Surds And Indices - Study Mode

[#161] If $$a = frac{{sqrt 5 + 1}}{{sqrt 5 - 1}}$$ xa0 and $$b{ ext{ = }}frac{{sqrt 5 - 1}}{{sqrt 5 + 1}}$$ xa0 then the value of $$left( {frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}}
ight)$$ xa0 xa0is = ?
Correct Answer

(B) $$frac{4}{3}$$

Explanation

Solution: $$eqalign{
& a + b = frac{{sqrt 5 + 1}}{{sqrt 5 - 1}} + frac{{sqrt 5 - 1}}{{sqrt 5 + 1}} cr
& = frac{{{{left( {sqrt 5 + 1}
ight)}^2} + {{left( {sqrt 5 - 1}
ight)}^2}}}{{left( {sqrt 5 - 1}
ight)left( {sqrt 5 + 1}
ight)}} cr
& = frac{{2left[ {{{left( {sqrt 5 }
ight)}^2} + 1}
ight]}}{{5 - 1}} cr
& = frac{{2left( {5 + 1}
ight)}}{4} cr
& = 3 cr
& a.b = frac{{sqrt 5 + 1}}{{sqrt 5 - 1}} imes frac{{sqrt 5 - 1}}{{sqrt 5 + 1}} = 1 cr
& { ext{Put value in expression}} cr
& frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}} cr
& = frac{{{{left( {a + b}
ight)}^2} - ab}}{{{{left( {a + b}
ight)}^2} - 3ab}} cr
& = frac{{{3^2} - 1}}{{{3^2} - 3}} cr
& = frac{{9 - 1}}{{9 - 3}} cr
& = frac{4}{3} cr} $$

[#162] $$2 + frac{6}{{sqrt 3 }} + frac{1}{{2 + sqrt 3 }} + frac{1}{{sqrt 3 - 2}}$$ xa0 xa0 equals to
Correct Answer

(D) 2

Explanation

Solution: $$eqalign{
& 2 + frac{6}{{sqrt 3 }} + frac{1}{{2 + sqrt 3 }} + frac{1}{{sqrt 3 - 2}} cr
& Rightarrow 2 + frac{{2 imes 3sqrt 3 }}{{sqrt 3 imes sqrt 3 }} + frac{1}{{2 + sqrt 3 }} - frac{1}{{2 - sqrt 3 }} cr
& Rightarrow 2 + 2sqrt 3 + left( {frac{{left( {2 - sqrt 3 }
ight) - left( {2 + sqrt 3 }
ight)}}{{left( {2 + sqrt 3 }
ight)left( {2 - sqrt 3 }
ight)}}}
ight) cr
& Rightarrow 2 + 2sqrt 3 + left( {frac{{2 - sqrt 3 - 2 - sqrt 3 }}{{4 - 3}}}
ight) cr
& Rightarrow 2 + 2sqrt 3 - 2sqrt 3 cr
& Rightarrow 2 cr} $$

[#163] The simplified value of the following expression is: $$frac{1}{{sqrt {11 - 2sqrt {30} } }} - frac{3}{{sqrt {7 - 2sqrt {10} } }} - frac{4}{{sqrt {8 + 4sqrt 3 } }}$$
Correct Answer

