)

, we divide both sides by

(x - 2)

g (x) = \frac{f ( x ) ^{4} - 6 ^{4}}{x - 2}

We now examine the limit as

x \to 2

. Substituting

x = 2

directly yields:

\frac{f ( 2 ) ^{4} - 6 ^{4}}{2 - 2} = \frac{6 ^{4} - 6 ^{4}}{0} = \frac{0}{0}

We have encountered the classic $0/0$ indeterminate form, which necessitates the use of L'Hopital's Rule.

The Power of L'Hopital

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

x

Using the chain rule for the numerator, the derivative of

f (x)^{4}

4 f (x)^{3} \cdot f^{'} (x)

. The derivative of the constant

6^{4}

0

, and the derivative of the denominator

(x - 2)

1

The limit expression becomes:

x \to 2 lim g (x) = x \to 2 lim \frac{4 f ( x ) ^{3} \cdot f ^{'} ( x )}{1}

Final Calculation

Now, we substitute the known values

f (2) = 6

and

f^{'} (2) = \frac{1}{48}

into the expression:

x \to 2 lim g (x) = 4 \cdot (f (2))^{3} \cdot f^{'} (2)

Substituting the constants:

x \to 2 lim g (x) = 4 \cdot (6)^{3} \cdot \frac{1}{48}

Since

6^{3} = 216

, we calculate:

x \to 2 lim g (x) = 4 \cdot 216 \cdot \frac{1}{48} = \frac{864}{48} = 18

The final result is $18$ .

Common Traps & FAQs

Wait, why do we reject the negative root here? Or do we even have a root?

Great question! In this problem, we are not solving for

x

f (x)

as an equation to find roots. We are evaluating a limit. The expression

f (x)^{4} - 6^{4}

becomes

0

x = 2

because

f (2) = 6

. We are not looking for roots of the function, but rather analyzing the behavior of the ratio as

x

approaches

2

. The indeterminate form

0/0

is exactly what we expect in limit problems involving derivatives.

Why can't we just use energy conservation or some other physics principle?

While physics problems often use energy conservation, this is a pure calculus problem. The 'energy' here is the structure of the limit. We must rely on the Fundamental Theorem of Calculus to evaluate the integral and L'Hopital's Rule to resolve the limit. Stick to the calculus toolkit!

Do I need to know the explicit form of

f (x)

to solve this?

Absolutely not! That is the beauty of this problem. We are only given the value of the function and its derivative at a single point,

x = 2

. Because the limit definition of a derivative relies only on the local behavior near that point, we have exactly enough information to solve it without ever knowing the full expression for

f (x)

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : R \to R$ be a continuously differentiable function such that $f (2) = 6$ and $f^{'} (2) = 1/48$ . If $\int_{6}^{f (x)} 4 t^{3} d t = (x - 2) g (x)$ , then $lim_{x \to 2} g (x)$ is equal to :

Select Answer:

Visualized Solution

Analyze the Integral Equation

Apply the Power Rule for Integration

Evaluate the Definite Integral

Equate to the Right-Hand Side

Isolate $g (x)$

Set Up the Limit

Identify the Indeterminate Form

Apply L'Hopital's Rule

Evaluate the Limit

Substitute the Given Values

Final Calculation

Conclusion

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

Analyzing the Setup

Unmasking the Integral

The Algebraic Bridge

The Power of L'Hopital

Final Calculation

Similar Questions

Let $f : R \to R$ be a function such that $f (2) = 4$ and $f^{'} (2) = 1$ . Then, the value of $lim_{x \to 2} \frac{x ^{2} f ( 2 ) - 4 f ( x )}{x - 2}$ is equal to :

Let $f (x) = 4$ and $f^{'} (x) = 4$ . Then $lim_{x \to 2} \frac{x f ( 2 ) - 2 f ( x )}{x - 2}$ is given by

Let $f (x)$ be a differentiable function at $x = a$ with $f^{'} (a) = 2$ and $f (a) = 4$ . Then $lim_{x \to a} \frac{x f ( a ) - a f ( x )}{x - a}$ equals :

Let $f : R \to R$ be a continuous function. Then $lim_{x \to \frac{π}{4}} \frac{\frac{π}{4} \int _{2}^{s e c^{2} x} f ( x ) d x}{x ^{2} - \frac{π ^{2}}{16}}$ is equal to :

$f (x)$ is the integral of $\frac{2 s i n x - s i n 2 x}{x ^{3}}, x \neq = 0$ , find $lim_{x \to 0} f^{'} (x)$

$lim_{h \to 0} \frac{f ( 2 h + 2 + h ^{2} ) - f ( 2 )}{f ( h - h ^{2} + 1 ) - f ( 1 )}$ , given that $f^{'} (2) = 6$ and $f^{'} (1) = 4$

If $f (a) = 2, f^{'} (a) = 1, g (a) = - 1, g^{'} (a) = 2$ , then the value of $lim_{x \to a} \frac{g ( x ) f ( a ) - g ( a ) f ( x )}{x - a}$ is

