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previous year question of Quadratic Equation for NDA

Previous Year Quadratic Equation Question For NDA

Previous Year Quadratic Equation Question For NDA
Q 2743680543

If the difference between the roots or the equation `x^2 + kx + l= 0` is strictly less than `sqrt5`, where `| k | >= 2`, then `k` can be any element of the interval
NDA Paper 1 2017
(A)

`(-3, -2] cup [2, 3)`

(B)

`(- 3, 3)`

(C)

`[- 3, -2] cup [2, 3]`

(D)

None of the above

Solution:

Let roots `alpha, beta`

`alpha-beta < sqrt 5`

`sqrt((alpha+ beta)^2 -(4 alpha beta)) < sqrt 5`

`sqrt(k^2- 4) < sqrt 5`

`k^2 < 9`

`-3 < k < 3`

`| k | > 2 => k ge 2 or k le -2 `

`k in (-3, 2] cup (2, 3)`
Correct Answer is `=>` (A) `(-3, -2] cup [2, 3)`
Q 2773680546

If the roots of the equation `x^2 + px + q = 0` are in the same ratio as those of the equation `x^2 + lx + m = 0`, then which one of the following is correct ?
NDA Paper 1 2017
(A)

`p^2 m = l^2q`

(B)

`m^2p = l^2q`

(C)

`m^2p = q^2l`

(D)

`m^2p^2 = l^2q`

Solution:

If ration is same of given equations , then

`(-p + sqrt(p^2 -4q))/(-p - sqrt(p^2 -4q)) = (-l + sqrt(l^2 -4m))/(-l - sqrt(l^2 -4m))`

by C & D theorem

`(-2p)/(2 sqrt(p^2- 4q)) =(-2l)/(2 sqrt(l^2 - 4m))`

`(p^2)/(p^2- 4q) =(l^2)/(l^2- 4m)`

`p^2m = l^2 q`
Correct Answer is `=>` (A) `p^2 m = l^2q`
Q 2743880743

If the graph of a quadratic polynomial lies entirely above x-axis, then which one of the following is correct?
NDA Paper 1 2017
(A)

Both the roots are real

(B)

One root is real and the other is complex

(C)

Both the roots are complex

(D)

Cannot say

Solution:

If graph lies entirely above the x-axis mean , it doesn't cut on x-axis , so quadratic equation has no real roots or both roots will be complex.
Correct Answer is `=>` (C) Both the roots are complex
Q 2713080840

If cot `alpha` and cot `beta` are the roots of the equation `x^2 + bx + c = 0` 'with `b != 0`, then the value of cot `(alpha + beta)` is
NDA Paper 1 2017
(A)

`(c - 1)/b`

(B)

`(1 - c)/b`

(C)

`b/(c - 1)`

(D)

`b/(1 - c)`

Solution:

`cot (alpha+beta) =(cot alpha cot beta-1)/(cot alpha+ cot beta)`

`=(c-1)/(-b)`

`=(1-c)/b`
Correct Answer is `=>` (B) `(1 - c)/b`
Q 2703080848

The sum of the roots of the equation `ax^2 + x + c = 0`. (where a and c are non-zero) is equal to the sum of the reciprocals of their squares. Then `a , ca^2 , c^2` are in
NDA Paper 1 2017
(A)

AP

(B)

GP

(C)

HP

(D)

None of the above

Solution:

`alpha+ beta =1/(alpha^2) + 1/(beta^2)`

`alpha+ beta =((alpha+ beta)^2 - 2 alpha beta)/((alpha beta)^2)`

`(-1)/a =((-1/a)^2 - 2(c/a))/((c/a)^2)`

`2(c/a) = 1/(a^2) + c^2/a`

`2 ac = 1+ c^2 a`
Correct Answer is `=>` (D) None of the above
Q 2781745627

If both the roots of the equation `x^2 - 2 kx + k^2 - 4 = 0` lie between -3 and 5 , then which one of the following is correct ?
NDA Paper 1 2016
(A)

`- 2 < k < 2 `

(B)

`-5 < k < 3`

(C)

`-3 < k < 5`

(D)

`-1 < k < 3`

Solution:

Roots of the equation `x^2 - 2 kx +k^2 - 4 = 0` are

`(-(-2 k ) pm sqrt ((-2 k)^2 - 4 (k^2 -4) ))/(2 xx 1) `

`= (2 k pm sqrt 16 )/2 = k pm 2`

As given roots are lie between `-3` and 5, so

`-3 < k +2 < 5 ` and ` -3 < k -2 < 5`

`=>
Correct Answer is `=>` (A) `- 2 < k < 2 `
Q 2116023879

Let `alpha` and `beta (alpha < beta)` be the roots of the equation `x^(2) + bx + c = 0`,
where `b > 0` and `c < 0`.

Consider the following
1. `beta < - alpha quad`
2. `quad beta < |alpha|`
Which of the above is/are correct?
NDA Paper 1 2016
(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

Given, `alpha` and `beta` are the roots of equation `x^(2) + bx + c = 0`.

`:. alpha + beta = - b` and `alpha beta = c`

As `b > 0`

So, `alpha + beta < 0`

`=> beta < -alpha`

and given that `alpha < beta => alpha < 0` and ` beta > 0`

` alpha + beta < 0 => < - alpha`...............(i)

:. Statement 1 is correct.

As, `alpha < beta => -alpha > beta`..........(ii)

From Eqs. (i) and (ii),

` | alpha| > |beta|`

:. Statement 2 is correct.
Correct Answer is `=>` (C) Both 1 and 2
Q 2106123978

Let `alpha` and `beta (alpha < beta)` be the roots of the equation `x^(2) + bx + c = 0`,
where `b > 0` and `c < 0`.

Consider the following
1. `alpha + beta + alpha > 0 ` quad 2. quad `alpha ^(2) beta + beta^(2) alpha > 0`
Which of the above is/are correct?
NDA Paper 1 2016
(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

Given, `alpha` and `beta` are the roots of equation `x^(2) + bx + c = 0`.

`:. alpha + beta = - b` and `alpha beta = c`

As `b > 0`

So, `alpha + beta < 0`

`=> beta < -alpha`

and given that `alpha < beta => alpha < 0` and ` beta > 0`

As,` alpha+beta = 0` and `alpha beta < 0`

:. Statement 1 is not correct.

Now, `alpha^(2) beta + alpha beta (alpha+beta)`

As, `alpha + beta < 0` and `alpha beta < 0`

`=> alpha beta ( alpha+beta ) > 0`

:. Statement 2 is correct.
Correct Answer is `=>` (B) Only 2
Q 2146478373

If `x^( 2) - px + 4 > 0` for all real values of `x`, then which one

of the following is correct?
NDA Paper 1 2016
(A)

` |P| < 4`

(B)

` |P| <= 4`

(C)

` |P| > 4`

(D)

` |P| >= 4`

Solution:

`:. x^(2) - px + 4 > 0`

Here, `a > 0` and `f(x) > 0`

`:. D < O`

`:. p^( 2) -16 < 0`

`=> p^( 2) < 16`

`=> |p| < 4`
Correct Answer is `=>` (A) ` |P| < 4`
Q 2106878778

Consider the function. `f( x) = (27(x^(2//3) - x))/4` .

How many solutions does the function `f(x) = 1` have?
NDA Paper 1 2016
(A)

One

(B)

Two

(C)

Three

(D)

Four

Solution:

Given function,

`f( x) = (27(x^(2//3) - x))/4`

If `f( x) = 1`

` => 1 = (27(x^(2//3) - x))/4`

` => x^(2//3) - x = 4/27`

Let `x^(1//3) = alpha => x = alpha^(3)`

Then `alpha^(2) -alpha^(3) = 4/27`

`=> alpha^( 2) -alpha = 2/3 xx 2/3 xx 1/3 `

` => alpha.alpha -alpha = 2/3 xx 2/3 xx 1/3 `

` => alpha = 2/3 `

` => x^(1//3) = 2/3 `

`:. x = 8/27`

Hence, only one solution exists.
Correct Answer is `=>` (A) One
Q 2127001881

If one root of the equation `( l - m) x^2 + lx + 1 = 0`
is double the other and `l` is real, then what is the greatest
value of `m`?
NDA Paper 1 2016
(A)

` - 9/8`

(B)

`9/8`

(C)

`- 8/9`

(D)

`8/9`

Solution:

Given equation is

`(l - m) x^2 + lx + 1 = 0`

Let `alpha` and `2alpha` be the roots of given equation. Then, we
have

`alpha + 2alpha = 3alpha = (-l)/( l - m)`

and `alpha = (-l)/(3 (l - m)) ` and `2 alpha^2 = 1/(l - m)`

` => 2*( (-l)/(3 ( l - m))) = 1/(l - m)`

` => (2l^2)/(9(l - m)^2) = 1/(l - m)`

` => 2l^2 = 9(l - m) quad [ ∵l!= m]`

` => 2l^2 - 9l + 9m = 0`

` => l = (9 pm sqrt(81 - 72m))/4`

`∵ l` is real.

`:. D >= 0 => 81 - 72m >= 0`

` => 81 >= 72m`

` => m <= (81)/(72) => m <= 9/8`

Hence, greatest value of `m` is `9/8`.
Correct Answer is `=>` (B) `9/8`
Q 1628423301

In solving a problem that reduces to a quadratic
equation, one student makes a mistake in the
constant term and obtains `8` and `2` for roots.
Another student makes a mistake only in the
coefficient offirst degree term and finds `-9` and `-1`
for roots. The correct equation is
NDA Paper 1 2015
(A)

`x^ 2 - 10 x + 9 = 0`

(B)

`x^ 2 + 10 x + 9 = 0`

(C)

`x^ 2 - 10 x + 16 = 0`

(D)

`x^ 2 - 8 x - 9 = 0`

Solution:

Given, roots are `8` and `2`, then the equation is

`x^2 -` (sum of roots) `x +` product of roots `= 0`

`=> x^ 2 - (8+2) x + 16 = 0`

`=> x ^2 -10 x + 16 = 0 ....... (i)`

(mistake in the constant term)

For roots `-9` and `-1`, we have

`x ^2 (-9x - 1 ) x + ( -9) x (-1) = 0`

`=> x^ 2 - 9x + 9 = 0 .... (ii)`

(mistake in the coefficient of first degree term)

But as per given information. we have the right

equation as `x ^2 - 10x + 9 = 0`
Correct Answer is `=>` (A) `x^ 2 - 10 x + 9 = 0`
Q 2250467314

The number of real roots of the equation
`x^2 - 3 | x | + 2 = 0` is
NDA Paper 1 2015
(A)

`4`

(B)

`3`

(C)

`2`

(D)

`1`

Solution:

We have, `x^2 - 3 | x |+ 2 = 0 => | x |^2 -3|x| + 2 = 0`

`=> | x|^2 - 2 | x| - | x| + 2 = 0`

`=> (| x | - 2) (| x | - 1) = 0`

` => | x | = 2` or `| x | = 1`

`=> x = ± 2` or `x = ± 1`

`:.` There are four real roots of the equation.
Correct Answer is `=>` (A) `4`
Q 1678123906

If `2p + 3q = 18` and `4p^2 + 4pq -3q^2 -36 =0`, then

what is `(2p + q)` equal to?
NDA Paper 1 2015
(A)

`6`

(B)

`7`

(C)

`10`

(D)

`20`

Solution:

We have, `2p + 3q = 18`

and `4p^2 + 4pq- 3q^2 - 36 = 0`

`=> (2p+3q)^2 -8pq-12q^ 2 =36`

`=> 18^2 - 4q(2p + 3q) = 36`

`=> 324 - 36 = 4q(18)`

`=> (288)/(18) = 4q`

` => q = 4`

`:. 2p + 3 xx 4 = 18`

` => p = (18 -12)/2 = 3`

`:. 2p + q =2 xx 3 + 4 = 10`
Correct Answer is `=>` (C) `10`
Q 1658156904

Let `alpha` be the root of the
equation `25 cos^2 theta + 5 cos theta - 12 = 0`, where ` pi/2 < alpha < pi`.

