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51.

The equation `(x^(2)+3x+4)^(2)+3(x^(2)+3x+4)+4=x` hasA. all its solutions real but not all positiveB. only two of its solutions realC. two of its solutions positive and negativeD. none of solutions real

Answer» Correct Answer - D
`(d)` `f(x)=x^(2)+3x+4`
`:.f(x)=x` has no real solution
`:.f(f(x))-f(x)=x` also has no real solution.
52.

In equation `x^4-2x^3+4x^2+6x-21=0`if two its roots are equal inmagnitude but opposite e in find the roots.

Answer» Given that `alpha + beta = 0 but alpha + beta + gamma + delta = 2`. Hence ,
`gamma + delta = 2`
Let `alpha beta = p and gamma delta = q.` Therefore, given equation is equivelent to `(x^(2) + P)(x^(2) -2x + q) = 0`.
Comparing the coefficients, we get
`p + q = 4, -2p = 6 , pq = -21` .Therefore, p = -3, q = 7 and they satisfy pq = -21. Hence,
`(x^(2) - 3)(x^(2) - 2x + 7) = 0`
Therefore, the roots are `pm i sqrt(6)(where i = sqrt(-1))`
53.

Solve `sqrt(5x^2-6x+8)+sqrt(5x^2-6x-7)=1.`

Answer» Let `5x^(2) - 6x = y`, Then,
`sqrt(5x^(2)-6x+8)-sqrt(5x^(2)-6x-7)=1`
or ` sqrt(y+8)-sqrt(y-7)=1`
or `(sqrt(y+8)-sqrt(y-7))^(2) = 1`
or ` y = sqrt(y^(2)+y-56)`
or `y^(2) = Y^(2) + y- 56`
or y = 56
`rArr 5x^(2)-6x = 56 [because y = 5x^(2) - 6x`]
or (5x+ 14) (x - 4) = 0
or `x = 4,(-14)/(5)`
Clearly, both the values satisfy the given equation. Hence, the roots of the given equation are 4 and `-14//5`.
54.

Solve the equation `(x+2)(x+3)(x+8)xx(x+12)=4x^2dot`

Answer» Since `(-2)(-12)=(-3)(-8)` so we can we write given equation as
`(x+2)(x+12)(x+3)(x+8)=4x^(2)`
`implies(x^(2)+14x+24)(x^(2)+11x+24)=4x^(2)`……I
Now `x=0` is not a root of given equation on dividing by `x^(2)` in both sides of Eq. (i) we get
`(x+24/x+14)(x+24/x+11)=4` ........ii
Put `x+24/x=y` then Eq. (ii) can be reduced in the form
`(y+14)(y+11)=4` or `y^(2)+25y+150=0`
`:.y_(1)=-15` and `y_(2)=-10`
Thus, the original equation is equivalent to the collection of equatiosn
`[(x+24/x=-15),(x+24/x=-10):]`
i.e. `[(x^(2)+15x+24=0),(x^(2)+10x+24=0):]`
On solving these collection we get
`x_(1)=(-15-sqrt(129))/2,x_(2)=(-15+sqrt(129))/2,x_(3)=-6,x_(4)=-4`
55.

Let `alpha,beta`be the roots of the equation `x^2-p x+r=0`and `alpha/2,2beta`be the roots of the equation `x^2-q x+r=0`, the value of `r`is (2007, 3M)`2/9(p-q)(2q-p)`(b) `2/9(q-p(2p-q)``2/9(q-2p)(2q-p)`(d) `2/9(2p-q)(2q-p)`A. `2/9(p-q)(2q-p)`B. `2/9(q-p)(2p-q)`C. `2/9(q-2p)(2q-p)`D. `2/9(2p-q)(2q-p)`

Answer» Correct Answer - D
The equation `x^(2)-px+r=0` has roots `(alpha, beta)` and the equation `x^(2)-qx+r` has roots `((alpha)/2, 2beta)`
`impliesr=alpha beta` and `alpha+beta=p` and `(alpha)/2+2beta=q`
`implies(beta=(2q-p)/3` and `alpha-(2(2p-q))/3`
`impliesalpha beta=r=2/9(2q-p)(2p-q)`
56.

The equation `x^(2)+bx+c=0` has distinct roots. If `2` is subtracted from each root the result are the reciprocal of the original roots, then `b^(2)+c^(2)` isA. `2`B. `3`C. `4`D. `5`

Answer» Correct Answer - D
`(d)` Let `r_(1)`, `r_(2)` be the roots, from given condition
`r-2=(1)/®impliesr^(2)-2r-1=0`, `b=-2`, `c=-1`, `b^(2)+c^(2)=5`
57.

Solve the equation `sqrt((6-4x-x^(2)))=x+4`

Answer» We have `sqrt((6-4x-x^(2)))=x+4`
This equation is equivalent to the system
`{(x+4ge0),(6-4x-x^(2)=(x+4)^(2)):}`
`implies{(xge-4),(x^(2)+6x+5=0):}`
On solving the equation `x^(2)+6x+5=0`
We find that `x_(1)=(-1)` and `x_(2)=(-5)` only `x_(1)=(-1)` satisfies the condition `xge-4`.
Consequently the number `-1` is the only solution of the given equation.
Form 3 An equation in the form
`root(3)(f(x))+root(3)(g(x))=h(x)` .........i
where `f(x),g(x)` are the functions of `x` but `h(x)` is a function of `x` or constant, can be solved as follows cubing both sides of the equation we obtain
`f(x)+g(x)=3root(3)((f(x)(g))(root(3)(f(x))+root(3)(g(x)))=h^(3)(x)`
`impliesf(x)+g(x)+3root(3)(f(x)g(x))(h(x))=h^(3)(x)`
[from Eq. (i)]
We find its roots and then substituting then into the original equation, we choose those which are the roots of the original equation.
58.

