Maths M.C.Q (1 Marks)

Clearly $x + 2 = 2x - 2$ ==> $x = 4$

$ \Rightarrow $$n = {m^{n - 1}} \Rightarrow m = {n^{{1 \over {n - 1}}}}$.

$= {2^{{{17} \over 3}}}{x^3}$.

${b^2} = ac \Rightarrow {({k^y})^2} = {k^x}.{k^z}$ $ \Rightarrow $ ${k^{2y}} = {k^{x + z}} \Rightarrow x + z = 2y$.

$ \Rightarrow $${a^{x - 2}}{b^{y - 2}}{c^{z - 2}} = 1 = {a^0}{b^0}{c^0}$

$\therefore x = y = z = 2$

$\therefore xyz = {2^3} = 8 = x + y + z + 2$.

$ \Rightarrow $$mn + m + 3n = mn + 2m + n$$ \Rightarrow $$m = 2n$.

$ \Rightarrow $${x^{{x^{3/2}}}} = {x^{(3/2)x}}$

$ \Rightarrow $${x^{3/2}} = {3 \over 2}x$ $ \Rightarrow $${x^{1/2}} = {3 \over 2}$

$ \Rightarrow $$x = {9 \over 4}$

Also $x = 1$ is an obvious solution.

$ = {{\sqrt {4 + 2\sqrt 3 } + \sqrt {4 - 2\sqrt 3 } } \over {2\sqrt 3 }} = {{(\sqrt 3 + 1)\, + \,(\sqrt 3 - 1)} \over {2\sqrt 3 }} = 1$.

$ = {{4(1 + \sqrt 2 + \sqrt 3 )} \over {3 + 2\sqrt 2 - 3}} + {{\sqrt 6 (\sqrt 3 - \sqrt 2 )} \over {3 - 2}}$

$ = \sqrt 2 (1 + \sqrt 2 + \sqrt 3 ) = 2 + \sqrt 2 + \sqrt 6 $.

$ = {{3\sqrt 2 (\sqrt 6 - \sqrt 3 )} \over {6 - 3}} - {{4\sqrt 3 \,(\sqrt 6 - \sqrt 2 )} \over {6 - 2}} + {{\sqrt 6 (\sqrt 3 - \sqrt 2 )} \over {3 - 2}}$

$ = \sqrt 2 \,(\sqrt 6 - \sqrt 3 ) - \sqrt 3 \,(\sqrt 6 - \sqrt 2 ) + \sqrt 6 (\sqrt 3 - \sqrt 2 )= 0.$

$ \Rightarrow $${x^2} + 1 = {(a - x)^2} = {x^2} - 2ax + {a^2}$

$ \Rightarrow $$x = {{1 - {a^2}} \over { - 2a}} = {{{a^2} - 1} \over {2a}} = {1 \over 2}\left( {a - {1 \over a}} \right)$.

Put $x = - 1$; $2( - 1) + 3 = a( - 1 - 3) \Rightarrow 1 = - 4a \Rightarrow a = {{ - 1} \over 4}$

Now put $x = 3$; $2(3) + 3 = b(3 + 1)$ ==> $9 = 4b$ ==> $b = {9 \over 4}$

Therefore, $a + b = {{ - 1} \over 4} + {9 \over 4} = 2$.

==> $(3x + a) = A(x - 1) - 10(x - 2)$

==> $3 = A - 10$, $a = - A + 20$

(On equating coefficients of $x$ and constant term)

==> $A = 13, a = 7.$

==> $3x + 4 = A{(x + 1)^2}$+$B(x + 1)\,(x - 1) + C(x - 1)$

Putting $x = 1$, we get $7 = A{(2)^2}$ ==> $A = {7 \over 4}$.

==> $(3x - 1) = (Ax + B)\,(x + 2)\, + \,C({x^2} - x + 1)$

Comparing the coefficient of like terms, we get $A + C = 0$, $2A + B - C = 3$, $2B + C = - 1$ ==> $A = 1$, $B = 0$, $C = - 1$

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

$ + ... + {A_r}x(x + 1)\,(x + 2)....(x + r - 1)\,(x + r + 1)\,(x + r + 2)$
$.......(x + n)$

Putting $x = - r$,

$1 = {A_r}( - r)\,( - r + 1)\,( - r + 2),\,.....( - 1).1.2....( - r + n)$

$ \Rightarrow $ $1 = {A_r}.{( - 1)^r}r!.(n - r)\,!$; ${A_r} = {{{{( - 1)}^r}} \over {r\,!\,(n - r)\,!}}$.

