Maths M.C.Q (1 Marks)

$\therefore \,\,\sqrt {134 + \sqrt {62\,92} }  = 11 + \sqrt {13} $

= ${\log _n}2 + {\log _n}3 + {\log _n}4 + ...... + {\log _n}1983$

= ${\log _n}(2.3.4....1983) = {\log _n}(1983!) = {\log _n}n = 1$.

= ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}........{\log _{{x_{n - 1}}}}x_{n - 1}^{x_{n - 2}^{{.^{{.^{{.^{{x_1}}}}}}}}}$

= ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}......{\log _{{x_{n - 2}}}}x_{n - 2}^{x_{n - 3}^{{.^{{.^{{.^{{x_1}}}}}}}}}$

= ${\log _{{x_1}}}{\log _{{x_2}}}x_2^{{x_1}} = {\log _{{x_1}}}{x_1} = 1$.

For logarithm to be defined,

${x^2} - 2x - 2 > 0$

$ \Rightarrow $ ${(x - 1)^2} > 3$

==> $x - 1 < - \sqrt 3 $ or$x - 1 > \sqrt 3 $

==> $x < 1 - \sqrt 3 $ or $x > 1 + \sqrt 3 $

i.e., $x < - (\sqrt 3 - 1)$ or $x > (\sqrt 3 + 1)$

Now from $(i),$ ${x^2} - 2x - 2 \le 1$

==> ${x^2} - 2x - 3 \le 0$

==> $(x - 3)\,(x + 1) \le 0$ $ \Rightarrow $ $ - 1 \le x \le 3$

$\therefore $ $x \in [ - 1,\, - (\sqrt 3 - 1)\,[\, \cup \,]\,\,\sqrt 3 + 1,\,3]$.

i.e., $x \in [ - 1,\,1 - \sqrt 3 )\,\, \cup (1 + \sqrt 3 ,\,3)$.

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

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

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

==> $61 - 46\sqrt 5 = {(a - \sqrt b )^3} = {a^3} + 3ab - (3{a^2} + b)\sqrt b $

==> $61 = {a^3} + 3ab,\,46\sqrt 5 = (3{a^2} + b)\,\sqrt b $

==> $61 = ({a^2} + 3b)\,a$, $23\sqrt {20} = (3{a^2} + b)\sqrt b $

So,$a = 1,\,\,b = 20$.

Therefore, $\sqrt[3]{{61 - 46\sqrt 5 }} = 1 - \sqrt {20} = 1 - 2\sqrt 5 $.

$ \Rightarrow $$x = \sqrt {a + x} $$ \Rightarrow $${x^2} - x - a = 0$$ \Rightarrow $$x = \frac{{1 \pm \sqrt {1 + 4a} }}{2}$

As $a > 0$, $x > 0$; $\therefore $ $+ve$ sign should be considered.

$2B + C = - 1$ $x = \frac{1}{2}(1 + \sqrt {4a + 1} )$.

==> $\frac{{4 + 3\sqrt 3 }}{{2 + \sqrt 3 }} = a + \sqrt b $

==> $\frac{{(4 + 3\sqrt 3 )\,(2 - \sqrt 3 )}}{{(2 + \sqrt 3 )\,(2 - \sqrt 3 )}} = a + \sqrt b $

==> $ - 1 + 2\sqrt 3 = a + \sqrt b $

==> $ - 1 + \sqrt {12} = a + \sqrt b $; $\therefore $ $(a,\,b) = ( - 1,\,12)$.

==> ${3^x} - \frac{{{3^x}}}{3} = 6$

Let ${3^x} = t$, then given equation can be written as

$t - \frac{t}{3} = 6$ ==> $3t - t = 18$ ==> $2t = 18$ ==> $t = 9$

$\therefore$  ${3^x} = {3^2}$ ==> $x = 2$.

Hence, ${x^x} = {2^2} = 4$.

