Maths M.C.Q (1 Marks)

$ = {{\log 7} \over {2\log 7}} + {{\log 4} \over {2\log 7}} = {1 \over 2} + {1 \over 2}{\log _7}4$

$ = {1 \over 2} + {1 \over 2}.2{\log _7}2 = {1 \over 2} + {\log _7}2 = {1 \over 2} + m = {{1 + 2m} \over 2}$

$ = {1 \over 2}{\log _e}(ab) = {\log _e}\sqrt {ab} $

$ \Rightarrow {{a + b} \over 2} = \sqrt {ab} \,\, \Rightarrow a + b = 2\sqrt {ab} $

==> $\,a = b$.

$ = {3^{4{{\log }_3}5}} + {3^{3.{1 \over 2}{{\log }_3}36}} + {3^{4{{\log }_9}7}}$

$ = {3^{{{\log }_3}{5^4}}} + {3^{{{\log }_3}{{36}^{3/2}}}} + {3^{{{\log }_3}{7^{4/2}}}}$

$ = {5^4} + {36^{3/2}} + {7^2} = 890$.

$ab = {1 \over 2}(1 + {\log _2}3) \Rightarrow 2ab - 1 = {\log _2}3$

$\therefore {\log _3}2 = {1 \over {2ab - 1}}$.

$ \Rightarrow $ ${\log _x}5 = {1 \over {{{\log }_x}5}}$ $\Rightarrow $ $1 = - 2B = 0$

$ \Rightarrow $ ${\log _x}5 = \pm 1$

$ \Rightarrow $ ${x^{ \pm \,1}} = 5$ $ \Rightarrow $ $x = 5,\,{1 \over 5}$.

$ \Rightarrow $${(a + 2b)^2} = 16ab$

$ \Rightarrow $$2\log (a + 2b) = \log 16 + \log a + \log b$

$\therefore $ $\log (a + 2b) = {1 \over 2}[\log a + \log b + 4\log 2]$

$ = {\log _2}{\log _2}{\log _4}{4^4} + 2 \times {1 \over {(1/2)}}{\log _2}2$

$ = {\log _2}{\log _2}4 + 4 = {\log _2}{\log _2}{2^2} + 4$

$ = {\log _2}2 + 4 = 1 + 4 = 5$.

$\therefore {(1 + x)^{ - 1}} = {\log _{abc}}a$

$\therefore {(1 + x)^{ - 1}} + {(1 + y)^{ - 1}} + {(1 + z)^{ - 1}} = {\log _{abc}}a + {\log _{abc}}b + {\log _{abc}}c$

$ = {\log _{abc}}abc = 1$.

==> $x = {{\log b} \over {\log a}} = {\log _a}b$

Similarly $y = {\log _b}c,\,z = {\log _c}a$

$\therefore xyz = {\log _a}b.{\log _b}c.{\log _c}a = 1$.

$ = 12 \times 0.47712 + 8 \times 0.30103$

$ = 5.72544 + 2.40824 = 8.13368$

$\therefore$ Number of digits in $y$ $= 8 + 1 = 9.$

$= {\log _a}2.{{n(n + 1)} \over 2} = {{n(n + 1)} \over 2}{\log _a}2$.

$ \Rightarrow $ ${({x^2} + 5 + x)^{1/2}} = 5$

$ \Rightarrow $ $({x^2} + x + 5) = 25$$ \Rightarrow $ ${x^2} + x - 20 = 0$

$ \Rightarrow $ $ (x - 4)\,(x + 5) = 0$ $ \Rightarrow $ $ x = \,4,\, - 5$ ==> $x = 4$.

${\log _6}16 = {{\log 16} \over {\log 6}} = {{4\log 2} \over {\log 2 + \log 3}}$

$ = {{4\log 2} \over {\log 2 + {{2a\log 2} \over {3 - a}}}} = {{4(3 - a)} \over {3 - a + 2a}} = 4.{{3 - a} \over {3 + a}}$.

$ \Rightarrow $$\log x = k(b - c),\,\log y = k(c - a),\,\log z = k(a - b)$

$ \Rightarrow $$x = {e^{k(b - c)}},\,y = {e^{k(c - a)}},\,z = {e^{k(a - b)}}$

$\therefore xyz = {e^{k(b - c) + k(c - a) + k(a - b)}} = {e^0} = 1$

${x^a}{y^b}{z^c} = {e^{k(b - c)a + k(c - a)b + k(a - b)c}} = {e^0} = 1 = xyz$

${x^{b + c}}{y^{c + a}}{z^{a + b}} = {e^{k({b^2} - {c^2}) + k({c^2} - {a^2}) + k({a^2} - {b^2})}} = {e^0} = 1$.

