Quadratic Equation and Inequalities

Given $\alpha + \beta = - \sin \theta $ and$\alpha \beta = - 2\sin \theta $

${{\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){\alpha ^{12}}{\beta ^{12}}} \over {\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){{\left( {\alpha - \beta } \right)}^{24}}}} = {{{{\left( {\alpha \beta } \right)}^{12}}} \over {{{\left( {\alpha - \beta } \right)}^{24}}}}$

$\left| {\alpha - \beta } \right| = \sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } = \sqrt {{{\sin }^2}\theta + 8\sin \theta } $

Hence required quantity

${{{{\left( {\alpha \beta } \right)}^{12}}} \over {{{\left( {\alpha - \beta } \right)}^{24}}}} = {{{{\left( {2\sin \theta } \right)}^{12}}} \over {{{\sin }^{12}}\theta {{\left( {\sin \theta + 8} \right)}^{12}}}} = {{{2^{12}}} \over {{{\left( {\sin \theta + 8} \right)}^{12}}}}$

Given quadratic equation

(m² + 1) x² – 3x + (m² + 1)² = 0

Let roots of the equation $\alpha $ and $\beta $.

$ \therefore $ Sum of roots = $\alpha $ + $\beta $ = ${3 \over {{m^2} + 1}}$

Product of roots = $\alpha $$\beta $ = m² + 1

${3 \over {{m^2} + 1}}$ is maximum when m = 0

Hence equation becomes x² – 3x + 1 = 0

$\alpha + \beta = 3$,    $\alpha \beta = 1$   

$\left| {\alpha - \beta } \right|$ = $\sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } $ = $\sqrt {{{\left( 3 \right)}^2} - 4.1} $ = $\sqrt 5 $

$\left| {{\alpha ^3} - {\beta ^3}} \right| = \left| {(\alpha - \beta )({\alpha ^2} + {\beta ^2} + \alpha \beta )} \right| $

= $\sqrt 5 \left| {{{\left( {\alpha + \beta } \right)}^2} - \alpha \beta } \right|$

$= \sqrt 5 (9 - 1) = 8\sqrt 5 $

If a quadratic equation with rational coefficient has one irrational root then other root will be the conjugate of the irrational root.

Here x² + px + q = 0 has one root 2 - $\sqrt 3$.

$ \therefore $ Other root will be 2 + $\sqrt 3$.

Sum of the roots = -p = 4

and product of the roots = q = 1

You can see p² – 4q – 12 = 0 satisfy value of p = -4 and q = 1.

(1 + m² )x² – 2(1 + 3m)x + (1 + 8m) = 0

Given equation has no real solution,

$ \therefore $ Discriminant (D) < 0

$ \Rightarrow $ 4(1 + 3m)² - 4(1 + m²)(1 + 8m) < 0

$ \Rightarrow $ 4[9m² + 6m + 1 - 8m - 1 - 8m³ - m²] < 0

$ \Rightarrow $ -8m³ + 8m² - 2m < 0

$ \Rightarrow $ -2m(4m² - 4m + 1) < 0

$ \Rightarrow $ m(2m - 1)² > 0

$ \therefore $ m > 0 and m $ \ne $ ${{1 \over 2}}$

So we can say number of integral values of m are infinitely many.

Case 1 : When $\sqrt x \ge 2$

then $\left| {\sqrt x - 2} \right| = \sqrt x - 2$

$ \therefore $ The given equation becomes,

$\left( {\sqrt x - 2} \right)$ + $\sqrt x \left( {\sqrt x - 4} \right) + 2$ = 0

$ \Rightarrow $ $\left( {\sqrt x - 2} \right)$ + $x - 4\sqrt x $ + 2 = 0

$ \Rightarrow $ $x - 3\sqrt x $ = 0

$ \Rightarrow $ $\sqrt x \left( {\sqrt x - 3} \right)$ = 0

$ \therefore $ $\sqrt x $ = 0 or 3

$\sqrt x $ = 0 is not possible as $\sqrt x \ge 2$.

So, $\sqrt x $ = 3

or $x$ = 9

Case 2 : When $\sqrt x < 2$

then $\left| {\sqrt x - 2} \right| = $$ - \left( {\sqrt x - 2} \right)$ = $2 - \sqrt x $

$ \therefore $ The given equation becomes,

$\left( {2 - \sqrt x } \right)$ + $\sqrt x \left( {\sqrt x - 4} \right) + 2$ = 0

$ \Rightarrow $ ${2 - \sqrt x }$ + $x - 4\sqrt x $ + 2 = 0

$ \Rightarrow $ $x - 5\sqrt x + 4$ = 0

$ \Rightarrow $ $x - 4\sqrt x - \sqrt x + 4$ = 0

$ \Rightarrow $ $\sqrt x \left( {\sqrt x - 4} \right)$$-\left( {\sqrt x - 4} \right)$ = 0

$ \Rightarrow $ $\left( {\sqrt x - 4} \right)$$\left( {\sqrt x - 1} \right)$ = 0

$ \therefore $ $\sqrt x $ = 4 or 1

$\sqrt x $ = 4 is not possible as $\sqrt x < 2$.

