Quadratic Equation and Inequalities

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

Given quadratic equation

${x^2} - x - 1 = 0$ having roots $\alpha $ and $\beta $, ($\alpha $ > $\beta $)

So, $\alpha = {{1 + \sqrt 5 } \over 2}$ and $\beta = {{1 - \sqrt 5 } \over 2}$

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

$ \because $ ${a_n} = {{{\alpha ^n} - {\beta ^n}} \over {\alpha - \beta }},\,n \ge 1$

So, ${a_{n + 1}} = {{{\alpha ^{n + 1}} - {\beta ^{n + 1}}} \over {\alpha - \beta }}$

${\alpha ^n} + {\alpha ^{n - 1}}\beta + {\alpha ^{n - 2}}{\beta ^2} + ... + \alpha {\beta ^{n - 1}} + {\beta ^n}$

${\alpha ^n} - {\alpha ^{n - 2}} - {\alpha ^{n - 3}}\beta - ... - {\beta ^{n - 2}} + {\beta ^n}$

[as $\alpha $$\beta $ = $ - $1]

= ${\alpha ^n} + {\beta ^n} - ({\alpha ^{n - 2}} + {\alpha ^{n - 3}}\beta + ... + {\beta ^{n - 2}})$

= ${\alpha ^n} + {\beta ^n} - {a_{n - 1}}$

$\left[ {as\,{a_{n - 1}} = {{{\alpha ^{n - 1}} - {\beta ^{n - 1}}} \over {\alpha - \beta }} = {\alpha ^{n - 2}} + {\alpha ^{n - 3}}\beta + ... + {\beta ^{n - 2}}} \right]$

$ \Rightarrow $ ${a_{n + 1}} + {a_{n - 1}} = {\alpha ^n} + {\beta ^n} = {b_n},\,\forall n \ge 1$

So, option (b) is correct.

Now, $\sum\limits_{n = 1}^\infty {{{{b_n}} \over {{{10}^n}}}} = \sum\limits_{n = 1}^\infty {{{{\alpha ^n} + {\beta ^n}} \over {{{10}^n}}}} $

[as, b_n = $\alpha $ⁿ + $\beta $ⁿ]

= $\sum\limits_{n = 1}^\infty {{{\left( {{\alpha \over {10}}} \right)}^n}} + \sum\limits_{n = 1}^\infty {{{\left( {{\beta \over {10}}} \right)}^n}} $

$ \because $ $\left[ {\left| {{\alpha \over {10}}} \right| < 1\,and\,\left| {{\beta \over {10}}} \right| < 1} \right]$

$ = {{{\alpha \over {10}}} \over {1 - {\alpha \over {10}}}} + {{{\beta \over {10}}} \over {1 - {\beta \over {10}}}} = {\alpha \over {10 - \alpha }} + {\beta \over {10 - \beta }}$

$ = {{10\alpha - \alpha \beta + 10\beta - \alpha \beta } \over {(10 - \alpha )(10 - \beta )}}$

$ = {{10(\alpha + \beta ) - 2\alpha \beta } \over {100 - 10(\alpha + \beta ) + \alpha \beta }}$

$ = {{10(1) - 2( - 1)} \over {100 - 10(1) - 1}}$

[as $\alpha $ + $\beta $ = 1 and $\alpha $$\beta $ = $ - $1]

$ = {{12} \over {89}}$

So, option (a) is not correct.

$ \because $ ${\alpha ^2} = \alpha + 1$ and ${\beta ^2} = \beta + 1$

$ \Rightarrow $${\alpha ^{n + 2}} = {\alpha ^{n + 1}} + {\alpha ^n}$ and ${\beta ^{n + 2}} = {\beta ^{n + 1}} + {\beta ^n}$

$ \Rightarrow $ $({\alpha ^{n + 2}} + {\beta ^{n + 2}}) = ({\alpha ^{n + 1}} + {\beta ^{n + 1}}) + ({\alpha ^n} + {\beta ^n})$

$ \Rightarrow {a_{n + 2}} = {a_{n + 1}} + {a_n}$

Similarly, ${a_{n + 1}} = {a_n} + {a_{n - 1}}$

${a_n} = {a_{n - 1}} + {a_{n - 2}}$

..............

