The region \(E\) is defined by the inequalities: \[ 3 - x \leq y \leq \sqrt{9 - x^2}, \quad 0 \leq x \leq 3. \] The point \(A = (p, p+1)\) lies inside the region. For \(A\) to lie above the line \(L: y = 3 - x\): \[ p + p + 1 - 3 > 0 \implies 2p > 2 \implies p > 1. \] For \(A\) to lie below the semicircle \(S: y = \sqrt{9 - x^2}\): \[ p^2 + (p+1)^2 - 9 < 0. \] Solving: \[ p^2 + p^2 + 2p + 1 - 9 < 0 \implies 2p^2 + 2p - 8 < 0 \implies p^2 + p - 4 < 0. \] The roots of \(p^2 + p - 4 = 0\) are: \[ p = \frac{-1 \pm \sqrt{17}}{2}. \] So, the valid interval is: \[ 0 < p < \frac{\sqrt{17} - 1}{2}. \] Combining with \(p > 1\): \[ 1 < p < \frac{\sqrt{17} - 1}{2}. \] Calculating: \[ b^2 + b - a^2 = 3. \] ✅ Therefore, the final answer is: \(3\).
The area enclosed by the closed curve $C$ given by the differential equation $\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$ is $4 \pi$.
Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $R S$ is
The statement
\((p⇒q)∨(p⇒r) \)
is NOT equivalent to
Let α, β(α > β) be the roots of the quadratic equation x2 – x – 4 = 0.
If \(P_n=α^n–β^n, n∈N\) then \(\frac{P_{15}P_{16}–P_{14}P_{16}–P_{15}^2+P_{14}P_{15}}{P_{13}P_{14}}\)
is equal to _______.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)