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Using only principal values of inverse functions, describe the set $A=\left\{x\geq0:\tan^{-1}(2x)+\tan^{-1}(3x)=\dfrac{\pi}{4}\right\}$.
Apply the addition formula: $\tan^{-1}(2x)+\tan^{-1}(3x)=\tan^{-1}\!\left(\dfrac{5x}{1-6x^2}\right)$ when $6x^2<1$.
Setting equal to $\pi/4$: $\dfrac{5x}{1-6x^2}=1 \Rightarrow 5x=1-6x^2 \Rightarrow 6x^2+5x-1=0$
$x=\dfrac{-5\pm\sqrt{25+24}}{12}=\dfrac{-5\pm7}{12}$
$x=\dfrac{2}{12}=\dfrac{1}{6}$ or $x=\dfrac{-12}{12}=-1$.
Since $x\geq0$, only $x=\dfrac{1}{6}$ is valid. Check: $6\cdot(1/6)^2=1/6<1$ ✓. So $A=\{1/6\}$ — a singleton.
Using mass ($M$), length ($L$), time ($T$), and electric current ($A$) as fundamental quantities, the dimensions of magnetic permeability are:
Magnetic permeability $\mu_0$ appears in: $F = \dfrac{\mu_0}{4\pi}\dfrac{I_1 I_2 l}{r}$
$[\mu_0] = \dfrac{[F][r]}{[I]^2[l]} = \dfrac{MLT^{-2} \cdot L}{A^2 \cdot L} = MLT^{-2}A^{-2}$
Alternatively, from $B = \mu_0 H$: $[H] = AL^{-1}$, $[B] = MT^{-2}A^{-1}$, so $[\mu_0] = \frac{[B]}{[H]} = \frac{MT^{-2}A^{-1}}{AL^{-1}} = MLT^{-2}A^{-2}$
Find all values of $\lambda$ for which the system $2x_1-2x_2+x_3=\lambda x_1$, $\ 2x_1-3x_2+2x_3=\lambda x_2$, $\ {-x_1+2x_2=\lambda x_3}$ has a non-trivial solution.
Rewrite as $(A-\lambda I)X=0$. For non-trivial solutions, $\det(A-\lambda I)=0$.
$A = \begin{pmatrix}2&-2&1\\2&-3&2\\-1&2&0\end{pmatrix}$
The characteristic equation simplifies to $\lambda^3 + \lambda^2 - 5\lambda + 3 = 0$, which factors as $(\lambda-1)^2(\lambda+3)=0$.
Roots: $\lambda=1$ (repeated) and $\lambda=-3$. So the set contains exactly two elements: $\{1, -3\}$.
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