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Percentage \(= \dfrac{21}{25} \times 100 = 84\%\)
Let $S=\left\{t\in\mathbb{R}: f(x)=|x-\pi|\cdot(e^{|x|}-1)\sin|x|\right.$ is not differentiable at $t\}$. Find $S$.
Check $x=0$: $f(x)=|x-\pi|(e^{|x|}-1)\sin|x|$. Near $x=0$, both $(e^{|x|}-1)$ and $\sin|x|$ have factors of $|x|$, so $f(x)\approx |x-\pi|\cdot|x|^2$ near 0, which is differentiable at 0.
Check $x=\pi$: near $\pi$, $(e^{|x|}-1)\sin|x|$ is smooth and non-zero ($e^\pi\sin\pi=0$!). Since $\sin\pi=0$, the product $(e^{|x|}-1)\sin|x|$ vanishes at $\pi$, making $f(x)=0$ near $\pi$ — wait, $\sin\pi=0$, so $f(\pi)=0$. Checking differentiability more carefully shows $f$ is differentiable at $\pi$ too.
Therefore $S=\boldsymbol{\emptyset}$.
- Spring constant $k$: $F = kx \Rightarrow [k] = \frac{[F]}{[x]} = \frac{MLT^{-2}}{L} = MT^{-2} = M^1L^0T^{-2}$ → (3)
- Pascal (pressure): $[P] = \frac{Force}{Area} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} = M^1L^{-1}T^{-2}$ → (4)
- Hertz (frequency): $[f] = T^{-1} = M^0L^0T^{-1}$ → (2)
- Joule (energy): $[E] = ML^2T^{-2} = M^1L^2T^{-2}$ → (1)
Correct match: A-3, B-4, C-2, D-1
Generally IE₁ increases across a period. However, N has a half-filled 2p³ configuration (extra stability), so its IE₁ is higher than O.
Correct order: B < C < O < N (not B < C < N < O)
So option (2) is wrong. The I < Br < Cl < F order for EGE is also wrong (Cl > F), but option (3) is listed incorrectly too — however, the key answer is (2).
Cause: The 4f electrons have a poor shielding effect (they cannot effectively shield each other or outer electrons from the increasing nuclear charge). As protons are added (Z increases from 58 to 71) with poor f-electron shielding, the effective nuclear charge experienced by outer electrons steadily increases, pulling them inward.
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