JEE Main 2024 (30 Jan Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let $f : R \to R$ be defined $f (x) = a e^{2 x} + b e^{x} + c x$ . If $f (0) = - 1, f^{'} (lo g_{e} 2) = 21$ and $\int_{0}^{l o g_{e} 4} (f (x) - c x) d x = \frac{39}{2}$ , then the value of $∣ a + b + c ∣$ equals:

Q: I keep messing up $e^{2\ln 2}$. How do I handle these logs?

Think of the power rule for logarithms: $n \ln x = \ln(x^n)$. So, $2 \ln 2 = \ln(2^2) = \ln 4$. Then, $e^{\ln 4}$ is simply $4$. Never try to calculate the decimal value of $\ln 2$; always keep it in logarithmic form until the very end. The beauty of exponential functions is that they are designed to 'undo' logarithms.

Q: Why do we need the absolute value $|a+b+c|$? Is it just to make the answer positive?

Yes, but it also serves as a final check. If you calculate $a+b+c$ and get a negative number, the modulus reminds you to re-read the question. It is a safeguard against sign errors in your final summation. In competitive exams, always treat the final requested format as a hint about the nature of your result.

Select Answer:

Progressive Hints

Hint 1

Look at the structure of the function

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

. You have three unknowns,

a

b

, and

c

. The problem provides three distinct conditions. Your first task is to translate these conditions into a system of three linear equations. Don't let the integral intimidate you; look for a way to simplify the integrand before you start calculating.

Hint 2

When you tackle the derivative condition

f^{'} (ln 2) = 21

, remember the chain rule for

e^{2 x}

. For the integral condition

\int_{0}^{l n 4} (f (x) - c x) d x = \frac{39}{2}

, notice that subtracting

c x

removes the linear term entirely. This is a deliberate design choice by the examiner to make your life easier—embrace it!

Hint 3

Your system of equations should look like this: (1)

a + b = - 1

, (2)

8 a + 2 b + c = 21

, and (3)

5 a + 2 b = 13

. Start by solving the system of (1) and (3) to isolate

a

and

b

. Once you have those, finding

c

from equation (2) is just a matter of careful arithmetic. Finally, don't forget the absolute value sign!

Visualized Solution

Introduction to the Function $f (x)$

Given function: $f (x) = a e^{2 x} + b e^{x} + c x$
Goal: Find the value of $∣ a + b + c ∣$
Three conditions are provided to find three unknowns $a, b, c$ .

Applying the First Condition $f (0) = - 1$

Condition 1: $f (0) = - 1$
Substitute $x = 0$ into $f (x)$ :
$f (0) = a e^{2 (0)} + b e^{0} + c (0)$

Simplifying to Equation 1

Recall that $e^{0} = 1$
$- 1 = a (1) + b (1) + 0$
Equation 1: $a + b = - 1$

Differentiating the Function $f (x)$

Condition 2 involves the derivative $f^{'} (x)$
Differentiate $f (x)$ with respect to $x$ :
$f^{'} (x) = \frac{d}{d x} (a e^{2 x} + b e^{x} + c x)$
$f^{'} (x) = 2 a e^{2 x} + b e^{x} + c$

Applying $f^{'} (lo g_{e} 2) = 21$

Substitute $x = ln 2$ into $f^{'} (x)$ :
$f^{'} (ln 2) = 2 a e^{2 l n 2} + b e^{l n 2} + c = 21$
Use logarithmic property: $e^{l n k} = k$ and $e^{2 l n k} = e^{l n k^{2}} = k^{2}$

Simplifying to Equation 2

$e^{l n 2} = 2$
$e^{2 l n 2} = e^{l n (2^{2})} = e^{l n 4} = 4$
Substitute these values:
$2 a (4) + b (2) + c = 21$
Equation 2: $8 a + 2 b + c = 21$

Setting up the Definite Integral

Condition 3: $\int_{0}^{l n 4} (f (x) - c x) d x = \frac{39}{2}$
Notice that $f (x) - c x = a e^{2 x} + b e^{x}$
Substitute this into the integral:
$\int_{0}^{l n 4} (a e^{2 x} + b e^{x}) d x = \frac{39}{2}$

Evaluating the Antiderivative

Find the antiderivative of $a e^{2 x} + b e^{x}$ :
$\int (a e^{2 x} + b e^{x}) d x = \frac{a}{2} e^{2 x} + b e^{x}$
Apply the limits from $0$ to $ln 4$ :
$[\frac{a}{2} e^{2 x} + b e^{x}]_{0}^{l n 4} = \frac{39}{2}$