$ \therefore $ $\sqrt x $ = 1

or $x$ = 1

So, Sum of all solutions = 9 + 1 = 10

Expression is always positive it

2m + 1 > 0 $ \Rightarrow $ m > $-$ ${1 \over 2}$ &

D < 0 $ \Rightarrow $ m² $-$ 6m $-$ 3 < 0

      3 $-$ $\sqrt {12} $ < m < 3 + $\sqrt {12} $ . . . . (iii)

$ \therefore $  Common interval is

3 $-$ $\sqrt {12} $ < m < 3 + $\sqrt {12} $

$ \therefore $  Intgral value of m {0, 1, 2, 3, 4, 5, 6}

3m²x² + m(m $-$ 4) x + 2 = 0

$\lambda + {1 \over \lambda } = 1,{\alpha \over \beta } + {\beta \over \alpha } = 1,{\alpha ^2} + {\beta ^2} = \alpha \beta $

($\alpha $ + $\beta $)² = 3$\alpha $$\beta $

${\left( { - {{m\left( {m - 4} \right)} \over {3{m^2}}}} \right)^2} = {{3\left( 2 \right)} \over {3{m^2}}},{{{{\left( {m - 4} \right)}^2}} \over {9{m^2}}} = {6 \over {3m}}$

${\left( {m - 4} \right)^2} = 18,m = 4 \pm \sqrt {18,} \,\,4 \pm 3\sqrt 2 $

D = (1 + sin$\theta $ cos$\theta $)² $-$ 4sin$\theta $cos$\theta $ = (1 $-$ sin$\theta $ cos$\theta $)²

$ \Rightarrow $  roots are $\beta $ = cosec$\theta $ and $\alpha $ = cos$\theta $

$\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + {{\left( { - {1 \over \beta }} \right)}^n}} \right)} = \sum\limits_{n = 0}^\infty {{{\left( {\cos \theta } \right)}^n}} + \sum\limits_{n = 0}^n {{{\left( { - \sin \theta } \right)}^n}} $

$ = {1 \over {1 - \cos \theta }} + {1 \over {1 + \sin \theta }}$

81x² + kx + 256 = 0 ; x = $\alpha $, $\alpha $³

$ \Rightarrow $  $\alpha $⁴ = ${{256} \over {81}}$ $ \Rightarrow $   $\alpha $ = $ \pm $ ${{4} \over {3}}$

Now   $-$ ${k \over {81}}$ = $\alpha $ + $\alpha $³ = $ \pm $ ${{100} \over {27}}$

$ \Rightarrow $  k = $ \pm $300

$\alpha $ + $\beta $ = $\lambda $ $-$ 3

$\alpha $$\beta $ = 2 $-$ $\lambda $

$\alpha $² + $\beta $² = ($\alpha $ + $\beta $)² $-$ 2$\alpha $$\beta $ = ($\lambda $ $-$ 3)² $-$ 2$\left( {2 - \lambda } \right)$

= $\lambda $² + 9 $-$ 6$\lambda $ $-$ 4 + 2$\lambda $

= $\lambda $² $-$ 4$\lambda $ + 5

= ($\lambda $ $-$ 2)² + 1

$ \therefore $  $\lambda $ = 2

Let f(x) = (c $-$ 5)x² $-$ 2cx + c $-$ 4

$ \therefore $  f(0)f(2) < 0      . . . . .(1)

& f(2)f(3) < 0      . . . . .(2)

from (1) and (2)

(c $-$ 4)(c $-$ 24) < 0

& (c $-$ 24)(4c $-$ 49) < 0

$ \Rightarrow $  ${{49} \over 4}$ < c < 24

$ \therefore $  s = {113, 14, 15, . . . . . 23}

Number of elements in set S = 11

x² $-$mx + 4 = 0

Case-I :

D > 0

m² $-$ 16 > 0

$ \Rightarrow $  m $ \in $ ($-$ $\infty $, $-$ 4) $ \cup $ (4, $\infty $)

Case-II :

$ \Rightarrow \,\,1 < {{ - b} \over {2a}} < 5$

$ \Rightarrow \,\,1 < {m \over 2} < 5 \Rightarrow \,m \in \left( {2,10} \right)$

Case-III :

f(1) > 0   and f(5) > 0

1 $-$ m + 4 > 0   and 25 $-$ 5m + 4 > 0

m < 5   and m < ${{29} \over 5}$

Case-IV :

Let one root is x = 1

1 $-$ m + 4 = 0

m = 5

Now equation x² $-$ 5x + 4 = 0

(x $-$ 1) (x $-$ 4) = 0

x = 1 i.e. m = 5 is also included

hence m $ \in $ (4, 5]

So given option is (4, 5)

For rational D must be perfect square

D = 121 $-$ 24$\alpha $

for 121 $-$ 24$\alpha $ to be perfect square a must be 3, 4, 5

So, ans $\alpha $ = 3

Here, 9x² + 27x + 20 = 0

$\therefore\,\,\,$ x = ${{ - b \pm \sqrt {{b^2} - 4ac} } \over {2a}}$

$ \Rightarrow $$\,\,\,$ x = ${{ - 27 \pm \sqrt {{{27}^2} - 4 \times 9 \times 20} } \over {2 \times 9}}$

$ \Rightarrow $$\,\,\,$ x = $-$ ${4 \over 3}$, $-$ ${5 \over 3}$

Given, cosA = $-$ ${3 \over 5}$

$\therefore\,\,\,$ sec A = ${1 \over {\cos A}}$ = $-$ ${5 \over 3}$

Here, A is an obtuse angle.

$\therefore\,\,\,$ tan A = $-$ $\sqrt {{{\sec }^2}A - 1} = - {4 \over 3}.$

Hence, roots of the equation are sec A and tan A.

