Mathematics Must Do Problems of Quadratic Equation for NDA

Must Do Problem For NDA

Must Do Problem For NDA
Q 1580645517

`sin A, sin B, cos A` are in `GP` . Roots of `x ^2 + 2x cot B + 1 = 0` are always:
BITSAT 2005
(A)

real

(B)

imaginary

(C)

greater than ` 1`

(D)

equal

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.
Correct Answer is `=>` (A) real
Q 1429623511

If `a, b, c` are real and for some real `x`,
`(a^2 + b^2)x^2 + 2 (ab +bc) x + (b^ 2 + c^2) <= 0,` then

(A)

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

(B)

`a, b, c` are in `AP`

(C)

`a x^2 + 2bx + c >= 0` for all `x`

(D)

`a x^2 + 2bx + c = 0` for all `x`

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.
Correct Answer is `=>` (A)
Q 2424780651

The number of roots of the equation

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

UPSEE 2012
(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

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`
Correct Answer is `=>` (D) `4`
Q 2355745664

The roots of the equation `x^2 + 2(3a + 5)x + 2(9a^2 + 25) = 0` are rational for
BITSAT Mock
(A)

` a in R`

(B)

`a= -4/5`

(C)

`a= 5/3`

(D)

no value of `a`

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`
Correct Answer is `=>` (C) `a= 5/3`
Q 1522780631

If `S` is the set containing values of `x` satisfying `[x]^2-5[x]+ 6 \le 0`, where `[x]` denotes greatest integer function then `S` contains :
BITSAT 2013
(A)

`(2, 4)`

(B)

`(2, 4]`

(C)

`[2, 3]`

(D)

`[2, 4)`

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)`
Correct Answer is `=>` (D) `[2, 4)`
Q 2570023816

If one root of the equation `x^2 - lamda x + 12 = 0` is
even prime while `x^2 + lamda x+ mu = 0` has equal
roots, then `mu` is equal to
BCECE Stage 1 2013
(A)

`8`

(B)

`16`

(C)

`24`

(D)

`32`

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`
Correct Answer is `=>` (B) `16`
Q 2474880756

If the roots of the equation `ax^2 + bx + c = 0` are

real and of the form `alpha/(alpha -1)` and `(alpha +1)/alpha`, then the value of `(a+ b + c)^2` is
UPSEE 2012
(A)

`b^2 - 4ac`

(B)

`b^2 - 2ac`

(C)

`2 b^2 - ac`

(D)

None of these

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`
Correct Answer is `=>` (A) `b^2 - 4ac`
Q 1458612504

Let `alpha , beta` be the roots of `ax^2 + bx + c = 0` and `gamma, delta` be the
roots of `px^2 + qx + r = 0`. Suppose `D_1` and `D_2` are the
discriminants of the equations `ax^2 + bx + c = 0` and
`px^2 + qx + r = 0`, respectively. If `alpha, beta , gamma , delta` are in AP, then
`D_1 : D_2 = ······`

(A)

`a^2 : p^2`

(B)

`a^2 : q^2`

(C)

`a^2 : r^2`

(D)

`b^2 : p^2`

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`
Correct Answer is `=>` (A) `a^2 : p^2`
Q 2351291124

If `alpha` and `beta` are the roots of the equation

`ax^2 + 2bx + c= 0, Delta = b^2 -ac` and `alpha + beta`,

`alpha^2 + beta^2 , alpha^3 + beta^3` are in G.P., then `(a ne 0)`
BITSAT Mock
(A)

`bc ne 0`

(B)

`Delta ne 0`

(C)

`Delta b =0`

(D)

`Delta c =0`

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`)
Correct Answer is `=>` (D) `Delta c =0`
Q 2218745600

The value of ‘a’ for which the roots `alpha, beta` of the equation `2x^2 + 6x + a = 0` satisfies
the condition` alpha/beta + beta/alpha < 2` is
BITSAT Mock
(A)

`a > 9/2`

(B)

`a < 0`

(C)

`a > 0`

(D)

None of these

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.
Correct Answer is `=>` (A) `a > 9/2`
Q 1448712603

The value of the parameter a for which there is at least one
`x` satisfying the conditions

`x^2 + (1- 3/(2 a)) x + a^2/2 - a/2 <0, x = a^2 -1/2 ` lie in the interval........

