| (i) \[\frac{{{a}^{2}}-{{b}^{2}}}{a(a-b)}-\frac{a{{b}^{2}}+{{a}^{2}}b}{a{{b}^{2}}}\] is equal to P . |
| (ii) \[\frac{64{{y}^{4}}+8{{y}^{3}}}{4{{y}^{3}}}\] is equal to Q . |
| (iii) When we divide \[(38{{a}^{3}}{{b}^{3}}{{c}^{2}}-19{{a}^{4}}{{b}^{2}}c)\] by \[19{{a}^{2}}bc\], the result is \[ka{{b}^{2}}c-{{a}^{2}}b\]. Then \[k=\underline{\,\,\,R\,\,\,}\]. |
A)
P Q R \[\frac{(a+b)(b-a)}{ab}\] \[3(8y+1)\] 1
B)
P Q R \[\frac{(a+b\,)(b-a)}{ab}\] \[3(8y+1)\] 1
C)
P Q R \[\frac{(a+b)(a-b)}{ab}\] \[2(8y+1)\] 1
D)
P Q R \[\frac{(a+b)(b-a)}{ab}\] \[2(8y+1)\] 2
Correct Answer: D
Solution :
(i) Reduce to lowest terms, \[\frac{{{a}^{2}}-{{b}^{2}}}{a(a-b)}-\left[ \frac{a{{b}^{2}}+{{a}^{2}}b}{a{{b}^{2}}} \right]\] \[=\frac{(a-b)(a+b)}{a(a-b)}-\left[ \frac{ab(b+a)}{(ab)b} \right]\] \[=\frac{a+b}{a}-\frac{(b+a)}{b}=\frac{(a+b)\times b-(a+b)\times a}{ab}\] \[=\frac{(a+b)(b-a)}{ab}\] (ii) \[\frac{64{{y}^{4}}+8{{y}^{3}}}{4{{y}^{3}}}=\frac{4{{y}^{3}}(16y+2)}{4{{y}^{3}}}\] \[=16y+2=2(8y+1)\] (iii) \[\frac{38{{a}^{3}}{{b}^{3}}{{c}^{2}}-19{{a}^{4}}{{b}^{2}}c}{19{{a}^{2}}bc}\] \[=\frac{38{{a}^{3}}{{b}^{3}}{{c}^{2}}}{19{{a}^{2}}bc}-\frac{19{{a}^{4}}{{b}^{2}}c}{19{{a}^{2}}bc}\] \[=2a{{b}^{2}}c-{{a}^{2}}b=ka{{b}^{2}}c-{{a}^{2}}b\] \[\therefore k=2\]
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