Given,

${1 \over {x + p}} + {1 \over {x + q}} = {1 \over r}$

$ \Rightarrow $$\,\,\,$ ${{x + p + x + q} \over {\left( {x + p} \right)\left( {x + q} \right)}} = {1 \over r}$

$ \Rightarrow $$\,\,\,$ (2x + p + q) r = x² + px + qx + pq

$ \Rightarrow $$\,\,\,$ x² + (p + q $-$ 2r) x + pq $-$ pr $-$ qr = 0

Let $\alpha $ and $\beta $ are the roots,

$\therefore\,\,\,$ $\alpha $ + $\beta $ = $-$ (p + q $-$ 2r)

and $\alpha $ $\beta $ = pq $-$ pr $-$ qr

Given that, $\alpha $ = $-$ $\beta $ $ \Rightarrow $ $\alpha $ + $\beta $ = 0

$\therefore\,\,\,$ $-$ (p + q $-$ 2r) = 0

Now, $\alpha $² + $\beta $²

= ($\alpha $ + $\beta $)² $-$ 2$\alpha $ $\beta $

= ($-$ (p + q $-$ 2r))² $-$ 2 (pq $-$ pr $-$ qr)

= p² +q² + 4r² + 2pq $-$ 4pr $-$ 4qr $-$ 2pq + 2pr + 2qr

= p² + q² + 4r² $-$ 2pr $-$ 2qr

= p² + q² $-$ 2r (p + q $-$ 2r)

= p² + q² $-$ 2r (0)

= p² + q²

Given,

$2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$

Case 1 :

When $\sqrt x - 3 \ge 0,$ then equation becomes

$2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$

$ \Rightarrow \,\,\,\,2\sqrt x - 6 + x - 6\sqrt x + 6 = 0$

$ \Rightarrow \,\,\,\,x\, - 4\sqrt x = 0$

$ \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 4} \right) = 0$

$\therefore\,\,\,$ $\sqrt x = 0,4$

but as $\sqrt x - 3 \ge 0$ or $\sqrt x \ge 3$ then $\sqrt x \ne 0$

$\therefore\,\,\,$ $\sqrt x = 4$ value is possible.

Case 2 :

When $\sqrt x - 3 < 0$ or $\sqrt x < 3.$ There equation becomes

$ - 2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$

$ \Rightarrow \,\,\,\,\, - 2\sqrt x + 6 + x - 6\sqrt x + 6 = 0$

$ \Rightarrow \,\,\,\,x - 8\sqrt x + 12 = 0$

$ \Rightarrow \,\,\,\,x - 6\sqrt x - 2\sqrt x + 12 = 0$

$ \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 6} \right) - 2\left( {\sqrt x - 6} \right) = 0$

$ \Rightarrow \,\,\,\,\left( {\sqrt x - 2} \right)\left( {\sqrt x - 6} \right) = 0$

$\therefore\,\,\,$ $\sqrt x = 2,6$

as $\sqrt x < 3$ so $\sqrt x \ne 6$

$\therefore\,\,\,$ $\sqrt x = 2$ is possible.

So, total possible value of $\sqrt x = 2,4$

or for x possible values are 4, 16.

$\therefore\,\,\,$ Set S contains exactly two elements 4 and 16.

Let $\alpha $ and $\beta $ = - 1 are the roots of the polynomial, then we get

f(x) = x² + (1 - $\alpha $)x - $\alpha $

$ \therefore $ f(1) = 2 - 2$\alpha $

and f(2) = 6 - 3$\alpha $

Also given,

f (1) + f (2) = 0

$ \therefore $ 2 - 2$\alpha $ + 6 - 3$\alpha $ = 0

$ \Rightarrow $ $\alpha $ = ${8 \over 5}$

Let $\alpha $, $\beta $ are the roots of the equation,

$ \therefore $ $\alpha $ + $\beta $ = $\lambda $ $-$ 2 and $\alpha $$\beta $ = 10 $-$ $\lambda $

${\alpha ^3} + {\beta ^3}$ = ($\alpha $ + $\beta $)³ $-$ 3$\alpha $$\beta $ ($\alpha $ + $\beta $)

= ($\lambda $ $-$ 2)³ $-$ 3(10 $-$ $\lambda $)($\lambda $ $-$ 2)

= $\lambda ^3$ $-$ 3$\lambda ^2$ $-$ 24$\lambda $ + 52

Let $f(\lambda $) = $\lambda ^3$ $-$ 3$\lambda ^2$ $-$ 24$\lambda $ + 52

$ \therefore $ ${{df(\lambda )} \over {d\lambda }}$ = 3$\lambda ^2$ $-$ 6$\lambda $ $-$ 24

$ \therefore $ at maximum of minimum ${{df(\lambda )} \over {d\lambda }}$ = 0

$ \therefore $ $\lambda ^2$ $-$ 2$\lambda $ $-$ 8 = 0

$ \Rightarrow $ ($\lambda $ + 2) ($\lambda $ $-$ 4) = 0

$ \Rightarrow $ $\lambda $ = $-$2, 4

${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$ = 2$\lambda $ $-$ 2

When $\lambda $ = $-$2

${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$ = $-$ 6 < 0

$ \therefore $ at $\lambda $ = $-$2, f($\lambda $) has maximum value.

When $\lambda $ = 4

${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$ = 6 > 0

$ \therefore $ at $\lambda $ = 4, f($\lambda $) has minimum value.