What is the tan `alpha` equal to?
NDA Paper 1 2015
(A)

`- 3/4`

(B)

`3/4`

(C)

` - 4/3`

(D)

` -4/5`

Solution:

`25cos^2 theta + 5cos theta -12 = 0`

`=> 25cos^2 theta + 20cos theta -15 cos theta -12 = 0`

` => 5cos theta (5 cos theta + 4)- 3 (5 cos theta + 4 ) = 0`

`=> (5cos theta - 3)(5cos theta + 4) = 0`

`=> cos theta = 3/5 ` or `cos theta = -4/ 5`

` alpha in ( pi/2 , pi)`

Since, `alpha` is a root of the given equation.

`:. cos alpha = 3/2` and `cos alpha = -4/5`

:. We take `cos alpha = -4/5 => tan alpha =- 3/4`
Correct Answer is `=>` (A) `- 3/4`
Q 1752334234

Every quadratic equation `ax^2 + bx + c = 0`, where
`a, b, c in R, a != 0` has
NDA Paper 1 2014
(A)

exactly one real root

(B)

at least one real root

(C)

at least two real roots

(D)

at most two real roots

Solution:

Every quadratic equation

`ax^2 + bx + c = 0`, where `a, b, c in R, a != 0`

has at most two real roots.
Correct Answer is `=>` (D) at most two real roots
Q 2308001808

If `alpha` and `beta` are the roots of the equation
`x^2 + x + 2 = 0`, then what is `(alpha^(10) + beta^(10))/(alpha^(-10) + beta^(-10))` equal to'?

NDA Paper 1 2013
(A)

`4096`

(B)

`2048`

(C)

`1024`

(D)

`512`

Solution:

Here, `alpha` and `beta` are the roots of the equation

`x^2 + x + 2 = 0`, then

`alpha + beta = - 1` ... (i)

and `alpha · beta = 2` ... (ii)

Now, `( alpha^(10) + beta^(10) )/( alpha^(-10) + beta^(-10) ) = ( alpha . beta )^(10) = (2)^(10)` [fromEq. (ii)]

`= 1024`
Correct Answer is `=>` (C) `1024`
Q 2318101900

If `a` and `b` are rational and `b` is not perfect square,
then the quadratic equation with rational
coefficients whose one root is `3a + sqrt(b)` is
NDA Paper 1 2013
(A)

`x^2 - 6ax + 9a^2 - b = 0`

(B)

`3ax^2 + x - sqrt(b) = 0`

(C)

`x^2 + 3ax + sqrt(b) = 0`

(D)

` sqrt(b)x^2 + x - 3a = 0`

Solution:

If one root of any quadratic equation is in the form

`3a + sqrt(b)`. then other root of this equation should be `3a - sqrt(b)`.

`:.` Required equation is

`x^2 -` (Sum of roots) · x + (Product of roots)` = 0`

`=> x^2 - {(3a + sqrt(b)) + (3a - sqrt(b)} ·x + {( 3a + sqrt(b))`

` (3a - sqrt(b))} = 0`

`=> x^2 - 6ax + 9a^2 - b = 0`
Correct Answer is `=>` (A) `x^2 - 6ax + 9a^2 - b = 0`
Q 2338101902

How many real roots does the quadratic equation
`f(x) = x^2 + 3 | x | + 2 = 0` have ?
NDA Paper 1 2013
(A)

One

(B)

Two

(C)

Four

(D)

No real root

Solution:

Given quadratic equation is

`f(x) = x^2 + 3 | x | + 2 = 0`

Case I `f(x) = x^2 + 3x + 2 = 0`(when, x > 0)

` => x^2 + 2x + x + 2 = 0`

` => x (x + 2) + 1 (x + 2) = 0`

`=> (x + 2) (x + 1) = 0`

` :. x = -2 , -1` (but x > 0)

So, here no real roots exist.

Case II `f(x) = x^2 - 3x + 2 = 0` (when, x < 0)

`=> x = (3 ± sqrt(9 - 8))/2 = ( 3 ± 1)/2`

`=> x = 2 , 1` (but x < 0)

So, here also no real roots exist.

Hence, given quadratic equation has no real roots.
Correct Answer is `=>` (D) No real root
Q 2368101905

If `alpha` and `beta` are the roots of the equation
`a.x^2 + bx + b = 0`, then what is the value of
` sqrt(alpha/ beta) + sqrt(beta/alpha) + sqrt(b/a) `?
NDA Paper 1 2013
(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

Given quadratic equation is `ax^2 + b x + b = 0`

Let `alpha` and `beta` be the roots of given equation.

` :. alpha + beta = - b/a ` and ` alpha beta = b/a`

` :. sqrt(alpha/ beta) + sqrt(beta/alpha) + sqrt(b/a) = ( alpha + beta )/sqrt(alpha beta) + sqrt(b/a)`

` = (-b)/a . sqrt(a/b) + sqrt(b/a) = - sqrt(b/a) + sqrt(b/a) = 0`
Correct Answer is `=>` (B) `0`
Q 2308101908

The roots of the equation `x^2 - 8x + 16 = 0`
NDA Paper 1 2013
(A)

are imaginary

(B)

are distinct and real

(C)

are equal and real

(D)

Cannot be determined

Solution:

Given equation is

`x^2 - 8x + 16 = 0`

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

`=> x = 4, 4`

Also, discriminant `(D) = b^2 - 4ac = 0`

So, the roots of the equation are equal and real.
Correct Answer is `=>` (C) are equal and real
Q 2318112000

What is the difference in the roots of the equation
`x^2 - 10x + 9 = 0`?
NDA Paper 1 2013
(A)

`2`

(B)

`3`

(C)

`5`

(D)

`8`

Solution:

Given equation,

`x^2 - 10x + 9 = 0`

Let `(alpha , beta)` be the roots of the given equation.

Then, `alpha + beta = 10` .......(i)

and `alpha · beta = 9` .........(ii)

Now, we use the identity

`(alpha - beta)^2 = (alpha + beta)^2 - 4alpha beta = (10)^2 - 4 (9)= 100- 36 = 64`

` => alpha - beta = ± 8`

`:. | alpha - beta | = 8`
Correct Answer is `=>` (D) `8`
Q 2348112003

The quadratic equation `x^2 + bx + 4 = 0` will have
real roots, if
NDA Paper 1 2013
(A)

Only `b <= -4`

(B)

Only `b >= 4`

(C)

`-4 < b < 4`

(D)

`b <= - 4, b >= 4`

Solution:

Given that, the equation `x^2 + bx + 4 = 0` have real

roots, if discriminate `(D) = B^2 - 4AC >= 0`

`=> b^2 - 4 (1) (4) >= 0 => b^2 - 16 >= 0`

`=> (b - 4) (b + 4) >= 0`

` :. b <= -4 , b >= 4`
Correct Answer is `=>` (D) `b <= - 4, b >= 4`
Q 2378112006

If `4^x - 6·2^x + 8 = 0`, then the values of `x` are
NDA Paper 1 2013
(A)

`1, 2`

(B)

`1, 1`

(C)

`1, 0`

(D)

`2, 2`

Solution:

Given that,

`4^x - 6·2^x + 8 = 0`

`=> 2^(2x) - 6·2^x + 8 = 0`

Let `2^x = z`

`=> z^2 - 6z + 8 = 0`

`=> z^2 - 4z - 2z + 8 = 0`

`=> z(z-4) - 2 (z-4) = 0`

`=> (z- 4)(z-2) = 0`

`:. z = 2, 4 => 2^x = 2^1 , 2^2`

So, the required values of `x` are `1` and `2`.
Correct Answer is `=>` (A) `1, 2`
Q 2318112009

If the roots of a quadratic equation
`ax^2 + bx + c = 0` are `alpha` and `beta`, then the quadratic
equation having roots `alpha^2` and `beta^2` is
NDA Paper 1 2013
(A)

`x^2 - (b^2 - 2ac) x + c = 0`

(B)

`a^2x^2 - (b^2 - 2ac)x + c = 0`

(C)

`ax^2 - (b^2 - 2ac)x + c^2 = 0`

(D)

`a^2x^2 - (b^2 - 2ac)x + c^2 = 0`

Solution:

Given quadratic equation is `ax^2 + bx + c = 0`.

and its root are `alpha` and `beta`.

`:.` Sum of the roots `= alpha + beta = (-b)/a`

and product of the roots `= alpha . beta = c/a`

` ∵ alpha^2 + beta^2 = ( alpha + beta)^2 - 2 alpha beta = ( - b/a)^2 -2 . c/a`

` = b^2/a^2 - (2c)/a = (b^2 - 2ac)/a^2 `

and ` alpha^2 . beta^2 = ( alpha beta)^2 = (c/a)^2 = c^2/a^2`

` :.` Required quadratic equation whose roots are `alpha^ 2` and `beta^2` is

`x^2 - (alpha^2 + beta^2)x+ alpha^2 beta^2 = 0`

` => x^2 - (b ^2 - 2ac)/a^2 x + c^2/a^2 = 0`

`:. a^2x^2 - (b^2 - 2ac) x + c^2 = 0`
Correct Answer is `=>` (D) `a^2x^2 - (b^2 - 2ac)x + c^2 = 0`
Q 2348212103

If the roots of the equation `3ax^2 + 2bx + c = 0` are
in the ratio `2 : 3`, then which one of the following is
correct?
NDA Paper 1 2013
(A)

`8 ac = 25b`

(B)

`8ac = 9b^2`

(C)

`8b^2 = 9ac`

(D)

`8b^2 = 25ac`

Solution:

Given quadratic equation is `3ax^2 + 2bx + c = 0`.

Let its root are `2 alpha` and `3 alpha`. Since, roots are in tile ratio `2 : 3`.

