Maths M.C.Q (1 Marks)

For log to be defined, ${{x + 2} \over x} > 0$ $ \Rightarrow $$x > 0$ or $x < - 2$

Now from $(i),$ ${\log _{0.2}}{{x + 2} \over x} \le {\log _{0.2}}0.2$

$ \Rightarrow $${{x + 2} \over x} \ge 0.2$ …..$(ii)$

Case $(i)$ $x > 0$

From $(ii),$ $x + 2 \ge 0.2x$

$ \Rightarrow $ $0.8x \ge - 2$

$ \Rightarrow $$x \ge - {5 \over 2}$.

Case $(ii)$ $x < - 2$

From $(ii),$ $x + 2 \le 0.2x$$ \Rightarrow $$0.8x \le - 2$$ \Rightarrow $$x \le - {5 \over 2}$

$ \Rightarrow $$x \in (0,\,\infty )\, \cup \,\left( { - \infty ,\, - {5 \over 2}} \right]$; 

$\therefore x \in \left( { - \infty ,\, - {5 \over 2}} \right] \cup (0,\,\infty )$.

$= {{7\,.\,({2^{1/4}} - 1)} \over {{2^{3/4}} - 1}} = A + B\,.\,{2^{1/4}} + C.\,{2^{1/2}} + D{.2^{3/4}}$

==> $7\,.\,{2^{1/4}} - 7 = (A - D)\,{2^{3/4}} + (2B - A) + (2C - B){.2^{1/4}}$$ + (2D - C){2^{1/2}}$

==> $(2B - A + 7) + (A - D){2^{3/4}} + (2C - B - 7){2^{1/4}}$ $+ (2D - C){2^{1/2}} = 0$

==> $2B - A + 7 = A - D = 2C - B - 7 = 2D - C = 0$

==> $A = D = 1,\,B = - 3,\,C = 2$.

==> ${3^{2x}} + {1 \over 3}{.3^{2x}} = {2.2^{x + {1 \over 2}}} + {2^{x + {1 \over 2} - 2}}$

==> $4\,.\,{3^{2x - 1}} = 3.\,{2^{x + {1 \over 2}}}$ ==> ${3^{2x - 2}} = {2^{x + {1 \over 2} - 2}}$

==> ${3^{2x - 2}} = {2^{x - {3 \over 2}}}$ ==> ${\left( {{9 \over 2}} \right)^{x - 1}} = {2^{ - 1/2}}$

==> $(x - 1)\,{\log _{9/2}}9/2 = - {1 \over 2}{\log _{9/2}}2$

==> $x - 1 = - {1 \over 2}{\log _{9/2}}2$

==> $x = 1 - {\log _{9/2}}\sqrt 2 = {\log _{9/2}}9/2 - {\log _{9/2}}\sqrt 2 $

==> $x = {\log _{9/2}}(9/2\sqrt 2 )$;

$\therefore x = {\log _{9/2}}(9/\sqrt 8 )$.

$6 + \sqrt {35} = y$, then $6 - \sqrt {35} = {1 \over y}$

$\therefore $ Given expression = ${{{x^{3/2}} + {1 \over {{x^{3/2}}}}} \over {{y^{3/2}} - {1 \over {{y^{3/2}}}}}} = {{{x^3} + 1} \over {{y^3} - 1}}.{\left( {{y \over x}} \right)^{3/2}}$

$ = {{{{(4 + \sqrt {15} )}^3} + 1} \over {{{(6 + \sqrt {35} )}^3} - 1}}\,.\,{\left( {{{6 + \sqrt {35} } \over {4 + \sqrt {15} }}} \right)^{3/2}}$

$ = {{(4 + \sqrt {15} + 1)\,\{ {{(4 + \sqrt {15} )}^2} - (4 + \sqrt {15} ) + 1\} } \over {(6 + \sqrt {35} - 1)\,\{ {{(6 + \sqrt {35} )}^2} + (6 + \sqrt {35} ) + 1\} }} \times {\left( {{{6 + \sqrt {35} } \over {4 + \sqrt {15} }}} \right)^{3/2}}$