............

${a_3} = {a_2} + {a_1}$

On adding, we get

${a_{n + 2}} = ({a_n} + {a_{n - 1}} + {a_{n - 2}} + ... + {a_2} + {a_1}) + {a_2}$

$ \because $ $\left[ {{a_2} = {{{\alpha ^2} - {\beta ^2}} \over {\alpha - \beta }} = \alpha + \beta = 1} \right]$

So, ${a_{n + 2}} - 1 = {a_1} + {a_2} + {a_3} + ...... + {a_n}$

So, option (c) is also correct.

And, now $\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{{10}^n}}}} = \sum\limits_{n = 1}^\infty {{{{\alpha ^n} - {\beta ^n}} \over {(\alpha - \beta ){{10}^n}}}} $

$ = {1 \over {\alpha - \beta }}\left[ {\sum\limits_{n = 1}^\infty {{{\left( {{a \over {10}}} \right)}^n}} - \sum\limits_{n = 1}^\infty {{{\left( {{\beta \over {10}}} \right)}^n}} } \right]$

$ = {1 \over {\alpha - \beta }}\left[ {{{{\alpha \over {10}}} \over {1 - {\alpha \over {10}}}} - {{{\beta \over {10}}} \over {1 - {\beta \over {10}}}}} \right]$,

$\left[ {as\left| {{\alpha \over {10}}} \right| < 1\,and\,\left| {{\beta \over {10}}} \right| < 1} \right]$

$ = {{10(\alpha - \beta )} \over {(\alpha - \beta )[100 - 10(\alpha + \beta ) + \alpha \beta ]}}$

$ = {{10} \over {100 - 10 - 1}} = {{10} \over {89}}$

Hence, options (b), (c) and (d) are correct.

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

$\alpha $² = $\alpha $ + 1

$\beta $² = $\beta $ + 1

a_n = p$\alpha $ⁿ + q$\beta $ⁿ

= p($\alpha $^n$-$1 + $\alpha $^n$-$2) + q($\beta $^n$-$1 + $\beta $^n$-$2)

= a_n$-$1 + a_n$-$2

$ \therefore $ a₁₂ = a₁₁ + a₁₀

$\alpha = {{1 + \sqrt 5 } \over 2}$,

$\beta = {{1 - \sqrt 5 } \over 2}$

${a_4} = {a_3} + {a_2}$

$ = 2{a_2} + {a_1}$

$ = 3{a_1} + 2{a_0}$


$28 = p(3\alpha + 2) + q(3\beta + 2)$

$28 = (p + q)\left( {{3 \over 2} + 2} \right) + (p - q)\left( {{{3\sqrt 5 } \over 2}} \right)$

$ \therefore $ p $-$ q = 0

and $(p + q) \times {7 \over 2} = 28$

$ \Rightarrow $ p + q = 8

$ \Rightarrow $ p = q = 4

$ \therefore $ p + 2q = 12

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, first equation $x^2-2 x \sec \theta+1=0$

Using quadratic equation formula we get,

$ \begin{aligned} x & =\frac{-(-2 \sec \theta) \pm \sqrt{(-2 \sec \theta)^2-4}}{2} \\\\ \Rightarrow & x=\frac{2 \sec \theta \pm \sqrt{4 \sec ^2 \theta-4}}{2} \\\\ \Rightarrow & x=\frac{2 \sec \theta \pm 2 \tan \theta}{2} \\\\ \Rightarrow & x=\sec \theta \pm \tan \theta \text { as } \theta \in\left(\frac{-\pi}{6}, \frac{-\pi}{2}\right) \\\\ \Rightarrow & \alpha_1=\sec \theta-\tan \theta \text { and } \beta_1=\sec \theta+\tan \theta \end{aligned} $