(A) 0

Explanation

Solution: $$eqalign{
& frac{1}{{sqrt {11 - 2sqrt {30} } }} cr
& = frac{1}{{sqrt {6 + 5 - 2 imes sqrt 6 imes sqrt 5 } }} cr
& = frac{1}{{sqrt {{{left( {sqrt 6 }
ight)}^2} + {{left( {sqrt 5 }
ight)}^2} - 2 imes sqrt 6 imes sqrt 5 } }} cr
& = frac{1}{{sqrt {{{left( {sqrt 6 - sqrt 5 }
ight)}^2}} }} cr
& = frac{1}{{sqrt 6 - sqrt 5 }} cr
& = frac{{left( {sqrt 6 + sqrt 5 }
ight)}}{{left( {sqrt 6 - sqrt 5 }
ight)left( {sqrt 6 + sqrt 5 }
ight)}} cr
& = sqrt 6 + sqrt 5 cr
& frac{3}{{sqrt {7 - 2sqrt {10} } }} cr
& = frac{3}{{sqrt {5 + 2 - 2 imes sqrt 5 imes sqrt 2 } }} cr
& = frac{3}{{sqrt 5 - sqrt 2 }} cr
& = frac{{3 imes left( {sqrt 5 + sqrt 2 }
ight)}}{{left( {sqrt 5 - sqrt 2 }
ight)left( {sqrt 5 + sqrt 2 }
ight)}} cr
& = frac{{3left( {sqrt 5 + sqrt 2 }
ight)}}{{5 - 2}} cr
& = sqrt 5 + sqrt 2 cr
& frac{4}{{sqrt {8 + 4sqrt 3 } }} cr
& = frac{4}{{sqrt {8 + 2sqrt {12} } }} cr
& = frac{4}{{sqrt {6 + 2 + 2 imes sqrt 6 imes sqrt 2 } }} cr
& = frac{4}{{sqrt {{{left( {sqrt 6 + sqrt 2 }
ight)}^2}} }} cr
& = frac{{4 imes left( {sqrt 6 - sqrt 2 }
ight)}}{{left( {sqrt 6 + sqrt 2 }
ight)left( {sqrt 6 - sqrt 2 }
ight)}} cr
& = frac{{4left( {sqrt 6 - sqrt 2 }
ight)}}{{6 - 2}} cr
& = sqrt 6 - sqrt 2 cr
& herefore { ext{Expression}} cr
& = left( {sqrt 6 + sqrt 5 }
ight) - left( {sqrt 5 + sqrt 2 }
ight) - left( {sqrt 6 - sqrt 2 }
ight) cr
& = sqrt 6 + sqrt 5 - sqrt 5 - sqrt 2 - sqrt 6 + sqrt 2 cr
& = 0 cr} $$

[#164] Which value among $$
oot 4 of 7 ,,
oot 3 of {11} $$ xa0 and $$
oot {12} of {1257} $$ xa0is the largest?
Correct Answer

(A) $$ oot 3 of {11} $$

Explanation

Solution: Given, $$
oot 4 of 7 ,,
oot 3 of {11} $$ xa0 and $$
oot {12} of {1257} $$ Take LCM of 4, 3 and 12 = 12 $${left( 7
ight)^{frac{1}{4}}},,{left( {11}
ight)^{frac{1}{3}}},,{left( {1257}
ight)^{frac{1}{{12}}}}$$ Multiplying the power by 12 = 7 3 , 11 4 , 1257 i.e., 343, 14641, 1257 Therefore, from above greater one is $$
oot 3 of {11} $$

[#165] If $$frac{{4 + 3sqrt 3 }}{{sqrt {7 + 4sqrt 3 } }} = { ext{A}} + sqrt { ext{B}} ,$$ xa0 xa0 then B - A is
Correct Answer

(C) 13

Explanation

Solution: $$eqalign{
& frac{{4 + 3sqrt 3 }}{{sqrt {7 + 4sqrt 3 } }} = { ext{A}} + sqrt { ext{B}} cr
& Rightarrow sqrt {7 + 4sqrt 3 } cr
& Rightarrow sqrt {{{left( 2
ight)}^2} + {{left( {sqrt 3 }
ight)}^2} + 2 imes 2sqrt 3 } cr
& Rightarrow sqrt {{{left( {2 + sqrt 3 }
ight)}^2}} cr
& Rightarrow left( {2 + sqrt 3 }
ight) cr
& Rightarrow frac{{4 + 3sqrt 3 }}{{2 + sqrt 3 }} = { ext{A}} + sqrt { ext{B}} cr
& Rightarrow frac{{4 + 3sqrt 3 }}{{2 + sqrt 3 }} imes frac{{2 - sqrt 3 }}{{2 - sqrt 3 }} = { ext{A}} + sqrt { ext{B}} cr
& Rightarrow frac{{left( {4 + 3sqrt 3 }
ight)left( {2 - sqrt 3 }
ight)}}{{4 - 3}} = { ext{A}} + sqrt { ext{B}} cr
& Rightarrow 8 - 4sqrt 3 + 6sqrt 3 - 9 = { ext{A}} + sqrt { ext{B}} cr
& Rightarrow 2sqrt 3 - 1 = { ext{A}} + sqrt { ext{B}} cr
& { ext{A}} = - 1{ ext{ and }}sqrt { ext{B}} = 2sqrt 3 cr
& { ext{B}} = 2sqrt 3 imes 2sqrt 3 = 12 cr
& { ext{B }} - { ext{ A}} = 12 - left( { - 1}
ight) = 13 cr} $$