$ = {{5 + \sqrt {15} } \over {5 + \sqrt {35} }}.\,{{\{ 31 + 8\sqrt {15} - 4 - \sqrt {15} + 1\} } \over {\{ 71 + 12\sqrt {35} + 6 + \sqrt {35} + 1\} }}\,.\,{\left( {{{6 + \sqrt {35} } \over {4 + \sqrt {15} }}} \right)^{3/2}}$

$ = {{\sqrt 5 + \sqrt 3 } \over {\sqrt 5 + \sqrt 7 }} \times \,{{28 + 7\sqrt {15} } \over {78 + 13\sqrt {35} }}\,{\left( {{{6 + \sqrt {35} } \over {4 + \sqrt {15} }}} \right)^{3/2}}$

$ = {{\sqrt 5 + \sqrt 3 } \over {\sqrt 5 + \sqrt 7 }}.{7 \over {13}}.\sqrt {{{6 + \sqrt {35} } \over {4 + \sqrt {15} }}} $

$ = {7 \over {13}}\,.\,{{\sqrt 3 + \sqrt 5 } \over {\sqrt 5 + \sqrt 7 }}\,.\,\sqrt {{{{{(\sqrt 5 + \sqrt 7 )}^2}} \over 2}\,.\,{2 \over {{{(\sqrt 3 + \sqrt 5 )}^2}}}} $

$ = {7 \over {13}}\,.\,{{\sqrt 3 + \sqrt 5 } \over {\sqrt 5 + \sqrt 7 }}\,.\,{{\sqrt 5 + \sqrt 7 } \over {\sqrt 3 + \sqrt 5 }} = {7 \over {13}}$.

$\therefore 3{x^2} + 4xy - 3{y^2} = 3\,(x - y)\,(x + y + 4)$

$ = 3\,.\,\left( {{{\sqrt 5 + \sqrt 2 } \over {\sqrt 5 - \sqrt 2 }} - {{\sqrt 5 - \sqrt 2 } \over {\sqrt 5 + \sqrt 2 }}} \right)\,\,\left( {{{\sqrt 5 + \sqrt 2 } \over {\sqrt 5 - \sqrt 2 }} + {{\sqrt 5 - \sqrt 2 } \over {\sqrt 5 + \sqrt 2 }}} \right)\, + 4$

$ = {{3\,[{{(\sqrt 5 + \sqrt 2 )}^2} - {{(\sqrt 5 - \sqrt 2 )}^2}]} \over {(5 - 2)\,(5 - 2)}}\,[{(\sqrt 5 + \sqrt 2 )^2} + {(\sqrt 5 - \sqrt 2 )^2}] + 4$

$ = {1 \over 3}.4\sqrt {10} \,.\,2\,(5 + 2) + 4 = {{56} \over 3}\sqrt {10} + 4 = {1 \over 3}(56\sqrt {10} + 12)$.

$ \Rightarrow $${x^3} = 9\sqrt 3 + 11\sqrt 2 $

$ = 6\sqrt 3 + 3\sqrt 3 + 9\sqrt 2 + 2\sqrt 2 $

$ = 3\sqrt 3 + 2\sqrt 2 + 6\sqrt 3 + 9\sqrt 2 $

$ = 3\sqrt 3 + 2\sqrt 2 + 3(2\sqrt 3 + 3\sqrt 2 )$

$ = 3\sqrt 3 + 2\sqrt 2 + 3\sqrt 2 \,.\,\sqrt 3 (\sqrt 2 + \sqrt 3 )$

$ = {(\sqrt 3 )^3} + {(\sqrt 2 )^3} + 3.\sqrt 2 .\sqrt 2 \,(\sqrt 3 + \sqrt 2 ) = {(\sqrt 3 + \sqrt 2 )^3}$

So, ${x^3} = {(\sqrt 3 + \sqrt 2 )^3}$

$x = \sqrt 3 + \sqrt 2 $.