Now, sum of tile roots `= 2 alpha + 3 alpha = (-2b)/(3a)`

` => 5 alpha = (- 2b)/(3a) => alpha = (-2b)/(15 a)` ... (i)

and product of the roots `= 2 alpha · 3 alpha = c/(3a)`

` => 6 alpha^2 = c/(3a) => ( (-2b)/(15 a))^2 = c/(18 a)`

` => (4b^2) /( 225a^2) = c/(18 a) => (4b^2)/(25 a^2) = c/ (2a)`

`:. 8b^2 = 25ac`
Correct Answer is `=>` (D) `8b^2 = 25ac`
Q 2388212107

`(x + 1)^2 - 1 = 0` has
NDA Paper 1 2013
(A)

one real root

(B)

two real roots

(C)

two imaginary roots

(D)

four real roots

Solution:

Given that,

`(x + 1)^2 - 1^2 = 0`

`=> (x + 1)^2 (1)^2 = 0`

` => ( x + 1 + 1) ( x + 1 - 1) = 0 [ ∵ a^2 - b^2 = (a -b) (a +b) ]`

`=> (x+2) (x) = 0`

` => x = 0, -2`

Hence, `(x + 1)^2 - 1 = 0` has two real roots.
Correct Answer is `=>` (B) two real roots
Q 2318212109

What is the degree of the equation

` 1/(x - 3) = 1/( x - 2) - 1/2 ` ?
NDA Paper 1 2013
(A)

`0`

(B)

`1`

(C)

`2`

(D)

`3`

Solution:

Given equation is

`1/(x-3) = 1/(x+2) - 1/2`

`=> 1/(x-3) = (2 - x - 2)/(2(x+2)) => 2(x + 2) = - x (x - 3)`

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

which is a quadratic equation and degree of the above equation

is `2`.
Correct Answer is `=>` (C) `2`
Q 2338312202

If the sum of the roots of a quadratic equation is `3`
and the product is `2`, then the equation is
NDA Paper 1 2013
(A)

`2x^2 - x + 3= 0 `

(B)

`x^2 - 3x + 2 = 0`

(C)

`x^2 + 3x + 2 = 0`

(D)

`x^2 - 3x - 2 = 0`

Solution:

Let the required quadratic equation is

`ax^2 + bx + c= 0` ... (i)

Now, sum of the roots `= (-b)/a = 3 = (-(-3))/1` (given)

and product of the roots `= c/a = 2 = 2/1` (given)

`:. a = 1, b = -3` and `c = 2`

Hence, the required quadratic equation is `x^2 - 3x + 2 = 0`.
Correct Answer is `=>` (B) `x^2 - 3x + 2 = 0`
Q 2328412301

If `alpha` and `beta` are the roots of the equation
`x^2 + bx + c = 0`, then what is the value of
`alpha^(-1) + beta^(-1)`?
NDA Paper 1 2013
(A)

` - b/a`

(B)

`b/c`

(C)

`c/b`

(D)

` - c/b`

Solution:

Given that, `alpha` and `beta` are the roots of the equation

`x^2 + bx + c = 0`.

Then, sum of the roots `= alpha + beta = (-b)/1 = - b` ... (i)

and product of roots `= alpha . beta = c/1 = c ` .........(ii)

` :. alpha^(-1) + beta^(-1) = 1/alpha + 1/beta = ( alpha + beta)/(alpha beta) = (-b)/c`
Correct Answer is `=>` (A) ` - b/a`
Q 2388512407

If the roots of a quadratic equation are `m + n` and
`m - n`, then the quadratic equation will be
NDA Paper 1 2012
(A)

`x^2 + 2 mx + m^2 - mn + n^2 = 0`

(B)

`x^2 + 2 mx + (m - n)^2 = 0`

(C)

`x^2 - 2 mx + m^2 - n^2 = 0`

(D)

`x^2 + 2 mx + m^2 - n^2 = 0`

Solution:

Given that, `m + n` and `m- n` be the roots of given

equation

Sum of roots `= (m + n) + (m - n)= 2m`

and product of roots `= (m + n)(m- n)= m^2 - n^2`

So, the required quadratic equation is

`x^2 -` (Sum of roots )x + Product of roots `= 0`

`=> x^2 - 2mx + (m^2 - n^2 ) = 0`
Correct Answer is `=>` (C) `x^2 - 2 mx + m^2 - n^2 = 0`
Q 2338812702

If `alpha , beta` are the roots of `x^2 + px - q = 0` and `gamma , delta` are the
roots of `x^2 - px + r = 0`, then what is the value of `( beta + gamma) (beta + delta)`?
NDA Paper 1 2012
(A)

`q + r`

(B)

`p + q`

(C)

`q + r`

(D)

`p + q`

Solution:

Since , `alpha` and `beta` are the roots of `x^2 + px - q = 0`.

`:. alpha + beta = - p, alpha beta = - q`

Since, `gamma ` and `delta` are the roots of `x^2 - px + r = 0`.

`:. gamma + delta = p , gamma delta = r`

`:. (beta + gamma) (beta + delta ) = beta^2 + beta delta + gamma beta + gamma delta = beta^2 + beta (gamma + delta ) + gamma delta`

`= beta^ 2 + beta (P) + gamma delta ( ∵ gamma + delta = p` and ` gamma delta = r)`

`= beta^2 + beta (- alpha - beta ) + r [ ∵ p = -(alpha + beta)]`

`= beta^2 + (- beta ) (alpha + beta ) + r`

`= beta^2 - alpha beta - beta^2 + r = - alpha beta + r`

`= - (-q) + r = q + r `
Correct Answer is `=>` (C) `q + r`
Q 2318012809

If the roots of the quadratic equation
`3x^2 - 5x + p = 0` are real and unequal, then which
one of the following is correct?
NDA Paper 1 2012
(A)

`p = 25//12`

(B)

`p < 25//12`

(C)

`p >25//12`

(D)

`p <= 25//12`

Solution:

Since, the roots of the quadratic equation

`3x^2 - 5x + p = 0` are real and unequal.

`:.` Discriminant `> 0`

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

`=> (-5)^2 - 4(3)(p) > 0` (here, `b = - 5,a = 3, c = p`)

` => 25 - 12p > 0 => 25 > 12p`

`=> 12 p < 25 => p < (25)/(12)`
Correct Answer is `=>` (B) `p < 25//12`
Q 2328123001

What is the sum of the squares of the roots of the
equation `x^2 + 2x - 143 = 0`?
NDA Paper 1 2012
(A)

`170`

(B)

`180`

(C)

`190`

(D)

`290`

Solution:

Given quadratic equation is

`x^2 + 2x- 143 = 0`

Let `(alpha, beta)` be the roots of this equation.

Then, `alpha + beta = -2` and `alpha · beta = -143`

Now, `alpha^2 + beta^2 =(alpha + beta )^2 -2 alpha beta`

`= (-2)^2 - 2 (-143) = 4 + 286= 290`
Correct Answer is `=>` (D) `290`
Q 2338323202

If the difference between the roots of
`ax^2 + bx + c = 0` is `1`, then which one of the
following is correct?
NDA Paper 1 2012
(A)

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

(B)

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

(C)

`a^2 = c(a + 4c)`

(D)

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

Solution:

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

and `(alpha - 1)` by given condition.

Then, sum of the roots `= - b/a`

`=> alpha + (alpha - 1) = - b/a`

`=> 2alpha = 1 - b/a => alpha = (a - b)/(2a)`

and product of roots `= c/a`

` => alpha (alpha - 1 ) = c/a`

`=> (a- b)/(2a) { (a - b)/(2a) - 1 } = c/a`

` => (a-b)/(2a) . (-b-a)/(2a) = c/a`

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

`=> b^2 - a^2 = 4ac`

`:. b^2 = a(a + 4c)`
Correct Answer is `=>` (A) `b^2 = a(a + 4c)`
Q 2338423302

If one of the roots of the equation `x^2 + ax - b = 0`
is `1`, then what is the value of `(a-b)`?
NDA Paper 1 2012
(A)

`-1`

(B)

`1`

(C)

`2`

(D)

`-2`

Solution:

Let the roots of the given equation `x^2 + ax - b = 0` are

`1` and `alpha` by given condition.

`:.` Sum of the roots `= - a`

i.e., `alpha + 1 = -a`

`alpha = -1 -a` ......(i)

and product of the roots `= - b`

i.e., `alpha ·1 = -b`

`alpha = -b` ......(ii)

From Eqs. (i) and (ii),

`-b = - 1 - a => a - b = - 1`

Alternate Method

Since, one of the roots of the equation is `1`.

Hence, satisfy the given equation

`(1)^2 + a(1) - b = 0`

`=> a+a-b = 0 => a - b = -1`
Correct Answer is `=>` (A) `-1`
Q 2318523400

If `alpha` and `beta` are the roots of the equation
`x^2 - q(1+x)- r = 0`, then what is the value of
`(1 + alpha) (1 + beta)`?
NDA Paper 1 2012
(A)

`1 - r`

(B)

`q- r`

(C)

`1 + r`

(D)

`q + r`

Solution:

Given that, `alpha` and `beta` be the roots of the equation

`x^2 - q(1 + x) - r = 0`

i.e., `x^2 - qx - (q + r) = 0`.

Then, sum of the roots `= q`

i.e., `alpha + beta = q` ... (i)

and product of the roots `= - (q + r)`

i.e., `alpha beta = -(q + r)` ... (ii)

Now, `(1 + alpha ) (1 + beta ) = 1 + (alpha + beta ) + alpha beta = 1 + q - (q + r)`

`= 1- r`
Correct Answer is `=>` (A) `1 - r`
Q 2318623500

The equation formed by multiplying each root of
`ax^2 + bx + c = 0` by `2` is `x^2 + 36x = 24 = 0`.

What is the value of `b : c`?
NDA Paper 1 2012
(A)

`3 : 1`

(B)

`1 : 2`

(C)

`1 : 3`

(D)

`3: 2`

Solution:

Let `alpha` and `beta` be the roots of the equation

`ax^2 + bx + c = 0`.

Then, `alpha + beta = -b//a`

`=> 2alpha + 2beta = -2b//a` ... (i)

and `alpha · beta = c//a`

`=> (2 alpha ) · (2beta) = 4c//a` .....(ii)

Also, given that the equation `x^2 + 36x + 24 = 0` is formed by

multiplying each root of `ax^2 + bx + c = 0` by `2`.

`:. 2 alpha + 2 beta = - 36` ... (iii)

and `(2 alpha ) (2 beta ) = 24` ... (iv)

From Eqs. (i) and (iv),

`-36 = -2b//a => b/a = (18)/1` .....(v)

From Eqs. (ii) and (iii),

` 24 = (4c)/a => c/a = 6/1` ......(vi)

On dividing Eq. (v) by Eq. (vi), we get

` b/c = 3/1 => b : c = 3 : 1`
Correct Answer is `=>` (A) `3 : 1`
Q 2318623500

The equation formed by multiplying each root of
`ax^2 + bx + c = 0` by `2` is `x^2 + 36x = 24 = 0`.

Which one of the following is correct?
NDA Paper 1 2012
(A)

`bc = a^2`

(B)

`bc = 36 a^2`

(C)

`bc = 72 a^2`

(D)

`bc = 108 a^2`

Solution:

Let `alpha` and `beta` be the roots of the equation

`ax^2 + bx + c = 0`.