Next, $x \pm 2\sqrt {x - 1} \ge 0$

==> ${x^2} \ge 4(x - 1) \Rightarrow {x^2} - 4x + 4 \ge 0 \Rightarrow {(x - 2)^2} \ge 0$,

which is true $\forall \,\,x$, $\therefore x \ge 1$.

For $1 \le x \le 2$, $\sqrt {x + 2\sqrt {x - 1} } + \sqrt {x - 2\sqrt {x - 1} } $

= $\sqrt {1 + (x - 1) + 2\sqrt {x - 1} } + \sqrt {1 + (x - 1) - 2\sqrt {x - 1} } $

= $(1 + \sqrt {x - 1} ) + (1 - \sqrt {x - 1} ) = 2$

For $x > 2$, $\sqrt {x + 2\sqrt {x - 1} } + \sqrt {x - 2\sqrt {x - 1} } $

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

$ + r(2x - 1)\,(x - 3) + s\,(2x - 1)\,(x + 2)$

Equating coefficient of ${x^3}$; $1 = 2p \Rightarrow p = \frac{1}{2}$

Equating coefficient of ${x^0}$i.e., constant term,

$0 = 6p - 6q + 3r - 2s \Rightarrow 6q - 3r + 2s = 3$.

$1 < \frac{{{{\log }_{0.09}}(x - 1)}}{{{{\log }_{0.3}}(x - 1)}}$

==> $1 < {\log _{0.3}}(0.09)$

==> $1 < {\log _{0.3}}{(0.3)^2}$

==> $1 < 2$

Which of true therefore it is true for every positive value of $2$

$\therefore$ $x \in (1,\infty )$.

$\left(10^{10^{10}}\right)^{\frac{1}{5}}$

$=10^{10^{10} \times \frac{1}{5}}=10^{10^9 \times \frac{10}{5}}=10^{2 \times 10^9}$

$\frac{p}{q}=\frac{\log _a x \cdot \log _x\left(\frac{b}{a}\right)}{\log _x b \cdot \log _{a b} x}$

$\frac{ p }{ q }=\frac{\frac{(\log b -\log a)}{\log a}}{\frac{\log b }{(\log a+\log b)}}$$\frac{ p }{ q }=\frac{(\log b )^2-(\log a)^2}{(\log a)(\log b )}$

Given $\log _{ a } b =10 \Rightarrow \log b =10 \log a$

$\frac{ p }{ q }=\frac{(100-1)(\log a)^2}{10(\log a)^2}=\frac{99}{10}$

$p+q=109$

$\frac{1}{ q }$ is terminating decimal

$\begin{array}{ll}\Rightarrow q =2^{ m } 5^{ n } \\ m , n \in w \\ 1 \leq 2^{ m } & 5^{ n } \leq 2021 \\ m =0 & n =0,1,2,3,4 \\ m =1 & n =0,1,2,3,4 \\ m =2 & n =0,1,2,3 \\ m =3 & n =0,1,2,3 \\ m =4 & n =0,1,2,3 \\ m =5 & n =0,1,2 \\ m =6 & n =0,1,2 \\ m =7 & n =0,1 \\ m =8 & n =0,1 \\ m =9 & n =0 \\ m =10 & n =0\end{array}$

as $\sqrt{ q } \in Q \Rightarrow m , n$ must be even

So total $11$ cases

Given, $\log (x-2 y)^2=\log x y$

$x^2+4 y^2-4 x y =x y$

$x^2+4 y^2-5 x y =0$

$\left(\frac{x}{y}\right)^2-5\left(\frac{x}{y}\right)+4 =0$

$\left(\frac{x}{y}-1\right)\left(\frac{x}{y}-4\right) =0 \Rightarrow \frac{x}{y}=1 \text { or } 4$

But if $\frac{x}{y}=1$ then, $x-2 y< 0$

Hence, only $\frac{x}{y}=4$

Given number,

$n=15^2 \times 5^{18}$

$n=3^2 \times 5^2 \times 5^{18}$

$n=9 \times 5^{20}$

Taking log base 10 both side

$\log _{10} n=\log _{10} 9+\log _{10} 5^{20}$

$=2 \log _{10} 3+20 \log _{10} 5$

$=2 \times 0.4771+20 \times(1-0.3010)$

$=14$ characters value

Hence, the number have $15$ digits $S=$ Sum of digits of the number Now, $n$ has last digit is $5 .$

