If $\int _{\alpha }^{2026-\alpha }\frac{(2026-x)^{2}}{x^{2}+(2026-x)^{2}}dx=100$, then $\alpha =$

Question

If the solution of \[ \left( 1 + 2e^\frac{x}{y} \right) dx + 2e^\frac{x}{y} \left( 1 - \frac{x}{y} \right) dy = 0 \] is \[ x + \lambda y e^\frac{x}{y} = c \quad \text{(where $c$ is an arbitrary constant), then $ \lambda $ is:} \]

Answer 1

Step 1: Observe the Symmetry
The two fractions $\dfrac{x^{2}}{x^{2}+(2026-x)^{2}}$ and $\dfrac{(2026-x)^{2}}{x^{2}+(2026-x)^{2}}$ add up to $1$.

Step 2: Use the Limits
The interval is symmetric about $x=1013$, so both integrals are equal and each is half the integral of $1$, i.e. $\dfrac{2026-2\alpha}{2}$.

Step 3: Solve
$1013-\alpha=100$, so $\alpha=913$.

Final Answer:
Solving 1013 - alpha = 100 gives alpha = 913. \[ \boxed{\text{(A) }913} \]

If \(\int _{\alpha }^{2026-\alpha }\frac{(2026-x)^{2}}{x^{2}+(2026-x)^{2}}dx=100\), then \(\alpha =\)

Show Hint

The Correct Option is A

Solution and Explanation

Top Questions on Some Properties of Definite Integrals

Questions Asked in MHT CET exam