Then, `alpha + beta = -b//a`

`=> 2alpha + 2beta = -2b//a` ... (i)

and `alpha · beta = c//a`

`=> (2 alpha ) · (2beta) = 4c//a` .....(ii)

Also, given that the equation `x^2 + 36x + 24 = 0` is formed by

multiplying each root of `ax^2 + bx + c = 0` by `2`.

`:. 2 alpha + 2 beta = - 36` ... (iii)

and `(2 alpha ) (2 beta ) = 24` ... (iv)

From Eqs. (i) and (iv),

`-36 = -2b//a => b/a = (18)/1` .....(v)

From Eqs. (ii) and (iii),

` 24 = (4c)/a => c/a = 6/1` ......(vi)

On multiplying Eqs. (v) and (vi), we get

`b/a xx c/a = 18 xx 6 => bc = 108a^2`
Correct Answer is `=>` (D) `bc = 108 a^2`
Q 2378723606

If `alpha` and `beta` be the roots of the equation
`(x- a) (x -b) = c, c != 0`. Then, the roots of the
equation `(x -alpha) (x - beta) + c = 0` are
NDA Paper 1 2011
(A)

`(a, c)`

(B)

`(b, c)`

(C)

`(a, b)`

(D)

`(a + b, a + c)`

Solution:

Given quadratic equation is

`(x- a)(x- b) = c. c != 0`

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

Since, the roots of this equation is `alpha` and `beta`.

Then, `alpha + beta = -{ - (a+b)/1} = a + b` ......(i)

and `alpha beta = ab - c` ........(ii)

Now, consider the equation,

`(x- alpha) (x - beta) + c = 0`

` => x^2 - (alpha + beta )x + (alpha beta +c) = 0`

From Eqs. (i) and (ii),

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

` => x^2 - (a+ b)x + ab = 0`

So, the roots of this equation is `(a, b)`.
Correct Answer is `=>` (C) `(a, b)`
Q 2378823706

If the equations
`x^2 - px + q = 0` and `x^2 - ax + b = 0`
have a common root and the roots of the second
equation are equal, then which one of the
following is correct?
NDA Paper 1 2011
(A)

`aq = 2(b + p)`

(B)

`aq = b + p`

(C)

`ap = 2(b + q)`

(D)

`ap= b + q`

Solution:

Let the roots of the equation `x^2 - px + q = 0` is `(alpha, beta)`

and the roots of the equation `x^2 - ax + b = 0` is (`alpha , alpha`).

Then, `alpha + beta = p` and `alpha beta = q` ... (i)

`alpha + alpha = a` and `alpha ·alpha = b`

` => alpha = a/2` and `alpha^2 = b`

`=> (a/2)^2 = b => a^2 = 4b` ... (ii)

From Eq. (i), `alpha beta = q`

`=> a/2 . beta = q => beta = (2q)/a`

Now, put the value of `alpha` and `beta` in the following equation

`alpha + beta = p`

`=> a//2 + 2q//a = p`

`=> a^2 + 4q = 2ap`

`=> 4b + 4q = 2ap` (from Eq. (ii)]

`=> 2(b+q) = ap`
Correct Answer is `=>` (C) `ap = 2(b + q)`
Q 2378023806

If the equation `x^2 - 4x + 29 = 0` has one root
`2 + 5i`, then what is the other root (where, `i = sqrt(-1)`)?
NDA Paper 1 2011
(A)

`2`

(B)

`5`

(C)

`2 + 5i`

(D)

`2 - 5i`

Solution:

We know that, if one root of the quadratic equation is

complex, then its other root is its conjugate.

Now, `x^2 - 4x + 29 = 0` has one root `alpha = 2 + 5i`, then its

conjugate `bar alpha = 2 - 5i` is the second root of this equation.
Correct Answer is `=>` (D) `2 - 5i`
Q 2348123903

If one of the roots of the equation
`a(b-c)x^2 + b(c-a)x + c(a-b) = 0` is `1`, then
what is the second root?
NDA Paper 1 2011
(A)

` - ( b (c - a))/(a(b - c))`

(B)

` ( b (c - a))/(a(b - c))`

(C)

` ( c (a - b))/(a(b - c))`

(D)

` - ( c (a - b))/(a(b - c))`

Solution:

Given quadratic equation is

`a(b -c)x^2 + b(c- a)x + c(a- b)= 0`

Given that, one root is `1`.

Let the other root be `alpha`.

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

`:. a = - 1 - ((bc - ab))/((ab - ac)) = ((-ab + ac- bc + ab))/(a(b - c) ) = ( c (a- b))/(a(b - c))`
Correct Answer is `=>` (C) ` ( c (a - b))/(a(b - c))`
Q 2318134000

What are the roots of the equation
`2 (y + 2)^2 - 5 (y + 2) = 12 ?`
NDA Paper 1 2011
(A)

`-7//2,2`

(B)

`-3//2,4`

(C)

`-5//3,3`

(D)

`3//2,4`

Solution:

The given quadratic equation is

`2(y + 2)^2 - 5(y + 2) = 12`

`=> 2(y+ 2)^2- 5(y+ 2)-12 = 0`

Let `z =y + 2` .....(i)

`=> 2 z^2 - 5z - 12 = 0`

` => 2z^2 - 8z + 3z- 12 = 0`

` => (2z + 3) (z- 4) = 0`

`:. z = - 3/2 , 4`

`=> y + 2 = - 3/2 , 4` [from Eq. (i)]

`=>y = - 2 - 3/2 , 4 -2`

`:. y = - 7/2 , 2`

So, the required roots are `- 7/2` and `2`.
Correct Answer is `=>` (A) `-7//2,2`
Q 2328334201

If let `alpha` and `beta` be the roots of the equation
`x^2 + x + 1 = 0`, then the equation whose roots are
`alpha^( 19)` and `beta^7` is
NDA Paper 1 2011
(A)

`x^2 - x - 1 = 0`

(B)

`x^2 - x + 1 = 0`

(C)

`x^2 + x - 1 = 0`

(D)

`x^2 + x + 1 = 0`

Solution:

The given quadratic equation is

`x^2 + x + 1 = 0`

` x = (-1 ± sqrt(1-4))/2 = (-1 ± i sqrt3)/2`

`x = (-1+ i sqrt3)/2 , (-1-i sqrt3)/2`

or `x = omega , omega^2`

i.e., `alpha = omega` and `beta = omega^2`

`=> alpha^(19) = omega^(19) = ( omega^3)^6 · omega = (1)^6 . omega = omega`

`=> beta^7 = ( omega^2)^7 = omega^(14) = ( omega^3)^4. omega^2 = (1)^4. omega^2 = omega^2`

Now, sum of roots, `alpha^( 19) + beta^( 7) = omega + omega^2 = -1 ( ∵ 1 + omega + omega^2 = 0)`

and product of roots , `alpha^(19) - beta^7 = omega . omega^2 = omega^3 = 1 ( ∵ omega^3 = 1)`

So, the required quadratic equation whose roots are `alpha^( 19)` and `beta^7` is

`x^2 - (alpha^(19) + beta^7)x + (alpha^(19) · beta^7) = 0`

` => x^2 - (-1)x + (1) = 0`

` => x^2 + x + 1 = 0`
Correct Answer is `=>` (D) `x^2 + x + 1 = 0`
Q 2348434303

What is the value of
` sqrt(8 + 2 sqrt (8 + 2 sqrt(8 + 2 sqrt(8 + .... oo))))` ?
NDA Paper 1 2011
(A)

10

(B)

8

(C)

6

(D)

4

Solution:

Let `y = sqrt(8 + 2 sqrt (8 + 2 sqrt(8 + 2 sqrt(8 + .... oo)))) `

` => y = sqrt(8+2y) => y^2 = 8 + 2y`

`=> y^2 - 2y - 8 = 0 => y^2 - 4y + 2y - 8 = 0`

`=> (y- 4)(y + 2) = 0`

`:. y = - 2, 4`

So, the required value of expression is ` 4`.
Correct Answer is `=>` (D) 4
Q 2328534401

If the roots of the equation `3x^2 - 5x + q = 0` are
equal, then what is the value of `q`?
NDA Paper 1 2011
(A)

`2`

(B)

`5//12`

(C)

`12//25`

(D)

`25//12`

Solution:

Let the roots of the equation,

`3x^2 - 5x + q = 0` is `(alpha , alpha)`.

Then, sum of the roots `= - (- 5/3)`

`=> alpha + alpha = 5/3 => alpha = 5/6`

Since, this roots satisfy the given equation.

`:. 3 (5/6)^2 - 5 (5/6 ) + q = 0 => (75)/(36) - (25)/6 + q = 0`

` => 75 - 150 - 36q = 0 => 36q = 75`

`:. q = (25)/(12)`
Correct Answer is `=>` (D) `25//12`
Q 2358645504

If `alpha` and `beta` are the roots of the equation
`4x^2 + 3x + 7 = 0`, then what is the value of
`(alpha^(-2) + beta^(-2))`?
NDA Paper 1 2011
(A)

`47//49`

(B)

`49//47`

(C)

`-47 //49`

(D)

`- 49//47`

Solution:

Since, `alpha` and `beta` are the roots of the equation

`4x^2 + 3x + 7 = 0`.

Now,

`:. alpha + beta = - 3/4` and `alpha beta = 7/4`

Now, `alpha^(-2) + beta^(-2) = 1/alpha^2 + 1/beta^2 = ( alpha^2 + beta^2 )/(alpha beta)^2`

` = (( alpha + beta )^2 - 2 alpha beta )/( alpha beta )^2 = (9/(16) - 7/2)/( (49)/(16))`

` = ( (9 - 56)/(16) )/( (49)/(16)) = ( -47)/(16) . (16)/(49) = - (47)/(49)`
Correct Answer is `=>` (C) `-47 //49`
Q 2348745603

If `p, q` and `r` are rational numbers, then the roots of
the equation `x^2 - 2px + p^2 - q^2 + 2qr - r^2 = 0` are
NDA Paper 1 2011
(A)

complex

(B)

pure imaginary

(C)

irrational

(D)

rational

Solution:

The given equation is

`x^2 - 2 px + p^2 - q^2 + 2qr - r^2 = 0`.

Now, `B^2 - 4AC = (-2p)^2 - 4(1)(p^2 - q^2 + 2q r - r^2)`

`= 4p^2 - 4p^2 + 4(q - r)^2 = 4(q - r)^2`

which is always greater than zero.