Solve `sqrt(x^2+4x-21)+sqrt(x^2-x-6)=sqrt(6x^2-5x-39.)`

Answer» Correct Answer - `x = 3`
We have
`sqrt(x^(2) + 4x - 21) + sqrt(x^(2) - x - 6) = sqrt(6x^(2) - 5x - 39)`
or `sqrt((x + 7)(x - 3)) + sqrt((x - 3)(x - 2)) = sqrt((x-3)((6x + 13))) (1)`
or ` sqrt(x - 3) + (sqrt(x - 7) = sqrt(x-2)-sqrt(6x + 13) ) `
`sqrt(x - 3) =0 or (sqrt(x - 7) = sqrt(x+2)-sqrt(6x + 13)= 0 `
`rArr x = 3 - or sqrt(x + 7) + sqrt(x + 2) = sqrt(6x = 13) `
Now , `sqrt(x + 7) + sqrt(x + 2) = sqrt(6x + 13)`
or `(sqrt(x + 7) + sqrt(x + 2))^(2) = sqrt(6x + 13)`
or `
or `x + 7 + x + 2+2 = sqrt((x - 7) = (x-2)) = 6x + 13`
or `2x + 9+2 = sqrt((x - 7) = (x-2)) = 6x + 13`
or `2sqrt((x - 7) = (x-2)) = 4x+ 4`
`sqrt((x - 7) = (x-2)) = 2(x+1)`
or `(x - 7) = (x-2) = 4(x+1)^(2)` (squraring both sides)
or `x^(2) + 9x + 14 = 4 (x^(2)+ 2x + 1)`
or ` 3x^(2) - x - 10 = 0`
or `(x - 2) (3x + 5) = 0`
`rArr x = 2 or x = (-5)/(3) ,` which does not satisfy (1).
Hence , x = 3 only.
59.

Solve `sqrt(x-2)(x^2-4x-5)=0.`

Answer» Correct Answer - `x = 2 or 5`
Given equation is solvable for x`in [2,infty)`. Now
`sqrt(x - 2)(x^(2) - 4x - 5) = 0`
or `sqrt(x - 2)(x - 5)(x - 1) = 0`
lt `rArr x = 2 or 5 (x ne -1, as x in [2, infty)`
60.

If `alpha,beta` are roots of `x^2-px+q=0` then find the quadratic equation whose roots are `((alpha^2-beta^2)(alpha^3-beta^3))` and `alpha^2beta^3+alpha^3beta^2`

Answer» Since `alpha, beta` are the roots of `x^(2)-px+q=0`
`:. alpha+beta=p,alphabeta=q`
`impliesalpha-beta=sqrt((p^(2)-4q))`
Now `(alpha^(2)-beta^(2))(alpha^(3)-beta^(3))`
`=((alpha+beta)(alpha-beta)(alpha-beta)(alpha^(2)+alpha beta+ beta^(2))`
`=(alpha+beta)(alpha-beta)^(2){(alpha+beta)^(2)-alpha beta}`
`=p(p^(2)-4q)(p^(2)-q)`
and `alpha^(3)beta^(2)+alpha^(2)beta^(3)=alpha^(2)beta^(2)(alpha+beta)=pq^(2)`
`S=` Sum of roots `=p(p^(2)-4q)(p^(2)-q)+pq^(2)`
`=p(p^(4)-5p^(2)q+5q^(2))`
`P=` Product of roots `=p^(2)q^(2)(p^(2)-4q)(p^(2)-q)`
`:.` Required equation is `x^(2)-Sx+P=0`
i.e. `x^(2)-p(p^(4)-5p^(2)q+5q^(2))x+p^(2)q^(2)(p^(2)-4q)(p^(2)-q)=0`
61.

If `a b+b c+c a=0,`then solve `a(b-2c)x^2+b(c-2a)x+c(a-2b)=0.`

Answer» As it is given that`ab + bc + ca = 0`, Putting x =1 in the equation, we get
`f(1) = a(b - 2c) + b(c - 2a)+ c(a - 2b)`
`= - (ab + bc + ca)`
=0
So x = 1 is a root of the equation.
Let the other root be `alpha`
Now product of the roots is
`1xxalpha = (c(a - 2b))/(a(b - 2c))`
or `alpha =(c)/(a) ((a - 2b)/(b - 2c))`
62.

If the equation `x^(2)-px+q=0` and `x^(2)-ax+b=0` have a comon root and the other root of the second equation is the reciprocal of the other root of the first, then prove that `(q-b)^(2)=bq(p-a)^(2)`.

Answer» Let `alpha` and `beta` be the roots of `x^(2)-px+q=0`.Then
`alpha+beta=p` …….i
`alpha beta=q` ……………..ii
And `alpha` and `1/(beta)` be the roots of `x^(2)-ax+b=0`. Then
`alpha+1/(beta)=a` ……….iii
`(alpha)/(beta)=b` ……..iv
Now LHS`=(q-b)^(2)`
`=(apha beta-(alpha)/(beta))^(2)` [from Eqs (ii) and(iv) ]
`=alpha^(2)(beta-1/(beta))^(2)=alpha^(2)[(alpha+beta)-(alpha+1/(beta))]^(2)`
`=alpha^(2)(p-a)^(2)` [from Eqs (i) and (iii)]
`=apha .beta . (alpha)/(beta)(p-a)^(2)`
`=pq(p-a)^(2)` [from Eqs (ii) and (iv) ]
`=`RHS
63.

Let f(x) be a quadratic polynomial satisfying f(2) + f(4) = 0. If unity is one root of f(x) = 0 then find the other root.

Answer» From the given information,
` f(x) = (x-1) (ax+b)`
So, `" " f(2) = 2a + b`
and " " `f(4) = 3(4a + b) = 12a + 3b`
Given that `f(2) + f(4) = 0`
`rArr " " 14a + 4b = 0`
`rArr " " b = - (7a)/(2)`
`therefore" " Equation is (x-1)(ax-(7a)/(2))=0`
Therefore, other root is `7//2`
64.