$ \Rightarrow $ ${\log _\pi }3 + {\log _\pi }4 > x$$ \Rightarrow $ ${\log _\pi }12 > x$

${\pi ^2} < 12 < {\pi ^3}$

$\therefore 12 > {\pi ^2}$;

$\therefore {\log _\pi }12 > {\log _\pi }{\pi ^2}$

i.e., ${\log _\pi }12 > 2$;

$\therefore x$ will be $2$.

$ = {\log _2}.{\log _3}....{\log _{98}}^{{{98}^{{{97}^{{.^{{.^{{.^{{2^1}}}}}}}}}}}}$

$ = {\log _2}\,\,2'{\log _3}3 = {\log _2}2 = 1$.

$={({x^{l - m}})^{({l^2} + lm + {m^2})}}{({x^{m - n}})^{{m^2} + nm + {n^2}}}$${({x^{n - l}})^{{n^2} + nl + {l^2}}}$

$={x^{{l^3} - {m^3}}}.{x^{{m^3} - {n^3}}}.{x^{{n^3} - {l^3}}}$=${x^{{l^3} - {m^3} + {m^3} - {n^3} + {n^3} - {l^3}}} = {x^0}=1$

Then $xyz = {{{k^3}} \over 6} = 288$, So $k = 12$

.$\therefore x = 12,y = 6,z = 4$ Therefore, ${1 \over {2x}} + {1 \over {4y}} + {1 \over {8z}} = {{11} \over {96}}$

$L.C.M.$ of $3, 4, 6$ is $12$

$\therefore \sqrt[3]{9} = {9^{1/3}} = {({9^4})^{1/12}} = {(6561)^{1/12}}$

$\root 4 \of {11} = {(11)^{1/4}}{({11^3})^{1/12}} = {(1331)^{1/12}}$,

$\root 6 \of {17} = {(17)^{1/6}} = {({17^2})^{1/2}} = {(289)^{1/12}}$

Hence, $\root 3 \of 9 $ is the greatest number.

$ = {{15} \over {\sqrt {10} + \sqrt {20} + \sqrt {40} - \sqrt 5 - \sqrt {80} }}$

$ = {{15} \over {\sqrt {10} + 2\sqrt 5 + 2\sqrt {10} - \sqrt 5 - 4\sqrt 5 }}$

$ = {{15} \over {3\sqrt {10} - 3\sqrt 5 }} = {5 \over {\sqrt {10} - \sqrt 5 }}\,.\,{{\sqrt {10} + \sqrt 5 } \over {\sqrt {10} + \sqrt 5 }}$

$ = \sqrt {10} + \sqrt 5 = \sqrt 5 (\sqrt 2 + 1)$

$3 + \sqrt 5 = \,x + y + 2\sqrt {xy} $. Obviously $x + y = 3$

and $4xy = 5$. So ${(x - y)^2} = 9 - 5 = 4$ or $(x - y) = 2$

After solving $x = {5 \over 2},y = {1 \over 2}$.

Hence, $\sqrt {3 + \sqrt 5 } = \sqrt {{5 \over 2}} + \sqrt {{1 \over 2}} = {{\sqrt 5 + 1} \over {\sqrt 2 }}$.

$\therefore \sqrt[4]{{(17 + 12\sqrt 2 )}} = \sqrt {(3 + 2\sqrt 2 )}  = \sqrt 2  + 1$

Squaring both sides, we get, $ - 2\sqrt {({x^2} - 1)} = 2x - 1$

Squaring again, we get, $x = {5 \over 4},$ which does not satisfy eq. $(i).$ 

Hence, there is no solution of the given equation.

$ \Rightarrow \,\,{a^{x + y + z}} = {(x + y + z)^{x + y + z}}$$ \Rightarrow $ $x + y + z = a$

Now, ${a^x} = {(x + y + z)^y} = {a^y}$$ \Rightarrow $ $x = y$, similarly $y = z$

$\therefore x = y = z = {a \over 3}$.

