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JEE Main 2023
Definite Integration
Definite Integration
Hard

Question

Let f:RR\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R} be defined as f(x)=ae2x+bex+cxf(x)=a e^{2 x}+b e^x+c x. If f(0)=1,f(loge2)=21f(0)=-1, f^{\prime}\left(\log _e 2\right)=21 and 0loge4(f(x)cx)dx=392\int_0^{\log _e 4}(f(x)-c x) d x=\frac{39}{2}, then the value of a+b+c|a+b+c| equals

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Solution

Key Concepts and Formulas

  1. Differentiation of Exponential Functions: The derivative of aekxae^{kx} is akekxake^{kx}.
  2. Properties of Logarithms and Exponentials: elogex=xe^{\log_e x} = x and enlogex=xne^{n \log_e x} = x^n. Also, e0=1e^0 = 1.
  3. Definite Integration: The definite integral of a function g(x)g(x) from aa to bb is found by evaluating its antiderivative G(x)G(x) at the limits: abg(x)dx=G(b)G(a)\int_a^b g(x) dx = G(b) - G(a).
  4. Solving Systems of Linear Equations: Using algebraic manipulation to find the values of unknown variables.

Step-by-Step Solution:

We are given the function f(x)=ae2x+bex+cxf(x) = a e^{2x} + b e^x + c x and three conditions that will allow us to determine the values of the constants aa, bb, and cc.

Step 1: Use the condition f(0)=1f(0) = -1 to form the first equation.

We are given f(x)=ae2x+bex+cxf(x) = a e^{2x} + b e^x + c x. Substitute x=0x=0 into the function: f(0)=ae2(0)+be0+c(0)f(0) = a e^{2(0)} + b e^0 + c(0) f(0)=ae0+be0+0f(0) = a e^0 + b e^0 + 0 Using the property e0=1e^0 = 1: f(0)=a(1)+b(1)f(0) = a(1) + b(1) f(0)=a+bf(0) = a + b We are given that f(0)=1f(0) = -1. Therefore, our first equation is: a+b=1(Equation 1)a + b = -1 \quad (\text{Equation 1}) Explanation: This step utilizes the definition of a function and the property of the exponential function at zero to establish a linear relationship between the constants aa and bb.

Step 2: Use the condition f(loge2)=21f'(\log_e 2) = 21 to form the second equation.

First, we need to find the derivative of f(x)f(x), denoted by f(x)f'(x). f(x)=ae2x+bex+cxf(x) = a e^{2x} + b e^x + c x Differentiating term by term with respect to xx:

  • The derivative of ae2xa e^{2x} is a(2e2x)=2ae2xa \cdot (2e^{2x}) = 2a e^{2x}.
  • The derivative of bexb e^x is bexb e^x.
  • The derivative of cxc x is cc. So, the derivative function is: f(x)=2ae2x+bex+cf'(x) = 2a e^{2x} + b e^x + c Now, we substitute x=loge2x = \log_e 2 into f(x)f'(x): f(loge2)=2ae2(loge2)+beloge2+cf'(\log_e 2) = 2a e^{2(\log_e 2)} + b e^{\log_e 2} + c Using the properties enlogex=xne^{n \log_e x} = x^n and elogex=xe^{\log_e x} = x:
  • e2(loge2)=eloge(22)=eloge4=4e^{2(\log_e 2)} = e^{\log_e (2^2)} = e^{\log_e 4} = 4.
  • eloge2=2e^{\log_e 2} = 2. Substituting these values: f(loge2)=2a(4)+b(2)+cf'(\log_e 2) = 2a (4) + b (2) + c f(loge2)=8a+2b+cf'(\log_e 2) = 8a + 2b + c We are given that f(loge2)=21f'(\log_e 2) = 21. Therefore, our second equation is: 8a+2b+c=21(Equation 2)8a + 2b + c = 21 \quad (\text{Equation 2}) Explanation: This step involves differentiating the function and then applying the given condition using properties of logarithms and exponentials to establish another linear relationship between aa, bb, and cc.

Step 3: Use the condition 0loge4(f(x)cx)dx=392\int_0^{\log_e 4}(f(x)-c x) d x=\frac{39}{2} to form the third equation.

First, let's find the expression for f(x)cxf(x) - cx: f(x)cx=(ae2x+bex+cx)cxf(x) - cx = (a e^{2x} + b e^x + c x) - cx f(x)cx=ae2x+bexf(x) - cx = a e^{2x} + b e^x Now, we need to evaluate the definite integral of this expression from 00 to loge4\log_e 4: 0loge4(ae2x+bex)dx\int_0^{\log_e 4} (a e^{2x} + b e^x) dx Let's find the antiderivative of (ae2x+bex)(a e^{2x} + b e^x):