Thequadratic equations `x^2""6x""+""a""=""0""a n d""x^2""c x""+""6""=""0`have one root in common. The other rootsofthe first and second equations are integers in the ratio 4 : 3. Then thecommon root is(1)1(2) 4(3) 3(4) 2A. 4B. 3C. 2D. 1

Answer» Correct Answer - C
Let `alpha, 4 beta` be roots of `x^(2)-6x+a=0` and `alpha, 3beta` be the roots of `x^(2)-cx+6=0`
Then `alpha+4beta=6` and `4alpha beta=a`…I
`alpha+3beta=c` and `3alpha beta=6`.ii
From Eq I and ii we get
`a=8, alpha beta=2`
Now first equation becomes
`x^(2)-6x+8=0`
`impliesx=2,4`
If `alpha-2, 4beta=4,` then `3 beta=3`
If `alpha=4, 4 beta=2` then `3beta=3/2`
`:.` Common root is `x=2`.
65.

Consider quadratic equations `x^2-ax+b=0 and x^2+px+q=0` If the above equations have one common root and the other roots are reciprocals of each other, then `(q-b)^2` equalsA. bq(p-a)^(2)`B. `b(p-a)^(2)`C. `q(p-a)^(2)`D. none of these

Answer» Correct Answer - A
`(a)` x^(2)-ax+b=0`……..`(i)`
x^(2)-px+q=0`……….`(ii)`
Let the roots of `(i)` be `alpha`, `beta` and that of `(ii)` be `alpha`, `(1)/(beta)`
`:. Alpha+beta=a`, `alphabeta=b`, `alpha+(1)/(beta)=p`, `(alpha)/(beta)=q`
`:.((q-b)^(2))/((p-a)^(2))=(alpha^(2)((1)/(beta)-beta)^(2))/(((1)/(beta)-beta)^(2))`
But `alpha^(2)=alphabeta*(alpha)/(beta)=bq`
`implies (q-b)^(2)=bq(p-a)^(2)`
66.

Consider quadratic equations `x^(2)-ax+b=0`……….`(i)` and `x^(2)+px+q=0`……….`(ii)` If for the equations `(i)` and `(ii)` , one root is common and the equation `(ii)` have equal roots, then `b+q` is equal toA. `-ap`B. `ap`C. `(1)/(2)ap`D. `2ap`

Answer» Correct Answer - C
`(c )` Let `alpha`, `beta` be the roots of `(i)` and `alpha`, `alpha` those of `(ii)`
`:.alpha+beta=a`, `alphabetb`, `2alpha=p`, `alpha^(2)=q`
`:. B+q=alpha(beta+alpha)=(1)/(2)p.a`
67.

Let `a,b in N, a != b` and the two quadratic equations `(a-1)x^2-(a^2+2)x+a^2+2a=0 and (b-1)x^2-(b^2+2)x+(b^2+2b)=0` have a common root. The value of `ab` isA. `4`B. `6`C. `8`D. `10

Answer» Correct Answer - C
`(c )` `a gt 1`, `b gt 1`, `a ne b`
Let `alpha `is the common root
`(a-1)alpha^(2)-(a^(2)+1)alpha+a^(2)+2a=0` ………`(i)`
`(b-1)alpha^(2)-(b^(2)+2)alpha+(b^(2)+2b)=0` ………..`(ii)`
`(i)xx(b-1)-(ii)xx(a-1)`, we get
`(a-b)(ab-a-b-2)(alpha-1)=0`..........`(iii)`
If `alpha=1`, then `a=b=1`
So, `a ne 1`
From `(iii)`, we get
`ab=a+b+2`
Let `a gt b gt 1`
`:. b=1+(b)/(a)+(2)/(a) lt 3 implies b=2`, `a=4`
`:. ab=8`
68.

If `[x]` is the greatest integer less than or equal to `x` and `(x)` be the least integer greater than or equal to `x` and `[x]^(2)+(x)^(2)gt25` then `x` belongs toA. `[3,4]`B. `(-oo,-4]`C. `[4,oo)`D. `(-oo,-4]uu[4,oo)`

Answer» Correct Answer - D
69.

If `0 lt x lt 1000 and [x/2]+[x/3]+[x/5]=31/30x`, (where `[.]` denotes the greatest integer function then number of possible values of x.A. 32B. 33C. 34D. none of these

Answer» Correct Answer - B
70.

If, for a positive integer `n ,`the quadratic equation,`x(x+1)+(x-1)(x+2)++(x+ n-1)(x+n)=10 n`has two consecutive integral solutions, then `n`is equal to :`10`(2) `11`(3) `12`(4) `9`A. 12B. 9C. 10D. 11

Answer» Correct Answer - D
Given quadratic equation is
`x(x+1)+(x+1)(x+2)+...+(x+bar(n-1))(x+n)=10m`
`implies(x^(2)+x^(2)+..+x^(2))+[(1+3+5+...+(2n-1)]x+[(1.2+2.3+...+(n-1)n]=10n`
`impliesnx^(2)+nx+(n(n^(2)-1))/(3)-10n=0`
`x^(2)+nx+(n^(2)-1)/(3)-10=0`
`implies3x^(2)+3nx+n^(2)-31=0`
Let `alpha and beta` be the roots.
Since, `alpha and beta` are consective.
`therefore|alpha-beta|=1implies(alpha-beta)^(2)=1`
Again, `(alpha-beta)^(2)=(alpha+beta)^(2)-4alpha beta`
`implies1=((-3n)/(3))^(2)-4((n^(2)-31)/(3))`
`implies1=n^(2)-4/3(n^(2)-31)implies3=3n^(2)-4n^(2)+124`
`impliesn^(2)=121=n=+-11`
`therefore" "n=11" "[becausengt0]`
71.

IF `[x^(2) - 2x + a]=0` has no solution, then find the values of a (where `[*] ` represents the greatest integer).

Answer» We have `[x^(2) - 2x + a]=0`
`rArr x^(2) - 2x + a in` [ 0, 1)
So, equation has no solution if `x^(2) - 2x a notin` [0, 1).
`therefore x^(2) - 2x + a lt AA x in` R, which is not possible.
So, `x^(2) - 2x + a lt 1 AA x in` R
`rArr x^(2) - 2x + a - 1 ge 0, AA x in` R
Now, `Dge 0`
`rArr 4-4 (a- 1) le 0`
`rArr 2 - a le 0`
`rArr a ge 2`
`rArr a in [ 2, infty)`
72.