$ \therefore $ When $\lambda $ = 4 equation is,

x² $-$ 2x + 6 = 0

$ \therefore $ ($\alpha $ $-$ $\beta $)² = ($\alpha $ + $\beta $)² $-$ 4$\alpha \beta$

$ \Rightarrow $ x² $-$ 4 $ \times $ 6

= $-$ 20

$ \Rightarrow $ ($\alpha $ $-$ $\beta $) = $2\sqrt 5 i$

$ \Rightarrow $ $\left| {\alpha - \beta } \right|$ = $2\sqrt 5$ (ans)

As tan A and tan B are the roots of 3x² $-$ 10x $-$ 25 = 0,

So, tan(A + B) = ${{\tan A + \tan B} \over {1 - \tan A\tan B}}$

= ${{{{10} \over 3}} \over {1 + {{25} \over 3}}}$ = ${{10/3} \over {28/3}}$ = ${5 \over {14}}$

Now, cos² (A + B) = $-$ 1 + 2 cos² (A + B)

= ${{1 - {{\tan }^2}(A + B)} \over {1 + {{\tan }^2}(A + B)}}\,$ $ \Rightarrow $ cos²(A + B) = ${{196} \over {221}}$

$\therefore\,\,\,$ 3sin²(A + B) $-$ 10sin(A + B)cos(A + B) $-$ 25 cos²(A + B)

= cos²(A + B) [ 3tan²(A + B) $-$ 10tan(A + B) $-$ 25]

= ${{75 - 700 - 4900} \over {196}} \times {{196} \over {221}}$

= $-$ ${{5525} \over {196}}$ $ \times $ ${{196} \over {221}}$ = $-$ 25

We know, 2^x = 1 only when x = 0.

Similarly, 2^{(x$-$1)(x2 + 5x $-$ 50)} = 1 when

(x$-$1)(x² + 5x $-$ 50) = 0

$ \Rightarrow $ (x - 1)(x + 10)(x - 5) = 0

$ \therefore $ x = 1, -10, 5

Sum of real values of x = 1 + (-10) + 5 = -4

Let, P(x) = ax² + bx + c

As, P(0) = 1,

$\therefore\,\,\,$ a(0)² + b(0) + c = 1

$ \Rightarrow $$\,\,\,$ c = 1

$\therefore\,\,\,$ P(x) = ax² + bx + 1

If   P(x) is divided by x $-$ 1, remainder = 4

$ \Rightarrow $$\,\,\,$ P$\left( 1 \right) = 4$

$\therefore\,\,\,$ a + b + 1 = 4 . . . . . (1)

If    P(x) is divided by x + 1, remainder = 6

$ \Rightarrow $$\,\,\,$ P($-$ 1) = 6

$\therefore\,\,\,$ a $-$ b + 1 = 6 . . . .(2)

By solving (1) and (2) we get,

a = 4, and b = $-$1

$\therefore\,\,\,$ P(x) = 4x² $-$ x + 1

P(2) = 4(2)² $-$ 2 + 1 = 15

P($-$ 2) = 4 ($-$2)² $-$ ($-$ 2) + 1 = 19

$\sum\limits_{r = 1}^n {\left( {x + r - 1} \right)\left( {x + r} \right)} = 10n$

$ \Rightarrow $ $\sum\limits_{r = 1}^n {\left( {{x^2} + xr + \left( {r - 1} \right)x + {r^2} - r} \right)} = 10n$

$ \Rightarrow $ $\sum\limits_{r = 1}^n {\left( {{x^2} + \left( {2r - 1} \right)x + r\left( {r - 1} \right)} \right)} = 10n$

$ \Rightarrow $ $n{x^2} + \left\{ {1 + 3 + 5 + .... + \left( {2n - 1} \right)} \right\}x$

$ + \left\{ {1.2 + 2.3 + ... + \left( {n - 1} \right)n} \right\}$ = 10n

$ \Rightarrow $ $n{x^2} + {n^2}x + {{n\left( {{n^2} - 1} \right)} \over 3} = 10n$

$ \Rightarrow $ ${x^2} + nx + {{\left( {{n^2} - 31} \right)} \over 3} = 0$

Let $\alpha $ and $\alpha $ + 1 be its two solutions

$ \therefore $ $\alpha $ + ($\alpha $ + 1) = -n

$ \Rightarrow $ $\alpha $ = ${{ - n - 1} \over 2}$ ....(1)

Also $\alpha $($\alpha $ + 1) = ${{\left( {{n^2} - 31} \right)} \over 3}$ ......(2)

Putting value of (1) in (2), we get

$ - \left( {{{n + 1} \over 2}} \right)\left( {{{1 - n} \over 2}} \right) = {{\left( {{n^2} - 31} \right)} \over 3}$

$ \Rightarrow $ n² = 121

$ \Rightarrow $ n = 11

Given,

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

$ \Rightarrow $   $\sqrt {2x + 1} = 1 + \sqrt {2x - 1} $

Squaring both sides, we get

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

$ \Rightarrow $   1 $=$ 2$\sqrt {2x - 1} $

$ \Rightarrow $   1 $=$ 4(2x $-$ 1)

$ \Rightarrow $   8x $-$ 4 $=$ 1

$ \Rightarrow $   x $=$ ${5 \over 8}$

So,    $\sqrt {4{x^2} - 1} $

$ = \sqrt {4\left( {{{25} \over {64}}} \right) - 1} $

$ = \sqrt {{{36} \over {64}}} $

$ = {6 \over 8}$

$ = {3 \over 4}$

Given,

x² + bx $-$ 1 = 0 . . . . .(1)

and x² + x + b = 0 . . . . . (2)

Performing (1) $-$ (2) we get,

bx $-$ 1 $-$ x $-$ b = 0

$ \Rightarrow $    x(b $-$ 1) = b + 1

$ \Rightarrow $   x = ${{b + 1} \over {b - 1}}$

putting value of x in equation (2),

${\left( {{{b + 1} \over {b - 1}}} \right)^2} + \left( {{{b + 1} \over {b - 1}}} \right) + b = 0$