(A)

`(1/2,1)`

(B)

`(-1/2,1)`

(C)

`(-1/2,-1)`

(D)

None of theses

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`
Correct Answer is `=>` (B) `(-1/2,1)`
Q 1555478364

If `|x^(2) - x - 6| = x + 2`, then the values of `x` are
VIT 2013
(A)

`-2, 2, - 4`

(B)

`-2, 2, 4`

(C)

`3, 2, - 2`

(D)

`4, 4, 3`

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`.
Correct Answer is `=>` (B) `-2, 2, 4`
Q 1514178050

If ` \alpha` and ` \beta ` are the roots of the equation ` x ^{ 2 }-2x+4=0`, then the value of ` \alpha ^{ n }+ \beta ^{ n }` will be:
BITSAT 2011
(A)

` i 2 ^{ n+1 }\sin ( { n\pi }/{ 3 } ) `

(B)

` 2 ^{ n+1 }\cos ( { n\pi }/{ 3 } )`

(C)

` i 2 ^{ n-1 }\sin ( { n\pi }/{ 3 } )`

(D)

`2 ^{ n-1 }\cos ( { n\pi }/{ 3 } )`

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 } )`
Correct Answer is `=>` (B) ` 2 ^{ n+1 }\cos ( { n\pi }/{ 3 } )`
Q 1407091888

If `p, q, r` are three distinct real numbers, `p ne 0` such that `x^ 2 + qx + pr = 0` and `x^2 + rx + pq = 0` have a common root, then the value of `p + q + r` is .....

(A)

`2`

(B)

`0`

(C)

`1`

(D)

`3`

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`
Correct Answer is `=>` (B) `0`
Q 2554123954

Find the values of `m` for which exactly one root of the equation `x^2 - 2mx + m^2 - 1 = 0`, lies in the interval `(- 2, 4)`.

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)`.
Q 1488756607

Let `S` be the set of real values for which the roots of
`x^2- 6ax + 2- 2a + 9a^2 = 0` exceed `3`. Then `S = (11/9, oo )`.

(A) True
(B) False
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 )`.
Correct Answer is `=>` (A)
Q 1412680530

The value of `a` for which the quadratic equation `3x^2 + 2(a^2 + 1)x + (a^2
- 3a + 2) = 0` possesses roots of opposite sign lies in

(A)

`(-oo, 0)`

(B)

`(-oo, -1)`

(C)

`(1, 2)`

(D)

`(3/2,2)`

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.`
Correct Answer is `=>` (C) `(1, 2)`
Q 1448623503

If `d, e, f` are in `GP` and the two quadratic equations `a x^ 2 + 2 bx + c = 0` and `d x^ 2 + 2ex + f = 0` have a common
root, then

(A)

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

(B)

`d/a, e/b , f/c ` are in `GP`

(C)

`2 dbf = aef + cde`

(D)

`b^2df = ace^2`

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`
Correct Answer is `=>` (A)
Q 1486556477

The value of `p` for which both the roots of the equation `4x^2- 20px + (25p ^2 + 15p- 66) = 0`, are less than `2`, lies in

(A)

`(4/5, 2)`

(B)

`(2, oo)`

(C)

`(-1, - 4/5)`

(D)

`(-oo, -1)`

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]`
Correct Answer is `=>` (D) `(-oo, -1)`
Q 1417680589

If the equation `ax^2 + bx + c = 0 (a> 0) ` has two roots `alpha` and `beta` such that `alpha < - 2` and `beta > 2`, then

(This question may have multiple correct answers)

(A) `b^2 - 4ac > 0`
(B) `c < 0`
(C) `a+ |b| +c <0`
(D) `4a + 2 | b |+ c < 0`
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`.
Correct Answer is `=>` (A)

 
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