If one root of the quadratic equation `ix^2-2(i+1)x +(2-i)=0,i =sqrt(-1)` is `2-i`, the other root isA. `3+i`B. `3+sqrt(-1)`C. `-1+i`D. `-1-i`

Answer» Correct Answer - D
Since all the coefficients of given equation are not real.
Therefore, other root `!=3+i`
Let other root be `alpha`.
Then sum of the roots`=(2(1+i))/i`
`impliesalpha+3-i=(2(1+i))/i`
`impliesalpha+3-i=2-2i`
`:.alpha=-1-i`
73.

A quadratic equations `p(x)=0` having coefficient `x^(2)` unity is such that `p(x)=0` and `p(p(p(x)))=0` have a common root, thenA. `p(0) p(1) gt 0`B. `p(0) p(1) lt 0`C. `p(0) p(1) = 0`D. `p(0)=0` and `p(1)=0`

Answer» Correct Answer - C
`(c )` let `p(x)=x^(2)+ax+b` and let `alpha` be the common root
`:. P(alpha)=0` and `p(p(p(alpha)))=0impliesp(p(0))=0`
`implies p(alpha)=0`
Now `p(0)=b`
`:. P(b)=0` ,brgt `implies b^(2)+ab+b=0`
`implies b(b+a+1)=0`
`implies p(0)p(1)=0`
74.

Let three quadratic equations ` ax^(2) - 2bx + c = 0, bx^(2) - 2 cx + a = 0` and `cx^(2) - ax + b = 0 `, all have only positive roots. Then ltbr. Which of these are always ture?A. `b^(2) = ac `B. `c^(2) = ab `C. each pair of equations has exactly one root commonD. each pair of equations has two roots common

Answer» Correct Answer - 1,2,4
For each equation , required condition
(i) Dicriminant ` ge 0 rArr b^(2) le ac, c^(2) ge ab , a^(2) ge bc `
(ii) ` f(0) gt 0 rArr c,a,b ge 0 `
(iii) Abscissa of vertex ` gt 0 rArr (b)/(a) gt 0 , (c)/(b) gt 0, (a)/(c) gt0 ` .
Now, `b^(2) ac , c^(2) ge ab rArr b^(2) c^(2) ge a^(2) bc rArr bc ge a^(2).`
But ` a^(2) ge bc. So, a^(2) = bc `
Similarly, ` b^(2) = ac and c^(2) = ab `
Therefore , a = b= c .
75.

The number of solutions of `|[x]-2x|=4, "where" [x]]` is the greatest integer less than or equal to x, isA. infiniteB. 4C. 3D. 2

Answer» Correct Answer - A
We have `|[x]-2x|=4`
`implies|[x]-2([x]+{x})|=4`
`implies|[x]+2{x}|=4`
which is possible only when
`2{x}=0,1`
If `{x}=0 then `[x]=+-4` and then `x=-4,4` and if `{x}=1/2`, then
`[x]+1=+-4`
`implies[x]=3,-5`
`:.x=3+1/2` and `-5+1/2`
`impliesx=7/2,-9/2impliesx=-4, -9/2,7/2,4`
76.

THe value of x which satisfy the equation `(sqrt(5x^2-8x+3))-sqrt((5x^2-9x+4))=sqrt((2x^2-2x))-sqrt((2x^2-3x+1))` isA. 3B. 2C. 1D. 0

Answer» Correct Answer - C
We have `sqrt((5x^(2)-8x+3))-sqrt((5x^(2)-9x+4))`
`=sqrt((2x^(2)-2x))-sqrt((2x^(2)-3x+1))`
`impliessqrt((5x-3)(x-1))-sqrt((5x-4)(x-1))`
`=sqrt(2x(x-1))-sqrt((2x-1)(x-1))`
`impliessqrt(x-1)(sqrt(5x-3)-sqrt(5x-4))=sqrt(x-1)(sqrt(2x)-sqrt(2x-1))`
`impliessqrt(x-1)=0`
`impliesx=1`
77.

Show that the equation `e^(sinx)-e^(-sinx)-4=0`has no real solution.A. infinite number of real rootsB. no real rootsC. exactly one real rootD. exactly four real roots

Answer» Correct Answer - 2
Let ` e^(sin x) =t`
` rArr t^(3) - 4t - 1 = 0 `
` t = (4 pmsqrt(16 + 4))/(2)`
`rArr t = e^(sin x ) = 2 pm sqrt(5)`
`rArr t = e^(sin x ) = 2 - sqrt(5),e^(sin x ) = 2 + sqrt(5)`
`rArr e^(sin x) = - sqrt(5) lt 0 ` , whihc is not possible
or ` sin x = In (2 + sqrt(5)) gt 1`, which is not possible
Hence no solution .
78.

If `ax^2+bx+c=0 and cx^2 + bx+a=0(a,b,c in R)` have a common non-real roots, then which of the following is not true ?A. `-2|a|lt |b| lt |a|`B. `-2|c|lt b lt 2|c|`C. `a=c`D. None of these

Answer» Correct Answer - D
`(d)` `D_(1)=b^(2)-4ac lt 0`, `D_(2)=b^(2)-4ac lt 0`, as the root is non-real
`implies` Both roots will be common.
`implies (a)/(c )=(b)/(b)=(c )/(a)=1 implies a=c`
Now, `b^(2)-4ac lt 0impliesb^(2)-4a^(2)` ( or `4c^(2)`) ` lt 0`
`implies |b| lt 2 |a| ("or" 2|c |)`.
79.

