From the condition $|f(x)-f(y)|\le 2|x-y|^{3/2}$, divide both sides by $|x-y|$:

$\left|\frac{f(x)-f(y)}{x-y}\right|\le 2|x-y|^{1/2}$

Taking $y\to x$: $|f'(x)|\le 2\cdot0=0$, so $f'(x)=0$ for all $x$.

Hence $f$ is constant. Since $f(0)=1$, we have $f(x)=1$ for all $x$.

$\int_0^1 f^2(x)\,dx = \int_0^1 1\,dx = \mathbf{1}$

Reverse saturation current is very small (μA range for Si).

A triangle is formed by choosing any 3 vertices from $n$ vertices, so $T_n = \binom{n}{3}$.

$T_{n+1} - T_n = \binom{n+1}{3} - \binom{n}{3} = \binom{n}{2}$

Setting $\binom{n}{2} = 10$: $\dfrac{n(n-1)}{2} = 10 \Rightarrow n(n-1) = 20 \Rightarrow n = 5$... wait: $5 \times 4 = 20$ ✓. But checking $n=5$: $T_6 - T_5 = \binom{6}{3}-\binom{5}{3}=20-10=10$ ✓. So $n = 5$.

Actually re-checking: $\binom{n}{2}=10 \Rightarrow n=5$. Answer: $n=\mathbf{5}$, but the JEE 2013 answer is $n=5$. Wait — options show 7 as (A). Let me verify: $T_{n+1}-T_n=\binom{n}{2}=10 \Rightarrow n(n-1)=20 \Rightarrow n=5$. The correct answer is $\mathbf{5}$ (option B, index 1).

Moles of H$_2$O $= \dfrac{0.72}{18} = 0.04\ \text{mol}$ → H atoms $= 0.08\ \text{mol}$

Moles of CO$_2 = \dfrac{3.08}{44} = 0.07\ \text{mol}$ → C atoms $= 0.07\ \text{mol}$

Ratio C : H $= 0.07 : 0.08 = 7 : 8$

Empirical formula: $\mathbf{C_7H_8}$ (like toluene)