$ \Rightarrow $ $a = 6 + 4 - 5 = 5$, $b = 21 + 2 - 5 = 18$,

$c = 18 - 6 + 10 = 22$.

$ \Rightarrow $ $1 = - 6 + B$, $2 = 9 - B$ $ \Rightarrow $ $B = 7$.

$ \Rightarrow $ $3=A-B,$ $4 = - A + 2B$

$ \Rightarrow $ $A = 10,\,\,\,B = 7$

$\therefore (A,\,B) = (10,\,7)$.

$ \Rightarrow $$1 = \sin \alpha ,\,\, - 1 = - 2 + \sin \alpha $$ \Rightarrow $$\alpha = {\pi \over 2}$.

$ \Rightarrow $${x^2} = k\,[({a^2} - {b^2}){x^2}] \Rightarrow 1 = k({a^2} - {b^2})$

$\therefore k = {1 \over {{a^2} - {b^2}}}$.

For $x = 1,\,\,9 = 9A \Rightarrow A = 1$

For $x = - 2,\,9 = - 3C \Rightarrow C = - 3$

Equating coefficient of ${x^2},\,0 = A + B \Rightarrow B = - A = - 1$

$\therefore A - B - C = 1 - ( - 1) - ( - 3) = 1 + 1 + 3 = 5$.

==> ${x^2} + 13x + 15 = A{(x + 3)^2} + B(2x + 3)\,(x + 3) + C(2x + 3)$

For $x = - 3,\,C = 5$ and for $x = - {3 \over 2};\,A = - 1$

Equating coefficient of ${x^2}$

$1 = A + 2B \Rightarrow B = {{1 - A} \over 2} = 1$

$\therefore $ Given expression = ${1 \over {x + 3}} - {1 \over {2x + 3}} + {5 \over {{{(x + 3)}^2}}}$.

==>$3{x^3} - 8{x^2} + 10 = A{(x - 1)^3} + B{(x - 1)^2} + C(x - 1) + D$

Equating coefficients of different powers of $x$, $3 = A$

$ - 8 = - 3A + B \Rightarrow B = 1$

$0 = 3A - 2B + C \Rightarrow C = - 7$

$10 = - A + B - C + D \Rightarrow D = 5$

$\therefore$ Given expression

= ${3 \over {x - 1}} + {1 \over {{{(x - 1)}^2}}} - {7 \over {{{(x - 1)}^3}}} + {5 \over {{{(x - 1)}^4}}}$.

For $x = i,\, - B + Ci = - 2i$ $ \Rightarrow $ $B = 0,\,C = - 2$

Equating coefficient of ${x^2}$,

$A + B = 1 \Rightarrow A = 1 - B = 1 - 0 = 1$;

$\therefore A = 1,\,B = 0,\,C = - 2$.

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

For $x = 1,$ $2 = 2C \Rightarrow C = 1$

For $x = - 1$, $2 = 6B \Rightarrow B = {1 \over 3}$

For $x = {1 \over 2}$, ${5 \over 4} = - {3 \over 4}A \Rightarrow $$A = - {5 \over 3}$

$\therefore $ Given expression = $ - {5 \over 3}{1 \over {(2x - 1)}} + {1 \over 3}{1 \over {x + 1}} + {1 \over {x - 1}}$

$\therefore a = 3$.

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

$ \Rightarrow $ $a = 3,\,a + b = 5 \Rightarrow b = 2$;

$\therefore (a,\,b) = (3,\,2)$.

Equating coefficient of ${x^2},\,C = 1$.