$ \Rightarrow $   (b + 1)² + (b + 1) (b $-$ 1) + b (b $-$ 1)² = 0

$ \Rightarrow $    b² + 2b + 1 + b² $-$ 1 + b (b² $-$ 2b + 1) = 0

$ \Rightarrow $    2b³ + 2b + b³ $-$ 2b² + b = 0

$ \Rightarrow $     b³ + 3b = 0

$ \Rightarrow $    b(b² + 3) = 0

b² = $-$ 3, b = 0

$ \therefore $    b = $ \pm \sqrt 3 i$

$ \Rightarrow $    $\left| b \right|$ = $\sqrt 3 $

Given equation, ${\left( {{x^2} - 5x + 5} \right)^{{x^2} + 4x - 60}} = 1$

Case 1 : When x² - 5x + 5 = 1 and x² + 4x - 60 is any real no then this equation satisfy.
Note : When we put any real number as a power of 1 the value stays always 1 (1 ^{any real no} = 1).
x² - 5x + 5 = 1
(x - 1)(x - 4) = 0
$\therefore$ x = 1, 4

Case 2 : When x² - 5x + 5 is a real no and x² + 4x - 60 = 0 then the given equation satisfy. As we know if power of any real no is zero then it will become 1((any real number)⁰ = 1).
For, x² + 4x - 60 = 0
(x - 6)(x + 10) = 0
$\therefore$ x = 6, -10

Case 3 : When x² - 5x + 5 = -1 and x² + 4x - 60 is even this equation satisfy. As we know (-1)^even = 1.
For, x² - 5x + 5 = -1
(x - 2)(x - 3) = 0
$\therefore$ x = 2, 3
But x can't be 3 because when x = 3 the value of x² + 4x - 60 becomes 3² + 4.3 - 60 = - 39 which is an odd number, then (-1)^-39 = -1. So for x = 3 equation does not satisfy.

$\therefore$ The sum of all the real values = 1 + 4 + 6 + (-10) + 2 = 3

Given equation, x² - 6x - 2 = 0

Roots are $\alpha $ and $\beta $.

So, $\alpha + \beta = 6$ and $\alpha \beta = - 2$

In the question given, ${a_n} = {\alpha ^n} - {\beta ^n}$

$\therefore$ ${a_8} = {\alpha ^8} - {\beta ^8}$

and ${a_9} = {\alpha ^9} - {\beta ^9}$

and ${a_{{10}}} = {\alpha ^{{10}}} - {\beta ^{10}}$

Now, the given equation

${{{a_{10}} - 2{a_8}} \over {2{a_9}}}$

= ${{{\alpha ^{10}} - {\beta ^{10}} - 2\left( {{\alpha ^8} - {\beta ^8}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$

=${{{\alpha ^{10}} - {\beta ^{10}} + \alpha \beta \left( {{\alpha ^8} - {\beta ^8}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$ (as $\alpha \beta = - 2$)

=${{{\alpha ^{10}} - {\beta ^{10}} + {\alpha ^9}\beta - \alpha {\beta ^9}} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$

= ${{{\alpha ^9}\left( {\alpha + \beta } \right) - {\beta ^9}\left( {\alpha + \beta } \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$

= ${{\left( {\alpha + \beta } \right)\left( {{\alpha ^9} - {\beta ^9}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$

= ${{\left( {\alpha + \beta } \right)} \over 2}$

= ${6 \over 2}$ (as ${ {\alpha + \beta } }$ = 6)

= 3

Let $p,q,r$ are in $AP$

$ \Rightarrow 2q = p + r\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

Given ${1 \over \alpha } + {1 \over \beta } = 4 \Rightarrow {{\alpha + \beta } \over {\alpha \beta }} = 4$

We have $\alpha + \beta = - q/p$ and $\alpha \beta = {r \over p}$

$ \Rightarrow {{ - {q \over p}} \over {{r \over p}}} = 4 \Rightarrow q = - 4r\,\,\,\,\,\,\,\,...\left( {ii} \right)$

From $(i),$ we have

$2\left( { - 4r} \right) = p + r \Rightarrow p = - 9r$

$q = - 4r$

Now $\left| {\alpha - \beta } \right| = \sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } $

$ = \sqrt {{{\left( {{{ - q} \over p}} \right)}^2} - {{4r} \over p}} $

$ = {{\sqrt {{q^2} - 4pr} } \over {\left| p \right|}}$

$ = {{\sqrt {16{r^2} + 36{r^2}} } \over {\left| { - 9r} \right|}}$

$ = {{2\sqrt {13} } \over 9}$

Given, $ - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$
As we know, $\left[ x \right] + \left\{ x \right\} = x$
where $\left[ x \right]$ is integral part and $\left\{ x \right\}$ is fractional part.
$\therefore$$\left\{ x \right\} = x - \left[ x \right]$
Now put $\left\{ x \right\}$ inplace of $x - \left[ x \right]$ in the equation.
The new equation is $ - 3{\left\{ x \right\}^2} + 2\left\{ x \right\} + {a^2} = 0$

[Note : Question says this equation has no integral solution, it means $\left\{ x \right\} \ne $ 0. So, $x$ is not a integer.]

$\therefore$ $\left\{ x \right\}$ = ${{ - 2 \pm \sqrt {4 - 4 \times \left( { - 3} \right){a^2}} } \over { - 6}}$
             = ${{ - 2 \pm \sqrt {4 + 12{a^2}} } \over { - 6}}$
As $\left\{ x \right\}$ is fractional part so it is lies between 0 to 1($0 \le \left\{ x \right\} < 1$).

By considering positive sign, we get

$0 \le {{ - 2 + \sqrt {4 + 12{a^2}} } \over { - 6}} < 1$

$ \Rightarrow $$0 \ge - 2 + \sqrt {4 + 12{a^2}} > - 6$

$ \Rightarrow $$2 \ge + \sqrt {4 + 12{a^2}} > - 4$

$\because$$ + \sqrt {4 + 12{a^2}} $ is always positive which is greater than any negative no. So can ignore the inequality $ + \sqrt {4 + 12{a^2}} > - 4$

Consider this inequality,
$2 \ge + \sqrt {4 + 12{a^2}} $
$ \Rightarrow $ $4 \ge 4 + 12{a^2}$
$ \Rightarrow $ $12{a^2} \le 0$
$ \Rightarrow $ ${a^2} \le 0$
$ \Rightarrow $ ${a^2} = 0$
$ \Rightarrow $ ${a} = 0$

If $a$ = 0 then $ - 3{\left\{ x \right\}^2} + 2\left\{ x \right\} = 0$ so $\left\{ x \right\}$ becomes 0 but question says $\left\{ x \right\}$ $ \ne $ 0.
So $a$ can't be 0.