$\sqrt x = \sqrt {3 - \sqrt 5 } = {1 \over {\sqrt 2 }}\,.\sqrt {6 - 2\sqrt 5 } = {1 \over {\sqrt 2 }}(\sqrt 5 - 1)$         $3x - 2 = 9 - 3\sqrt 5 - 2 = 7 - 3\sqrt 5 = {{14 - 6\sqrt 5 } \over 2}$

= ${{{{(3 - \sqrt 5 )}^2}} \over 2}$;

$ \Rightarrow $ $\sqrt 2 + \sqrt {3x - 2} = \sqrt 5 \,.\,\sqrt x $;

$\therefore {{\sqrt x } \over {\sqrt 2 + \sqrt {3x - 2} }} = {1 \over {\sqrt 5 }}$.

= $(\sqrt {21} - \sqrt {18} ) - (\sqrt {20} - \sqrt {17} )$

= ${{(\sqrt {21} - \sqrt {18} )(\sqrt {21} + \sqrt {18} )} \over {\sqrt {21} + \sqrt {18} }} - {{20 - 17} \over {\sqrt {20} + \sqrt {17} }}$

= $3\,\left[ {{1 \over {\sqrt {21} + \sqrt {18} }} - {1 \over {\sqrt {20} + \sqrt {17} }}} \right]$

= ${{3\,[\sqrt {20} + \sqrt {17} - \sqrt {21} - \sqrt {18} ]} \over {(\sqrt {21} + \sqrt {18} )\,(\sqrt {20} + \sqrt {17} )}}$

= ${{3\,[(\sqrt {20} - \sqrt {21} ) + (\sqrt {17} - \sqrt {18)} ]} \over {(\sqrt {21} + \sqrt {18} )\,(\sqrt {20} + \sqrt {17} )}}$

= ${{ - 3\,[(\sqrt {21} - \sqrt {20} ) + (\sqrt {18} - \sqrt {17} )} \over {(\sqrt {21} + \sqrt {18} )\,(\sqrt {20} + \sqrt {17} )}} < 0$,

$\therefore a < b$.

and ${{f(x)} \over {x + 2}} = {\phi _3}(x) + {{15} \over {x + 2}}$

${{f(x)} \over {(x + 1)\,(x + 2)\,(x - 2)}} = \phi (x) + {{Q(x)} \over {(x + 1)\,(x + 2)\,(x - 2)}}$

We have to find $Q(x)$, which will be a second degree polynomial. When $Q(x)$ is divided by $(x + 1)$, we should get the same remainder as being obtained by dividing $f(x)$ by $(x + 1)$ i.e., $6$. Similarly when $Q(x)$ is divided by $(x - 2)$, remainder should be $3$ and when $f(x)$ is divided by $x + 2,$ the remainder should be $15.$

$\therefore Q( - 1) = 6$

$Q(2) = 3$, $Q( - 2) = 15$

Let $Q(x) = \alpha {x^2} + \beta x + \gamma $,

$\therefore \alpha - \beta + \gamma = 6$…..(i)

$4\alpha + 2\beta + \gamma = 3$.....(ii); $4\alpha - 2\beta + \gamma = 15$ …..(iii)

$ \Rightarrow $$\alpha = 2,\,\beta = - 3,\,\gamma = 1$;

$\therefore Q(x) = 2{x^2} - 3x + 1$.

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

$ - 2A + B = 0$, $1 = - 2B + 4C$

$\therefore A = {3 \over 8},\,B = {3 \over 4},\,C = {5 \over 8}$

$\therefore {{{x^2} + 1} \over {({x^2} + 4)\,\,(x - 2)}} = {{{3 \over 8}x + {3 \over 4}} \over {{x^2} + 4}} + {{{5 \over 8}} \over {x - 2}}$

$ = \,{1 \over 4}\,\left( {{3 \over 8}x + {3 \over 4}} \right)\,{\left( {1 + {{{x^2}} \over 4}} \right)^{ - 1}} + {5 \over 8}\,{1 \over {( - 2)}}\,{\left( {1 - {x \over 2}} \right)^{ - 1}}$