If a, b, c are positive real numbers such that the equations `ax^(2) + bx + c = 0 and bx^(2) + cx + a = 0`, have a common root, thenA. `1: 2: 3`B. `3: 2: 1`C. `1: 3 : 2`D. `3 : 1 : 2 `

Answer» Correct Answer - 1
` x^(2) + 2x + 3 = 0 `
` D = 2^(2) - 4 .1.3 lt 0 `
i.e., both roots are complex.
Hence , both roots are common of ` x^(2) + 2x + 3 = 0 `
and ` ax ^(2) + bx + c = 0`
`therefore (a)/(1) = (b)/(2) = (c)/(3)` .
80.

if `x^2+x+1` is a factor of `ax^3+bx^2+cx+d` then the real root of `ax^3+bx^2+cx+d=0 ` is : (a) `-d/a ` (B) `d/a ` (C) `a/b` (D)none of theseA. `-d/a`B. `d/a`C. `a/d`D. none of these

Answer» Correct Answer - A
We know that `x^(2)+x+1` is factor of `ax^(3)+bx^(2)+cx+d`
Hence roots of `x^(2)+x+1=0` are also roots of
`ax^(3)+bx^(2)+cx+d=0`. Since `omega` and `omega^(2)`
(where `omega=-1/2+(3i)/2`) are two complex roots of `x^(2)+x+1=0`
Therefore `omega` and `omega^(2)` are two complex roots fo `ax^(3)+bx^(2)+cx+d=0`
We know that a cubic equation has atleast one real root. Let real root be `alpha`. Then
`alpha. omega. omega^(2)=-d/aimpliesalpha=-d/a`
81.

If `a ,b ,c ,d`are four consecutive terms of an increasing A.P., then the roots of theequation `(x-a)(x-c)+2(x-b)(x-d)=0`area. non-real complexb. real and equalc. integers d. real and distinctA. non real complexB. real and equalC. integersD. real and distinct

Answer» Correct Answer - D
82.

Statement 1 If one root of `Ax^(3)+Bx^(2)+Cx+D=0 A!=0`, is the arithmetic mean of the other two roots, then the relation `2B^(3)+k_(1)ABC+k_(2)A^(2)D=0` holds good and then `(k_(2)-k_(1))` is a perfect square. Statement -2 If a,b,c are in AP then `b` is the arithmetic mean of a and c.A. Statement -1 is true, Statement -2 is true, Statement -2 is a correct explanation for Statement-1B. Statement -1 is true, Statement -2 is true, Statement -2 is not a correct explanation for Statement -1C. Statement -1 is true, Statement -2 is falseD. Statement -1 is false, Statement -2 is true

Answer» Correct Answer - A
Let roots of `Ax^(3)+Bx^(2)+Cx+D=0`……….i
are `alpha-beta, alpha, alpha+beta` (in AP)
Then `(alpha-beta)+alpha+(alpha+beta)=-B/A`
`impliesalpha=-B/(3A)`, which is a root of Eq. (i)
Then `A alpha^(3)+B alpha^(2)+C alpha +D=0`
`impliesA(-B/(3A))^(3)+B(-B/(3A))^(2)+C(-B/(3A))+D=0`
`implies-(B^(3))/(27A^(2))+(B^(3))/(9A^(2))-(BC)/(3A)+D=0`
`implies2B^(3)-9ABC+27A^(2)D=0`
Now comparing with `2B^(3)+k_(1)ABC+k_(2)A^(2)D=0` we get
`k_(1)=-9,k_(2)=27`
`:.k_(2)-k_(1)=27-(-9)=36=6^(2)`
Hence both statement are true and Statement 2 is a correct explanation of Statement -1.
83.

If `alpha, beta, gamma` are the roots of the cubic `x^(3)-px^(2)+qx-r=0` Find the equations whose roots are (i) `beta gamma +1/(alpha), gamma alpha+1/(beta), alpha beta+1/(gamma)` (ii)`(beta+gamma-alpha),(gamma+alpha-beta),(alpha+beta-gamma)` Also find the valueof `(beta+gamma-alpha)(gamma+alpha-beta)(alpha+beta-gamma)`

Answer» Correct Answer - (i) `ry^(3)-q(r+1)y^(2)+p(r+1)^(2)y-(r+1)^(3)=0`
(ii) `y^(3)-py^(2)+(4q-p^(2))y+(8r-4pq+p^(3))=0` and `4pq-p^(3)-8r`
Given `alpha, beta` and `gamma` are the roots of the cubic equation
`x^(3)-px^(2)+qx-r=0` ………..i
`:. alpha +beta+gamma=p,alpha beta+beta gamma+gamma alpha=q,alpha beta gamma =r`
(i) Let `y=beta gamma +1/(alpha)`
`impliesy=(alpha beta gamma +1)/(alpha)=(r+1)/(alpha)`
`:.alpha=(r+1)/y`
From Eq. (i) we get
`alpha^(3)-palpha^(2)+q alpha-r=0`
`implies((r+1)^(3))/(y^(3))-(p(r+1)^(2))/(y^(2))+(q(r+1))/y-r=0`
or `ry^(3)-q(r+1)y^(2)+p(r+1)^(2)y-(r+1)^(3)=0`
(ii) Let `y=beta+gamma -alpha=(alpha+beta+gamma)-2alpha=p-2alpha`
`alpha=(p-y)/2`
From Eq. (i) we get
`alpha^(3)-palpha^(2)+q alpha-r=0`
`implies((p-y)^(3))/8-(p(p-y)^(2))/4+(q(p-y))/2-r=0`
or `y^(3)-py^(2)+(4q-p^(2))y+(8r-4pq+p^(3))=0`
Also product of roots `=(8r-4pq+p^(3))`
84.

If `c ,d`are the roots of the equation `(x-a)(x-b)-k=0`, prove that a, b are roots of the equation `(x-c)(x-d)+k=0.`

Answer» Since c and d are the roots of the equation
(x-a) (x-b) - K=0, we have
(x - a)(x-b)-K = (x - c) (x - d)
or (x - a)(x-b) = (x - c) (x - d) + K
or (x - c)(x-d)+ K = (x - a) (x - b)
Clearly, a and b are roots of the equation (x - a ) (x - b) = 0 .Hence, a and b are roots of (x-c)(x-d) + k = 0
85.

