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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
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. |
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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))` |
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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`. |
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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` |
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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)` |
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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` |
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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. |
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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. |
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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)` |
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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` |
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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))` |
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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 |
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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` |
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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`. |
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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)` |
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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` |
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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` |
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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]` |
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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)` |
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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` |
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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` |
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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 . |
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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` |
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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` |
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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 . |
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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 |)`. |
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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)` . |
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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` |
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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. |
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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))` |
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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 |
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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. |
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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. |
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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` |
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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. |
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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` |
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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` |
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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` |
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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` |
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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.` |
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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` |
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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 ` |
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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))` |
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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` |
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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. |
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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` |
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