$\therefore$ Minimum value of $S=1+5=6$

Maximum value of $S=9 \times 14+5$

$=126+5=131$

$\therefore 6 \leq S < 140$

We know,

$\frac{\log _e(1+x)}{x} < 1$

$\frac{\log _e\left(1+\frac{1}{x}\right)}{1 / x}-1$

$\log _e\left(\frac{x+1}{x}\right) < \frac{1}{x}$

$\log _e(x+1)-\log _e x < \frac{1}{x}$

$\log _e 2-\log _e 1 < 1$

$\log _e 3-\log _e 2 < \frac{1}{2}$

$\log _e(n+1)-\log _e n < \frac{1}{n}$

$\log _e(n+1)-\log _e 1<1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}$

$\log _e(n+1) \leq 4$

${\left[\because 1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n} \geq 4\right] }$

$n+1 < e^4$

$n < e^4-1$

$n < (2.7)^4-1$

$n < 5459-1$

$n < 53.59$

$\quad[\because e=2718]$

$\therefore$ Hence, option $(a)$ is correct.

We have,

$\log _{1 / 3}(x+y)+\log _3(x-y)=2$

$2 y^2=512^{x+1}$

$\Rightarrow \log _{3^{-1}}(x+y)+\log _3(x-y)=2$

$\Rightarrow-\log _3(x+y)+\log _3(x-y)=2$

$\Rightarrow \quad \frac{x-y}{x+y}=3^2=9$

$\Rightarrow \quad x-y=9 x+9 y$

$\Rightarrow \quad-8 x=10 y \Rightarrow-4 x=5 y$

and $\quad 2^{y^2}=2^{9(x+1)}$

$\Rightarrow \quad y^2=9(x+1)$

$\begin{array}{cc}\Rightarrow & 16 x^2=225 x+225 \\ \Rightarrow & 16 x^2-225 x-225=0 \\ \Rightarrow & (16 x+15)(x-15)=0 \\ \Rightarrow & x=15,-\frac{15}{16}\end{array}$

$x=-\frac{15}{16}, y=\frac{3}{4}$ (not possible)

$\therefore$ Only one solution $(15,-12)$.

Let $x=0.75$

According to the question, $\frac{x^3}{1-x}+\left(x+x^2+1\right)$

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

$=\frac{x^3+1-x^3}{1-x}=\frac{1}{1-x}$

Now, put the value of $x$

$\frac{1}{1-0.75} =\frac{1}{0.25}$

$=\frac{100}{25}=4$

So, square root of the equation $=\sqrt{4}=2$

We have, $\log _a b=4$, $\log _c d=2, a, b, c, d \in N$

$\Rightarrow b=a^4, d=c^2$

$\Rightarrow b-d=7=a^4-c^2$

$\Rightarrow 7=\left(a^2+c\right)\left(a^2-c\right)$

$\Rightarrow 7 \times 1=\left(a^2+c\right)\left(a^2-c\right)$

$\therefore a^2+c=7 \text { and } a^2-c=1$

$\text { On solving, we get } a=2 \text { and } c=3$

$\therefore c-a=3-2=1$

$=\sum \limits_{n=0}^{\infty} \frac{1}{(n-3) !}+\sum \limits_{n=0}^{\infty} \frac{3}{(n-2) !}$

$+\sum \limits_{n=0}^{\infty} \frac{1}{(n-1) !}+\sum \limits_{n=0}^{\infty} \frac{1}{(2 n-1) !}-\sum \limits_{n=0}^{\infty} \frac{1}{(2 n) !}$

$=e+3 e+e+\frac{1}{2}\left(e-\frac{1}{e}\right)-\frac{1}{2}\left(e+\frac{1}{e}\right)$