$\therefore {a^x} = {b^y} = {c^z} = abc = k\,({\rm{say}})$

$ \Rightarrow $$a = {k^{1/x}} \Rightarrow {1 \over x} = {\log _k}a$; $\sum\limits_{}^{} {{1 \over x} = {{\log }_k}a + {{\log }_k}b + {{\log }_k}c} = {\log _k}abc = {\log _{abc}}abc = 1$.

Now, ${\left( {{x \over y}} \right)^{x/y}} = {\left( {{x \over {{x^{y/x}}}}} \right)^{x/y}} = {\left( {{x^{1\, - \,{y \over x}}}} \right)^{x/y}}$

$={x^{(x/y) - 1}} = {x^{(x/y) - k}} \Rightarrow k = 1$.

==> ${x^{{x^{4/3}}}} = {\left( {{x^{4/3}}} \right)^x}$= ${x^{{x^{4/3}}}} = {x^{{4 \over 3}x}} \Rightarrow {x^{4/3}} = {4 \over 3}x$

$\Rightarrow {x^{\frac{4}{3} - 1}} = \frac{4}{3} \Rightarrow {x^{1/3}} = \frac{4}{3}$

 $\therefore x = {\left( {{4 \over 3}} \right)^3} = {{64} \over {27}}$

Also $x=1$ is an obvious solution .

$ \Rightarrow $$x\ln a = y\ln b = xy\ln (ab) = k\,{\rm{(say)}}$

$\ln a = {k \over x},\,\,\ln b = {k \over y}$

$\ln (a\,b) = {k \over {xy}}$$ \Rightarrow $ $\ln a + \ln b = {k \over {xy}}$

$ \Rightarrow$ ${k \over x} + {k \over y} = {k \over {xy}}$

$ \Rightarrow $ ${1 \over x} + {1 \over y} = {1 \over {xy}}$ $ \Rightarrow $ ${{x + y} \over {xy}} = {1 \over {xy}}$;

$\therefore x + y = 1$.

$ \Rightarrow $${x^3} = 2 - {2^{ - 1}} - {3.2^{1/3}}{.2^{ - 1/3}}({2^{1/3}} - {2^{ - 1/3}})$

$ \Rightarrow $${x^3} = 2 - {1 \over 2} - 3x$ ==> ${x^3} + 3x = {3 \over 2}$

$\therefore 2{x^3} + 6x = 3$.

==>${5^{x - 1}} + {5^{3 - x}} = 26$$ \Rightarrow $${5^{x - 1}} + 25\,.\,{5^{ - (x - 1)}} - 26 = 0$

==>${5^{2(x - 1)}} - 26.\,{5^{(x - 1)}} + 25 = 0$

==>${5^{2(x - 1)}} - {5^{x - 1}} - {25.5^{x - 1}} + 25 = 0$

==>${5^{x - 1}}({5^{x - 1}} - 1) - 25({5^{x - 1}} - 1) = 0$

==>$({5^{x - 1}} - 25)({5^{x - 1}} - 1) = 0$ ==>$({5^{x - 1}} - {5^2})\,({5^{x - 1}} - {5^0}) = 0$

==> ${5^{x - 1}} = {5^2}$or ${5^{x - 1}} = {5^0}$$ \Rightarrow $$x = 3,\,1$.

$ \Rightarrow $ ${3^{2x - 3}} = {2^{{{2x - 3} \over 2}}}$ $ \Rightarrow $ ${2^{{{2x - 3} \over 2}}} = {\left( {{3^{{{2x - 3} \over 2}}}} \right)^2}$
 $ \Rightarrow $$2x - 3 = 0$,

$\therefore x = {3 \over 2}$.

$ = {{12\,(3 - 2\sqrt 2 - \sqrt 5 )} \over {{{(3 - 2\sqrt 2 )}^2} - 5}} = {{12\,(3 - 2\sqrt 2 - \sqrt 5 )} \over {17 - 12\sqrt 2 - 5}}$

$ = {{(3 - 2\sqrt 2 - \sqrt 5 )} \over {1 - \sqrt 2 }} = {{(\sqrt 5 + 2\sqrt 2 - 3)\,(\sqrt 2 + 1)} \over {(\sqrt 2 - 1)\,(\sqrt 2 + 1)}}$

$ = {{\sqrt {10} + 4 - 3\sqrt 2 + \sqrt 5 + 2\sqrt 2 - 3} \over {2 - 1}} = 1 + \sqrt 5 + \sqrt {10} - \sqrt 2 $.