Let $f (x) = \int_{0}^{x} (t + sin (1 - e^{t})) d t, x \in R$ . Then, $lim_{x \to 0} \frac{f ( x )}{x ^{3}}$ is equal to

Let $f : R \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18$ , $f^{'} (3) = 0$ and $f^{''} (3) = 4$ . Then $lim_{x \to 1} (lo g_{e} (\frac{f ( 2 + x )}{f ( 3 )}))^{\frac{18}{( x - 1 ) ^{2}}}$ is equal to :

Let $f : [- \frac{π}{2}, \frac{π}{2}] \to R$ be a differentiable function such that $f (0) = \frac{1}{2}$ . If $lim_{x \to 0} \frac{x \int _{0}^{x} f ( t ) d t}{e ^{x^{2}} - 1} = α$ , then $8 α^{2}$ is equal to :

Select Answer:

Visualized Solution

Analyze the Integral Equation

Apply the Power Rule for Integration

Evaluate the Definite Integral

Equate to the Right-Hand Side

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

Set Up the Limit

Identify the Indeterminate Form

Apply L'Hopital's Rule

Evaluate the Limit

Substitute the Given Values

Final Calculation

Conclusion

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

Analyzing the Setup

Unmasking the Integral

The Algebraic Bridge

The Power of L'Hopital

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a function such that f(2)=4 and f′(2)=1. Then, the value of limx→2​x−2x2f(2)−4f(x)​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=4 and f′(x)=4. Then limx→2​x−2xf(2)−2f(x)​ is given by

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a differentiable function at x=a with f′(a)=2 and f(a)=4. Then limx→a​x−axf(a)−af(x)​ equals :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a continuous function. Then limx→4π​​x2−16π2​4π​∫2sec2x​f(x)dx​ is equal to :

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limh→0​f(h−h2+1)−f(1)f(2h+2+h2)−f(2)​, given that f′(2)=6 and f′(1)=4

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=∫0x​(t+sin(1−et))dt,x∈R. Then, limx→0​x3f(x)​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→(0,∞) be a twice differentiable function such that f(3)=18, f′(3)=0 and f′′(3)=4. Then limx→1​(loge​(f(3)f(2+x)​))(x−1)218​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:[−2π​,2π​]→R be a differentiable function such that f(0)=21​. If limx→0​ex2−1x∫0x​f(t)dt​=α, then 8α2 is equal to :

Isolate $g (x)$

Let $f : R \to R$ be a function such that $f (2) = 4$ and $f^{'} (2) = 1$ . Then, the value of $lim_{x \to 2} \frac{x ^{2} f ( 2 ) - 4 f ( x )}{x - 2}$ is equal to :

Let $f (x) = 4$ and $f^{'} (x) = 4$ . Then $lim_{x \to 2} \frac{x f ( 2 ) - 2 f ( x )}{x - 2}$ is given by

Let $f (x)$ be a differentiable function at $x = a$ with $f^{'} (a) = 2$ and $f (a) = 4$ . Then $lim_{x \to a} \frac{x f ( a ) - a f ( x )}{x - a}$ equals :

Let $f : R \to R$ be a continuous function. Then $lim_{x \to \frac{π}{4}} \frac{\frac{π}{4} \int _{2}^{s e c^{2} x} f ( x ) d x}{x ^{2} - \frac{π ^{2}}{16}}$ is equal to :

$f (x)$ is the integral of $\frac{2 s i n x - s i n 2 x}{x ^{3}}, x \neq = 0$ , find $lim_{x \to 0} f^{'} (x)$

$lim_{h \to 0} \frac{f ( 2 h + 2 + h ^{2} ) - f ( 2 )}{f ( h - h ^{2} + 1 ) - f ( 1 )}$ , given that $f^{'} (2) = 6$ and $f^{'} (1) = 4$

If $f (a) = 2, f^{'} (a) = 1, g (a) = - 1, g^{'} (a) = 2$ , then the value of $lim_{x \to a} \frac{g ( x ) f ( a ) - g ( a ) f ( x )}{x - a}$ is

Let $f (x) = \int_{0}^{x} (t + sin (1 - e^{t})) d t, x \in R$ . Then, $lim_{x \to 0} \frac{f ( x )}{x ^{3}}$ is equal to

Let $f : R \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18$ , $f^{'} (3) = 0$ and $f^{''} (3) = 4$ . Then $lim_{x \to 1} (lo g_{e} (\frac{f ( 2 + x )}{f ( 3 )}))^{\frac{18}{( x - 1 ) ^{2}}}$ is equal to :

Let $f : [- \frac{π}{2}, \frac{π}{2}] \to R$ be a differentiable function such that $f (0) = \frac{1}{2}$ . If $lim_{x \to 0} \frac{x \int _{0}^{x} f ( t ) d t}{e ^{x^{2}} - 1} = α$ , then $8 α^{2}$ is equal to :