$=5 e-\frac{1}{e}$

$a^2-b+c=26$

$\frac{1}{2} \sum \limits_{ n =1}^{\infty} \frac{2 n (2 n -1)+8 n +8}{(2 n ) !}$

$\frac{1}{2} \sum \limits_{ n =1}^{\infty} \frac{1}{(2 n -2) !}+2 \sum \limits_{ n =1}^{\infty} \frac{1}{(2 n -1) !}+4 \sum \limits_{ n =1}^{\infty} \frac{1}{(2 n ) !}$

$e =1+1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots \ldots$

$e ^{-1}=1-1+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}+\ldots \ldots$

$\left( e +\frac{1}{ e }\right)=2\left(1+\frac{1}{2 !}+\frac{1}{4 !}+\ldots \ldots .\right)$

$e -\frac{1}{ e }=\left(1+\frac{1}{3 !}+\frac{1}{5 !}+\ldots . .\right)$

$\text { Now }$

$\frac{1}{2}\left(\sum \limits_{ n =1}^{\infty} \frac{1}{(2 n -2) !}\right)+2 \sum \limits_{ n =1}^{\infty} \frac{1}{(2 n -1) !}+4 \sum \limits_{ n =1}^{\infty} \frac{1}{(2 n ) !}$

$=\frac{1}{2}\left[\frac{ e +\frac{1}{ e }}{2}\right]+2\left[\frac{ e -\frac{1}{ e }}{2}\right]+4\left(\frac{ e +\frac{1}{ e }-2}{2}\right)$

$=\frac{\left( e +\frac{1}{ e }\right)}{4}+ e -\frac{1}{ e }+2 e +\frac{2}{ e }-4$

$=\frac{13}{4} e +\frac{5}{4 e }-4$

Feasible region : $x+\frac{7}{2}>0 \Rightarrow x > -\frac{7}{2}$

And $x+\frac{7}{2} \neq 1 \Rightarrow x \neq-\frac{5}{2}$

And $\frac{x-7}{2 x-3} \neq 0 \quad$ and $2 x-3 \neq 0$

$\Downarrow$

$x \neq 7$

$x \neq \frac{3}{2}$

Taking intersection : $x \in\left(\frac{-7}{2}, \infty\right)-\left\{-\frac{5}{2}, \frac{3}{2}, 7\right\}$

Now $\log _a b \geq 0$ if $a > 1$ and $b \geq 1$

Or

$a \in(0,1)$ and $b \in(0,1)$

$\text { C-I; } \quad x+\frac{7}{2}>1 \text { and }\left(\frac{x-7}{2 x-3}\right)^2 \geq 1$

$x > -\frac{5}{2} \quad(2 x-3)^2-(x-7)^2 \leq 0$

$(2 x-3+n-7)(2 x-3-x+7) \leq 0$

$(3 x-10)(x+4) \leq 0$

$\quad x \in\left[-4, \frac{10}{3}\right]$

$\text { Intersection : } x \in\left(\frac{-5}{2}, \frac{10}{3}\right]$

$\text { C-II } x+\frac{7}{2} \in(0,1) \text { and }\left(\frac{x-7}{2 x-3}\right)^2 \in(0,1)$

$0 < x+\frac{7}{2} < 1 \quad \quad\left(\frac{x-7}{2 x-3}\right)^2 < 1$

$-\frac{7}{2} < x < \frac{-5}{2} \quad(x-7)^2<(2 x-3)^2$

$x \in(-\infty,-4) \cup\left(\frac{10}{3}, \infty\right)$

No common values of $x$.