$ \Rightarrow $ ${x^2} - 5 = A(x - 2) + B(x - 1) + C(x - 1)(x - 2)$

$ \Rightarrow $ $C = 1,\,A + B - 3C = 0,\, - 2A - B + 2C = - 5$

$\therefore A = 4,\,B = - 1,\,C = 1$

$\therefore $ Given expression = $1 + {4 \over {x - 1}} - {1 \over {x - 2}}$

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

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

$ + C(\sin x - 1)\,(2\sin x - 3)$

$ \Rightarrow $ $1 = 2C \Rightarrow C = {1 \over 2}$

$0 = A + 2B - 5C,\,1 = - A - 3B + 3C$.

$\therefore A = {{13} \over 2}$, $B = - 2$, $C = {1 \over 2}$, $A + B + C = 5$.

$\therefore {1 \over 2}{\log _{0.3}}(x - 1) < 0$

or ${\log _{0.3}}(x - 1) < \,0 = \log 1$ or $(x - 1) > 1$ or $x > 2$

As base is less than $1$, therefore the inequality is reversed, now $x > 2$ $ \Rightarrow $$x$ lies in $(2,\infty )$.

Let $y = x + 5$, $y = {64.2^{ - x}}$ will intersect at one point.

Number of solutions $= 1.$

$\therefore {\log _{20}}3 \in \left( {{1 \over 3},\,{1 \over 2}} \right)$.

==>${x^4} + 24{x^2} + 28 = ({A_1}x + {B_1})\,{({x^2} + 1)^2}$

$ + ({A_2}x + {B_2})\,({x^2} + 1) + ({A_3}x + {B_3})$

Putting $x = i,\,5 = {A_3}i + {B_3}$$ \Rightarrow $ ${A_3} = 0,\,{B_3} = 5$

Equating different powers of $x$,

$0 = {A_1},\,{B_1} = 1,\,2{A_1} + {A_2} = 0 \Rightarrow {A_2} = 0$

$2{B_1} + {B_2} = 24 \Rightarrow {B_2} = 22$.

$\therefore$ Partial fraction = ${1 \over {{x^2} + 1}} + {{22} \over {{{({x^2} + 1)}^2}}} + {5 \over {{{({x^2} + 1)}^3}}}$.

$b = {\log _{36}}24 = {{3\log 2 + \log 3} \over {2(\log 2 + \log 3)}}$

$c = {\log _{48}}36 = {{2(\log 2 + \log 3)} \over {4\log 2 + \log 3}}$

$\therefore abc = {{2\,\,\log 2 + \log 3} \over {4\log 2 + \log 3}}$

==> $1 + abc = {{6\log 2 + 2\log 3} \over {4\log 2 + \log 3}} = 2.{{3\log 2 + \log 3} \over {4\log 2 + \log 3}} = 2bc$.

$ = {20^{ - 2{{\log }_{20}}(1/9)}} = {20^{2{{\log }_{20}}9}} = {20^{{{\log }_{20}}{9^2}}} = {9^2} = 81$.

==> ${1 \over {\ln a.\ln b.\ln c}}[{(\ln a)^3} + {(\ln b)^3} + {(\ln c)^3} - 3\ln a.\ln b.\ln c] = 0$

==> ${(\ln a)^3} + {(\ln b)^3} + {(\ln c)^3} - 3\ln a.\ln b.\ln c = 0$

==> $\ln a + \ln b + \ln c = 0$

==> $\ln (abc) = ln 1$, $[{a^3} + {b^3} + {c^3} - 3abc = 0$

==> $a + b + c = 0]$,

$\therefore abc = 1$.

This quadratic equation will have exactly one solution if its discriminant vanishes.

$\therefore {( - 1)^2} - 4.1.{\log _{16}}k = 0 \Rightarrow 1 = {\log _{16}}{k^4}$

$ \Rightarrow $${k^4} = 16$ $ \Rightarrow $ ${k^2} = 4$ $ \Rightarrow $ $k = \pm 2$.

But ${\log _{16}}k$ is not defined $k < 0$, $k = 2$.

$\therefore$ Number of real values of $k = 1$.

There is a possibility of a solution $x = 3$

For this value, $LHS =$ ${3^{{3 \over 4}{{.1}^2} + 1 - \left( {{5 \over 4}} \right)}} = {3^{{2 \over 4}}} = {3^{{1 \over 2}}} = {\rm{RHS}}$.

