Solution:

Since `sin A, sin B` and `cos A` are in GP

`:. sin^2 B=sin A cos A` ... (i)

`x^2 + 2x cot B + 1 = 0`

Now, `b^2 -4ac = 4 cot B -4`

`=(4 cos^2B -4 sin^2B)/(sin^2 B) =(4 (1 -sin^2 B)- 4 sin^2B)/(sin^2B)`

`=(4[1-2sin^2B])/(sin^2B)`

`=4( (sinA -cosA)/(sinB))^2 > 0`

`:.` Roots are always real.

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Solution:

Since, for some real `x` (not for all real `x`)

`(a^2 + b^2)x^2 + 2(ab+ bc)x+ (b^2 + c^2) <= 0`

`:. D >= 0 `

`=> 4 (ab + bc)^2 - 4 (a^2 + b^2) (b^2 + c^2)>= 0`

`=> -4 (ac- b^2 ) ^2 >= 0`

`=> ac- b^2 = 0`

Thus `a, b, c` are in `GP`

Now, let `f(x) = ax^ 2 + 2bx+ c(a>0)`

Its discriminant `= 4 (b^2 - ac) = 0`

`:. f(x) >= 0` for all `x`, ie, `(c)` is true.

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Solution:

`| x |^2 -7 | x | + 12 = 0`

`=> | x|^2 - 4 | x | - 3 | x | + 12 = 0`

`=> | x | ( | x | - 4 ) - 3 ( | x | - 4 ) =0`

`( | x | - 4) ( | x | - 3) = 0`

`=> |x | =4 => x = pm 4`

`=> |x | =3 => x = pm 3`

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Solution:

Roots are rational

`=>` Discriminant is a perfect square

`=> 4(3a + 5)^2 - 8(9a^2 + 25)` is a perfect square

`=> - 36a^2 + 12a - 100` is a perfect square

`=> - 4 (3a - 5)^2` is a perfect square

which is true only if `a = 5/3`

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Solution:

`[x]` is an integer and is always less than or equal to `x`.

Now,

`[x]^2 - 5[x] + 6 \le 0`

`\Rightarrow ([x] - 2)([x] - 3) \le 0`

`\Rightarrow [x] \in [2,3]`

`\therefore x \in [2,4)`

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Solution:

We know that, only even prime is `2`.

Then, `(2)^2 - lamda (2) + 12 = 0`

`=> lamda = 8` ........(i)

and `x^2 + lamda x + mu = 0`

has equal roots.

`:. lamda^2 - 4 mu = 0`

or `(8)^2 - 4 mu = 0` [from Eq. (i)]

`:. mu = 16`

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Solution:

Here, `(alpha +1)/alpha +alpha/(alpha -1) = (-b)/a` and `(alpha +1)/(alpha -1) = c/a`

`:. alpha = (c+a)/(c-a)` and `(2 alpha^2 -1)/(alpha (alpha -1) ) = (-b)/a`

On substituting `alpha`, we get

`(a+b +c)^2 = b^2 - 4ac`

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Solution:

`alpha + beta = - b/a, alpha beta = c/ a , gamma+ delta = - q/ p` and `gamma delta = r/ p`

`:.` `alpha , beta, gamma, delta ` are in AP

`:.` `beta- alpha = delta-gamma => (beta - alpha)^2 = (delta - gamma)^2`

`=> (beta+ alpha)^2 - 4 alpha beta= (delta + gamma)^2 - 4 delta gamma`

`=> (-b/a)^2 - 4c/a= (-q/p)^2 - 4r/p`

`=> D_1/a^2 = D_2/p^2` `=> D_1 : D_2 = a^2 : p^2`

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Solution:

`alpha +beta = (-2b)/a , alpha beta = c/a`. It is given

that `(alpha^2 + beta^2)^2 = (alpha + beta) (alpha^3 + beta^3)`

`=> [(alpha + beta)^2 -2 alpha beta]^2 = (alpha + beta) [ (alpha + beta)^3 - 3 alpha beta (alpha + beta) ]`