$= {{\sqrt 5 + (3 - \sqrt {5)} } \over {\sqrt 7 + (5 - \sqrt 7 )}} = {3 \over 5}$, which is rational.

$= \sqrt {12 - \sqrt {{{(6 + 4\sqrt 2 )}^2}} } $= $\sqrt {12 - 6 - 4\sqrt 2 } = \sqrt {6 - 4\sqrt 2 } $

$= \sqrt {{2^2} + {{(\sqrt 2 )}^2} - 2\,.\,2\sqrt 2 } = 2 - \sqrt 2 $.

$\therefore \sqrt {\sqrt {50} + \sqrt {48} } = {2^{1/4}}(\sqrt 3 + \sqrt 2 )$.

$ = {1 \over {\sqrt 2 }}\,[(1 + \sqrt 5 ) - (1 + \sqrt 3 )] = {1 \over {\sqrt 2 }}(\sqrt 5 - \sqrt 3 ) = \sqrt {{5 \over 2}} - \sqrt {{3 \over 2}} $.

$ = {5^{1/4}}\sqrt {11 + 1 + 2\sqrt {11} } = {5^{1/4}}(\sqrt {11} + 1)$

$ \Rightarrow $$y = {4 \over x} = {4 \over {\sqrt 7 + \sqrt 3 }} = {{4(\sqrt 7 - \sqrt 3 )} \over {7 - 3}} = \sqrt 7 - \sqrt 3 $

${x^4} + {y^4} = {({x^2} + {y^2})^2} - 2{x^2}{y^2}$

$ = {[{(x + y)^2} - 2xy]^2} - 2{(xy)^2} = {[{(2\sqrt 7 )^2} - 8]^2} - 2\,.\,{4^2} = 368$.

$ \Rightarrow $ $x + 10 = 36 + x - 2 - 12\sqrt {x - 2} $

$ \Rightarrow $ $2 = \sqrt {x - 2} $ $ \Rightarrow $ $4 = x - 2$ $ \Rightarrow $ $x = - 6$

This value satisfies the given equation.

$\therefore x=6$.

= ${{(1 + \sqrt 2 + \sqrt 3 ) - 1} \over {(\sqrt 3 + \sqrt 2 )}}$= ${{\sqrt 3 + \sqrt 2 } \over {\sqrt 3 + \sqrt 2 }} = 1$.

==> ${(x + 1)^2} = A({x^2} + 1) + (Bx + C)\,x$

==> $A + B = 1$, $C = 2$,$A = 1$ ==> $B=0$

${\sin ^{ - 1}}\left( {\frac{A}{C}} \right) = {\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = 30^\circ  = \frac{\pi }{6}$

==>${x \over {(x - 1)\,{{({x^2} + 1)}^2}}} = {1 \over 4}\left[ {{1 \over {(x - 1)}} - {{x + 1} \over {{x^2} + 1}}} \right] + {{Ax + B} \over {{{({x^2} + 1)}^2}}}$

==> $4x = {({x^2} + 1)^2} - (x + 1)\,(x - 1)\,({x^2} + 1) + 4(Ax + B)\,(x - 1)$

==> $4A + 2 = 0$,$4B - 4A = 4$ ==> $A = {{ - 1} \over 2}$, $B = {1 \over 2}$

$\therefore y = {{Ax + B} \over {{{({x^2} + 1)}^2}}} = {1 \over 2}{{(1 - x)} \over {{{({x^2} + 1)}^2}}}$

Rewriting the denominators for expressions, we get

= ${{{{ - 4} \over 3}} \over {2\left( {1 + {x \over 2}} \right)}} + {{{{11} \over 3}} \over {1 - x}} = {{ - 2} \over 3}{\left( {1 + {x \over 2}} \right)^{ - 1}} + {{11} \over 3}{(1 - x)^{ - 1}}$

= ${{ - 2} \over 3}\left[ {1 - {x \over 2} + {{{x^2}} \over 4} - {{{x^3}} \over 8} + ...... + {{( - 1)}^n}{{{x^n}} \over {{2^n}}} + ......} \right]\,$

$ + {{11} \over 3}[1 + x + {x^2} + ....... + {x^n} + .....]$

The coefficient of ${x^n}$ in the given expression is

${{ - 2} \over 3}{( - 1)^n}{1 \over {{2^n}}} + {{11} \over 3}$.