Hence intersection with feasible region

We get $x \in\left(\frac{-5}{2}, \frac{10}{3}\right]-\left\{\frac{3}{2}\right\}$

Integral value of $x$ are $\{-2,-1,0,1,2,3\}$

No. of integral values $=6$

$=\left(\frac{4}{25}\right)^{\log _{\left(\frac{5}{2}\right)}\left(\frac{1}{2}\right)}$

$=\left(\frac{1}{2}\right)^{\log _{\left(\frac{5}{2}\right)}\left(\frac{4}{25}\right)}=\left(\frac{1}{2}\right)^{-2}=4$

$\Rightarrow \quad 6 x+4 y=5$     $. . . (1)$

$2 x - y =\log _{\frac{1}{5^{1 / 6} \sqrt{6}}}\left(5^{\frac{1}{6}} \sqrt{6}\right)^3=-3$

$\Rightarrow \quad 2 x - y =-3$      $. . . (2)$

$\text { equation (1) - (2) }$

$\Rightarrow \quad 4 x+5 y=8$

Take log to the base 5 on both sides and put $\log _5 x=t$

$16 t^4-68 t^2+16=0$

$\Rightarrow 4 t^4-17 t^2+4=0\left\{\begin{array}{l}t_1 \\ t_2 \\ t_3 \\ t_4\end{array}\right.$

$t_1+t_2+t_3+t_4=0$

$\log _5 x_1+\log _5 x_2+\log _5 x_3+\log _5 x_4=0$

$x_1 x_2 x_3 x_4=1$

$x=(x-1) \log _3 4 $

$x\left(1-2 \log _3 2\right)=-2 \log _3 2 $

$x=\frac{2 \log _3 2}{2 \log _3 2-1} \quad\quad$ Ans. $(A)$

Again

$x \log _2 3=(x-1) \cdot 2 $

$x\left(\log _2 3-2\right)=-2 $

$x=\frac{2}{2-\log _2 3}\quad\quad$ Ans. $(B)$

$x=\frac{1}{1-\frac{1}{2} \log _2 3}=\frac{1}{1-\log _4 3}\quad\quad$ Ans.$(C)$

$\sqrt{4-\frac{1}{3 \sqrt{2}} t}=t $

$4-\frac{1}{3 \sqrt{2}} t=t^2 \Rightarrow $

$t^2+\frac{1}{3 \sqrt{2}} t-4=0 \Rightarrow 3 \sqrt{2} t^2+t-12 \sqrt{2}=0 $

$t=\frac{-1 \pm \sqrt{1+4 \times 3 \sqrt{2} \times 12 \sqrt{2}}}{2 \times 3 \sqrt{2}}=\frac{-1 \pm 17}{2 \times 3 \sqrt{2}}$

$t=\frac{16}{6 \sqrt{2}}, \frac{-18}{6 \sqrt{2}}$

$t =\frac{8}{3 \sqrt{2}}, \frac{-3}{\sqrt{2}}$ and $\frac{-3}{\sqrt{2}}$ is rejected

$\text { so } 6+\log _{3 / 2}\left(\frac{1}{3 \sqrt{2}} \times \frac{8}{3 \sqrt{2}}\right)=6+\log _{3 / 2}\left(\frac{4}{9}\right)=6+\log _{3 / 2}\left(\left(\frac{2}{3}\right)^2\right)=6-2=4$

Taking $\ln$ both sides,

$\Rightarrow(\ln 2)(\ln 2 x)=(\ln 3)(\ln 3 y)$

$\Rightarrow(\ln 2)(\ln 2+\ln x)=(\ln 3)(\ln 3+\ln y) \rightarrow(1)$

Now, we will take the second equation,

$3^{\ln x}=2^{\ln y}$

Taking $\ln$ both sides,

$\Rightarrow(\ln x)(\ln 3)=(\ln y)(\ln 2)$

$\Rightarrow(\ln y)=\frac{(\ln x)(\ln 3)}{\ln 2}$

Putting value of $\ln y$ in $(1)$,

$(\ln 2)(\ln 2+\ln x)=(\ln 3)\left(\ln 3+\frac{(\ln x)(\ln 3)}{\ln 2}\right)$

$\Rightarrow(\ln x)\left(\ln 2-\left(\frac{(\ln 3)^2}{\ln 2}\right)\right)=(\ln 3)^2-(\ln 2)^2$

$\Rightarrow \frac{\ln x}{\ln 2}\left((\ln 2)^2-(\ln 3)^2\right)=(\ln 3)^2-(\ln 2)^2$

$\Rightarrow \frac{\ln x}{\ln 2}=-1$

$\Rightarrow(\ln x)=\ln (2)^{-1}$

$\Rightarrow x=2^{-1} \Rightarrow x=\frac{1}{2}$, which is the required value of $x_0$.