Substituting the Limits

Upper limit ( $x = ln 4$ ): $\frac{a}{2} e^{2 l n 4} + b e^{l n 4}$
Since $e^{2 l n 4} = 16$ and $e^{l n 4} = 4$ , this becomes $\frac{a}{2} (16) + b (4) = 8 a + 4 b$
Lower limit ( $x = 0$ ): $\frac{a}{2} e^{0} + b e^{0} = \frac{a}{2} + b$

Simplifying to Equation 3

Difference: $(8 a + 4 b) - (\frac{a}{2} + b) = \frac{39}{2}$
Simplify: $\frac{15 a}{2} + 3 b = \frac{39}{2}$
Multiply the entire equation by $2$ : $15 a + 6 b = 39$
Divide by $3$ :
Equation 3: $5 a + 2 b = 13$

Solving for $a$ and $b$

We have a system of two linear equations:
1) $a + b = - 1$
3) $5 a + 2 b = 13$
Multiply Equation 1 by $2$ : $2 a + 2 b = - 2$
Subtract this from Equation 3:
$(5 a + 2 b) - (2 a + 2 b) = 13 - (- 2)$
$3 a = 15 \Rightarrow a = 5$
Substitute $a = 5$ into Equation 1:
$5 + b = - 1 \Rightarrow b = - 6$

Solving for $c$

Recall Equation 2: $8 a + 2 b + c = 21$
Substitute $a = 5$ and $b = - 6$ :
$8 (5) + 2 (- 6) + c = 21$
$40 - 12 + c = 21$
$28 + c = 21$
$c = 21 - 28 \Rightarrow c = - 7$

Final Calculation: $∣ a + b + c ∣$

We have $a = 5$ , $b = - 6$ , $c = - 7$
Calculate the sum: $a + b + c = 5 + (- 6) + (- 7)$
$a + b + c = 5 - 13 = - 8$
Find the absolute value:
$∣ a + b + c ∣ = ∣ - 8∣ = 8$
Final Answer: $8$

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The function is defined as

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

. We are tasked with finding the constants

a

b

, and

c

using three given conditions.

In the JEE Advanced arena, encountering three unknowns with three conditions signals a System of Equations. We will solve this systematically.

Phase 1

The First Key

We begin with the condition

f (0) = - 1

. Substituting

x = 0

into the function:

f (0) = a e^{2 (0)} + b e^{0} + c (0) = a + b = - 1

This gives us our first linear equation: Equation 1:

a + b = - 1

Phase 2

The Derivative Dance

Next, we utilize the condition

f^{'} (ln 2) = 21

. First, we find the derivative

f^{'} (x)

f^{'} (x) = 2 a e^{2 x} + b e^{x} + c

Substituting

x = ln 2

, and noting that

e^{l n 2} = 2

and

e^{2 l n 2} = 4

, we obtain:

2 a (4) + b (2) + c = 21

Equation 2:

8 a + 2 b + c = 21

Phase 3

The Integral's Grace

Finally, we evaluate the integral

\int_{0}^{l n 4} (f (x) - c x) d x = \frac{39}{2}

. Since

f (x) - c x = a e^{2 x} + b e^{x}

, the integral simplifies significantly:

\int_{0}^{l n 4} (a e^{2 x} + b e^{x}) d x = [\frac{a}{2} e^{2 x} + b e^{x}]_{0}^{l n 4} = \frac{39}{2}

Evaluating at the limits

ln 4

and

0

(\frac{a}{2} (16) + b (4)) - (\frac{a}{2} (1) + b (1)) = \frac{15 a}{2} + 3 b = \frac{39}{2}

Multiplying by

2

and dividing by

3

yields: Equation 3:

5 a + 2 b = 13

Phase 4

The Final Resolution

We now solve the system consisting of Equation 1 and Equation 3: 1)

a + b = - 1

5 a + 2 b = 13

Multiplying Equation 1 by

2

gives

2 a + 2 b = - 2

. Subtracting this from Equation 3:

(5 a + 2 b) - (2 a + 2 b) = 13 - (- 2) \Rightarrow 3 a = 15 \Rightarrow a = 5

Substituting

a = 5

into Equation 1:

5 + b = - 1 \Rightarrow b = - 6

Finally, substituting

a = 5

and

b = - 6

into Equation 2:

8 (5) + 2 (- 6) + c = 21 \Rightarrow 40 - 12 + c = 21 \Rightarrow 28 + c = 21 \Rightarrow c = - 7

The values are

a = 5

b = - 6

, and

c = - 7

. The final result is:

∣ a + b + c ∣ = ∣5 - 6 - 7∣ = ∣ - 8∣ = 8

Common Traps & FAQs

Why does the

c x

term disappear in the integral? Is that a coincidence?