Now, for the equation $x^2+2 x \tan \theta-1=0$

$ \begin{aligned} & x=\frac{-2 \tan \theta \pm \sqrt{4 \tan ^2 \theta+4}}{2} \\\\ \Rightarrow & x=\frac{-2 \tan \theta \pm 2 \sec \theta}{2} \\\\ \Rightarrow & x=-\tan \theta \pm \sec \theta \\\\ \Rightarrow & x=(\sec \theta-\tan \theta),-(\sec \theta+\tan \theta) \end{aligned} $

Given, that $\alpha_2$ and $\beta_2$ are the roots of equation and $\alpha_2>\beta_2$

$ \begin{aligned} &\Rightarrow \alpha_2 =\sec \theta-\tan \theta, \beta_2=-(\sec \theta+\tan \theta) \\\\ &\Rightarrow \alpha_1+\beta_2 =(\sec \theta-\tan \theta)-(\sec \theta+\tan \theta) \\\\ & =\sec \theta-\tan \theta-\sec \theta-\tan \theta \\\\ &\Rightarrow \alpha_1+\beta_2 =-2 \tan \theta \end{aligned} $

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

Given, x₁ and x₂ are roots of

$\alpha {x^2} - x + \alpha = 0$

$\therefore$ ${x_1} + {x_2} = {1 \over \alpha }$ and ${x_1}{x_2} = 1$

Also, $\left| {{x_1} - {x_2}} \right| < 1$

$ \Rightarrow {\left| {{x_1} - {x_2}} \right|^2} < 1 \Rightarrow {({x_1} - {x_2})^2} < 1$

or, ${({x_1} + {x_2})^2} - 4{x_1}{x_2} < 1$

$ \Rightarrow {1 \over {{\alpha ^2}}} - 4 < 1$ or ${1 \over {{\alpha ^2}}} < 5$

$ \Rightarrow 5{\alpha ^2} - 1 > 0$

or, $(\sqrt 5 \alpha - 1)(\sqrt 5 \alpha + 1) > 0$

$\therefore$ $\alpha \in \left( { - \infty , - {1 \over {\sqrt 5 }}} \right) \cup \left( {{1 \over {\sqrt 5 }},\infty } \right)$ .....(i)

Also, $D > 0$

$ \Rightarrow 1 - 4{\alpha ^2} > 0$ or $\alpha \in \left( { - {1 \over 2},{1 \over 2}} \right)$ ...... (ii)

$\alpha \in \left( { - {1 \over 2},{{ - 1} \over {\sqrt 5 }}} \right) \cup \left( {{1 \over {\sqrt 5 }},{1 \over 2}} \right)$

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, a quadratic equation $p(x)=0$ with real coefficients has purely imaginary roots.

Let $i \lambda$ and $-i \lambda$ are the roots of $p(x)=0$ where $i=\sqrt{-1}$ and $\lambda$ is a real number except zero.

$\begin{aligned} & \therefore p(x)=a(x-i \lambda)(x+i \lambda) \\ & \Rightarrow p(x)=a\left(x^2+\lambda^2\right) \end{aligned}$

Now, $\quad p(p(x))=0$

$\Rightarrow \quad a\left((p(x))^2+\lambda^2\right)=0$

$\Rightarrow a\left[a^2\left(x^2+\lambda^2\right)^2+\lambda^2\right]=0$

$\Rightarrow a^2\left(x^2+\lambda^2\right)^2+\lambda^2=0$

$\Rightarrow \quad\left(x^2+\lambda^2\right)^2=-\frac{\lambda^2}{a^2}$

$\Rightarrow \quad x^2+\lambda^2= \pm i \frac{\lambda}{a}$

$\Rightarrow \quad x^2= \pm i \frac{\lambda}{a}-\lambda^2$

$\Rightarrow \quad x= \pm \sqrt{ \pm i \frac{\lambda}{a}-\lambda^2}$

Hence, $p(p(x))=0$ has four roots but all the roots are neither purely real nor purely imaginary.