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

$ - {5 \over {16}}\left( {1 + {x \over 2} + {{\left( {{x \over 2}} \right)}^2} + .....} \right)$

Coefficient of ${x^5}$= ${3 \over {32}}\,.\,{1 \over {{4^2}}} + {3 \over {16}} \times 0 - {5 \over {16}}\,{\left( {{1 \over 2}} \right)^5}$

= ${3 \over {{2^9}}} - {5 \over {{2^9}}} = - {1 \over {{2^8}}} = - {1 \over {256}}$ .

$ \Rightarrow {\left( {{2^{({x^2} + 2)}}} \right)^2} - {9.2^{({x^2} + 2)}} + 8 = 0$

Put ${2^{({x^2} + 2)}}^2 = y$. Then ${y^2} - 9y + 8 = 0$, which gives $y = 8,y = 1$.

when $y = 8\,\, \Rightarrow \,\,{2^{{x^2} + 2}} = 8$ ==> ${2^{{x^2} + 2}} = {2^3}$ ==> ${x^2} + 2 = 3$

==> ${x^2} = 1$ ==>$x = 1, - 1$.

when $y = 1$ ==> ${2^{{x^2} + 2}} = 1$ ==> ${2^{{x^2} + 2}} = {2^o}$

==> ${x^2} + 2 = 0$ ==>${x^2} = - 2$, which is not possible.

$ = a + b + c - 2\sqrt {ab} - 2\sqrt {bc} + 2\sqrt {ca} $

$a,b,c > 0$. Then $a + b + c = 10,$

$ab = 6$, $bc = 10,$$ca = 15$

${a^2}{b^2}{c^2} = 900$==> $abc = 30$ $( \ne \pm 30)$.

So, $a = 3,\,\,\,b = 2,\,\,c = 5$

Therefore, $\sqrt {(10 - \sqrt {24} - \sqrt {40} + \sqrt {60} )} = \pm (\sqrt 3 + \sqrt 5 - \sqrt 2 )$

= ${1 \over {{x^{b + c}} + {x^{c + a}} + {x^{a + b}}}}\sum\limits_{}^{} {{x^{b + c}}} $

= ${1 \over {{x^{b + c}} + {x^{c + a}} + {x^{a + b}}}}\,({x^{b + c}} + {x^{c + a}} + {x^{a + b}}) = 1$.

$\therefore {x \over {\sqrt 2 + \sqrt x }} + {y \over {\sqrt 2 - \sqrt y }}$=${{x(\sqrt 2 - \sqrt x )} \over {2 - x}} + {{y(\sqrt 2 + \sqrt y )} \over {2 - y}}$

==> ${{x(\sqrt x - \sqrt 2 )} \over {x - 2}} + {{y\,(\sqrt y + \sqrt 2 )} \over {2 - y}} = {{x\,(\sqrt x - \sqrt 2 )} \over {\sqrt 3 }} + {{y\,(\sqrt y + \sqrt 2 )} \over {\sqrt 3 }}$

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

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

= ${1 \over {\sqrt 3 }}\left[ {{1 \over {{2^{3/2}}}}{{(4 + 2\sqrt 3 )}^{3/2}} + {1 \over {{2^{3/2}}}}{{(4 - 2\sqrt 3 )}^{3/2}} - 2\sqrt 6 } \right]$

= ${1 \over {\sqrt 3 }}\,\left[ {{1 \over {2\sqrt 2 }}\{ {{(\sqrt 3 + 1]}^3} + {{(\sqrt 3 - 1)}^3}\} - 2\sqrt 6 } \right]$

= ${1 \over {\sqrt 3 }}\left[ {{1 \over {2\sqrt 2 }}\{ 2.3\sqrt 3 + 6.\sqrt 3 \} - 2\sqrt 6 } \right]$

= ${1 \over {\sqrt 3 }}(3\sqrt 6 - 2\sqrt 6 ) = \sqrt 2 $.