It is absolutely not a coincidence! In JEE Advanced problems, look for symmetry or cancellations. The examiner designed the integral

\int_{0}^{l n 4} (f (x) - c x) d x

specifically to isolate the exponential terms. It is a hint to simplify before you integrate. Always look for these 'gifts' in the problem statement; they are there to reward your observation skills.

I keep messing up

e^{2 l n 2}

. How do I handle these logs?

Think of the power rule for logarithms:

n ln x = ln (x^{n})

. So,

2 ln 2 = ln (2^{2}) = ln 4

. Then,

e^{l n 4}

is simply

4

. Never try to calculate the decimal value of

ln 2

; always keep it in logarithmic form until the very end. The beauty of exponential functions is that they are designed to 'undo' logarithms.

Why do we need the absolute value

∣ a + b + c ∣

? Is it just to make the answer positive?

Yes, but it also serves as a final check. If you calculate

a + b + c

and get a negative number, the modulus reminds you to re-read the question. It is a safeguard against sign errors in your final summation. In competitive exams, always treat the final requested format as a hint about the nature of your result.

Animated Solution for Mathematics - Definite Integration: Let $f : R \to R$ be defined $f (x) = a e^{2 x} + b e^{x} + c x$ . If $f (0) = - 1, f^{'} (lo g_{e} 2) = 21$ and $\int_{0}^{l o g_{e} 4} (f (x) - c x) d x = \frac{39}{2}$ , then the value of $∣ a + b + c ∣$ equals:

Select Answer:

Visualized Solution

Introduction to the Function $f (x)$

Applying the First Condition $f (0) = - 1$

Simplifying to Equation 1

Differentiating the Function $f (x)$

Applying $f^{'} (lo g_{e} 2) = 21$

Simplifying to Equation 2

Setting up the Definite Integral

Evaluating the Antiderivative

Substituting the Limits

Simplifying to Equation 3

Solving for $a$ and $b$

Solving for $c$

Final Calculation: $∣ a + b + c ∣$

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $f (x) = 2 + ∣ x ∣ - ∣ x - 1∣ + ∣ x + 1∣, x \in R$ . Consider (S1): $f^{'} (- \frac{3}{2}) + f^{'} (- \frac{1}{2}) + f^{'} (\frac{1}{2}) + f^{'} (\frac{3}{2}) = 2$ , (S2): $\int_{- 2}^{2} f (x) d x = 12$ . Then,

Let $5 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} + 3, x > 0$ . Then $18 \int_{1}^{2} f (x) d x$ is equal to

If $\int_{0}^{1} \frac{1}{3 + x + 1 + x} d x = a + b 2 + c 3$ , where $a, b, c$ are rational numbers, then $2 a + 3 b - 4 c$ is equal to :

Let $f (x)$ be a differentiable function defined on $[0, 2]$ such that $f^{'} (x) = f^{'} (2 - x)$ for all $x \in (0, 2)$ , $f (0) = 1$ and $f (2) = e^{2}$ . Then the value of $\int_{0}^{2} f (x) d x$ is:

Let the function $f : [0, 2] \to R$ be defined as $f (x) = {e^{m i n {x^{2}, x - [x]}}, e^{[x - l n x]}, x \in [0, 1) x \in [1, 2]$ , where $[t]$ denotes the greatest integer less than or equal to $t$ . Then the value of the integral $\int_{0}^{2} x f (x) d x$ is

Let $[t]$ denote the largest integer less than or equal to $t$ . If $\int_{0}^{3} ([x^{2}] + [\frac{x ^{2}}{2}]) d x = a + b 2 - 3 - 5 + c 6 - 7$ , where $a, b, c \in Z$ , then $a + b + c$ is equal to

The value of the integral $\int_{- 2}^{2} \frac{∣ x ^{3} + x ∣}{( e ^{x ∣ x ∣} + 1 )} d x$ is equal to

If $\int_{0}^{1} 4 cot^{- 1} (1 - 2 x + 4 x^{2}) d x = a tan^{- 1} (2) - b lo g_{e} (5)$ , where $a, b \in N$ , then $(2 a + b)$ is equal to .........