Hint:

(i) In a quadratic equation imaginary roots are always in conjugate pair i.e, if one root is $p+i q$, then other root must be $p-i q$

(ii) A quadratic equation with real coefficients has purely imaginary roots, then consider $\pm i \lambda$ are the roots of the given quadratic, where $\lambda \in \mathrm{R}-\{0\}$.

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, $3^x=4^{x-1}$

Taking $\log$ on both side with base 3

$\begin{aligned} & \Rightarrow \log _3 3^x=\log _3\left(4^{x-1}\right) \\ & \Rightarrow x \log _3 3=(x-1) \log _3 4 \\ & \Rightarrow \quad x .1=(x-1) \log _3 4 \\ & \Rightarrow \quad x=\frac{\log _3 4}{\log _3 4-1} \\ & \Rightarrow \quad x=\frac{1}{1-\frac{1}{\log _3 4}} \\ \end{aligned}$

$\text { According to base changing rule } \frac{1}{\log _3 4}=\log _4 3$

$\begin{array}{ll} \therefore & x=\frac{1}{1-\log _4 3} \quad \text{... (i)}\\ \Rightarrow & x=\frac{1}{1-\log _{2^2} 3} \\ \Rightarrow & x=\frac{1}{1-\frac{1}{2} \log _2 3} \quad\left[\log _{a^m} x=\frac{1}{m} \log _a x\right] \\ \Rightarrow & x=\frac{2}{2-\log _2 3} \quad \text{... (ii)} \end{array}$

$\begin{aligned} & \Rightarrow \quad x=\frac{2}{2-\frac{1}{\log _3 2}} \\ & \Rightarrow \quad x=\frac{2 \log _3 2}{2 \log _3 2-1} \quad \text{... (iii)} \end{aligned}$

From equation (i), (ii) and (iii), it is clear that option (A), (B) and (C) are correct.

Hints:

(i) Base changing Rule $\log _b^a=\frac{1}{\log _a^b}$

(ii) $\log _a x^n=n \log _a x$

(iii) $\log _{a^m} x=\frac{1}{m} \log _a x$

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.

Let a + 1 = t⁶. Thus, when a $\to$ 0, t $\to$ 1.

$\therefore$ $({t^2} - 1){x^2} + ({t^3} - 1)x + (t - 1) = 0$

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

as t $\to$ 1

$2{x^2} + 3x + 1 = 0$

$ \Rightarrow 2{x^2} + 2x + x + 1 = 0$

$ \Rightarrow (2x + 1)(x + 1) = 0$

Thus, x = $-$1, $-$1/2

or, $\mathop {\lim }\limits_{a \to {0^ + }} \alpha (a) = - {1 \over 2}$

and $\mathop {\lim }\limits_{a \to {0^ + }} \beta (a) = - 1$

We have,

${(2x)^{\ln 2}} = {(3y)^{\ln 3}}$ ...... (1)

${3^{\ln x}} = {2^{\ln y}}$ ....... (2)

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

$ \Rightarrow \log y = {{(\log x)(\log 3)} \over {\log 2}}$ ........ (3)

Taking log both sides of Eq. (1), we get

$(\log 2)\{ \log 2 + \log x\} = \log 3\{ \log 3 + \log y\} $

${(\log 2)^2} + (\log 2)(\log x) = {(\log 3)^2} + {{{{(\log 3)}^2}(\log x)} \over {\log 2}}$ from Eq. (3)

$ \Rightarrow {(\log 2)^2} - {(\log 3)^2} = {{{{(\log 3)}^2} - {{(\log 2)}^2}} \over {\log 2}}(\log x)$

$ \Rightarrow - \log 2 = \log x$

$ \Rightarrow x = {1 \over 2} \Rightarrow {x_0} = {1 \over 2}$.

We have, ${a_n} = {\alpha ^n} - {\beta ^n}$

${\alpha ^2} - 6\alpha - 2 = 0$

Multiplying with $\alpha$⁸ on both sides, we get

${\alpha ^{10}} - 6{\alpha ^9} - 2{\alpha ^8} = 0$ ..... (1)

Similarly, ${\beta ^{10}} - 6{\beta ^9} - 2{\beta ^8} = 0$ ..... (2)

From Eqs. (1) and (2), we get

${\alpha ^{10}} - {\beta ^{10}} - 6({\alpha ^9} - {\beta ^9}) = 2({\alpha ^8} - {\beta ^8})$

$ \Rightarrow {a_{10}} - 6{a_9} = 2{a_8} \Rightarrow {{{a_{10}} - 2{a_8}} \over {2{a_9}}} = 3$.