$ = {{\log 4} \over {\log 3}}.{{\log 5} \over {\log 4}}.{{\log 6} \over {\log 5}}.{{\log 7} \over {\log 6}}.{{\log 8} \over {\log 7}}.{{\log 9} \over {\log 8}} = {{\log 9} \over {\log 3}}$

$ = {\log _3}9 = {\log _3}{3^2} = 2$.

$ \Rightarrow $$x + 1 = A\,(x - 2)\,(x - 3)\, + B(x - 1)\,(x - 3) + C(x - 1)\,(x - 2)$

Putting $x = 1,\,A = 1$ ; $x=2$ gives $B = - 3$,

For $x = 3,\,C = 2$

$\therefore$ Given expression = ${1 \over {x - 1}} - {3 \over {x - 2}} + {2 \over {x - 3}}$.

For $x = 0,\,A = 1$ and for $x = i$, $ - B + Ci = 1$

$ \Rightarrow B = - 1,\,C = 0$ $ \Rightarrow $ $(A,B,C) = (1,\, - 1,\,0)$.

Taking log both the sides, $\log \,y = \log {3^{40}}$

==> $\log \,\,y = 40\,\log 3$$ \Rightarrow \,\,\log \,y = 19.08$

$\therefore$  Number of digits in $y = 19 + 1 = 20$ .

$(x - 1) = y \Rightarrow x = y + 1$

$\therefore {{{x^2}} \over {{{(x - 1)}^3}(x - 2)}} = {{{{(1 + y)}^2}} \over {{y^3}(y - 1)}} = {{1 + 2y + {y^2}} \over {{y^3}( - 1 + y)}}$

Dividing the numerator,

$(1 + 2y + {y^2})$ by $( - 1 + y)$ till ${y^3}$ appears as factor,

we get ${{1 + 2y + {y^2}} \over { - 1 + y}} = ( - 1 - 3y - 4{y^2}) + {{4{y^3}} \over { - 1 + y}}$

Given expression = ${{ - 1} \over {{y^3}}} - {3 \over {{y^2}}} - {4 \over y} + {4 \over { - 1 + y}}$

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

Then, ${p \over q} = {\log _2}7\,\,\,\,\, \Rightarrow 7 = {2^{p/q}}\,\,\, \Rightarrow {2^p} = {7^q}$,

which is false since $L.H.S$ is even and $R.H.S$ is odd. Obviously ${\log _2}7$ is not an integer and hence not a prime number.

${(2\sqrt 2 )^x} = 32\root 5 \of 4 $

==> ${({2.2^{1/2}})^x} = {2^5}{.2^{2/5}}$;  ${2^{{{3x} \over 2}}} = {2^{5 + {2 \over 5}}}$

Here, by equating the indices, ${3 \over 2}x = {{27} \over 5}$

$\therefore x = \frac{{18}}{5} = 3.6$
.

Obviously, ${\log _{10}}\alpha ,\,{\log _3}\alpha ,\,{\log _e}\alpha ,\,{\log _2}\alpha $ are in increasing order.

$ = {\log _7}7 - {\log _7}8 = 1 - {\log _7}{2^3} = 1 - 3{\log _7}2$.

$ \Rightarrow $ $x = {5^2} = 25$.

$ = {\log _2}3 + {1 \over 2},$ which is irrational.

$ = {{{{\log }_e}a} \over {{{\log }_e}b}} \times {{{{\log }_e}b} \over {{{\log }_e}c}} \times {{{{\log }_e}c} \over {{{\log }_e}a}} = 1$.