For $x = 1$, $2 = 3A$ $ \Rightarrow $ $A = {2 \over 3}$

For $x = \omega ,\,2\omega = A(1 + \omega + {\omega ^2}) + B{\omega ^2} + (C - B)\,\omega - C$

$ \Rightarrow $ $2\omega = A.0 + B{\omega ^2} + (C - B)\,\omega - C$

$\omega = {{ - 1 + \sqrt 3 i} \over 2},\,\,{\omega ^2} = {{ - 1 - \sqrt 3 i} \over 2}$

$ \therefore - 1 + \sqrt 3 i = B\left( { - {1 \over 2} - {{\sqrt 3 } \over 2}i} \right) + (C - B)\,\left( { - {1 \over 2} + {{\sqrt 3 } \over 2}i} \right) - C$

$ \Rightarrow $$ - 1 + \sqrt 3 i = \left( { - {B \over 2} - {C \over 2} + {B \over 2} - C} \right) + {{i\sqrt 3 } \over 2}(C - 2B)$

$ \Rightarrow $$ - 1 = - {3 \over 2}C,\,\sqrt 3 = {{\sqrt 3 } \over 2}(C - 2B)$

$C = {2 \over 3},\,B = {{C - 2} \over 2} = - {2 \over 3}$

$\therefore A = C \ne B$ $ \Rightarrow $ $A \ne B \ne C$.

$(1 - x + {x^2} - {x^3} + {x^4} - ..)\,\left( {1 + {x \over 2} + {{\left( {{x \over 2}} \right)}^2} + {{\left( {{x \over 2}} \right)}^3} + ...} \right)$

Coefficient of ${x^4} = - {3 \over 2}\left[ { - 1.1 + 1.{1 \over 2} - 1.{1 \over 4} + 1.{1 \over 8}} \right]$

$ = - {3 \over 2}\,\left[ { - 1 + {1 \over 2} - {1 \over 4} + {1 \over 8}} \right] = {{15} \over {16}}$.

$\therefore {1 \over y} = {1 \over 2}{\log _5}17$

${1 \over x} = {\log _5}3 = {1 \over 2}{\log _5}9$.

Clearly, ${1 \over y} > {1 \over x}$; $\therefore x > y$

$ \Rightarrow $ ${{\log x} \over {y - z}} = {{\log y} \over {z - x}} = {{\log z} \over {x - y}} = k\,{\rm{(say)}}$

$ \Rightarrow $ $\log x = k(y - z),\,\log y = k(z - x),\,\log z = k(x - y)$

$\therefore \log x + \log y + \log z = 0$$ \Rightarrow $$x + y = 1$ $ \Rightarrow $$xyz = 1$.

$x\log x + y\log y + z\log z$

=$x.k.(y - z) + y.k.(z - x) + z.k(x - y) = 0$

$ \Rightarrow $$\log ({x^x}.{y^y}.{z^z}) = \log 1$

$\therefore x^xy^yz^z=1 $.

${1 \over 2} \le {\log _{0.1}}\,x \Rightarrow {\log _{0.1}}{(0.1)^{1/2}} \le {\log _{0.1}}x$

$ \Rightarrow $${(0.1)^{1/2}} \ge x$ $ \Rightarrow $$x \le {1 \over {\sqrt {10} }}$

${\log _{0.1}}x \le 2 \Rightarrow {\log _{0.1}}x \le {\log _{0.1}}{(0.1)^2}$

$x \ge {(0.1)^2} \Rightarrow x \ge {1 \over {100}}$, ${1 \over {100}} \le x \le {1 \over {\sqrt {10} }}$.

Hence, ${x_{{\rm{max}}}} = {1 \over {\sqrt {10} }},{x_{{\rm{min}}{\rm{.}}}} = {1 \over {100}}$.

= ${{\sqrt {11 + 2\sqrt {30} } } \over {\sqrt 1 }} - {{3\,(7 + 2\sqrt {10} )} \over {\sqrt 9 }} - {{2\sqrt {2 - \sqrt 3 } } \over {\sqrt 1 }}$

= $\sqrt {11 + 2\sqrt {30} } - \sqrt {7 + 2\sqrt {10} } - 2\sqrt {2 - \sqrt 3 } $

= $(\sqrt 6 + \sqrt 5 ) - (\sqrt 5 + \sqrt 2 ) - \sqrt {8 - 4\sqrt 3 } $

= $(\sqrt 6 - \sqrt 2 ) - (\sqrt 6 - \sqrt 2 ) = 0$.