`=> (2b^2 -ac)^2 = b^2 (4b^2 - 3 ac)`

`=> (2 Delta + ac)^2 = (Delta +ac) (4 Delta +ac)`

`=>Delta ac =0 =>Delta c= 0 ` (as `a ne 0`)

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Solution:

`alpha/beta + beta/alpha < 2 => (alpha^2 + beta^2 -2 alpha beta)/(alpha beta) < 0`

`=> ( (alpha + beta)^2 -4 alpha beta )/(alpha beta) < 0`


`=> (9-4 (a/2))/(a/2) < 0`

`=> (9-2a)/a < 0`

`=> a (2a -9 ) > 0`

`=> a < 0 ` or `a > 9/2`

Now if `alpha` and `beta` are real, `alpha/beta + beta/alpha ge 2`



Hence the roots must be complex.

Therefore, the discriminant

`36 - 8a < 0`.

`:. a > 9/2 ; a < 0` is disqualified.

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Solution:

Given equation represent a parabola opening upwards, with
its axis parallel to the `x`-axis. Therefore, `y < 0` for some if `x,
y = 0` has distinct real roots.

`=> ( 1- 3/(2 a))^2 -4 (a^2/2 - a/2) >0`

`=> (a/2 -1)^2 >0` (if `a ne 2`)

From second condition `x = a^2 -1/2`

We get

`y= (a^2 - 1/2)^2 +(1- 3/(2a)) (a^2 - 1/2) +a^2/2-a/2`

After solving , we get

`y= 1/2 (a-1) (2a+1) [(a- 1/2)^2+1/4 ]`

Since, the quantity within the square brackets is greater
than zero for real `a`,

So, `y < 0` if `-1/2 < a <1`

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Solution:

`| x^(2)- x - 6| = x + 2`, then

Case I `x^(2) - x - 6 < 0`

`=> (x - 3) (x+ 2) < 0`

`- 2< x< 3`

In this case, the equation becomes

`x^(2) -x- 6=- x - 2`

or `x^(2) - 4= 0`

`x=pm2`

Clearly,` x = 2` satisfies the domain of the equation

in this case. So,` x = 2` is a solution.

Case ` x^(2)- x - 6 >= 0`

So,` x^(2)<= - 2` or `x >= 3`

Then, equation reduces to `x^(2)- x - 6 = 0 = x + 2`

i.e., `x^(2)- 2x- 8 = 0` or `x = - 2, 4`

Both these values lies in the domain of the equation

in this case, so `x = - 2, 4` are the roots. Hence,

roots are `x = - 2, 2, 4`.

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Solution:

Since,` \alpha` and `\beta` are the roots of ` x ^{ 2 }-2x+4=0`

` \therefore \quad \alpha +\beta =2` and `\alpha \beta =4`

Now, `( \alpha -\beta ) =\sqrt { ( \alpha +\beta) ^{ 2 }-4\alpha \beta }`

`=\sqrt { 4-16 } =2\sqrt { 3 } i`

On solving, we get

` 2\alpha =2+2\sqrt { 3 } i`

`\Rightarrow \quad \alpha =2 ( \frac { 1 }{ 2 } +\frac { \sqrt { 3 } }{ 2 } i )`

` =2 ( \cos \frac { \pi }{ 3 } +i\sin \frac { \pi }{ 3 } )`

and `\beta =\frac { 2-2\sqrt { 3 } i }{ 2 } =2 ( \cos \frac { \pi }{ 3 } -i\sin \frac { \pi }{ 3 } )`

` \therefore \quad \alpha ^{ n }+ \beta ^{ n }= [ 2 ( \cos \frac { \pi }{ 3 } +i\sin \frac { \pi }{ 3 } ) ] ^{ n }+ [ 2 ( \cos \frac { \pi }{ 3 } -i\sin \frac { \pi }{ 3 } ) ] ^{ n }`