$ = \ln {{1 + x} \over {1 + x + {{{x^2}} \over {2\,!}} + {{{x^3}} \over {3\,!}} + ...}} < {\rm{ }}0,\,{\rm{as }}1 + x < 1 + x + {{{x^2}} \over {2\,!}} + .... + $

$\therefore {\log _e}(1 + x) < x$, for $x > 0$.

${x \over {1 + x}} - \log (1 + x) = 1 - {1 \over {1 + x}} - \log (1 + x)$

= $1 - \left[ {{1 \over {1 + x}} + \log (1 + x)} \right]\, < 0$, for $x > 0$

$\therefore {x \over {1 + x}} < \log (1 + x)$, $\therefore(b)$ is true

${e^x} - (1 + x) = 1 + x + {{{x^2}} \over {2\,!}} + {{{x^3}} \over {3\,!}} + ..... - (1 + x)$

= ${{{x^2}} \over {2\,!}} + {{{x^3}} \over {3\,!}} + ..... > 0$, for $x > 0$

$\therefore {e^x} > 1 + x$, for $x > 0$; $\therefore (c)$ is true

${e^x} - (1 - x) = 1 + x + {{{x^2}} \over {2\,!}} + ...... - 1 + x$

= $2x + {{{x^2}} \over {2\,!}} + {{{x^3}} \over {3\,!}} + ....... > 0$, for $x > 0$

$\therefore {e^x} > 1 - x$, for $x > 0$

Thus, ${e^x} < (1 - x),$ for $x > 0$ is not true.

$\Rightarrow$ $x^3 + y^3 + z^3 = 3xyz$

$\Rightarrow 3$

$\Rightarrow \frac{2|z|^{2}+2|z|-3}{|z|+1}>\frac{2|z|^{2}+2|z|-3}{|z|+1}>\left(\frac{1}{\sqrt{3}}\right)^{-2}$

$\frac{2|z|^{2}+2|z|-3}{|z|+1}>3 $

$\Rightarrow 2|z|^{2}-|z|-6>0$

$\Rightarrow|z|>2$

Taking logarithm on base $5,$ then

$\frac{1}{4}\left(\log _{5}^{2} x\right) \geq 1+\frac{1}{5}\left(\log _{5} x\right)\left(\log _{5} x\right)$

$\Rightarrow \frac{1}{20} \log _{5}^{2} x \geq 1$

or $\left(\log _{5}^{2} x\right) \geq 20$

or $\log _{5} x \geq 2 \sqrt{5} $ and $ \log _{5} x \leq-2 \sqrt{5}$

or $x \geq 5^{2 \sqrt{5}} $ and $ x \leq 5^{-2 \sqrt{5}}$

But $x>0$

$\therefore $ $x \in\left(0,5^{-2 \sqrt{5}}\right] \cup\left[5^{2 \sqrt{5}}, \infty\right)$

$x y=100$

$\frac{x+y}{2} \geq \sqrt{x y}$

$(x+y) \geq 20$

when $a, b, c$ has same sign.

$\log _{x} z=6$

$\Rightarrow \mathrm{x} \in(-\infty, 5) \cup(9, \infty)$     $\ldots(1)$

$4-x>0 $ and $ 4-x \neq 1 \Rightarrow x<4 $ and $ x \neq 3 \dots(2)$

from $( 1)$ $\cap(2)$

$x \in(-\infty, 4)-\{3\}$

$\log _{7}(t-1)+\log _{7}(x->)=1$

$\log [(t-1)(x-7)]=1$

$(x-1)(t-2)=7$

$k^{2}-7 t-x+7=7$

$t^{2}-8 t=0$

$t_{1}=0$ or $t=8$

$2^{x}=0 \quad$ or $\quad 2^{x}=8$

$2^{x}=2^{3}$

$\Rightarrow x=3$

$y=\frac{c a b-c}{990}$

$y=\frac{100 c+10 a+b-c}{990} $

$\Rightarrow y=\frac{99 c+10 a+b}{990}$