If $\int_{0}^{π /4} \frac{s i n ^{2} x}{1 + s i n x c o s x} d x = \frac{1}{a} lo g_{e} (\frac{a}{3}) + \frac{π}{b 3}$ , where $a, b \in N$ , then $a + b$ is equal to

Let $f (x) = {- 2, x - 2, - 2 \leq x \leq 0 0 < x \leq 2$ and $h (x) = f (∣ x ∣) + ∣ f (x) ∣$ . Then $\int_{- 2}^{2} h (x) d x$ is equal to :

Select Answer:

Visualized Solution

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

Applying the First Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0)=−1

Simplifying to Equation 1

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

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(loge​2)=21

Simplifying to Equation 2

Setting up the Definite Integral

Evaluating the Antiderivative

Substituting the Limits

Simplifying to Equation 3

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

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

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣a+b+c∣

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=2+∣x∣−∣x−1∣+∣x+1∣,x∈R. Consider (S1): f′(−23​)+f′(−21​)+f′(21​)+f′(23​)=2, (S2): ∫−22​f(x)dx=12. Then,

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let 5f(x)+4f(x1​)=x1​+3,x>0. Then 18∫12​f(x)dx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫01​3+x​+1+x​1​dx=a+b2​+c3​, where a,b,c are rational numbers, then 2a+3b−4c 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 differentiable function defined on [0,2] such that f′(x)=f′(2−x) for all x∈(0,2), f(0)=1 and f(2)=e2. Then the value of ∫02​f(x)dx is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let [t] denote the largest integer less than or equal to t. If ∫03​([x2]+[2x2​])dx=a+b2​−3​−5​+c6​−7​, where a,b,c∈Z, then a+b+c is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of the integral ∫−22​(ex∣x∣+1)∣x3+x∣​dx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫01​4cot−1(1−2x+4x2)dx=atan−1(2)−bloge​(5), where a,b∈N, then (2a+b) is equal to .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫0π/4​1+sinxcosxsin2x​dx=a1​loge​(3a​)+b3​π​, where a,b∈N, then a+b is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={−2,x−2,​−2≤x≤00<x≤2​ and h(x)=f(∣x∣)+∣f(x)∣. Then ∫−22​h(x)dx is equal to :

Introduction to the Function $f (x)$

Applying the First Condition $f (0) = - 1$

Differentiating the Function $f (x)$

Applying $f^{'} (lo g_{e} 2) = 21$

Solving for $a$ and $b$

Solving for $c$

Final Calculation: $∣ a + b + c ∣$

Let $f (x) = 2 + ∣ x ∣ - ∣ x - 1∣ + ∣ x + 1∣, x \in R$ . Consider (S1): $f^{'} (- \frac{3}{2}) + f^{'} (- \frac{1}{2}) + f^{'} (\frac{1}{2}) + f^{'} (\frac{3}{2}) = 2$ , (S2): $\int_{- 2}^{2} f (x) d x = 12$ . Then,

Let $5 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} + 3, x > 0$ . Then $18 \int_{1}^{2} f (x) d x$ is equal to

If $\int_{0}^{1} \frac{1}{3 + x + 1 + x} d x = a + b 2 + c 3$ , where $a, b, c$ are rational numbers, then $2 a + 3 b - 4 c$ is equal to :

Let $f (x)$ be a differentiable function defined on $[0, 2]$ such that $f^{'} (x) = f^{'} (2 - x)$ for all $x \in (0, 2)$ , $f (0) = 1$ and $f (2) = e^{2}$ . Then the value of $\int_{0}^{2} f (x) d x$ is:

Let $[t]$ denote the largest integer less than or equal to $t$ . If $\int_{0}^{3} ([x^{2}] + [\frac{x ^{2}}{2}]) d x = a + b 2 - 3 - 5 + c 6 - 7$ , where $a, b, c \in Z$ , then $a + b + c$ is equal to

The value of the integral $\int_{- 2}^{2} \frac{∣ x ^{3} + x ∣}{( e ^{x ∣ x ∣} + 1 )} d x$ is equal to

If $\int_{0}^{1} 4 cot^{- 1} (1 - 2 x + 4 x^{2}) d x = a tan^{- 1} (2) - b lo g_{e} (5)$ , where $a, b \in N$ , then $(2 a + b)$ is equal to .........

If $\int_{0}^{π /4} \frac{s i n ^{2} x}{1 + s i n x c o s x} d x = \frac{1}{a} lo g_{e} (\frac{a}{3}) + \frac{π}{b 3}$ , where $a, b \in N$ , then $a + b$ is equal to

Let $f (x) = {- 2, x - 2, - 2 \leq x \leq 0 0 < x \leq 2$ and $h (x) = f (∣ x ∣) + ∣ f (x) ∣$ . Then $\int_{- 2}^{2} h (x) d x$ is equal to :