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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.
| 251. |
If `{x}` and `[x]` represent fractional and integral part of x respectively, find the value of `[x]+sum_(r=1)^(2000)({x+r})/2000` |
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Answer» `[x]+sum_(r=1)^(2000)({x+r})/2000=[x]+sum_(r=1)^(2000)({x})/2000` [from property (i)] `=[x]+({x})/2000sum_(r=1)^(2000)1=[x]+({x})/2000xx2000=[x]+{x}=x` |
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| 252. |
Prove that the following equations has no solutions. (i) `sqrt((2x+7))+sqrt((x+4))=0` (ii) `sqrt((x-4))=-5` (iii) `sqrt((6-x))-sqrt((x-8))=2` (iv) `sqrt(-2-x)=root(5)((x-7))` (v) `sqrt(x)+sqrt((x+16))=3` (vi) `7sqrt(x)+8sqrt(-x)+15/(x^(3))=98` (vii) `sqrt((x-3))-sqrt(x+9)=sqrt((x-1))` |
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Answer» (i) We have `sqrt((2x+7))+sqrt((x+4))=0` This equation is defined for `2x+7ge0` and `x+4ge0implies{(xge-7/2),(xge-4):"` `:.xge-7/2` For `xge-7/2` , the left hand side of the original equation is positive, but right hand side is zero. Therefore, the equation has no roots. (ii) We have `sqrt(x-4))=-5` The equation is defined for `x-4ge0` `:.xge4` For `xge4`, the left hand side of the original equation is positive but right hand side is negative. Therefore, the equation has no roots. (iii) We have `sqrt((5-x))-sqrt(x-8)=2` The equation is defined for `6-xge0` and `x-8ge0` `:.{(xle6),(xge8):}` Consequently, there is no `x` for which both expressions would have sense. Therefore, the equation has no roots. (iv) We have `sqrt((-2-x))=root(5)((x-7))` This equation is defined for `-2-xge0impliesxle-2` For `xle-2` the left hand side is positive, but right hand side is negative. Therefore, the equation has no roots. (v) We have `sqrt(x)+sqrt((x+16))=3` The equation is defined for `xge0` and `x+16ge0implies{(xge0),(xge-16):}` Hence `xge0` For `xge0` the left hand side `ge4`, but right hand side is 3. Therefore, the equation has no roots. (vi) We have `7sqrt(x)+8sqrt(-x)+15/(x^(3))=98` For `xlt0` the expression `7sqrt(x)` is meaningless, For `xgt0`, the expressionn `8sqrt(-x)` is meaningles and for `x=0` the expression `15/(x^(3))` is meaningless. Consequently the left hand side of the original equation ils meaningless for any `x epsilonR`. Therefore the equation has no roots. (vii) We have `sqrt((x-3))-sqrt((x+9))=sqrt(x-1)` This equation is defined for `{(x-3ge0),(x+9ge0),(x-1ge0):}implies{(xge3),(xge-9),(xge1):}` Hence `xge3` For `xge3, sqrt(x-3)ltsqrt(x+9)` i.e. `sqrt((x-3))-sqrt((x+9))lt0` Hence for `xge3`,the left hand side of the original equation is negative and right hand side is positive. Therefore, the equation has no roots. |
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| 253. |
The equation `(x)^(2)=[x]^(2)+2x` where `[x]` and `(x)` are the integers just less than or equal to `x` and just greater than or equal to `x` respectively, then number of values of `x` satisfying the given equation |
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Answer» Case I If `x epsilonI` then `x=[x]=(x)` The given equation reduces to `x^(2)=x^(2)+2x` `implies2x=0` or `x=0`….i Case II If `x!inI`, then `(x)=[x]+1` The given equation reduces to `([x]+1)^(2)=[x]^(2)+2x` `implies1=2(x-[x])` or `{x}=1/2` `:.x=[x]+1/2=n+1/2,n epsilonI`............ii ,brgt Hence the solution of the original equation is `x=0, n+1/2, n epsilon I`. |
