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Let $I_n=\displaystyle\int\tan^n x\,dx$ for $n>1$. If $I_4+I_6 = a\tan^5x + bx^5 + C$, find the ordered pair $(a,b)$.
Use the reduction: $I_n+I_{n-2}=\displaystyle\int\tan^{n-2}x\cdot\tan^2x\,dx+\int\tan^{n-2}x\,dx=\int\tan^{n-2}x\sec^2x\,dx=\dfrac{\tan^{n-1}x}{n-1}+C$
$I_4+I_6$: set $n=5$: $I_4+I_6=\dfrac{\tan^5x}{5}+C$
Comparing with $a\tan^5x+bx^5+C$: $a=\dfrac{1}{5}$, $b=0$.
$(a,b) = \left(\dfrac{1}{5},0\right)$
A compound with molecular mass 180 is acylated with CH$_3$COCl to yield a compound with molecular mass 390. How many amino ($-\text{NH}_2$) groups does the original compound contain?
Acylation: each $-\text{NH}_2$ reacts with CH$_3$COCl to introduce a $-\text{COCH}_3$ group ($M = 42$) and release HCl.
Net mass added per amino group $= 42 - 1 = 42$ (the $-H$ is replaced by $-\text{COCH}_3$)
Wait — each NH$_2$ + CH$_3$COCl → NH$\cdot$COCH$_3$ + HCl. Mass change per group $= 42 - 1 = +41$... more precisely: $-NH_2$ (16) becomes $-NHCOCH_3$ (58): increase = 42.
Increase in mass $= 390 - 180 = 210$
Number of amino groups $= \dfrac{210}{42} = \mathbf{5}$
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