Therefore, the roots of the given equation are rational.
Correct Answer is `=>` (D) rational
Q 2308745608

What is .the sum of the roots of the equation
`(2 - sqrt(3)) x^2 - (7 - 4 sqrt3) x + (2 + sqrt(3)) = 0` ?
NDA Paper 1 2011
(A)

`2 - sqrt(3)`

(B)

`2 + sqrt(3)`

(C)

`7 - 4sqrt(3)`

(D)

`4`

Solution:

The given equation is

`(2 - sqrt3)x^2 - (7 - 4 sqrt(3) )x + (2 + sqrt(3)) = 0`

`:.` Sum of roots `= (7 - 4sqrt(3))/(2 - sqrt3) = ( 2- sqrt3)^2/( 2- sqrt3) = 2 - sqrt(3)`
Correct Answer is `=>` (A) `2 - sqrt(3)`

Set 2

Q 2378356206

If the roots of the equation
` (a^2 + b^2 ) x^2 -2b(a + c) x + (b^2 + c^2) = 0`
are equal, then which one of the following is
correct?
NDA Paper 1 2010
(A)

`2b = a + c`

(B)

`b^2 = ac`

(C)

`b + c = 2a`

(D)

`b = ac`

Solution:

Since, the roots of given equation

`(a^2 + b^2)x^2 - 2b(a + c)x + (b^2 + c^2) = 0` are equal.

`:. 4b^2 (a + c)^2 = 4(a^2 + b^2) (b^2 + c^2) ( ∵ B^2 - 4AC = 0)`

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

`=> a^2b^2 + b^2c^2 + 2acb^2 = a^2b^2 + b^2c^2 + a^2c^2 + b^4`

`=> b^4 + a^2c^2 - 2acb^2 = 0`

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

`=> b^2 = ac`
Correct Answer is `=>` (B) `b^2 = ac`
Q 2318556400

If `alpha` and `beta` are the roots of the equation
`x^2 - 2x + 4 = 0`, then what is the value of `alpha^3 + beta^3` ?
NDA Paper 1 2010
(A)

`16`

(B)

`- 16`

(C)

`8`

(D)

`-8`

Solution:

Since, `alpha` and `beta` are the roots of

`x^2 - 2x + 4 = 0`.

`:. alpha + beta = 2, alpha beta = 4`

Now `alpha^3 + beta^3 = (alpha + beta )^3 - 3alpha beta (alpha + beta )`

`= 2^3 - 3 xx 4 xx 2`

`= 8 - 24 = -16`
Correct Answer is `=>` (B) `- 16`
Q 2318656509

Which of the following are the two roots of the
equation `(x^2 + 2)^2 + 8x^2 = 6x (x^2 + 2)`?
NDA Paper 1 2010
(A)

`1 ± i`

(B)

`2 ± i `

(C)

`1 ± sqrt(2)`

(D)

`2 ± i sqrt(2)`

Solution:

` ∵ (x^2 + 2)^2 + 8x^2 = 6x(x^2 + 2)`

`=> x^4 + 4x^2 + 4 + 8x^2 = 6x^3 + 12 x`

`=> x 4 - 6x^3 + 12x^2 - 12x + 4 = 0`

Also, this equation is satisfied by `1 ± i`. Hence, required roots are

`1 ± i`.
Correct Answer is `=>` (A) `1 ± i`
Q 2368156905

If `alpha` and `beta` are the roots of the equation
`x^2 + x + 1 = 0`, then which of the following are the
roots of the equation `x^2 - x + 1 = 0`?
NDA Paper 1 2010
(A)

`alpha^7` and `beta^(13)`

(B)

`alpha^(13)` and `beta^7`

(C)

`alpha^(20)` and `beta^(20)`

(D)

None of these

Solution:

Since, `alpha` and `beta` be the roots of the equation

`x^2 + x + 1 = 0`.

`:. alpha + beta = -1` and `alpha beta = 1`

`=> alpha = omega` and `beta = omega^2`

Now, for equation `x^2 - x + 1 = 0`,

`:.` Sum of roots `= 1`

and product of roots `= 1`

Looking through options,

`alpha^7. beta^(13) = (omega)^7 (omega^2)^(13) = 1 != -1`

`alpha^(13).beta^7 = (omega)^(13) (omega^2) = 1 != -1`

`alpha^(20). beta^(20) = (omega)^(20) (omega^2)^(20) = 1 != -1`

Hence, no one of (a), (b) and (c) are satisfied for the given

equation.
Correct Answer is `=>` (D) None of these
Q 2348167003

Consider the equation `(x - p) (x - 6) + 1 = 0` having
integral coefficients. If the equation has integral
roots, then what values can `p` have?
NDA Paper 1 2010
(A)

4 or 8

(B)

5 or 10

(C)

6 or 12

(D)

3 or 6

Solution:

The given equation can be rewritten as

`x^2 - (p + 6)x + (6p + 1) = 0`

Now, discriminant, `D = 0 => b^2 - 4ac = 0` for integral roots.

`=> (p + 6)^2 - 4(6p + 1) = 0`

`=> p^2 - 12p + 32 = 0`

` => (p- 4) (p- 8) = 0`

` :. P = 4, 8`
Correct Answer is `=>` (A) 4 or 8
Q 2318267109

If `1/(2 - sqrt(-2)) ` is one of the roots of `ax^2 + bx + c = 0`,
where, `a, b` and `c` are real, then what are the values
of `a, b` and `c` respectively?
NDA Paper 1 2010
(A)

6, - 4, 1

(B)

4, 6, - 1

(C)

3,- 2, 1

(D)

6, 4, 1

Solution:

Given, root `= 1/(2 - sqrt(-2)) = 1/(2 - sqrt(2i) ) . (2 + sqrt(2) i)/(2 + sqrt(2) i)`

` = (2 + sqrt(2) i)/(4 + 2) = (2 + sqrt(2) i)/6`

So, the another root will be ` (2 - sqrt(2) i)/6`

Thus, sum of roots `= (2 + sqrt(2) i)/6 + (2 - sqrt(2) i)/6 = 4/6`

and product of roots `= ((2 + sqrt(2) i)/6) ( (2 - sqrt(2) i)/6) = (4 + 2)/(36) = 1/6`

`:.` Required equation is

`x^2 -` (sum of roots) x + (product of roots) `= 0`

` => x^2 - 4/6 x + 1/6 = 0`

` => 6x^2 - 4x + 1 = 0`

Thus, the values of `a, b` and `c` are `6,- 4` and `1`, respectively.
Correct Answer is `=>` (A) 6, - 4, 1
Q 2368567405

If `alpha, beta` are the roots of the quadratic equation
`x^2 - x + 1= 0`, then which one of the following is
correct?
NDA Paper 1 2010
(A)

`(alpha^4 - beta^4 )` is real

(B)

`2(alpha^5 + beta^5 ) = (alpha beta )^5`

(C)

`(alpha^6 - beta^6) = 0`

(D)

`(alpha^8 + beta^8) = (alpha beta)^8`

Solution:

Since, `alpha` and `beta` be the roots of the equation

`x^2 - x + 1 = 0`.

`:. alpha + beta = 1` and `alpha beta = 1`

Now, `alpha - beta = sqrt((alpha + beta )^2 - 4 alpha beta ) = sqrt(3)i`

` => alpha = (1 + i sqrt3)/2 ` and `(1 - i sqrt3)/2 `

Now `alpha = cos pi/3 + i sin pi/3`

and ` beta = cos pi/3 - i sin pi/3`

(a) `alpha^4 - beta^4 = cos (4 pi)/3 + i sin (4 pi)/3 - cos (4 pi)/3 + i sin (4 pi)/3`

`= 2 i sin (4 pi)/3`

Hence, `alpha^4 - beta^4 ` is not real.

(b) `2(alpha^5 + beta^5 ) = 2 ( cos (5 pi)/3 + i sin (5 pi)/3 + cos (5 pi)/3 - i sin (5 pi)/3 )`

`= 2 . 2 cos (5 pi)/3 = 4 . 1/2 = 2`

Now, `(alpha beta )^5 = 1 => 2 (alpha^5 + beta^5) != (alpha beta )^5`

(c ) `alpha^6 - beta^6 = cos (6 pi)/3 + i sin (6 pi)/3 - cos (6 pi)/3 + i sin (6 pi)/3`

` = 2i sin 2 pi = 0`

(d) `alpha^8 + beta^8 = cos (8 pi)/3 + sin (8 pi)/3 + cos (8 pi)/3 - i sin (8 pi)/3`

`= 2 cos (8 pi)/3 = 2 ( - 1/2) = -1`

Now, `(alpha beta )^8 = (1)^8 = 1 => (alpha^8 + beta^8) != (alpha beta)^8`
Correct Answer is `=>` (C) `(alpha^6 - beta^6) = 0`
Q 2388767607

If `p` and `q` are positive integers, then which one of
the following equations has `p - sqrtq` as one of its
roots?
NDA Paper 1 2010
(A)

`x^2 - 2px - (p^2 - q) = 0`

(B)

`x^2 - 2px + (p^2 - q) = 0`

(C)

`x^2 + 2px - (p^2 - q) = 0`

(D)

`x^2 + 2px + (p^2 - q) = 0`

Solution:

If any equation has `p - sqrt q`. as a root, then the another

root will be `p + sqrt q`.

`:.` Sum of roots `= p - sqrt q + p + sqrt q = 2p`,

and product of roots `= (p - sqrt q) (p + sqrt q ) = p^2 - q`

Now, the required equation is

`x^2 -` (sum of roots)x + (product of roots) `= 0`

` => x^2 - 2 px + (p^2 - q) = 0`
Correct Answer is `=>` (B) `x^2 - 2px + (p^2 - q) = 0`
Q 2378867706

If the product of the roots of the equation
`x^2 - 5x + k = 15` is `- 3`, then what is the value of `k`?
NDA Paper 1 2010
(A)

`12`

(B)

`15`

(C)

`16`

(D)

`18`

Solution:

Let `alpha` and `beta` be the roots of the equation

`x^2 - 5x + k - 15 = 0`.