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| 254. |
Let `x^2+y^2+x y+1geqa(x+y)AAx ,y in R ,`then the number of possible integer (s) in the range of `a`is___________. |
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Answer» Correct Answer - 3 `rArr (y-a)^(2)=4(y^(2)-ay+1)le0` `rArr -3y^(2)+2ay+a^(2)-4le0AAy inR` `therefore 3y^(2)-2ay+4-a^(2)ge0AAy in R` `Dle0` `rArr4a^(2)-4 " . "3(4-a^(2))le0` `rArra^(2)-3(4-a^(2))le0rArr4a^(2)-12le0` `therefore " Range of " a in [-sqrt3, sqrt3]` `rArr " number of integer " {-1,0,1}` |
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| 255. |
If the roots of the equation `(b-c)x^2+(c-a)x+(a-b)=0` are equal then `a,b,c` will be in |
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Answer» Give equation is `a(b-c)x^(2)+b(c-a)x+c(a-b)=0`…………i Here, coefficient of `x^(2)+` coefficient of `x+` constant term `=0` i.e. `a(b-c)+b(c-a)+c(a-b)=0` The, 1 is a root of Eq. (i) Since, its roots are equal. Therefore its other root will be also equal to 1. Then product of roots `=1xx1=(c(a-b))/(a(b-c))` `impliesab-ac=ca-bc` `:.b=(2ac)/(a+c)` Hence a,b,c are in HP. |
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| 256. |
If one roots of the equation `x^(2)-sqrt(5)x-19=0` is `(9+sqrt(5))/2` then find the other root. |
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Answer» All coefficients of the given equation are not rationa, then other root `!=(9-sqrt(5))/2` Let other root be `alpha` sum of roots `=sqrt(5)` `=(9+sqrt(5))/2+alpha=sqrt(5)impliesalpha=(-9+sqrt(5))/2` Hence other root is `(-9+sqrt(5))/2` |
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| 257. |
Find all roots of the equation `x^4+2x^3-16x^2-22x+7=0`, if one root is `2+sqrt3` |
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Answer» All coefficients are teal, irrational roots will occur in congugate pairs. Hence another root is `2-sqrt(3)`. `:.` Product of these roots `=(x-2-sqrt(3))(x-2+sqrt(3))` `=(x-2)^(2)-3=x^(2)-4x+1` On dividing `x^(4)+2x^(3)-16x^(2)-22x+7` by `x^(2)-4x+1`, then the other quadratic factor is `x^(2)+6x+y`. Then the given equation reduce in the form `(x^(2)-4x+1)(x^(2)+6x+7)=0` `:.x^(2)+6x+7=0` Then `x=(-6+-sqrt(36-28))/2=-3+-sqrt(2)` Hence the other roots are `2-sqrt(3),-3+-sqrt(2)` |
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| 258. |
Let `alpha,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`a ,c`(b) `b ,c``a ,b`(d) `a+c ,b+c` |
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Answer» Since `alpha, beta` are the roots fo `(x-a)(x-b)=c` or `(x-a)(x-b)-c=0` Then `(x-a)(x-b)-c-(x-alpha)(x-beta)` `implies(x-alpha)(x-beta)+c=(x-a)(x-b)` Hence roots of `(x-alpha)(x-beta)+c=0` are a,b. |
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| 259. |
Solve for `x(5+2sqrt(6))^x^(2-3)+(5-2sqrt(6))^x^(2-3)=10`(1985,5M) |
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Answer» `(5+2sqrt(6))(5-2sqrt(6))=1` `:.(5-2sqrt(6))=1/((5+2sqrt(6)))` `:.(5+2sqrt(6))^(x^(2)-3)+(5-2sqrt(6))^(x^(2)-3)=10` reduces to `(5+2sqrt(6))^(x^(2)-3)+(1/(5+2sqrt(6))))^(x^(2)-3)=10` Put `(5+2sqrt(6))^(x^(2)-3)=t`, then `t+1/t=10` `impliest^(2)-10t+1=0` or `t=(10+-sqrt((100-4)))/2=(5+-2sqrt(6))` `implies(5+2sqrt(6))^(x^(2)-3)=(5+-2sqrt(6))=(5+2sqrt(6))^(+-1)` `:.x^(2)-3=+-1` `impliesx^(2)-3=1` or `x^(2)-3=-1` `impliesx^(2)=4` or `x^(2)=2` Hence `x=+-2, +-sqrt(2)`. |