  • The antiderivative of ae2xa e^{2x} is a(e2x2)=a2e2xa \left(\frac{e^{2x}}{2}\right) = \frac{a}{2} e^{2x}.
  • The antiderivative of bexb e^x is bexb e^x. So, the antiderivative of (ae2x+bex)(a e^{2x} + b e^x) is a2e2x+bex\frac{a}{2} e^{2x} + b e^x. Now, we apply the Fundamental Theorem of Calculus: [a2e2x+bex]0loge4=392\left[\frac{a}{2} e^{2x} + b e^x\right]_0^{\log_e 4} = \frac{39}{2} Evaluate the expression at the upper limit (x=loge4x = \log_e 4) and subtract the evaluation at the lower limit (x=0x = 0): At x=loge4x = \log_e 4: a2e2(loge4)+beloge4=a2eloge(42)+b(4)=a2eloge16+4b=a2(16)+4b=8a+4b\frac{a}{2} e^{2(\log_e 4)} + b e^{\log_e 4} = \frac{a}{2} e^{\log_e (4^2)} + b(4) = \frac{a}{2} e^{\log_e 16} + 4b = \frac{a}{2}(16) + 4b = 8a + 4b At x=0x = 0: a2e2(0)+be0=a2e0+be0=a2(1)+b(1)=a2+b\frac{a}{2} e^{2(0)} + b e^0 = \frac{a}{2} e^0 + b e^0 = \frac{a}{2}(1) + b(1) = \frac{a}{2} + b Now, subtract the lower limit evaluation from the upper limit evaluation: (8a+4b)(a2+b)=392(8a + 4b) - \left(\frac{a}{2} + b\right) = \frac{39}{2} 8a+4ba2b=3928a + 4b - \frac{a}{2} - b = \frac{39}{2} Combine like terms: (8aa2)+(4bb)=392(8a - \frac{a}{2}) + (4b - b) = \frac{39}{2} (16aa2)+3b=392(\frac{16a - a}{2}) + 3b = \frac{39}{2} 15a2+3b=392\frac{15a}{2} + 3b = \frac{39}{2} To eliminate the fractions, multiply the entire equation by 2: 15a+6b=3915a + 6b = 39 We can simplify this equation by dividing by 3: 5a+2b=13(Equation 3)5a + 2b = 13 \quad (\text{Equation 3}) Explanation: This step involves simplifying the integrand, finding its antiderivative, and then applying the definite integral formula by evaluating at the limits. Properties of logarithms and exponentials are crucial here, as is algebraic simplification to obtain a linear equation in terms of aa and bb.

Step 4: Solve the system of linear equations to find the values of aa, bb, and cc.

We have the following system of equations:

  1. a+b=1a + b = -1
  2. 8a+2b+c=218a + 2b + c = 21
  3. 5a+2b=135a + 2b = 13

Let's first solve for aa and bb using Equations 1 and 3. From Equation 1, we can express bb in terms of aa: b=1ab = -1 - a Substitute this expression for bb into Equation 3: 5a+2(1a)=135a + 2(-1 - a) = 13 5a22a=135a - 2 - 2a = 13 3a2=133a - 2 = 13 3a=153a = 15 a=5a = 5 Now, substitute the value of aa back into the expression for bb: b=1a=15b = -1 - a = -1 - 5 b=6b = -6 Now that we have the values of aa and bb, we can substitute them into Equation 2 to find cc: 8a+2b+c=218a + 2b + c = 21 8(5)+2(6)+c=218(5) + 2(-6) + c = 21 4012+c=2140 - 12 + c = 21 28+c=2128 + c = 21 c=2128c = 21 - 28 c=7c = -7 So, the values of the constants are a=5a=5, b=6b=-6, and c=7c=-7. Explanation: This step involves a standard procedure for solving a system of linear equations. We used substitution to solve for two variables first and then used those values to find the third variable.

Step 5: Calculate a+b+c|a+b+c|.

We have found a=5a=5, b=6b=-6, and c=7c=-7. Now, we calculate the sum a+b+ca+b+c: a+b+c=5+(6)+(7)a+b+c = 5 + (-6) + (-7) a+b+c=567a+b+c = 5 - 6 - 7 a+b+c=17a+b+c = -1 - 7 a+b+c=8a+b+c = -8 Finally, we need to find the absolute value: a+b+c=8|a+b+c| = |-8| a+b+c=8|a+b+c| = 8 Explanation: This final step involves substituting the determined values of the constants into the expression a+b+c|a+b+c| and computing its absolute value.

Common Mistakes & Tips

  • Logarithm and Exponential Properties: Ensure correct application of rules like elogex=xe^{\log_e x} = x and enlogex=xne^{n \log_e x} = x^n. Mistakes here can lead to incorrect values in Steps 2 and 3.
  • Algebraic Errors: Be meticulous with algebraic manipulations, especially when solving systems of equations or simplifying expressions with fractions. Double-checking calculations can prevent errors.
  • Differentiation/Integration Errors: Carefully apply the rules of differentiation and integration for exponential functions. The constants of integration are not needed for definite integrals.

Summary

The problem was solved by systematically using the three given conditions to form a system of linear equations for the unknown constants aa, bb, and cc. The first condition, f(0)=1f(0)=-1, provided a direct relation between aa and bb. The second condition, involving the derivative f(loge2)=21f'(\log_e 2)=21, yielded another linear equation in aa, bb, and cc. The third condition, a definite integral, was evaluated to produce a third linear equation involving aa and bb. Solving this system of equations yielded a=5a=5, b=6b=-6, and c=7c=-7. Finally, the absolute value of the sum of these constants, a+b+c|a+b+c|, was calculated to be 8.

The final answer is 8\boxed{8}.

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