The given equations are

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

${x^2} + x + b = 0$ ....... (1)

Common root is $(b - 1)x - 1 - b = 0$.

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

This value of x satisfies Eq. (1), we get

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

$ \Rightarrow b = i\sqrt 3 ,\, - i\sqrt 3 ,\,0$

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

$ \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$

Sum of roots = ${{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}$ and product = 1

Given, $\alpha$ + $\beta$ = $-$ p and $\alpha$³ + $\beta$³ = q

$ \Rightarrow (\alpha + \beta )({\alpha ^2} - \alpha \beta + {\beta ^2}) = q$

$\therefore$ ${\alpha ^2} + {\beta ^2} - \alpha \beta = {{ - q} \over p}$ ..... (i)

and ${(\alpha + \beta )^2} = {p^2}$

$ \Rightarrow {\alpha ^2} + {\beta ^2} + 2\alpha \beta = {p^2}$ ..... (ii)

From Eq. (i) and (ii), we get

${\alpha ^2} + {\beta ^2} = {{{p^3} - 2q} \over {3p}}$

and $\alpha \beta = {{{p^3} + q} \over {3p}}$

$\therefore$ Required equation is

${x^2} - {{({p^2} - 2q)x} \over {({p^3} + q)}} + 1 = 0$

$ \Rightarrow ({p^3} + q){x^2} - ({p^3} - 2q)x + ({p^3} + q) = 0$

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 $-1.$

As a, b, c, p, q $\in$ R and the two given equations have exactly one common root.

$\Rightarrow$ Either both equations have real roots.

Or both equations have imaginary roots.

$\Rightarrow$ Either $\Delta_1\ge0$ and $\Delta_2 > 0$ or

$\Delta_1 < 0$ and $\Delta_2 < 0$

$\Rightarrow~p^2-q\ge0$ and $b^2-ac\ge0$

or $p^2-q\le0$ and $b^2-ac\le0$

$\Rightarrow~(p^2-q)(b^2-ac)\ge0$

$\therefore$ Statement 1 is true.

Also, we have $\alpha\beta=q$ and ${\alpha \over \beta } = {c \over a}$

$\therefore$ ${{\alpha \beta } \over {{\alpha \over \beta }}} = {q \over c} \times a \Rightarrow {\beta ^2} = {{qa} \over c}$

As $\beta \ne 1$ or $-1$

$ \Rightarrow {\beta ^2} \ne 1$

$ \Rightarrow {{qa} \over c} \ne 1$ or $c \ne qa$

Again as exactly one root $\alpha$ is common and $\beta\ne1$

$\therefore$ $\alpha + \beta \ne \alpha + {1 \over \beta } \Rightarrow {{ - 2b} \over a} \ne - 2P$

$ \Rightarrow b \ne ap$

$\therefore$ Statement 2 is correct.

But Statement 2 is not 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)$

Since $\alpha,\beta$ are the roots of

${x^2} - px + r = 0$

$\therefore$ $\alpha + \beta = p$ and $\alpha \beta = r$

We know that

${\alpha \over 2}$ and 2$\beta$ are the roots of the equation ${x^2} - qx + r = 0$

$\therefore$ ${\alpha \over 2} + 2\beta = q$ and ${\alpha \over 2} \times 2\beta = r$

Solving $\alpha + \beta = p$ and ${\alpha \over 2} + 2\beta = q$

We get $\alpha = {2 \over 3}(2p - q)$ and $\beta = {1 \over 3}(2q - p)$

$\alpha \beta = r$

$r = {2 \over 9}(2p - q)(2q - p)$

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

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

$D \ge 0$ 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 $41$