Now by considering negative sign, we get


$0 \le {{ - 2 - \sqrt {4 + 12{a^2}} } \over { - 6}} < 1$

$ \Rightarrow $$0 \ge - 2 - \sqrt {4 + 12{a^2}} > - 6$

$ \Rightarrow $$2 \ge - \sqrt {4 + 12{a^2}} > - 4$

As 2 is always greater than ${ - \sqrt {4 + 12{a^2}} }$. Ignore this inequality.

Now consider this inequality,

$ - \sqrt {4 + 12{a^2}} > - 4$

$ \Rightarrow $ $\sqrt {4 + 12{a^2}} < 4$

$ \Rightarrow $ $4 + 12{a^2} < 16$

$ \Rightarrow $ $12{a^2} < 12$

$ \Rightarrow $ ${a^2} < 1$

$ \Rightarrow $ $\left( {{a^2} - 1} \right) < 0$

$ \Rightarrow $ $\left( {a + 1} \right)\left( {a - 1} \right) < 0$

$ \Rightarrow $ $ - 1 < a < 1$

But earlier we found that $a$ $ \ne $ 0.

So, the range of $a$ is = $\left( { - 1,0} \right) \cup \left( {0,1} \right)$

Given equations are

$\,\,\,\,\,\,\,\,\,\,\,\,{x^2} + 2x + 3 = 0\,\,\,\,\,...\left( i \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,a{x^2} + bx + c = 0\,\,\,...\left( {ii} \right)$

Roots of equation $(i)$ are imaginary roots.

According to the question $(ii)$ will also have both roots same as $(i).$

Thus ${a \over 1} = {b \over 2} = {c \over 3} = \lambda \left( {say} \right)$

$ \Rightarrow a = \lambda ,b = 2\lambda ,c = 3\lambda $

Hence, required ratio is $1:2:3$

Given equation is ${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$

Put ${e^{{\mathop{\rm sinx}\nolimits} \,}} = t$ in the given equation,

we get ${t^2} - 4t - 1 = 0$

$ \Rightarrow t = {{4 \pm \sqrt {16 + 4} } \over 2}$

$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm \sqrt {20} } \over 2}$

$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm 2\sqrt 5 } \over 2}$

$\,\,\,\,\,\,\,\,\,\,\, = 2 \pm \sqrt 5 $

$ \Rightarrow {e^{\sin x}} = 2 \pm \sqrt 5 $ $\,\,\,\,\,$ (as $t = {e^{\sin x}}$)

$ \Rightarrow {e^{\sin x}} = 2 - \sqrt 5 $ and

$\,\,\,\,\,\,\,\,\,\,\,\,\,$ ${e^{\sin x}} = 2 + \sqrt 5 $

$ \Rightarrow {e^{\sin x}} = 2 - \sqrt 5 < 0$

and $\,\,\,\,\,\,\sin x = \ln \left( {2 + \sqrt 5 } \right) > 1$ So, rejected

Hence given equation has no solution.

$\therefore$ The equation has no real roots.

${x^2} - x + 1 = 0$

$ \Rightarrow x = {{1 \pm \sqrt {1 - 4} } \over 2}$

$x = {{1 \pm \sqrt 3 i} \over 2}$

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

$\beta = {1 \over 2} - {{i\sqrt 3 } \over 2} = - \omega $

${\alpha ^{2009}} + {\beta ^{2009}} = {\left( { - {\omega ^2}} \right)^{2009}} + {\left( { - \omega } \right)^{2009}}$

$ = - {\omega ^2} - \omega = 1$

Given that roots of the equation

$b{x^2} + cx + a = 0$ are imaginary

$\therefore$ ${c^2} - 4ab < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

Let $y = 3{b^2}{x^2} + 6bc\,x + 2{c^2}$

$ \Rightarrow 3{b^2}{x^2} + 6bc\,x + 2{c^2} - y = 0$

As $x$ is real, $D \ge 0$

$ \Rightarrow 36{b^2}{c^2} - 12{b^2}\left( {2{c^2} - y} \right) \ge 0$

$ \Rightarrow 12{b^2}\left( {3{c^2} - 2{c^2} + y} \right) \ge 0$

$ \Rightarrow {c^2} + y \ge 0$

$ \Rightarrow y \ge - {c^2}$

But from eqn. $(i),$ ${c^2} < 4ab$

or $ - {c^2} > - 4ab$

$\therefore$ we get $y \ge - {c^2} > - 4ab$

$y > - 4ab$

Let the roots of equation ${x^2} - 6x + a = 0$ be $\alpha $

and $4$ $\beta $ and that of the equation

${x^2} - cx + 6 = 0$ be $\alpha $ and $3\beta .$ Then

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + 4\beta = 6;\,\,\,\,\,\,\,4\alpha \beta = a$

and $\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + 3\beta = c;\,\,\,\,\,\,\,3\alpha \beta = 6$

$ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 8$

$\therefore$ The equation becomes

${x^2} - 6x + 8 = 0$

$ \Rightarrow \left( {x - 2} \right)\left( {x - 4} \right) = 0$

$ \Rightarrow $ roots are $2$ and $4$

$ \Rightarrow \alpha = 2,\beta = 1$

$\therefore$ Common root is $2.$

Statements $2$ is $\sqrt {n\left( {n + 1} \right)} < n + 1,n \ge 2$

$ \Rightarrow \sqrt n < \sqrt {n + 1} ,n \ge 2$ which is true

$ \Rightarrow \sqrt 2 < \sqrt 3 < \sqrt 4 < \sqrt 5 < - - - - - - \sqrt n $

Now $\sqrt 2 < \sqrt n \Rightarrow {1 \over {\sqrt 2 }} > {1 \over {\sqrt n }}$

$\sqrt 3 < \sqrt n \Rightarrow {1 \over {\sqrt 3 }} > {1 \over {\sqrt n }};$

$\sqrt n \le \sqrt n \Rightarrow {1 \over {\sqrt n }} \ge {1 \over {\sqrt n }}$

Also ${1 \over {\sqrt 1 }} > {1 \over {\sqrt n }}$

$\therefore$ Adding all, we get

${1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + ....... + {1 \over n} > {n \over {\sqrt n }} = \sqrt n $

Hence both the statements are correct and statement $2$ is a correct explanation of statement $-1.$

Let $\alpha $ and $\beta $ are roots of the equation ${x^2} + ax + 1 = 0$

So, $\alpha + \beta = - a$ and $\alpha \beta = 1$

given $\left| {\alpha - \beta } \right| < \sqrt 5 $

$ \Rightarrow \sqrt {{{\left( {\alpha - \beta } \right)}^2} - 4\alpha \beta } < \sqrt 5 $

(as ${\left( {\alpha - \beta } \right)^2} = {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta $ )

$ \Rightarrow \sqrt {{a^2} - 4} < \sqrt 5 $

$ \Rightarrow {a^2} - 4 < 5$

$ \Rightarrow {a^2} - 9 < 0 \Rightarrow {a^2} < 9$

$ \Rightarrow - 3 < a < 3$

$ \Rightarrow a \in \left( { - 3,3} \right)$

$y = {{3{x^2} + 9x + 17} \over {3{x^2} + 9x + 7}}$

$3{x^2}\left( {y - 1} \right) + 9x\left( {y - 1} \right) + 7y - 17 = 0$

$D \ge 0$ as $x$ is real

$81{\left( {y - 1} \right)^2} - 4 \times 3\left( {y - 1} \right)\left( {7y - 17} \right) \ge 0$

$ \Rightarrow \left( {y - 1} \right)\left( {y - 41} \right) \le 0$

$ \Rightarrow 1 \le y \le 41$

$\therefore$ Max value of $y$ is $41$

${x^2} + px + q = 0$

Sum of roots $ = \tan {30^ \circ } + \tan {15^ \circ } = - p$

Products of roots $ = \tan {30^ \circ }.\tan {15^ \circ } = q$

$\tan {45^ \circ } = {{\tan {{30}^ \circ } + \tan {{15}^ \circ }} \over {1 - \tan {{30}^ \circ }.\tan {{15}^ \circ }}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{ - p} \over {1 - q}} = 1$

$ \Rightarrow - p = 1 - q \Rightarrow q - p = 1$

$\therefore$ $2 + q - p = 3$

Equation ${x^2} - 2mx + {m^2} - 1 = 0$

${\left( {x - m} \right)^2} - 1 = 0$

or $\left( {x - m + 1} \right)\left( {x - m - 1} \right) = 0$

$x = m - 1,m + 1$

$m - 1 > - 2$ and $m + 1 < 4$

$ \Rightarrow m > - 1$ and $m<3$

or $\,\,\, - 1 < m < 3$

$\angle $R = 90^o $ \therefore $ $\angle $P + $\angle $Q = 90^o

$ \Rightarrow $ ${P \over 2} + {Q \over 2} = {{90} \over 2} = 45$^o

$\tan \left( {{P \over 2}} \right),\tan \left( {{Q \over 2}} \right)$ are the roots of $a{x^2} + bx + c = 0$

$ \therefore $ $\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right) = - {b \over a},\,\,$

and $\tan \left( {{P \over 2}} \right).\tan \left( {{Q \over 2}} \right) = {c \over a}$

${{\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right)} \over {1 - \tan \left( {{P \over 2}} \right)\tan \left( {{Q \over 2}} \right)}}$

$ = \tan \left( {{P \over 2} + {Q \over 2}} \right) $= tan 45^o = 1

$ \Rightarrow {{ - {b \over a}} \over {1 - {c \over a}}} = 1$

$ \Rightarrow - {b \over a} = {a \over a} - {c \over a}$

$ \Rightarrow - b = a - c$ or $c = a + b.$

Given quadratic equation,

${x^2} - \left( {a - 2} \right)x - a - 1 = 0$

Let $\alpha $ and $\beta $ are the roots of the equation.

$ \therefore $ $\alpha $ + $\beta $ = $a - 2$

and $\alpha $$\beta $ = $ - a - 1$

Now ${\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta $

$ \Rightarrow $ ${\alpha ^2} + {\beta ^2} = {\left( {a - 2} \right)^2} + 2\left( {a + 1} \right)$

$ \Rightarrow $ ${\alpha ^2} + {\beta ^2} = {a^2} - 2a + 6$

$ \Rightarrow $ ${\alpha ^2} + {\beta ^2} = {\left( {a - 1} \right)^2} + 5$

$ \Rightarrow $ The value of ${\alpha ^2} + {\beta ^2}$ will be minimum, when ${a - 1}$ = 0

$ \Rightarrow $ ${a = 1}$

Let n and (n + 1) be the roots of x² $-$ bx + c = 0.