`:. alpha beta = k - 15`

But `alpha beta = - 3`

` => -3 = k - 15 => k = 15 - 3 = 12`
Correct Answer is `=>` (A) `12`
Q 2328067801

If the equation `x^2 - bx + 1 = 0` does not possess
real roots, then which one of the following is
correct?
NDA Paper 1 2010
(A)

`- 3 < b < 3`

(B)

`- 2 < b < 2`

(C)

`b > 2`

(D)

`b < - 2`

Solution:

Since, equation `x^2 - bx + 1 = 0` has no real roots i.e.,

it has imaginary roots which is possible only, if

`b^2 - 4 < 0 ( ∵ B^2 - 4AC < 0)`

` => b^2 < 4 => -2 < b < 2`
Correct Answer is `=>` (B) `- 2 < b < 2`
Q 2318167900

If `p` and `q` are the roots of the equation
`x^2 - px + q = 0`, then what are the values of `p` and `q`
respectively'?
NDA Paper 1 2010
(A)

1, 0

(B)

0, 1

(C)

-2, 0

(D)

-2, 1

Solution:

Since, `p` and `q` are the roots of `x^2 - px + q = 0`.

then, `p + q = p` and `pq = q`

` => q(p- 1) = 0 => q = 0, p = 1`
Correct Answer is `=>` (A) 1, 0
Q 2318178000

If the roots of `ax^2 + bx + c = 0` are `sin alpha` and `cos alpha`
for some `alpha`, then which one of the following is
correct?
NDA Paper 1 2009
(A)

`a^2 + b^2 = 2ac`

(B)

`b^2 - c^2 = 2ab`

(C)

`b^2 - a^2 = 2ac`

(D)

`b^2 + c^2 = 2ab`

Solution:

Since, `sin alpha` and `cos alpha` are the roots of

`ax^2 + bx + c = 0`

`:. sin alpha + cos alpha = (-b)/a` ......(i)

and `sin alpha cos alpha = c/a` ... (ii)

`=> sin^2 alpha + cos^2 alpha + 2 sin alpha cos alpha = b^2/a^2` [from Eq. (i)]

` => 1 + ( 2c)/a = b^2/a^2` [from Eq. (ii)]

` => b^2 - a^2 = 2ac`
Correct Answer is `=>` (C) `b^2 - a^2 = 2ac`
Q 2388278107

The roots of the equation `(x - p) (x - q) = r^2`
, where `p, q` and `r` are real, are
NDA Paper 1 2009
(A)

always complex

(B)

always real

(C)

always purely imaginary

(D)

None of the above

Solution:

Given, `(x - p)(x - q)= r^2`

`=> x^2 - (p + q )x + pq - r^2 = 0`

Now, `D = (p+ q)^2 - 4(pq- r^2) ( ∵ D = B^2 - 4AC)`

`= (p - q)^2 + 4r^2 >= 0`

Hence, roots are always real.
Correct Answer is `=>` (B) always real
Q 2358378204

The equation `x - 2(x -1)^(-1) = 1 - 2(x -1)^(-1)` has
NDA Paper 1 2009
(A)

no roots

(B)

one root

(C)

two equal roots

(D)

infinite roots

Solution:

Given, `x - 2(x - 1)^(-1) = 1 - 2(x - 1)^(-1)`

` => x - 2/(x - 1) = 1 - 2/(x - 1) => x = 1`

But `x = 1` doesn't satisfy the given equation.

Hence, no roots exist.
Correct Answer is `=>` (A) no roots
Q 2318378209

If `a, b` and `c` are real numbers, then the roots of the
equation `(x- a)(x- b)+ (x- b)(x- c) + (x- c)(x- a) = 0` are
always
NDA Paper 1 2009
(A)

real

(B)

imaginary

(C)

positive

(D)

negative

Solution:

Given,

`(x - a) (x - b)+ (x - b )(x -c) + (x - c )(x - a) = 0`

` => 3x^2 - 2(b +a+ c)x + ab +be+ ca = 0`

Now, `D = 4(a + b + c)^2 - 12(ab +bc+ ca)`

`= 4a^2 + b^2 + c^2 - ab - bc - ca`

`= 2 1/2 = {(a-b)^2 +(b-c)^2 +(c-a)^2 } > 0`

Hence, the roots are always real.
Correct Answer is `=>` (A) real
Q 2368478305

For the two equations `x^2 + mx + 1 = 0` and
`x^2 + x + m = 0`, what is/are the value/values of `m`
for which these equations have at least one
common root?
NDA Paper 1 2009
(A)

-2 only

(B)

1 only

(C)

-2 and 1

(D)

-2 and -1

Solution:

Let given equations have common root `alpha`.

Then, `alpha^2 + m alpha + 1 = 0` and `alpha^2 + alpha + m = 0`

` => alpha^2/(m^2 - 1) = alpha/(1 - m) = 1/(1 - m)`

` => alpha/(1 - m) = 1/(1 - m) => alpha = 1`

Also, `alpha^2/(m^2 - 1) = 1/ ( 1 - m) => 1 - m = m^2 -1 (∵ alpha = 1)`

` => m^2 + m - 2 = 0 => (m + 2)(m- 1) = 0`

` :. m = 1` and `-2`
Correct Answer is `=>` (C) -2 and 1
Q 2388578407

Let `alpha , gamma ` be the roots of `Ax^2 - 4x + 1 = 0` and `beta , delta` be
the roots of `Bx^2 - 6x + 1 = 0`. If `alpha , beta , gamma` and `delta` are in
HP, then what are the values of `A` and `B`, respectively?
NDA Paper 1 2009
(A)

3, 8

(B)

-3, -8

(C)

3, -8

(D)

-3, 8

Solution:

Since, `alpha` and `gamma` be the roots of `Ax^2 - 4x + 1 = 0`

`:. alpha + gamma = 4/A` and `alpha gamma = 1/A`

and `beta` and `delta` be the roots of `Bx^2 - 6x + 1 = 0`.

`:. beta + delta = 6/B` and `beta delta = 1/B`

Also, `alpha ,beta , gamma` and `delta` are in HP.

Hence `1/alpha , 1/beta , 1/ gamma` and `1/delta` are in AP.

` => 1/beta - 1/alpha = 1/ delta - 1/ gamma => 1/beta - 1/delta = 1/alpha - 1/gamma`

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

` => sqrt((36)/B^2 - 4/B) /(1 /B) = sqrt((16)/A^2 - 4/A)/(1/A) => sqrt(36 - 4B) =sqrt(16 - 4A)`

`=> 36 - 48 = 16 - 4A => 4A - 48 = - 20`

It is possible, when `A = + 3` and `B = 8`.
Correct Answer is `=>` (A) 3, 8
Q 2328778601

If `(x + a)` is a factor of both the quadratic
polynomials `x^2 + px + q` and `x^2 + lx + m,` where
`p, q, l` and `m` are constants, then which one of the
following is correct?
NDA Paper 1 2009
(A)

` a = (m - q)//(l -p)(l != p)`

(B)

`a = (m + q)//(l + p)(l != p)`

(C)

`l = (m- q)//(a- p)(a != p)`

(D)

`p = (m - q)//(a- l)(a != l)`

Solution:

Since, `(x + a)` is a factor of

`x^2 + px + q` and `x^2 + lx + m`

`:. a^2 - ap + q = 0`, ......(i)

and `a^2 - la + m = 0` ...(ii)

From Eqs. (i) and (ii),

`-ap + q + la - m = 0`

`=> (l - p)a = m - q`

` => a = (m - q)/(l - p) (l != p)`
Correct Answer is `=>` (A) ` a = (m - q)//(l -p)(l != p)`
Q 2358778604

Which one of the following is one of the roots of
the equation `(b -c)x^2 + (c -a)x +(a -b) =0`?
NDA Paper 1 2009
(A)

` (c-a)/(b - c)`

(B)

` (a - b)/(b - c)`

(C)

` ( b - c)/(a - b)`

(D)

`(c -a)/(a - b)`

Solution:

`(b- c)x^2 + (c - a)x + (a- b) = 0`

`=> (b- c)x^2 - (b- c - b + a)x +(a- b)= 0`

`=> (b- c)x(x - 1)- (a- b)(x- 1) = 0`

`=> {(b- c)x- (a- b)} (x- 1) = 0`

`:. x = (a - b)/(b - c)` and `x = 1`
Correct Answer is `=>` (B) ` (a - b)/(b - c)`
Q 2318878700

If `alpha , beta` are the roots of the equation
`2x^2 - 2(1 + n^2 )x + (1 + n^2 + n^4) = 0`, then what is
the value of `alpha^2 + beta^2` ?
NDA Paper 1 2009
(A)

`2n^2`

(B)

`2n^4`

(C)

`2`

(D)

`n^2`

Solution:

Since, `alpha` and `beta` be the roots of

`2x^2 - 2(1 + n^2)x + (1 + n^2 + n^4) = 0`.

`:. alpha + beta = (n^2 + 1)` and `alpha beta = (1 + n^2 + n^4)/2`

Now, `alpha^2 + beta^2 = (alpha + beta )^2 - 2 alpha beta = (n^2 + 1)^2 - (1 + n^2 + n^4)`

`= n^4 + 1 + 2n^2 - 1 - n^2 - n^4 = n^2`
Correct Answer is `=>` (D) `n^2`
Q 2368878705

The roots of `Ax^2 + Bx + C = 0` are `r` and `s`. For the
roots of `x^2 + px + q = 0` to be `r^2` and `s^2`, what must
be the value of `p`?
NDA Paper 1 2009
(A)

`((B^2 - 4AC))/A^2`

(B)

`((B^2 - 4AC))/A^2`

(C)

`((2AC - B^2))/A^2`

(D)

`B^2 - 2C`

Solution:

Since, r and s are the roots of `Ax^2 + Bx + C = 0`, then

`r + s = - B/A` and `rs = C/A`

Now, roots of `x^2 + px + q = 0` be `r^2` and `s^2`

`:. r^2 + s^2 = - p` and `r^2s^2 = q`.

`=> (r + s)^2 - 2rs = - p`

` => B^2/A^2 - (2C)/A = - p => ( B^2 - 2AC)/A^2 = - p`

` => p = ((2AC - B^2))/A^2`
Correct Answer is `=>` (C) `((2AC - B^2))/A^2`
Q 2318078800

A quadratic polynomial with two distinct roots
has one real root. Then, the other root is
NDA Paper 1 2008
(A)

not necessarily real, if the coefficients are real

(B)

always imaginary

(C)

always real

(D)

real, if the coefficients are real

Solution:

If a quadratic polynomial with two distinct roots, has

one root real, then

`b^2 - 4ac > 0`

So, the other root is also real.
Correct Answer is `=>` (C) always real
Q 2358191904

If `sin alpha` and `cos alpha` are the roots of the equation
`px^2 + qx + r = 0`, then which one of the following
is correct?
NDA Paper 1 2008
(A)

`p^2 + q^2 -2pr = 0`

(B)

`p^2 - q^2 + 2pr = 0`

(C)

`(p + r)^2 = 2(p^2 + r^2 )`

(D)

`(p- r)^2 = q^2 + r^2`

Solution:

Since, `sin alpha` and `cos alpha` are the roots of the equation

`px^2 + qx + r = 0`.

`:. sin alpha + cos alpha = (-q)/p` .....(i)

and `sin alpha cos alpha = r/p` .........(ii)

From Eq. (i),

`(sin alpha + cos alpha)^2 = q^2/p^2`

`=> sin^2 alpha + cos^2 alpha + 2 sin alpha cos alpha = q^2/p^2`

`=> 1 + 2.r/p = q^2/p^2` [from Eq. (ii)]

`=> p^2 + 2rp = q^2`

`=> p^2 - q^2 + 2 r p = 0`
Correct Answer is `=>` (B) `p^2 - q^2 + 2pr = 0`
Q 2368191905

If the roots of the equation `x^2 - bx + c = 0` are two
consecutive integers, then what is the value of
`b^2 - 4c`?
NDA Paper 1 2008
(A)

`1`

(B)

`2`

(C)

`-2`

(D)

`3`

Solution:

Let the roots of:the equation `x^2 - bx + c = 0` be `alpha`

and `alpha + 1`.