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| 260. |
If `alpha,beta ` be the roots of the equation `3x^2+2x+1=0,` then find value of `((1-alpha)/(1+alpha))^3+((1-beta)/(1+beta))^3` |
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Answer» Let `(1-alpha)/(1+alpha)=ximpliesa=(1-x)/(1+x)` So replacing `x` by `(1-x)/(1+x)` in the given equation we get `3((1-x)/(1+x))^(2)+2((1-x)/(1+x))|+1=0impliesx^(2)-2x+3=0` …….i It is clear that `(1-alpha)/(1+alpha)` and `(1-beta)/(1+beta)` are the roots fo Eq. (i) `:.((1-alpha)/(1+alpha))+((1-beta)/(1+beta))=2`..........ii and `((1-alpha)/(1+alpha))((1-beta)/(1+beta))=3`.......iii `:.((1-alpha)/(1+alpha))^(3)+((1-beta)/(1+beta))^(3)=((1-alpha)/(1+alpha) +(1-beta)/(1+beta))^(3)-3` `((1-alpha)/(1+alpha))((1-beta)/(1+beta))((1-alpha)/(1+alpha)+(1-beta)/(1+beta))=2^(3)-3.3.2=8-18=-10` |
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| 261. |
Solve the equation `log_(3)(5+4log_(3)(x-1))=2` |
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Answer» We have `log_(3)(5+4log_(3)(x-1))=2` is equivalent ot the equation(here base `!=1,gt0`) `:.5+4log_(3)(x-1)=3^(2)` `implieslog_(3)(x-1)=1impliesx-1=3^(1)` `:.x=4` Hence `x_(1)=4` is the solution of the original equation. |
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| 262. |
Fill in the blanksIf the product of the roots of the equation `x^2-3k x+2e^(21 nk)-1=0`is 7, then the roots are real for____________. |
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Answer» Correct Answer - `k=2` Since, `x^(2)-3kx+2e^(2logk)-1=0` has product of roots 7. `implies2e^(2logh)-1=7` `impliese^(2log_(e)k)=4` `impliesk^(2)=4` `impliesk=2" "["neglectig-2"]` |
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| 263. |
Solve the equation `log_((x^(3)+6))(x^(2)-1)=log_((2x^(2)+5x))(x^(2)-1)` |
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Answer» The equation is equivalent to `{(x^(2)-1gt0),(2x^(2)+5xgt0),(2x^(2)+5x!=1),(x^(3)+6=2x^(2)+5x):}` `implies{(xlt-1 "and"xgt1),(xlt-5/2"and"xgt0),(x!=(-5+-sqrt(3))/4),(x=-2,1,3):}` Hence `x_(1)=3` is only root of the original equation. |
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| 264. |
If one root of the equation `(x-1)(7-x)=m` is three times the other, then `m` is equal toA. `-5`B. `0`C. `2`D. `5` |
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Answer» Correct Answer - C `(c )` `-x^(2)+8x-7=m implies x^(2)-8x+7+m=0` `alpha+3alpha=8 implies alpha=2` `alpha*3alpha=7+mimplies12=7+mimpliesm=5` |
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| 265. |
Solve the equationi `log_((x^(2)-1))(x^(3)+6)=log_((x^(2)-1))(2x^(2)+5x)` |
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Answer» This equation is equivalent to the system `{(2x^(2)+5xgt0),(x^(2)-1gt0),(x^(2)-1!=1),(x^(3)+6=2x^(2)+5x):}`impliesP(xlt-5/2` "and" xgt0),(xlt-1 "and "xgt1),(x!=+-sqrt(2)),(x=-2,1,3):}` Hence `x_(1)=3` is only root of the original equation.` |
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| 266. |