$\therefore x = 3$ is a solution, which is a $ + ve$ integer.

Next, $\left[ {{3 \over 4}{{({{\log }_3}x)}^2} + {{\log }_3}x - {5 \over 4}} \right]\,{\log _3}x = {1 \over 2}$

$ \Rightarrow $$[3\,{({\log _3}x)^2} + 4{\log _3}x - 5]\,{\log _3}x - 2 = 0$

$ \Rightarrow $$3{t^3} + 4{t^2} - 5t - 2 = 0$, [$t = {\log _3}x$]

$ \Rightarrow $$3{t^3} - 3{t^2} + 7{t^2} - 7t + 2t - 2 = 0$

$ \Rightarrow $$(3{t^2} + 7t + 2)\,(t - 1) = 0$$ \Rightarrow $ $(3t + 1)\,(t + 2)\,(t - 1) = 0$

$ \Rightarrow $$t = 1,\, - 2,\, - {1 \over 3}$$ \Rightarrow $${\log _3}x = 1,\, - 2,\, - {1 \over 3}$

$ \Rightarrow $$x = {3^1},\,{3^{ - 2}},{3^{ - 1/3}}$; $x = 3,\,{1 \over 9},\,{1 \over {\root 3 \of 3 }}$

Thus, there is  $one +ve$ integral value, one irrational value, two positive rational values.

$y = {\log _7}2058 = {\log _7}({7^3}.6) = 3 + {\log _7}6$

As ${\log _5}8 > {\log _5}5$ i.e., ${\log _5}8 > 1$. $x > 4$

And ${\log _7}6 < {\log _7}7$ i.e., ${\log _7}6 < 1$

$\therefore y < 4$;

$\therefore x > y$.

${\log _{1/\sqrt 2 }}\sin x > 0$, $x \in [0,\,4\pi ]$$ \Rightarrow $ $0 < \sin x < 1$

$\therefore$ Integral multiple of ${\pi \over 4}$ will be ${\pi \over 4},\,{{3\pi } \over 4},\,{{9\pi } \over 4},{{11\pi } \over 4}$

Number of required values $= 4.$

For log to be defined, ${x^2} - 6x + 12 > 0$

$ \Rightarrow $${(x - 3)^2} + 3 > 0$, which is true $\forall x \in R$.

From $(i),$ ${x^2} - 6x + 12 \le {\left( {{1 \over 2}} \right)^{ - 2}}$

$ \Rightarrow $${x^2} - 6x + 12 \le 4$ $ \Rightarrow $ ${x^2} - 6x + 8 \le 0$

$ \Rightarrow $ $(x - 2)(x - 4) \le 0$ $ \Rightarrow $ $2 \le x \le 4$;

$\therefore x \in [2,\,4]$.

$ \Rightarrow $ ${(x - 1)^2} > x + 5$$ \Rightarrow $${x^2} - 3x - 4 > 0$

$ \Rightarrow $ $(x - 4)\,(x + 1) > 0$$ \Rightarrow $$x > 4$ or $x < - 1$

But for ${\log _{\sqrt 2 }}(x - 1)$ to be defined, $x - 1 > 0$ i.e., $x > 1$

$\therefore x > 4 \Rightarrow x \in (4,\,\infty )$.

For log to be defined $x - 1 > 0 \Rightarrow x > 1$

From $(i),$ ${\log _{{{(0.2)}^2}}}(x - 1) \ge {\log _{0.2}}(x - 1)$

$ \Rightarrow $${1 \over 2}{\log _{0.2}}(x - 1) \ge {\log _{0.1}}(x - 1)$

$ \Rightarrow $$\sqrt {x - 1} \le (x - 1)$

$ \Rightarrow $$\sqrt {x - 1} (1 - \sqrt {x - 1} ) \le 0$ $ \Rightarrow $ $1 - \sqrt {x - 1} \le 0$

$ \Rightarrow $$\sqrt {x - 1} \ge 1$ $ \Rightarrow $ $x \ge 2$,

$\therefore \,\,x \in [2,\,\infty )$ .