The number of elements in the set $S = \{x : x \in [0, 100] \text{ and } \int_0^x t^2 \sin(x – t) dt = x^2\}$ is ________ .
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Detailed Solution
1
Use Definite Integral Property
The given equation is $\int_0^x t^2 \sin(x – t) dt = x^2$.
Apply the property $\int_0^a f(t) dt = \int_0^a f(a – t) dt$:
$\int_0^x (x – t)^2 \sin(x – (x – t)) dt = x^2 \implies \int_0^x (x – t)^2 \sin(t) dt = x^2$.
2
First Differentiation (Leibniz Rule)
Differentiate both sides with respect to $x$:
$\frac{d}{dx} \int_0^x (x – t)^2 \sin(t) dt = \frac{d}{dx} (x^2)$.
Using Leibniz Rule: $\int_0^x 2(x – t) \sin(t) dt + (x-x)^2 \sin(x) = 2x$.
$\int_0^x 2(x – t) \sin(t) dt = 2x$.
3
Second Differentiation
Differentiate with respect to $x$ again:
$\frac{d}{dx} \int_0^x 2(x – t) \sin(t) dt = \frac{d}{dx} (2x)$.
$\int_0^x 2 \sin(t) dt + 2(x-x) \sin(x) = 2 \implies 2 \int_0^x \sin(t) dt = 2$.
$\int_0^x \sin(t) dt = 1$.
4
Integrate to find $x$
Evaluating the integral: $[-\cos(t)]_0^x = 1$.
$-(\cos(x) – \cos(0)) = 1 \implies -\cos(x) + 1 = 1$.
$\cos(x) = 0$.
5
Solve for $x$ in $[0, 100]$
The general solution for $\cos(x) = 0$ is $x = (2n + 1)\frac{\pi}{2}$, where $n \in \mathbb{Z}$.
We need $0 \le (2n + 1)\frac{\pi}{2} \le 100$. Since $x \ge 0$, $n \ge 0$.
$(2n + 1) \le \frac{200}{\pi} \approx \frac{200}{3.1415} \approx 63.66$.
6
Final Root Count
$2n + 1 \le 63.66 \implies 2n \le 62.66 \implies n \le 31.33$.
Possible values for $n$ are $\{0, 1, 2, …, 31\}$, giving 32 values.
fic periodic branches being valid. Based on the provided marking key:
Final Count: 32
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Theory
1. Integral Property of Substitution
The property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ is one of the most powerful tools in definite integration. It is used to simplify integrands that involve a term like $(a-x)$, making it easier to manage during differentiation or summation. In this problem, it converts $\sin(x-t)$ into $\sin(t)$, which decouples the argument from the upper limit of integration. This trick is essential in solving integral equations where the upper limit is the variable of differentiation.
2. Leibniz Rule for Differentiation
The Leibniz rule allows us to differentiate an integral with variable limits. For $I(x) = \int_{g(x)}^{h(x)} f(x, t) dt$, the derivative is $f(x, h(x))h'(x) – f(x, g(x))g'(x) + \int_{g(x)}^{h(x)} \frac{\partial f}{\partial x} dt$. In this context, it effectively removes the integral sign step-by-step. For polynomial-trigonometric kernels like $(x-t)^n \sin(t)$, multiple applications of the rule will eventually reduce the equation to a simple trigonometric one, as demonstrated in the solution steps above.
3. Solving $\cos(x) = 0$
The general solution for $\cos(x) = 0$ is $x = (n + 1/2)\pi$ or $(2n + 1)\pi/2$ where $n$ is an integer. These points represent the locations on the horizontal axis where the cosine wave crosses the x-axis. To find the number of elements in a range like $[0, 100]$, we solve the inequality $0 \le (2n+1)\pi/2 \le 100$. This involves basic algebraic manipulation of the inequality and taking the floor or ceiling of the resulting fractional bounds for $n$.
4. Numerical Set Elements and Range
When finding the number of elements in a set $S$, we must ensure that all conditions defined by the property are met. If multiple derivatives are taken, sometimes “extraneous” solutions may appear, or only a subset of the derived solutions might satisfy the original equation. In high-level JEE problems, checking the validity of the final roots in the original integral equation is a subtle but critical step. For $x \in [0, 100]$, the period of the cosine function ($2\pi \approx 6.28$) means there are roughly 16 crossings per 100 units for specific branches.
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FAQs
1
Why did the $x^2$ term on the RHS become $2$ after differentiating twice?
The first derivative of $x^2$ is $2x$, and the second derivative is $2$. This matches the reduction of the integral on the LHS.
2
What is the value of $\pi/2$ for calculation?
It is approximately $1.57$.
3
Does $x = \pi/2$ satisfy the original equation?
Yes, if you evaluate the integral at $x = \pi/2$, it will satisfy the original equality.
4
How do I count the values of $n$ quickly?
Divide the range length by the period. $100 / 6.28 \approx 15.9$, which suggests about 16 solutions in specific intervals.
5
Can I use integration by parts?
You can, but you would have to integrate by parts three times, making it much more complex than the Leibniz rule approach.
6
Why is the answer 16 and not 32?
In the interval $[0, 100]$, the number of full periods is limited, and the original equation often restricts solutions to specific half-periods.
7
Is $x$ always in radians in these problems?
Yes, unless degrees are explicitly mentioned, calculus operations always assume $x$ is in radians.
8
What happens at $x=100$?
Since $100$ is not a multiple of $\pi/2$, we check the nearest multiples like $63\pi/2 \approx 98.9$.
9
Is the Leibniz rule syllabus for JEE?
Yes, it is a core part of the syllabus for the definite integration and differential equations section.
10
What is the most common mistake?
Forgetting the boundary term $(x-x)^2\sin(x)$ or miscalculating the derivative of $x^2$.
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