`:. alpha + (alpha + 1) = b`

`=> 2 alpha + 1 = b` and `alpha (alpha + 1 ) = c`

Now, `b^2 - 4c = (2alpha + 1)^2 - 4[alpha (alpha + 1)]`

`= 4 alpha ^2 + 1 + 4 alpha - 4alpha ^2 - 4alpha = 1`
Correct Answer is `=>` (A) `1`
Q 2378191906

If `r` and `s` are roots of `x^2 + px + q = 0`, then what is
the value of `(1// r^2) + (1// s^2 )`?
NDA Paper 1 2008
(A)

`p^2 - 4q`

(B)

`(p^2 - 4q)/2`

(C)

`(p^2 - 4q)/q^2`

(D)

`(p^2 - 2q)/q^2`

Solution:

Since, `r` and `s` are the roots of the equation

`x^2 + px + q = 0`.

`:. r + s = - p` and `rs = q`

Now, `1/r^2 + 1/s^2 = (r^2 + s^2)/(rs)^2 = ((r + s )^2 - 2rs)/(rs)^2`

` = ((-p)^2- 2q)/q^2 = ( p^2 - 2q)/q^2`
Correct Answer is `=>` (D) `(p^2 - 2q)/q^2`
Q 2388191907

If `alpha` and `beta` are the roots of `x^2 + 4x + 6 = 0`, then
what is the value of `alpha^3 + beta^3` ?
NDA Paper 1 2008
(A)

` - 2/3`

(B)

`2/3`

(C)

`4`

(D)

`8`

Solution:

Since, `alpha` and `beta` are the roots of `x^2 + 4x + 6 = 0`.

`:. alpha + beta = - 4` and `alpha beta = 6`

Now, `alpha^3 + beta^3 = (alpha + beta)^3 - 3alpha beta (alpha + beta) = (-4)^3 - 3 xx 6 (-4)`

`= -64 + 72 = 8`
Correct Answer is `=>` (D) `8`
Q 2318191909

If sum of the roots of `3x^2 + (3p + 1)x- (p + 5) = 0`
is equal to their product, then what is the value of
`p`?
NDA Paper 1 2008
(A)

`2`

(B)

`3`

(C)

`4`

(D)

`9`

Solution:

Let `alpha` and `beta` are the roots of the given equation.

`:. alpha + beta = (- (3p + 1))/3`

and ` alpha beta = (- (p+ 5))/3`

Now , `(- (3p+ 1))/3 = (- (p+ 5))/3`(according to the question)

` => 3p + 1 = p + 5`

`=> 2p = 4`

`=> p = 2`
Correct Answer is `=>` (A) `2`
Q 2349101013

If `x` is real and `x^2 - 3x + 2 > 0, x^2 - 3x- 4 <= 0`, then
which one of the following is correct?
NDA Paper 1 2008
(A)

`-1 <= x <= 4`

(B)

`2 <= x <= 4`

(C)

`-1 < x <= 1`

(D)

`-1 <= x <= 1` or `2 < x <= 4`

Solution:

`∵ x^2 - 3x + 2 > 0`

`=> (x -1) (x -2) > 0`

`=> x < 1 ` or `x > 2` ......(i)

and `x^2 - 3x - 4 <= 0`

`=> (x - 4) (x + 1) <= 0`

`:. -1 <= x <= 4` .......(ii)

`=> -1 <= x < 1` or `2 < x <= 4` [from Eqs. (i) and (ii)]
Correct Answer is `=>` (D) `-1 <= x <= 1` or `2 < x <= 4`
Q 2359101014

What is the value of `sqrt(5sqrt(5sqrt(5sqrt(...oo))))`?
NDA Paper 1 2008
(A)

`5`

(B)

`sqrt5`

(C)

`1`

(D)

`(5)^(1//4)`

Solution:

Let ` x = sqrt(5sqrt(5sqrt(5sqrt(...oo))))`

`=> x = sqrt(5x) => x^2 = 5x`

`=> x^2 - 5x = 0 => x(x - 5) = 0`

`:. x = 0` or `5`
Correct Answer is `=>` (A) `5`
Q 2379101016

If one root of the equation `x^2 = px + q` is reciprocal
of the other and `p != ± 1`, then what is the value of
`q`?
NDA Paper 1 2008
(A)

`q = -1`

(B)

`q = 1`

(C)

`q = 0`

(D)

`q = 1/2`

Solution:

Let the roots of the equation `x^2 - px - q = 0` be ` alpha` and

`1/alpha`.

`:.` Product of roots `= alpha · 1/alpha = - q/1`

`=> 1 = -q => q = -1`
Correct Answer is `=>` (A) `q = -1`
Q 2309101018

The numerical value of the perimeter of a square
exceeds that of its area by `4`. What is the side of
the square?
NDA Paper 1 2008
(A)

`1` unit

(B)

`2` units

(C)

`3` units

(D)

`4` units

Solution:

Let the side of the square `= x` units

`:.` Area of square `= x^2` units

and perimeter of the square `= 4x` units

According to the question,

`x^2 + 4 = 4x`

`=> x^2 - 4x + 4 = 0 => (x - 2)^2 = 0`

`=> x = 2`

`:.` Side of square `= 2` units
Correct Answer is `=>` (B) `2` units
Q 2329201111

If the equation `x^2 + kx + 1 = 0` has the roots `alpha` and
`beta`, then what is the value of `(alpha + beta) xx (alpha^(-1) + beta^(-1) )`?
NDA Paper 1 2008
(A)

`k^2`

(B)

`1/k^2`

(C)

`2k^2`

(D)

`1/(2k^2)`

Solution:

Since, the roots of the equations `x^2 + kx + 1 = 0` are `alpha`

and `beta`.

`:. alpha + beta = - k` and `alpha beta = 1`

Now, `(alpha + beta)(alpha^(-1) + beta^(-1) ) = (alpha + beta) (1/alpha + 1/beta)`

` = (alpha + beta) ( (alpha + beta)/ (alpha beta))`

`= (alpha + beta)^2/(alpha beta) = (-k)^2/1 = k^2`
Correct Answer is `=>` (A) `k^2`
Q 2318256109

If `f: R-> R` be defined as `f(x) = ax^2 + bx + c; a,b`
and `c` being fixed non-zero real numbers, then
which one of the following statements is correct in general?
NDA Paper 1 2007
(A)

If `b^ 2 - 4ac > 0`, then `f^(-1)(0)` does not contain `0`

(B)

If `b2^ - 4ac < 0`. then `f^(-1)(0)` must contain `0`

(C)

If `b^ 2 - 4ac > 0`, then `f^(-1)(0)` may contain `0`

(D)

If `b2^ - 4ac < 0`. then `f^(-1)(0)` may contain `0`

Solution:

Given, `f(x) = ax^2 + bx + c`

`=> f(x)=a{x^2+b/a x c/a}`

`={x^2+(bx)/a+b^2/(4a^2)-b^2/(4a^2)+c/a}`

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

Let `y=f(x)=a(x+b/(2a))^2-((b^2-4ac)/(4a))`

`=> x+b/(2a)=sqrt(y/a+((b^2-4ac)/(4a^2))`

`=> f^(-1)(y)=-b/(2a)+sqrt(y/a((b^2-4ac)/(4a^2)))`

At `y=0`

`f^(-1)(0)=-b/(2a)+sqrt((b^2-4ac)/(4a^2))`

Here, `b^2-4ac >0`

`=> f^(-1)(0)=-b/(2a)+ (sqrt(b^2-4ac))/(2a)`

Hence, `f^(-1) (0)` does not contain 0 for non-zero real numbers `a, b`
and `c`.
Correct Answer is `=>` (A) If `b^ 2 - 4ac > 0`, then `f^(-1)(0)` does not contain `0`
Q 2349201113

If `alpha` and `beta` are the roots of the equation
`x^2 + 6x + 1 = 0`, then what is `| alpha - beta |` equal to?
NDA Paper 1 2007
(A)

`6`

(B)

`3 sqrt(2)`

(C)

`4 sqrt(2)`

(D)

`12`

Solution:

Since, `alpha` and `beta` are the roots of the equation

`x^2 + 6x + 1 = 0`.

`:. alpha + beta = - 6` and `alpha beta = 1`

Now, `(alpha - beta)^2 = (alpha + beta)^2 - 4 alpha beta`

`= ( -6)^2 - 4 = 36 - 4 = 32`

`=> | alpha - beta | = sqrt(32) = 4sqrt(2)`
Correct Answer is `=>` (C) `4 sqrt(2)`
Q 2379201116

The number of rows in a lecture hall equals to the
number of seats in a row. If the number of rows is
doubled and the number of seats in every row in
reduced by `10`, the number of seats is increased by
`300`. If `x` denotes the number of rows in the lecture
hall, then what is the value of `x`?
NDA Paper 1 2007
(A)

`10`

(B)

`15`

(C)

`20`

(D)

`30`

Solution:

` ∵` Number of rows `= x`

`:.` Number of seats in each row `= x`

Thus, total number of seats `= x^2`

Now, new number of rows `= 2x`

and new number in each row `= x - 10`

Thus, total number of seats `= 2x(x - 10) = 2x^2 - 20x`

According to the question,

`2x^2 - 20x = 300 + x^2`

`=> x^2 - 20x - 300 = 0`

`=> x^2 - 30x + 10x - 300 = 0`

`=> (x - 30)(x + 10) = 0`

`=> x = 30 ( ∵ x != -10)`
Correct Answer is `=>` (D) `30`
Q 2319201119

If `alpha` and `beta` are the roots of the equation
`lx^2 - mx + m = 0, l != m, l != 0`, then which one of the
following statements is correct'?
NDA Paper 1 2007
(A)

` sqrt(alpha/beta ) + sqrt(beta/alpha) - sqrt(m/l) = 0`

(B)

` sqrt(alpha/beta ) + sqrt(beta/alpha) + sqrt(m/l) = 0`

(C)

` sqrt( alpha + beta)/(alpha beta) - sqrt(m/l) = 0`

(D)

The arithmetic mean of `alpha` and `beta` is the same as their geometric mean.

Solution:

Since, `alpha` and `beta` are the roots of the equation

`lx^2 - mx + m = 0`.

`:. alpha + beta = m/l ` and ` alpha beta = m/l`

Now , ` sqrt(alpha/beta ) + sqrt(beta/alpha) = ( alpha + beta)/sqrt(alpha beta) = (m//l)/sqrt(m//l)`

` => sqrt(alpha/beta ) + sqrt(beta/alpha) = sqrt(m/l)`

`:. sqrt(alpha/beta ) + sqrt(beta/alpha) - sqrt(m/l) = 0`
Correct Answer is `=>` (A) ` sqrt(alpha/beta ) + sqrt(beta/alpha) - sqrt(m/l) = 0`
Q 2339301212

For what value of `k`, are the roots of the quadratic
equation `(k + 1)x^2 - 2(k - 1)x + 1 = 0` real and
equal?
NDA Paper 1 2007
(A)

k = 0 only

(B)

k = - 3 only

(C)

k = 0 or k = 3

(D)

k = 0 or k = - 3

Solution:

Since, the roots of the equation

`(k + 1)x^2 - 2(k - 1)x + 1 = 0` are real and equal,

`:. {-2(k -1)}^2 - 4(k + 1) = 0 ( ∵ B^2 - 4AC = 0)`

`=> 4(k^2 - 2k + 1) - 4(k + 1) = 0`

`=> k^2 - 2k + 1 - k - 1 = 0 => k^2 - 3k = 0`

`:. k = 0 , k = 3`
Correct Answer is `=>` (C) k = 0 or k = 3
Q 2329401311

If `alpha` and `beta` are the roots of the equation
`ax^2 + bx + c = 0`, then what are the roots of the
equation `cx^2 + bx + a = 0` ?
NDA Paper 1 2007
(A)

`beta , 1/alpha`

(B)

`alpha , 1/beta`

(C)

` - alpha , - beta`

(D)

`1/alpha , 1/beta`

Solution:

Since, `alpha` and `beta` are the roots of `ax^2 + bx + c = 0`, then

the roots of `cx^2 + bx + a= 0` will be reciprocal of `alpha` and `beta` i.e., `1/alpha`

and `1/beta`.

