Q.19. Let y = y(x) be the solution of the differential equation : , , satisfying . If , than is equ

Q: What is the correct answer for JEE Main Mathematics Q19?

The correct answer is D. Sol. I.F. = $e^{\int \frac{6x^2+e^{-2x}(3x^2-2x-4)+(4x^3+8)e^{-2x}}{(x^2+2)(2+e^{-2x})}dx}$ Let $(x^2+2)(2+e^{-2x}) = t$ $(6x^2+e^{-2x}(3x^2-2x-4))dx = dt$ I.F. = $e^{\int \frac{d((x^2+2)(2+e^{-2x}))}{(x^2+2)(2+e^{-2x})} - \int \frac{4dx}{2+e^{-2x}}}$ = $e^{\ln((x^2+2)(2+e^{-2x})) - \int \frac{4e^{2x}}{2e^{2x}+1}dx}$ = $(x^2+2)(2+e^{-2x}) \cdot e^{-\ln(2e^{2x}+1)^2}$ = $(x^2+2)(2+e^{-2x}) \cdot \f

Q: Q.19. Let y = y(x) be the solution of the differential equation : , , satisfying . If , than is equal to :

The correct answer is D. Sol. I.F. = $e^{\int \frac{6x^2+e^{-2x}(3x^2-2x-4)+(4x^3+8)e^{-2x}}{(x^2+2)(2+e^{-2x})}dx}$ Let $(x^2+2)(2+e^{-2x}) = t$ $(6x^2+e^{-2x}(3x^2-2x-4))dx = dt$ I.F. = $e^{\int \frac{d((x^2+2)(2+e^{-2x}))}{(x^2+2)(2+e^{-2x})} - \int \frac{4dx}{2+e^{-2x}}}$ = $e^{\ln((x^2+2)(2+e^{-2x})) - \int \frac{4e^{2x}}{2e^{2x}+1}dx}$ = $(x^2+2)(2+e^{-2x}) \cdot e^{-\ln(2e^{2x}+1)^2}$ = $(x^2+2)(2+e^{-2x}) \cdot \f

Q: What are the options for this JEE Main Mathematics question?

A. $6/13$ · B. $13/8$ · C. $13/12$ · D. $12/13$

Question

Q.19. Let y = y(x) be the solution of the differential equation : , , satisfying . If , than  is equal to :

FirstInTest · Accepted Answer

Sol. I.F. = $e^{\int \frac{6x^2+e^{-2x}(3x^2-2x-4)+(4x^3+8)e^{-2x}}{(x^2+2)(2+e^{-2x})}dx}$
Let $(x^2+2)(2+e^{-2x}) = t$
$(6x^2+e^{-2x}(3x^2-2x-4))dx = dt$
I.F. = $e^{\int \frac{d((x^2+2)(2+e^{-2x}))}{(x^2+2)(2+e^{-2x})} - \int \frac{4dx}{2+e^{-2x}}}$
= $e^{\ln((x^2+2)(2+e^{-2x})) - \int \frac{4e^{2x}}{2e^{2x}+1}dx}$
= $(x^2+2)(2+e^{-2x}) \cdot e^{-\ln(2e^{2x}+1)^2}$
= $(x^2+2)(2+e^{-2x}) \cdot \frac{1}{(2e^{2x}+1)^2}$
This solution provided in the image is incomplete and seems to be for a different problem or a very complex step. Re-evaluating the problem statement and the provided solution steps, it seems the solution provided in the image is for a different differential equation or has errors. The given solution for Q.19 is: 
Sol. I.F. = $e^{\int \frac{6x^2+(3x^2-2x-4)e^{-2x}}{(x^2+2)(2+e^{-2x})}dx}$ 
Let $(x^2+2)(2+e^{-2x}) = t$ 
$dt = (2x(2+e^{-2x}) + (x^2+2)(-2e^{-2x}))dx = (4x+2xe^{-2x}-2x^2e^{-2x}-4e^{-2x})dx$ 
This does not match the numerator. The provided solution seems to be for a different problem. Assuming the question is solvable with standard methods and the answer is D based on the key.

Let's re-examine the integral factor from the image:
I.F. = $e^{\int \frac{6x^2+e^{-2x}(3x^2-2x-4)+(4x^3+8)e^{-2x}}{(x^2+2)(2+e^{-2x})}dx}$ 
This is a very complex integral. The solution provided in the image is for a different form of the differential equation. The image's solution for I.F. is simplified to $e^{\ln(x^2+2) - 2\ln(2+e^{-2x})} = \frac{x^2+2}{(2+e^{-2x})^2}$. This is not directly derivable from the given P(x).

Let's assume the given P(x) is $\frac{6x^2+e^{-2x}(3x^2-2x-4)}{(x^2+2)(2+e^{-2x})}$ and the solution is for a simpler P(x) or there's a typo in the question or solution. Given the complexity, I'll transcribe the provided solution as is, acknowledging its potential discrepancy.

From the image's solution:
I.F. = $e^{\ln(x^2+2) - 2\ln(2+e^{-2x})} = \frac{x^2+2}{(2+e^{-2x})^2}$
Then $y \cdot \frac{x^2+2}{(2+e^{-2x})^2} = \int (2+e^{-2x}) \cdot \frac{x^2+2}{(2+e^{-2x})^2} dx + C = \int \frac{x^2+2}{2+e^{-2x}} dx + C$
This step is incorrect. The RHS should be $\int Q(x) \cdot I.F. dx + C$. Here $Q(x) = 2+e^{-2x}$.
So, $y \cdot \frac{x^2+2}{(2+e^{-2x})^2} = \int (2+e^{-2x}) \cdot \frac{x^2+2}{(2+e^{-2x})^2} dx + C = \int \frac{x^2+2}{2+e^{-2x}} dx + C$
The solution in the image then states:
$y \frac{x^2+2}{2+e^{-2x}} = \frac{x^3}{3} + 2x + C$ (This implies I.F. was $\frac{x^2+2}{2+e^{-2x}}$ and $Q(x) = 2+e^{-2x}$)
At $y(0) = 3/2 \Rightarrow \frac{3}{2} \frac{0+2}{2+1} = 0+0+C \Rightarrow C = 1$
So $y \frac{x^2+2}{2+e^{-2x}} = \frac{x^3}{3} + 2x + 1$
At x = 1:
$y(1) \frac{1+2}{2+e^{-2}} = \frac{1}{3} + 2 + 1 = \frac{10}{3}$
$y(1) \frac{3}{2+e^{-2}} = \frac{10}{3}$
$y(1) = \frac{10}{9}(2+e^{-2})$
Given $y(1) = \alpha(2+e^2)$. This does not match. There is a clear discrepancy between the question and the provided solution steps. The solution seems to be for a different problem or has significant errors. However, following the provided solution's final step for $\alpha$:
So $\alpha = \frac{13}{12}$ (This is from the image's final step, which is not derived from the preceding steps.)
Given the discrepancy, I will use the final answer from the image and mark confidence lower.

Answer

$6/13$

Answer

$13/8$

Answer

$13/12$

Answer

$12/13$

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