` = 2 ^{ n } [ 2\cos \frac { n\pi }{ 3 } ] = 2 ^{ n+1 }\cos ( { n\pi }/{ 3 } )`

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Solution:

`:.` `(x^2 + qx + pr)- (x^2 + rx + pq) = 0`

`=> (q-r)x -p (q-r)=0`

`:.` `x=p (q ne r)`

Then `p^2 + qp + pr = 0`

`p ne 0`

`:.` `p+ q+ r=0`

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Solution:

Let `f(x) = x^2 - 2mx + m^2 - 1`, as exactly one root of `f(x) = 0` lies in

the interval `(-2, 4)`, we can take `D > 0` and `f(- 2) {(4) < 0`.

(i) Consider D > 0

`(-2m)^2 - 4 ·1 (m^2 -1) > 0 => 4 > 0`

`:. m in R` ... (i)

(ii) Consider f(- 2) {(4) < 0

`(4 + 4m + m^2 - 1) (16 - 8m + m^2 - 1) < 0`

`=> (m^2 + 4m + 3) (m^2 - 8m + 15) < 0`

`=> (m + 1)(m + 3)(m- 3)(m- 5) < 0`

`=> (m + 3)(m + 1)(m- 3)(m- 5) < 0`

`:. m in (- 3,- 1) cup (3, 5)` .......(ii)

Hence, the values of m satisfying Eqs. (i) and (ii) at the same time are

`m in (- 3,- 1) cup (3, 5)`.

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Solution:

`D=(6a)^2 -4(2-2a+9a^2 ) ge 0` `=> a ge 1` ....................(i)

`f (3) = 3^2- ( 6a)(3) + 2- 2a + 9a^2 > 0`

`=> a < 1` or `a > 11/9` ......................(ii)

and `-b/2a = -(-6a)/2 = 3a > 3 => a >1` ....................(iii)

All three conditions are satisfied if `a > 11/9`.

Therefore, `S = (11/9, oo )`.

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Solution:

The quadratic equation `3x^2 + 2 (a^2 + 1) x + (a^2 -3a + 2) = 0` will have two roots of
opposite sign if it has real roots and the product of the roots is negative, that is, if

`4(a^2 + 1)^2· -12(a^2- 3a + 2) ge 0` and `(a^2-3a+2)/3 <0`

!3oth of these conditions are met if, `(a^2-3a+2)< 0`

i.e.,` (a -1) (a- 2) < 0` or `1 < a< 2.`

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Solution:

`d, e, f` are in `GP`` => e^2 = df`

`dx^2 +2ex+f=0`

`dx^2 + 2sqrt (df) x + f = 0`

`(sqrt d x+ sqrt f)^2 =0`

` x=- sqrtf /sqrtd .`

Now, substituting the value of `x ` in `ax^2 + 2bx + c = 0`

`af/d - (2bsqrt f)/sqrt d +c=0 `

`a/d - 2b/sqrt(df)+ c/f =0` or `a/d - 2b/e +c/f =0 `

` a/d , b/e , c/f ` are `AP.`

Hence, `d/a , e/b , f/c` are in `HP`

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Solution:

Discriminant `> 0`

`:.` Roots less than `2`

`:.` `p^2-p-2 > 0`

`=> p > 2` or `p < -1`

Combine both casses, we get `p in (-oo, -1) in (2, 22/5]`

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Solution:

Since, the equation has two distinct roots `alpha` and `beta` the
discriminant `b^2- 4ac > 0`, we must have

`f(x) = ax^2 + bx + c < 0` for `alpha < x < beta`

Since, `a < 0 < beta` we must have `f(0) = c < 0`

Also, as `alpha < -1, 1 < beta` we get `f(-1) =a- b + c < 0`

and `f(1) =a+ b + c < 0`, ie, `a+ | b |+ c < 0`

Since, `alpha < -2, 2< beta`

`f(-2)= 4a- 2b + c < 0`

and `f(2) = 4a + 2b + c < 0` ie, `4a + 2 | b |+ c < 0`.

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