Statement -1 If the equation `(4p-3)x^(2)+(4q-3)x+r=0` is satisfied by `x=a,x=b` nad `x=c` (where a,b,c are distinct) then `p=q=3/4` and `r=0` Statement -2 If the quadratic equation `ax^(2)+bx+c=0` has three distinct roots, then a, b and c are must be 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 |
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Answer» Correct Answer - D If quadratic equation `ax^(2)+bx+c=0` is satisfied by more than two values of `x` then it must be an identity. Therefore `a=b=c=0` `:.` Statement -2 is true. But in Statement -1 `4p-3=4q-3=r=0` Then `p=q=3/4,r=0` which is false Since at one valueof `p` or `q` or `r` al coefficients at a time `!=0` `:.` Statement -1 is false. |
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| 267. |
Let `a`, `b`, `c` and `m in R^(+)`. The possible value of `m` (independent of `a`, `b` and `c`) for which atleast one of the following equations have real roots is `{:(ax^(2)+bx+cm=0),(bx^(2)+cx+am=0),(cx^(2)+ax+bm=0):}}`A. `(1)/(2)`B. `(1)/(8)`C. `(1)/(12)`D. `(1)/(4)` |
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Answer» Correct Answer - B::C::D `(b,c,d)` If at least one of the equations has real roots, then `D_(1)+D_(2)+D_(3) ge 0` `(b^(2)-5acm)+(c^(2)-4bam)+a^(2)-4cbm ge 0` `a^(2)+b^(2)+c^(2) ge 4(ab+bc+ca)m` `4m ge (a^(2)+b^(2)+c^(2))/(ab+bc+ca)`………`(1)` `AA a,b,c in R+` but `a^(2)+b^(2) ge 2ab` etc. `:. a^(2)+b^(2)+c^(2) ge ab+bc+ca` `(a^(2)+b^(2)+c^(2))/(ab+bc+ca) ge 1` `:.(a^(2)+b^(2)+c^(2))/(ab+bc+ca)|_(min)=1` , Hence `4m` must be less than or equal to the minimum value. `:. 4m le 1` gtbrgt `implies m le (1)/(4)` `implies m in (0,(1)/(4)]` |
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| 268. |
Fill in the blanksThe coefficient of `x^(99)`in the polynomial `(x-10(x-2)(x-100)i s______dot` |
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Answer» Correct Answer - `-5050` The coefficient of `x^(99)"in" (x-1)(x-2)...(x-100)` `=-(1+2+3...100)` `=-(100)/(2)(1+100)=-50(101)=-5056` |
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| 269. |
Solve the equation `log_(((2+x)/10))7=log_((2/(x+1)))7`. |
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Answer» The given equation is equivalent to `{(2/(x+1)gt0),(2/(x+1)!=1),((2+x)/10=2/(x+1)):}implies{(x+1gt0),(x!=1),(x=-6,3):}` `:.x_(1)=3` is root of the original equation. |
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| 270. |
Column I contains rational algebraic expressions and Column II contains possible integers of a. |
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Answer» Correct Answer - `(A)to(q,r,s,t),(B)to(q,r),(C)to(p,q)` `Ato(q,r,s,t), Bto(q,r,):Cto(p,q)` (A) We have `y=(ax^(2)+3x-4)/(3x-4x^(2)+a)` `impliesx^(2)(a+4y)+3(1-y)x-(ay+4)=0` As `x epsilon R ` we get `Dge0` `implies9(1-y)^(2)+4(a+4y)(ay+4)ge0` `implies(9+16a)y^(2)+(4a^(2)+46)y+(9+16a)ge0,AA y espsilon R` `implies` If `9+16agt0` then `Dle0` Now `D le0` `implies(4a^(2)+46)^(2)-4(9+16a)^(2)le0` `implies4[(2a^(2)+23)^(2)-(9+16a)^(2)]le0` `implies[(2a^(2)+23)+(9+16)][(2a^(2)+23)-(9+16a)]le0` ltbRgt `implies(2a^(2)+16a+32)(2a^(2)-16a+14)le0` `implies4(a+4)^(2)(a^(2)-8a+8)le0` `impliesa^(2)-8a+8le0` `implies(a-1)(a-8)le0` ltbr `implies1leale7` `:.9+16ag0` and 1leale7` `implies1leale7` (B) we have `y=(ax^(2)+x-2)/(a+x-2x^(2))` `impliesx^(2)(a+2y)+x(1-y)-(2+ay)=0` As `x epsilonR` we get `Dge0` `implies(1-y)^(2)+4(2+ay)(a+2y)ge0` `implies(1+8a)y^(2)+(4a^(2)+14)y+(1+8a)ge0` `implies` If `1+8agt0` then `Dle0` `implies(4a^(2)+14)^(2)-4(1+8a)^(2)le0` `implies4[(2a^(2)+7)^(2)-(1+8a)^(2)]le0` `implies[(2a^(2)+7)+(1+8a)][(2a^(2)+7)-(1+8a)]le0` `implies(2a^(2)+8a+8)(2a^(2)-8a+6)le0` `implies4(a+2)^(2)(a^(2)-4a+3)le0` `impliesa^(2)-4a+3le0` `implies(a-1)(a-3)le0` `implies1leale3` Thus `1+8agt0` and `1leale3` `implies1leale3` (C)We have `y=(x^(2)+2x+a)/(x^(2)+4x+3a)` `impliesx^(2)(y-1)+2(2y-1)x+a(3y-1)=0` As `x epsilon R` We get `Dge0` ltbRgt `implies4(2y-1)^(2)-4(y-1)a(3y-1)ge0` `implies(4-3a)y^(2)-(4-4a)y+(1-a)ge0` `implies` If `4-3agt0` then `Dle0` `implies(4-4a)^(2)-4(4-3a)(1-a)le0` `implies4(2-2a)^(2)-4(4-3a)(1-a) le 0` `implies4+4a^(2)-8a-(4-8a+3a^(2))le0` `implies a^(2)-ale0` `impliesa(a-1)le0` ltbRgt `implies0leale1` |
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| 271. |
If `alpha,beta`are the roots of the equation `a x^2+b x+c=0,`then find the roots of the equation `a x^2-b x(x-1)+c(x-a)^2=0`in term of `alphaa n dbetadot` |
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Answer» `ax^(2) - bx (x - 1) + c (x -1)^(2) = 0` `rArr (ax)/((1-x)^(20)) + (bx)/( 1- x) + c = 0` (1) Now , `alpha` is a root of `ax^(2) +bx + c = 0 `. Then let `alpha = (x) /(1-x)` `rArr x = (alpha)/(alpha + 1)` Hence, the roots of (1) are `alpha //(1 + alpha ),beta//(1 + beta)`. |
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| 272. |
Let a, b, c be real numbers with a = 0 and let `alpha,beta ` be the roots of the equation `ax^2 + bx + C = 0`. Express the roots of `a^3x^2 + abcx + c^3 = 0` in terms of `alpha,beta` |
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Answer» `alpha + beta = (-b)/(a), alpha beta = (c) /(a)` Roots of the equation `a^(3) x^(2) + abcx + c^(3) = 0` are : `x=(-abcpmsqrt((abc)^(2) - 4a^(2) c^(3)))/(2a^(3))` `= (-(b)/(a))((c)/(a)) pm(sqrt(((b)/(a))^(2)((c)/(a))^(2) - 4 ((c)/(a))^(3)))/(2)` `=(alpha beta) (((alpha + beta)pmsqrt((alpha - beta)^(2))))/(2)` `=alpha beta (((alpha + beta)(alpha - beta)))/(2)` `alpha^(2) beta, alphabeta^(2)` |
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| 273. |
If `P(x)=a x^2+b x+c&Q(x)=-a x^2+dx+c ,a c!=0,`then the equation `P(x)dotQ(x)=0`hasExactly two real rootsAtleast two real rootsExactly four real roots(d) No real roots |
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Answer» Correct Answer - 1 `P(x).Q(x)=(ax^(2)+bx+c)(-ax^(2)+bx+c)` Now, `D_(1)=b^(2)-4ac and D_(2)=b^(2)+4ac` Clearly, ` D_(1)+D_(2)=2b^(2)ge0` `therefore` Atleast one of `D_(1) and D_(2)` is (+ve). Hence, atleast two real roots. Hene, stamement is true. |
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| 274. |
Find the set of all solutions of the equation `2^|y| -|2^(y -1) -1| = 2^(y -1) +1 ` |
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Answer» Correct Answer - `y in{-1}uu[1,oo)` Given, ` 2^(|y|)-|2^(y-1)-1|=2^(y-1)+1` Case I When `y in (-oo, 0]` `therefore 2^(-y)+(2^(y-1)-1)2^(y-1)+1` `implies 2^(-y)=2` `impliesy=-1 in (-oo,0]" "...(i)` Case II When `y in (0,1]` `therefore2^(y)+(2^(y-1)-1)=2^(y-1)+1` `implies 2^(y)=2` `impliesy=1 in (0,1] " "...(ii)` Case III When `y in (1, oo)` `therefore2^(y)-2^(y-1)+1=2^(y-1)+1` `implies 2^(y)-2.2^(y-1)=0` implies 2^(y)-2^(y)=0 "true for all" y gt 1 " "...(iii)` From Eqs. (i), (ii), (iii), we get `y in{-1}uu[1,oo).` |
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| 275. |