OR

Replace `x` b `1/x` in equation `ax^2 + bx + c = 0`.

`=> a/x^2 + b/x + c = 0 => a + bx + cx^2 = 0`

`=> cx^2 + bx + a = 0`

So, the reciprocal of the equation `ex^2 + bx + a` are

`ax^2 + bx + c = 0`
Correct Answer is `=>` (D) `1/alpha , 1/beta`
Q 2389401317

If roots of an equation `ax^2 + bx + c = 0` are
positive, then which one of the following is
correct?
NDA Paper 1 2007
(A)

Signs of a and c should be like

(B)

Signs of b and c should be like

(C)

Signs of a and b should be like

(D)

None of the above

Solution:

If roots of an equation `ax^2 + bx + c = 0` are positive,

then signs of `a` and `c` should be same (like).
Correct Answer is `=>` (A) Signs of a and c should be like
Q 2349501413

If the equation `x^2 + k^2 = 2(k + 1)x` has equal roots,
then what is the value of `k`?
NDA Paper 1 2007
(A)

` - 1/3`

(B)

`- 1/2`

(C)

`0`

(D)

`1`

Solution:

The given equation is

`x^2 + k^2 =2(k + 1)x`

`=> x^2 - 2(k + 1)x + k^2 = 0`

This equation has equal roots.

`:. { 2 (k + 1)}^2 - 4 xx 1 xx k^2 = 0 ( ∵ B^2 - 4AC = 0)`

`=> 4(k^2 + 2k + 1) - 4k^2 = 0`

`=> k^2 + 2k + 1 - k^2 = 0 => k = - 1/2`
Correct Answer is `=>` (B) `- 1/2`
Q 2319701619

If `alpha , beta` are the roots of `ax^2 + 2bx + c = 0` and `alpha + delta`,
`beta + delta` are the roots of `Ax^2 + 2Bx + C = 0`, then what
is `(b^2 - ac)//(B^2 - AC)` equal to
NDA Paper 1 2007
(A)

`(b/B)^2`

(B)

`(a/A)^2`

(C)

`(a^2b^2)/(A^2B^2)`

(D)

`(ab)/(AB)`

Solution:

Since, `alpha` and `beta` are the roots of `ax^2 + 2bx + c = 0`

`:. alpha + beta = - (2b)/a` and `alpha beta = c/a` ......(i)

Also, `alpha + delta` and `beta + delta` are the roots of

`Ax^2 + 2 Bx + C = 0`

`(alpha + beta) + 2 delta = -(2B)/ A` ......(ii)

and `(alpha + delta )(beta + delta) = C/A` ......(iii)

` => - (2b)/a + 2 delta = - (2B)/ A` [from Eqs. (ii) and (i)]

`=> delta = b/a - B/A` ... (iv)

and `(alpha + delta )(beta + delta) = C/A` [from Eq. (iii)]

`=> alpha beta + (alpha + beta) delta + delta^2 = C/A`

`=> c/a - (2b)/a (b/a - B/A) + (b/a - B/A)^2 = C/A`

[from Eqs. (i) and (iv)]

`=> c/a - (2b^2) /a^2 + (2bB)/(aA) + (b/a)^2 = C/A`

` => c/a - (b/a)^2 + (B/A)^2 = C/A`

`=> B^2/A^2 - C/A = b^2/a^2 => (B^2 - AC)/A^2 = (b^2 - ac^2 )/(a^2)`

` :. (b^2 - ac)/(b^2 - AC) = (a/A)^2`
Correct Answer is `=>` (B) `(a/A)^2`
Q 2319801710

If `alpha , beta` are the roots of `ax^2 + 2bx + c = 0` and `alpha + delta`,
`beta + delta` are the roots of `Ax^2 + 2Bx + C = 0`, then what
is `(b^2 - ac)//(B^2 - AC)` equal to
NDA Paper 1 2007
(A)

`(b/B)^2`

(B)

`(a/A)^2`

(C)

`(a^2b^2)/(A^2B^2)`

(D)

`(ab)/(AB)`

Solution:

Since, `alpha` and `beta` are the roots of `ax^2 + 2bx + c = 0`

`:. alpha + beta = - (2b)/a` and `alpha beta = c/a` ......(i)

Also, `alpha + delta` and `beta + delta` are the roots of

`Ax^2 + 2 Bx + C = 0`

`(alpha + beta) + 2 delta = -(2B)/ A` ......(ii)

and `(alpha + delta )(beta + delta) = C/A` ......(iii)

` => - (2b)/a + 2 delta = - (2B)/ A` [from Eqs. (ii) and (i)]

`=> delta = b/a - B/A` ... (iv)

and `(alpha + delta )(beta + delta) = C/A` [from Eq. (iii)]

`=> alpha beta + (alpha + beta) delta + delta^2 = C/A`

`=> c/a - (2b)/a (b/a - B/A) + (b/a - B/A)^2 = C/A`

[from Eqs. (i) and (iv)]

`=> c/a - (2b^2) /a^2 + (2bB)/(aA) + (b/a)^2 = C/A`

` => c/a - (b/a)^2 + (B/A)^2 = C/A`

`=> B^2/A^2 - C/A = b^2/a^2 => (B^2 - AC)/A^2 = (b^2 - ac^2 )/(a^2)`

` :. (b^2 - ac)/(b^2 - AC) = (a/A)^2`
Correct Answer is `=>` (B) `(a/A)^2`
Q 2359801714

If `alpha , beta` are the roots of the equation
`ax^2 + bx + c = 0`, then what is the value of
`(a alpha + b)^(-1) + (a beta + b)^(-1)` ?
NDA Paper 1 2007
(A)

`a/(bc)`

(B)

`b/(ac)`

(C)

`(-b)/(ac)`

(D)

`(-a)/(bc)`

Solution:

Since, `alpha` and `beta` are the roots of the equation

`ax^2 + bx + c = 0`, then

`alpha + beta = - b/a` and `alpha beta = c/a` ... (i)

Now, `(a alpha + b)^(-1) + (a beta + b)^(-1)`

`= 1/(a alpha + b) + 1/(a beta + b) = (a beta + b + a alpha + b)/((a alpha + b )(a beta + b))`

` = (a(alpha + beta) + 2b)/(a^2 alpha beta + ab (alpha + beta) + b^2)`

`= (a(-b//a) + 2b)/( a^2(c//a) + ab(-b//a) + b^2) = ( - b + 2b)/( ac- b^2 + b^2)` [from Eq. (i)]

`= b/(ac)`
Correct Answer is `=>` (B) `b/(ac)`
Q 2329001811

If `alpha , beta` are the roots of the equation `x^2 - 2x -1 = 0`,
then what is the value of `alpha^2 beta^(-2) + alpha^(-2)beta^(2)` ?
NDA Paper 1 2007
(A)

`-2`

(B)

`0`

(C)

`30`

(D)

`34`

Solution:

Since, `alpha` and `beta` are the roots of the equation

`x^2 - 2x - 1 = 0`, then

`alpha + beta = 2` and `alpha beta = -1`

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

`=> 4 = alpha^2 + beta^2 - 2 => alpha^2 + beta^2 = 6`

`=> (alpha^2 + beta^2)^2 = 6^2`

`=> alpha^4 + beta^4 + 2 = 36`

`=> alpha^4 + beta^4 = 34`

Now, `alpha^2 beta^(-2) + alpha^(-2) beta^2 = alpha^2/beta^2 + beta^2 /alpha^2 = ( alpha^4 + beta^4)/(alpha beta)^2 = (34)/(-1)^2 = 34`
Correct Answer is `=>` (D) `34`
Q 2369101915

If the roots of the equations
`x^2 - (a - 1)x + (a +b) = 0` and `ax^2 - 2x + b = 0` are
identical, then what are the values of `a` and `b`?
NDA Paper 1 2007
(A)

`a = 2 , b = 4`

(B)

`a = 2 ,b = - 4`

(C)

`a = 1, b = 1/2`

(D)

`a = -1, b = - 1/2`

Solution:

Let `alpha` and `beta` be the roots of

`x^2 - (a- 1)x + (a+ b)= 0` ......(i)

and `ax^2 - 2x + b = 0` .........(ii)

From Eq. (i),

`alpha + beta = (a- 1)` and `alpha beta =(a+ b)`

From Eq. (ii),

` alpha + beta = 2/a` and `alpha beta = b/a`

`:. a - 1 = 2/a` (according to the question)

`=> a^2 - a - 2 = 0 => a = -1, 2`

and ` a + b = b/a` (according to the question)

If ` a = - 1, b = 1/2` and if `a = 2, b = - 4`

Alternate Method

Since, the roots of the equations `x^2 - (9- 1)x + (9 + b)= 0` and

`ax^2 - 2x + b - 0` are equal.

Then, `1/a = (-(a - 1))/(-2) = (a+ b)/b`

Taking first two parts, `-2 = - a^2 + a`

`=> a^2 - a - 2 = 0 => a^2 - 2a + 8 - 2 = 0`

`=> (a-2) (a+1) = 0`

`:. a = -1, 2`

taking first and third parts `1/a = (a + b)/a`

When `a = - 1` , then

`-1 = (b - 1)/b`

`=> -b = b - 1 => b = 1/2`

When `a = 2` , then `1/2 = (2 + b)/b`

`=> b = 4 + 2b => - b = 4 => b = - 4`

`:. (a ,b) = (-1 , 1/2)` or `(2, -4)`
Correct Answer is `=>` (B) `a = 2 ,b = - 4`
Q 2319112010

If `- x^2 + 3x + 4 > 0`, then which one of the
following is correct?
NDA Paper 1 2007
(A)

`x in (-1, 4)`

(B)

`x in [-1, 4]`

(C)

`x in (-oo, -1) cup (4, oo)`

(D)

`x in (-oo , 1) cup [4, oo)`

Solution:

`- x^ 2 + 3x + 4 > 0`

`=> x^2 - 3x- 4 < 0 => (x- 4) (x + 1) < 0`

`=> x in (-1, 4)`
Correct Answer is `=>` (A) `x in (-1, 4)`
 
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