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If $\displaystyle\sum_{r=0}^{25}\left\{{}^{50}C_r \cdot {}^{50-r}C_{25-r}\right\} = K\binom{50}{25}$, find the value of $K$.
Note: ${}^{50}C_r \cdot {}^{50-r}C_{25-r} = \dfrac{50!}{r!(50-r)!} \cdot \dfrac{(50-r)!}{(25-r)!(25)!} = \dfrac{50!}{r!(25-r)!25!} = {}^{50}C_{25}\cdot{}^{25}C_r$
So the sum $= {}^{50}C_{25}\displaystyle\sum_{r=0}^{25}{}^{25}C_r = {}^{50}C_{25}\cdot 2^{25}$
Therefore $K = \mathbf{2^{25}}$.
Expected trend: EA increases up the group (smaller atom = stronger attraction for incoming electron). However, F is an exception — its very small size causes high electron–electron repulsion in the compact 2p subshell, reducing its EA.
Actual order: Cl > F > Br > I
\(\text{F} > \text{I}\) is correct ✓. The correct option here is (4): F > I.
The synthesis of ATP using a proton gradient across a membrane is called:
Oxidative phosphorylation occurs in the inner mitochondrial membrane (or bacterial plasma membrane) via the electron transport chain and ATP synthase. It couples chemiosmosis with ATP synthesis.
Let $f:[0,1]\to\mathbb{R}$ satisfy $f(xy)=f(x)\cdot f(y)$ for all $x,y\in[0,1]$ with $f(0)\neq 0$. If $y=y(x)$ solves $\dfrac{dy}{dx} = f(x)$ with $y(0)=1$, find $y\!\left(\dfrac{1}{4}\right)+y\!\left(\dfrac{3}{4}\right)$.
From the functional equation $f(xy)=f(x)f(y)$ with $f(0)\neq 0$, putting $x=y=0$ gives $f(0)=f(0)^2 \Rightarrow f(0)=1$. Putting $y=0$: $f(0)=f(x)f(0) \Rightarrow f(x)=1$ for all $x$.
So $\frac{dy}{dx}=1$, giving $y=x+C$. With $y(0)=1$: $C=1$, so $y=x+1$.
$y(1/4)+y(3/4)=(1/4+1)+(3/4+1)=5/4+7/4=12/4=\mathbf{3}$
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