Statement -1 `ax^(3)+bx+c=0` where `a,b,c epsilonR` cannot have 3 non-negative real roots. Statement 2 Sum of roots is equal to zero.A. Statement -1 is true, Statement -2 is true, Statement -2 is a correct explanation for Statement-1B. Statement -1 is true, Statement -2 is true, Statement -2 is not a correct explanation for Statement -1C. Statement -1 is true, Statement -2 is falseD. Statement -1 is false, Statement -2 is true

Answer» Correct Answer - A
Let `y=ax^(3)+bx+c`
`:.(dy)/(dx)=3ax^(2)+b`
For maximum or minimum `(dy)/(dx)=0` we get
`x=+-sqrt(-b/(3a))`
Case If `a gt0, bgt0` then `(dy)/(dx)gt0`
In this case function is increasing so it has exactly one root
Case II If `alt0, blt0` ten `(dy)/(dx)lt0`
In this case function is decreasing so it has exactly one root.
Case III `agt0, blt0` or `alo0, bgt0` then `y=ax^(3)+bx+c` is maximum at one point and minimum at other point.
Hence all roots can never be non -netative.
`:.` Statement -1 is false. But ltgtbrgt Sum of roots `=-("Coefficient of" x^(2))/("Coefficient of"x^(3))=0`
i.e. Statement -2 is true.
86.

If `A_(1),A_(2),A_(3),………………,A_(n),a_(1),a_(2),a_(3),……a_(n),a,b,c epsilonR` show that the roots of the equation `(A_(1)^(2))/(x-a_(1))+(A_(2)^(2))/(x-a_(2))+(A_(3)^(2))/(x-a_(3))+…+(A_(n)^(2))/(x-a_(n))` `=ab^(2)+c^(2) x+ac `are real.

Answer» Assume `alpha + i beta` is a complex root of the given equation, then conjugate of this root i.e. `alpha-ibeta` is also root of this equation.
On putting `x=alpha+ibeta` and `x=alpha-ibeta` in the given equation we get
`(A_(1)^(2))/(alpha+ibeta-a_(1))+(A_(2)^(2))/(alpah+i beta-a_(2))+(A_(3)^(2))/(alpha+i beta-a_(3))+....+(A_(n)^(2))/(alpha+ibeta-a_(n))`
`=ab^(2)+c^(2)(alpha+ibeta)+ac`...i
and `(A_(1)^(2))/(alpha -ibeta-a_(1))+(A_(2)^(2))/(alpha-ibeta-a_(2))+(A_(3)^(2))/(alpha-ibeta-a_(3))+..........+(A_(n)^(2))/(alpha-i beta-a_(n))`
`=ab^(2)+c^(2)(alpha-i beta)+ac` ...........ii
On subtracting Eq. i from Eq ii we get
`2i beta[(A_(1)^(2))/((alpha-a_(1)^(2)+beta^(2))+(A_(2)^(2))/((alpha-a_(2))^(2)+beta^(2))+(A_(3)^(2))/((alpha-a_(3))^(2)+beta^(2))`
`+...........+(A_(n)^(2))/((alpha-a_(n))^(2)+beta^(2))+c^(2)]=0`
The expression in bracket `!=0`
`:.2ibeta=0impliesbeta=0`
Hence all roots of the given equation are real.
87.

Let `f(x)= x^(3) + x + 1` and P(x) be a cubic polynomial such that P(0) = -1 and roots of f(0) = 1 ; P(x) = 0 are the squares of the roots of f(x) = 0 . Then find the value of P(4).

Answer» According to the question,
`f(x) = x^(3) + x + 1 = (x - alpha ) (x- beta) (x - gamma) " "`…(1)
and `P(x) = k(x - alpha ^(2))(x - beta^(2))(x - gamma^(2))`
Given `P(0) = - 1 rArr - kalpha^(2)beta^(2)gamma^(2) = - 1`
Also `f(0) = 1 rArr - k alpha beta gamma = 1`
`therefore -K(-1)^(2) = - 1 rArr k = 1`
`therefore P(x) = (x - alpha ^(2))(x - beta^(2)) (x - gamma^(2))`
Now , P(4) = `(2 - alpha) (2 - beta) (2 - gamma) (2 + alpha ) (2+beta) (2+gamma)`
`= f (2) (-f(-2))`
`= (2^(3) + 2 + 1 ) (-(-2)^(3) - (-2) - 1)`
`11xx9 = 99`
88.

If the equation `(1+m)x^(2)-2(1+3m)x+(1-8m)=0` where `m epsilonR~{-1}`, has atleast one root is negative, thenA. `m epsilon(-oo,-1)`B. `m epsilon (-1/8,oo)`C. `m epsilon(-1-1/8)`D. `m epsilonR`

Answer» Correct Answer - C
89.

If G and L are the greatest and least values of the expression`(2x^(2)-3x+2)/(2x^(2)+3x+2), x epsilonR` respectively. The least value of `G^(100)+L^(100)` isA. `2^(100)`B. `3^(100)`C. `7^(100)`D. none of these

Answer» Correct Answer - D
Let `y=(2x^(2)-3x+2)/(2x^(2)+3x+2)`
`implies2x^(2)y+3xy+2y=2x^(2)-3x+2`
`implies2(y-1)x^(2)+3(y+1)x+2(y-1)=0`
As `x epsilonR`
`:.Dge0`
`=9(y+1)^(2)-4.2(y-1).2(y-1)ge0`
`implies9(y+1)^(2)-16(y-1)^(2)ge0`
`implies(3y+3)^(2)-(4y-4)^(2)ge0`
`implies(7y-1)(7-y)ge0`
`implies(7y-1)(y-7)le0`
`:.1/7leyle7`
`:.G=7` and `L=1/7`
`:.GL=1`
NOw `(G^(100)+L^(100))/2ge(GL)^(100)implies(G^(100)+L^(100))/2ge1`
`impliesG^(100)+L^(100)ge2`
Least value of `G^(100)` +L^(100)` is 2.
90.