Then, n + (n + 1) = b and n(n + 1) = c

$\therefore$ b² $-$ 4c = (2n + 1)² $-$ 4n(n + 1)

= 4n² + 4n + 1 $-$ 4n² $-$ 4n = 1

Let $\alpha ,\,\,\alpha + 1\,\,$ be roots

Then $\alpha + \alpha + 1 = b = $ sum of -

roots $\alpha \left( {\alpha + 1} \right) = c$

$=$ product of roots

$\therefore$ ${b^2} - 4c $

$ = {\left( {2\alpha + 1} \right)^2} - 4\alpha \left( {\alpha + 1} \right)$

$ = 1.$

${x^2} - \left( {a - 2} \right)x - a - 1 = 0$

$ \Rightarrow \alpha + \beta = a - 2;\,\,\alpha \beta = - \left( {a + 1} \right)$

${\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta $

$ = {a^2} - 2a + 6 = {\left( {a - 1} \right)^2} + 5$

For min. value of ${\alpha ^2} + {\beta ^2}$ where $\alpha $ is an integer

$ \Rightarrow \,\,a = 1.$

both roots are less than $5,$

then $(i)$ Discriminant $ \ge 0$

$\left( {ii} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p\left( 5 \right) > 0$

$\left( {iii} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{Sum\,\,of\,\,roots} \over 2} < 5$

Hence $\left( i \right)\,\,\,\,\,\,4{k^2} - 4\left( {{k^2} + k - 5} \right) \ge 0$

$4{k^2} - 4{k^2} - 4k + 20 \ge 0$

$4k \le 20 \Rightarrow k \le 5$

$\left( {ii} \right)\,\,\,\,\,f\left( 5 \right) > 0;25 - 10k + {k^2} + k - 5 > 0$

or ${k^2} - 9k + 20 > 0$

or $k\left( {k - 4} \right) - 5\left( {k - 4} \right) > 0$

or $\left( {k - 5} \right)\left( {k - 4} \right) > 0$

$ \Rightarrow k \in \left( { - \infty ,4} \right) \cup \left( { - \infty ,5} \right)$

$\left( {iii} \right)\,\,\,\,\,\,{{Sum\,\,of\,\,roots} \over 2}$

$ = - {b \over {2a}} = {{2k} \over 2} < 5$

The intersection of $(i)$, $(ii)$ & $(iii)$ gives

$k \in \left( { - \infty ,4} \right).$

$4$ is a root of ${x^2} + px + 12 = 0$

$ \Rightarrow 16 + 4p + 12 = 0$

$ \Rightarrow p = - 7$

Now, the equation ${x^2} + px + q = 0$

has equal roots.

$\therefore$ ${p^2} - 4q = 0$ $ \Rightarrow q = {{{p^2}} \over 4} = {{49} \over 4}$

Let the second root be $\alpha .$

Then $\alpha + \left( {1 - p} \right) = - p \Rightarrow \alpha = - 1$

Also $\alpha .\left( {1 - p} \right) = 1 - p$

$ \Rightarrow \left( {\alpha - 1} \right)\left( {1 - p} \right) = 0$

$ \Rightarrow p = 1$ [as $\alpha = - 1$]

$\therefore$ Roots are $\alpha = - 1$ and $p-1=0$

Let two numbers be a and b then ${{a + b} \over 2} = 9$

and $\sqrt {ab} = 4$

$\therefore$ Equation with roots $a$ and $b$ is

${x^2} - \left( {a + b} \right)x + ab = 0$

$ \Rightarrow {x^2} - 18x + 16 = 0$

Let the roots of given equation be $\alpha $ and $2$$\alpha $ then

$\alpha + 2\alpha = 3\alpha = {{1 - 3a} \over {{a^2} - 5a + 3}}$

and $\alpha .2\alpha = 2{\alpha ^2} = {2 \over {{a^2} - 5a + 3}}$

$ \Rightarrow \alpha = {{1 - 3a} \over {3\left( {{a^2} - 5a + 3} \right)}}$

$\therefore$ $2\left[ {{1 \over 9}{{{{\left( {1 - 3a} \right)}^2}} \over {{{\left( {{a^2} - 5a + 3} \right)}^2}}}} \right]$

$ = {2 \over {{a^2} - 5a + 3}}$

${{{{\left( {1 - 3a} \right)}^2}} \over {\left( {{a^2} - 5a + 3} \right)}} = 9$

or $9{a^2} - 6a + 1$

$ = 9{a^2} - 45a + 27$

or $39a = 26$ or $a = {2 \over 3}$

${x^2} - 3\left| x \right| + 2 = 0$

$ \Rightarrow {\left| x \right|^2} - 3\left| x \right| + 2 = 0$

$\left( {\left| x \right| - 2} \right)\left( {\left| x \right| - 1} \right) = 0$

$\left| x \right| = 1,2$ or $x = \pm 1, \pm 2$

$\therefore$ No. of solution $=4$

$a{x^2} + bx + c = 0,$ $\alpha + \beta = {{ - b} \over a},\alpha \beta = {c \over a}$

As for given condition, $\alpha + \beta = {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}}$

$\alpha + \beta = {{{\alpha ^2} + {\beta ^2}} \over {{\alpha ^2}{\beta ^2}}} - {b \over a}$

$ = {{{{{b^2}} \over {{a^2}}} - {{2c} \over a}} \over {{{{c^2}} \over {{a^2}}}}}$

On simplification $2{a^2}c = a{b^2} + b{c^2}$

$ \Rightarrow {{2a} \over b} = {c \over a} + {b \over c}$

$ \Rightarrow {c \over a},{a \over b},{b \over c}$ are in $A.P.$

$\therefore$ ${a \over c},{b \over a},\,\,\& \,\,$ are in $H.P.$