The quadratic `x^2+a x=b+1=0`has roots which are positive integers, then `(a^2+b^2)`can be equal to`50`b. `37`c. `61`d. `19`A. 50B. 37C. 61D. 19 |
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Answer» Correct Answer - 1 `x^(2) + ax + b + 1 = 0` has positive integral roots `alpha and beta`. Now, `(alpha + beta) = -a and alphabeta = b + 1` `rArr (alpha + beta)^(2) + (alphabeta - 1)^(2) = a^(2) + b^(2)` `rArr a^(2) +b^(2) = (alpha^(2) + 1)(beta^(2) + 1).` `rArr a^(2) + b^(2)` can be equal to 50 (since other options have prime numbers) |
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| 276. |
Solve the equation `log_(1//3)[2(1/2)^(x)-1]=log_(1//3)[(1/4)^(x)-4]` |
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Answer» The given equation is equivalent to `{(2(1/2)^(x)-1gt0),(2(1/2)^(x)-1=(1/4)^(x)-4):}` `implies{((1/2)^(x)gt1/2),((1/2)^(2x)-2(1/2)^(x)-3=0):}` `implies{(xlt1),([(1/2)^(x)-3][(1/2)^(x)+1]):}=0` `implies{(xlt1),((1/2)^(x)=3,(1/2)^(x)+1!=0):}implies{(xlt1),(x=(-log_(2)3)):}` Hence `x_(1)=-log_(2)3` is the root of the original equation. |
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| 277. |
Let `alpha and beta` be the roots of `x^2 - 5x - 1 = 0` then the value of `(alpha^15 + alpha^11 + beta^15 + beta^11)/(alpha^13 + beta^13)` is |
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Answer» `(alpha ^(15)+ alpha^(11) + alpha^(15) + beta^(11))/(alpha^(13) + beta^(13))`. = `(alpha ^(15)+ alpha^(15) + alpha^(2) beta^(2)(alpha^(11)+beta^(2)))/(alpha^(13) + beta^(13))" "( because alpha beta = -1)` `(alpha ^(15)+ alpha^(13) + beta^(2)beta^(15)+alpha^(2)beta^(13))/(alpha^(13) + beta^(13))` `((alpha ^(13)+ alpha^(13)) ( alpha^(2)+beta^(2)))/(alpha^(13) + beta^(13))` `alpha^(2) + beta^(2) = (alpha + beta)^(2) - 2alpha beta = 27` |
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| 278. |
If `alpha,beta` are the roots of `ax^2 + c = bx`, then the equation `(a + cy)^2 =b^2y` in y has the rootsA. `alpha beta^(-1), alpha^(-1)beta`B. `alpha^(-2), beta_(-2)`C. `alpha^(-1), beta^(-1)`D. `alpha^(2), beta^(2)` |
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Answer» Correct Answer - 2 `ax^(2) - bx + c = 0` `alpha + beta = b/a, abeta = c/a` Also, `(a + cy)^(2) = b^(2)y` `rArr c^(2)y^(2) - (b^(2) - 2ac)y + a^(2) = 0` `rArr (c/a)^(2)y^(2) - ((b/a)^(2) - 2(c/a))y + 1 = 0` `rArr (alphabeta)^(2)y^(2) - (alpha^(2) + beta^(2))y + 1 = 0` `rArr y^(2) - (alpha^(-2) + beta^(-2))y + alpha^(-2)beta^(-2) = 0` `rArr (y - alpha^(-2)) (y - beta^(-2)) = 0` Hence, the roots are `alpha^(-2), beta^(-2)`. |
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| 279. |
The equation `2x^(2)+3x+1=0` has an irrational root. |
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Answer» Given `2x^(2)+3x+1=0` Here, `D=(3)^(2)-4.2.1=1` which is a perfect square. `therefore` Roots are rational. Hence, statement is false. |
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| 280. |
Solve the equation `(1-2(2logx)^(2))/(logx-2(logx)^(2))=1` |
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Answer» The given equation can rewrite in the form `(1-2(2logx)^(2))/(logx-2(logx)^(2))=1` `implies(1-8(logx)^(2))/(logx-2)(logx)^(2))-1=0` Let `logx=t` then `(1-8t^(2))/(5-2t^(2))-1=0implies(1-8t^(2)-t+2t^(2))/(t-2t^(2))=0` `implies(1-t-6t^(2)/((t-2t^(2))=0implies((1+2t)(1-3t))/(t(1-2t))=0` `implies{(t=-1/2),(t=1/3):}implies{(logx=-1/2),(logx=1/3):}implies{:(x_(1)=10^(-1//2)),(x_(2)=10^(1//3)):}` Hence `x_(1)=1/(sqrt(10))` and `x_(2)=root(3)(10)` are the roots of the original equation. |