If the equation whose roots are the squares of the roots of the cubic `x^3-a x^2+b x-1=0`is identical with the given cubic equation, then`a=0,b=3`b. `a=b=0`c. `a=b=3`d. `a ,b ,`are roots of`x^2+x+2=0`A. `a=b=0`B. `a=0,b=3`C. `a=b=3`D. `a,b` are roots of `x^(2)+x+2=0`

Answer» Correct Answer - A::C::D
We have `x^(3)-ax^(2)+bx-1=0`……….i
Then `alpha^(2)+beta^(2)+gamma^(2)=(alpha+beta=gamma)^(2)-2(alpha beta+beta gamma+gamma alpha)`
`=a^(2)-2b`
`alpha^(2) beta^(2)+beta^(2)gamma^(2)+gamma^(2) alpha^(2)=(alpha beta+beta gamma+gamma alpha)^(2)`
`=-2 alpha beta gamma( alpha +beta+gamma)=b^(2)-2a`
and `alpha^(2) beta^(2) gamma^(2)=1`
Therefore, the equation whose roots are`alpha^(2),beta^(2)` and `gamma^(2)` is
`x^(3)-(a^(2)-2b)x^(2)+(b^(2)-2a)x-1=0`...........ii
`a^(2)-2b=a` and `b^(2)-2a=b`
Eliminating `b` we have
`((a^(2)-a)^(2))/4-2a=(a^(2)-a)/2`
`impliesa{a(a-1)^(2)-8-2(a-1)}=0`
`impliesa(a^(3)-2a^(2)-a-6)=0`
`impliesa(a-3)(a^(2)+a+2)=0`
`impliesa=0` or `a=3` or `a^(2)=a+2=0`
`impliesb=0` or `b=3`
or `b^(2)+b+2=0`
`:.a=b=0`
or `a=b=3`
or a and b are roots of `x^(2)+x+2=0`
91.

If the equation whose roots are the squares of the roots of the cubic `x^3-a x^2+b x-1=0`is identical with the given cubic equation, then`a=0,b=3`b. `a=b=0`c. `a=b=3`d. `a ,b ,`are roots of`x^2+x+2=0`A. ` a = 0, b = 3`B. a = b = 0C. a = b = 3D. a, b are roots of ` x^(2) + x + 2 = 0 `

Answer» Correct Answer - 2,3,4
Given equation is `x^(3) - ax^(3) + bx - 1 = 0 ` . If roots of the equationo
be ` alpha, beta , gamma`, then
`alpha^(2) + beta^(2) + gamma^(2) = (alpha + beta + gamma)^(2) - 2 (alpha beta + beta gamma + gamma alpha)`
` = alpha ^(2) - 2a `
`alpha^(2)beta^(2) + beta^(2) gamma^(2)+ gamma^(2)alpha ^(2) = (alpha beta + beta gamma+ gammaalpha )^(2) - 2 alpha beta gamma (alpha + beta + gamma )`
` b^(2) - 2a `
` alpha ^(2) beta^(2) gamma^(2) = 1`
So, the equations whose roots are ` alpha ^(2) , beta^(2) , gamma^(2)` is given by
`x^(3) - ax^(2) + bx - 1= 0 `
`rArr a^(2) - 2b = a and b^(2) - 2a = b`
Eliminatng b, we get
`((a^(2) - a)^(2))/(4) - 2a = (a^(2) - a)/(2)`
or ` a{a (a - 1)^(2) - 8 - 2a (a - 1)} = 0 `
or ` a (a^(3) - 2a ^(2) + a + 2) = 0`
or ` a(a - 3) (a^(2) + a + 2) = 0`
`rArr a = 0 or a = 3 or a^(2) + a + 2 = 0`
Which gives b = 0 or b = 3 `b^(2) + b+ 2 = 0 `.So a = b - 0 or
a = b =3 or a a, b are roots of ` x^(2) + x + 2 = 0`
92.

If G and L are the greatest and least values of the expression`(2x^(2)-3x+2)/(2x^(2)+3x+2), x epsilonR` respectively. If `L^(2)ltlamdaltG^(2), lamda epsilon N` the sum of all values of `lamda` isA. 1035B. 1081C. 1225D. 1176

Answer» Correct Answer - D
Let `y=(2x^(2)-3x+2)/(2x^(2)+3x+2)`
`implies2x^(2)y+3xy+2y=2x^(2)-3x+2`
`implies2(y-1)x^(2)+3(y+1)x+2(y-1)=0`
As `x epsilonR`
`:.Dge0`
`=9(y+1)^(2)-4.2(y-1).2(y-1)ge0`
`implies9(y+1)^(2)-16(y-1)^(2)ge0`
`implies(3y+3)^(2)-(4y-4)^(2)ge0`
`implies(7y-1)(7-y)ge0`
`implies(7y-1)(y-7)le0`
`:.1/7leyle7`
`:.G=7` and `L=1/7`
`:.GL=1`
NOw `(G^(100)+L^(100))/2ge(GL)^(100)implies(G^(100)+L^(100))/2ge1`
`impliesG^(100)+L^(100)ge2`
We have `L^(2)lt lamdalt G^(2)`
`(1/7)^(2)ltlamdalt 7^(2)`
`implies1/49lt lamdalt 49`
`implies lamda=1,2,3,..48` as `lamda epsilon N`
`:.` Sum of all values of `lamda=1+2+3+.............+48=(48xx49)/2=1176`
93.

If the sum of squares of roots of equation `x^(2)-(sin alpha-2)x-(1+sin alpha)=0` is the least, then `alpha` is equal toA. `pi//4`B. `pi//3`C. `pi//2`D. `pi//6`

Answer» Correct Answer - C
`(c )` Let `p`, `q` are roots.
`p^(2)+q^(2)=(p+q)^(2)-2pq`
`=(sinalpha-2)^(2)+2(1+sin alpha)`
`=(sin alpha-1)^(2)=5`
For least value `sin alpha=1`
`implies alpha=pi//2`
94.

