12 multiple-choice questions, progressively harder.
For which value of kkk does kx+1=3x+kkx + 1 = 3x + kkx+1=3x+k have no solution?
Solution
Correct answer: D
Reduce the equation.
kx−3x=k−1 ⇒ (k−3)x=k−1kx - 3x = k - 1 \;\Rightarrow\; (k - 3)x = k - 1kx−3x=k−1⇒(k−3)x=k−1
No solution needs the coefficient zero, the constant nonzero. At k=3k = 3k=3 the coefficient is 000 and the constant is k−1=2≠0k - 1 = 2 \neq 0k−1=2=0.
For which value of kkk does kx+1=3x+kkx + 1 = 3x + kkx+1=3x+k have infinitely many solutions?
Correct answer: C
The reduced form is (k−3)x=k−1(k - 3)x = k - 1(k−3)x=k−1. Infinitely many need both parts to vanish together.
k−3=0 and k−1=0 ⇒ k=3 and k=1k - 3 = 0 \text{ and } k - 1 = 0 \;\Rightarrow\; k = 3 \text{ and } k = 1k−3=0 and k−1=0⇒k=3 and k=1
No single kkk satisfies both, so infinitely many solutions never occur.
For which value of kkk does kx+10=5x+2kkx + 10 = 5x + 2kkx+10=5x+2k have infinitely many solutions?
Correct answer: B
kx−5x=2k−10 ⇒ (k−5)x=2(k−5)kx - 5x = 2k - 10 \;\Rightarrow\; (k - 5)x = 2(k - 5)kx−5x=2k−10⇒(k−5)x=2(k−5)
Both the coefficient k−5k - 5k−5 and the constant 2(k−5)2(k - 5)2(k−5) vanish at k=5k = 5k=5, giving 0⋅x=00 \cdot x = 00⋅x=0.
For k≠5k \neq 5k=5, what is the solution xxx of kx+10=5x+2kkx + 10 = 5x + 2kkx+10=5x+2k?
The reduced form is (k−5)x=2(k−5)(k - 5)x = 2(k - 5)(k−5)x=2(k−5). For k≠5k \neq 5k=5 cancel the common factor.
x=2(k−5)k−5=2x = \frac{2(k - 5)}{k - 5} = 2x=k−52(k−5)=2
The solution is x=2x = 2x=2 for every k≠5k \neq 5k=5.
Solve mx+n=px+qmx + n = px + qmx+n=px+q for xxx, with the correct condition.
Collect the xxx terms, factor, and divide by the factor m−pm - pm−p.
(m−p)x=q−n ⇒ x=q−nm−p,m≠p(m - p)x = q - n \;\Rightarrow\; x = \frac{q - n}{m - p}, \quad m \neq p(m−p)x=q−n⇒x=m−pq−n,m=p
The condition is on the difference m−pm - pm−p, not on mmm and ppp separately.
Solve the thin-lens equation 1f=1u+1v\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}f1=u1+v1 for uuu.
Correct answer: A
Isolate 1u\frac{1}{u}u1, combine over a common denominator, then take reciprocals.
1u=1f−1v=v−ffv ⇒ u=fvv−f\frac{1}{u} = \frac{1}{f} - \frac{1}{v} = \frac{v - f}{fv} \;\Rightarrow\; u = \frac{fv}{v - f}u1=f1−v1=fvv−f⇒u=v−ffv
The rearrangement u=fvv−fu = \dfrac{fv}{v - f}u=v−ffv requires which condition?
The divisor is the factor v−fv - fv−f.
v−f≠0 ⇒ v≠fv - f \neq 0 \;\Rightarrow\; v \neq fv−f=0⇒v=f
The rearrangement R=E−IrIR = \dfrac{E - Ir}{I}R=IE−Ir requires which condition?
The divisor is III.
R=E−IrI needs I≠0R = \frac{E - Ir}{I} \text{ needs } I \neq 0R=IE−Ir needs I=0
A zero current leaves RRR undetermined by this relation.
Solve ax=bax = bax=b for bbb (so bbb is now the unknown).
With bbb chosen as the unknown, aaa and xxx are parameters, and bbb already stands alone after reading the equation right to left.
ax=b ⇒ b=axax = b \;\Rightarrow\; b = axax=b⇒b=ax
No division is needed at all.
The family (k−6)x=k−6(k - 6)x = k - 6(k−6)x=k−6 has the solution x=1x = 1x=1 for every kkk except which value?
For k≠6k \neq 6k=6 divide both sides by k−6k - 6k−6 to get x=1x = 1x=1. At k=6k = 6k=6 the equation collapses.
0⋅x=00 \cdot x = 00⋅x=0
There every real number is a solution, so x=1x = 1x=1 is no longer the single answer.
Solve the trapezoid-area formula A=h(a+b)2A = \dfrac{h(a + b)}{2}A=2h(a+b) for bbb, with the correct condition.
Multiply by 222, divide by hhh, then subtract aaa.
2A=h(a+b) ⇒ 2Ah=a+b ⇒ b=2Ah−a2A = h(a + b) \;\Rightarrow\; \frac{2A}{h} = a + b \;\Rightarrow\; b = \frac{2A}{h} - a2A=h(a+b)⇒h2A=a+b⇒b=h2A−a
The division is by hhh, so the condition is h≠0h \neq 0h=0.
Which statement about the equation ax=bax = bax=b is correct?
The roles of unknown and parameter are assigned by the question, not carried by the symbols.
solve for x: x=ba;solve for a: a=bx\text{solve for } x: \; x = \frac{b}{a}; \qquad \text{solve for } a: \; a = \frac{b}{x}solve for x:x=ab;solve for a:a=xb
The same equation yields two different problems depending on which letter you choose to isolate.
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