The value of `lamda` such that sum of the squares of the roots of the quadratic equation, `x^(2)+(3-lamda)x+2=lamda` had the least value isA. `5/4`B. 1C. `15/8`D. 2

Answer» Correct Answer - D
Given, quadratic equation is
`x^(2)+(3+ - lamda)x+2=lamda`
`x^(2)+(3-lamda)x+(2-lamda)=0" "...(i)`
Let Eq. (1) has roots `alpha and beta,` then `alpha+beta,=lamda-3and alpha beta=2-lamda`
`" ""["because"For" ax^(2)+bx+c=0, "sum of roots"=-b/a"and product of roots"=c/a"]"`
Now, `alpha^(2)+beta^(2)=(alpha+beta)^(2)-2alphabeta`
`=(lamda-3)^(2)-2(2-lamda)`
`=lamda^(2)-4lamda+5=(lamda^(2)-4lamda+4)+1=(lamda-2)^(2)+1`
Clearly, `almda^(2)+beta^(2)` will be least when `lamda=2.`
95.

Find the values of a for which the expression `(ax^2+3x-4)/(3x-4x^2+a)` assumes all real values for all real values of x

Answer» Let `f(x) = (ax^(2) + 3x - 4)/(3x - x^(2) +a)`
`rArr (a + 4y) x^(2) + (3-3y) x - 4 - ay = 0`
Now, x is real. So
D `ge` 0
`rArr 9 (1 -y)^(2) + 4 (a + 4y) (4 + ay )ge`0
`rArr 9 ( + 16 a)y^(2) + (-18 + 4a^(2) + 64)y + (9 + 16a )ge 0, AA y in R (because y` all real values)
`rArr 9 + 16a gt 0 and (4a^(2) + 46)^(2) - 4 (9 + 16 a)^(2) le 0`
`rArr a gt - (9)/(16)`
and `(4a^(2) + 46 - 18 - 32 a) (4a^(2) + 46 + 18 32a )le 0`
`rArr a gt - (9(/(16) and (a^(2) - 8a + 7) (a^(2) + 8a + 16) le 0`
`rArr a gt - (9)/(16) and 1 le a le 7 or a = - 4`
`rArr 1 le a le 7`
96.

For real x, the function `(x-a)(x-b)/(x-c)` will assume all real values providedA. `a gt b gt c`B. `a lt b lt c`C. `a gt c lt b`D. `a le c le b`

Answer» Correct Answer - D
Let `y=(x^(2)-(a+b)x+ab)/(x-c)`
`impliesyx-cy=x^(2)-(a+b)x+ab`
`impliesx^(2)-(a+b+y)x+(ab+cy)=0`
For real roots, `D ge 0`
`implies(a+b+y)^(2)-4(ab+cy)ge0`
`implies(a+b)^(2)+y^(2)+2(a+b)y-4ab-4cyge0`
`impliesy^(2)+2(a+b-2x)y+(a-b)^(2)ge0`
which is true for all real values of y.
`therefore" "D le0`
`4(a+b-2a)^(2)-4(a-b)^(2)le0`
`implies4(a+b-2c+a-b)(a+b-2c-a+)le0`
`implies(2a-2c)(2b-c)le0`
`implies(a-c)(b-c)le0`
`implies(c-a)(c-b)le0`
`implies` c mule lie between a and b
`i.e. a le c le b or b le c lea `
97.

If a continous founction of defined on the real line R, assumes positive and negative values in R, then the equation f(x)=0 has a root in R. For example, if it is known that a continuous function f on R is positive at some point and its minimum values is negative, then the equation `f(x)=0` has a root in R. Considetr `f(x)=ke^(x)-x` for all real x where k is real constant. For `k gt 0,` the set of all values of k for which `ke^(x)-x=0` has two distinct, roots, isA. `(0,(1)/(e))`B. `((1)/(e),1)`C. `((1)/(e),oo)`D. `(0,1)`

Answer» Correct Answer - A
For two distinct roots, `1+ ln k lt 0(k gt0)`
`ln k lt-1impliesklt1/e`
Hence,` k in (0,(1)/(e))`
98.

Set of all real value of a such that `f(x)=((2a-1)+x^2+2(a+1)x+(2a-1))/(x^2-2x+40)`is always negative is`-oo,0`b. `0,oo`c. `-oo,1//2`d. noneA. `(-oo, 0)`B. `(0, oo)`C. `(-oo, 1//2)`D. None

Answer» Correct Answer - 1
`f(x) = ((2a - 1)x^(2) + 2(a + 1)x + (2a - 1))/(x^(2) - 2x + 40)`
Since `x^(2) - 2x + 40 gt 0` for all real x.,
`(2a - 1)x^(2) + 2(a + 1)x + (2a - 1) lt 0 AA x in R`
`therefore 2a - 1 lt 0` or `a lt 1/2" "`(1)
and `D lt 0`
`rArr 4(a + 1)^(2) - 4(2a - 1)^(2) lt 0`
`rArr a(a - 2) gt 0`
`rArr a lt 0` or `a gt 2" "`(2)
From (1) and (2), `a lt 0`
99.

If he expression `[m x-1+(1//x)]`is non-negative for all positive real `x ,`then the minimum value of `m`must be`-1//2`b. `0`c. `1//4`d. `1//2`A. `-1//2`B. 0C. `1//4`D. `1//2`

Answer» Correct Answer - 3
We know that `ax^(2) + bx c ge 0, AA x in R`,
if `a gt 0 and b^(2) - 4ac le 0.` So,
`mx - 1 + 1/x ge 0` or `(mx^(2) - x + 1)/(x) ge 0`
or `mx^(2) - x + 1 ge 0` as `x gt 0`.
Now,
`mx^(2) - x + 1 ge 0` if `m gt 0` and `1 - 4m le 0`
`rArr m gt 0 and m ge 1//4`
Thus, the minimum value of m is 1/4.
100.

If `a ,b in R ,a!=0`and the quadratic equation `a x^2-b x+1=0`has imaginary roots, then `(a+b+1)`isa. positiveb. negativec. zero d. Dependent on thesign of `b`A. positiveB. negativeC. zeroD. dependent on the sign of b

Answer» Correct Answer - 1
`D = b^(2) - 4a lt 0 rArr a gt 0`
Therefore the graph is concave upwards.
`f(x) gt 0, AA x in R`
`rArr f(-1) gt 0`
`rArr a + b + 1 gt 0`