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| 281. |
If `alphaa n dbeta`are roots of the equation `a x^2+b x+c=0,`then the roots of the equation `a(2x+1)^2-b(2x+1)(3-x)+c(3-x)^2=0`are`(2alpha+1)/(alpha-3),(2beta+1)/(beta-3)`b. `(3alpha+1)/(alpha-2),(3beta+1)/(beta-2)`c. `(2alpha-1)/(alpha-2),(2beta+1)/(beta-2)`d. none of theseA. `(2alpha + 1)/(alpha - 3), (2beta + 1)/(beta - 3)`B. `(3alpha + 1)/(alpha - 2), (2beta + 1)/(beta - 2)`C. `(2alpha - 1)/(alpha - 2), (2beta + 1)/(beta - 2)`D. none of these |
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Answer» Correct Answer - 2 `a((2x + 1)^(2))/((x - 3)^(2)) + b((2x + 1))/((x - 3)) + c = 0` `rArr (2x + 1)/(x - 3) = alpha` or `(2x + 1)/(x - 3) = beta` or `2x + 1 = alphax - 3alpha` or `x(alpha - 2) = 1 + 3alpha` or `x = (1 + 3alpha)/(alpha - 2), (1 + 3beta)/(beta - 2)` |
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| 282. |
Solve the equation `log_((log_(5)x))5=2` |
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Answer» We have `log_((log_(5)x))5=2` Base of logarithm `gt0` and `!=1` `:.log_(4)xgt0` and `log_(5)x!=1` `impliesxgt1` and `x!=5` `:.` The original equation is equivalent to `log_(5)x=5^(1//2)=sqrt(5)` ltrgt `:.x_(1)=5^(sqrt(5))` Here `5^(sqrt(5))` is the only root of the original equation. |
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| 283. |
The number of irrational roots of the equation `4x//(x^2+x+3)+5x//(x^2-5x+3)=-3//2`is`4`b. `0`c. `1`d. `2`A. 4B. 0C. 1D. 2 |
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Answer» Correct Answer - 4 Here, x = 0 is not a root, Divide both the numerator and denominator by x and put x + 3/x = y to obtain `(4)/(y + 1) + (5)/(y - 5) = - 3/2 rArr y = -5, 3` x + 3/x = -5 has two irrational roots and x + 3/x = 3 ahs imaginary roots. |
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| 284. |
Solve the equation `2x^(log_(4)^(3))+3^(log_(4)^(x)=27` |
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Answer» The domain of the admissible value of the equation is `xgt0`. The given equation is equivalent to `2.3^(log_(4)x)+3^(log_(4)^(x))=27` [from above result (v)] `=3.3^(log_(4)^(x)=27` `=3^(log_(4)x)=9` `implies3^(log_(4)x)=3^(2)` `implieslog_(4)x=2` `impliesx_(1)=4^(2)=16` is its only root. |
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| 285. |
If `a gt 0 and b^(2) - 4 ac = 0` then solve `ax^(3) + (a + b) x^(2) + (b + c) x + c gt 0` . |
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Answer» We must have `ax^(3) + (a + b)x^(2) + (b + c ) x + c gt 0` `rArr ax^(2) (x-1) + bx (x-1) + c (x + 1) gt 0` `rArr (x + 1) (ax^(2) + bx + c ) gt 0` `rArr a(x + 1 ) (x + (b)/(2a))^(2) gt 0 as b^(2) = 4ac` `rArr x gt -1 and x ne - (b)/(2a)` |
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| 286. |
Solve the inequation `log_(((x^2-12x+30)/10))(log_2((2x)/5))gt0` |
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Answer» This inequation is equivalent to the collection of two systems `{((x^(2)-12x+30)/10gt1),(log_(2)((2x)/5)gt1):}` `{(0lt(x^(2)-12x+30)/10lt1),(0ltlog_(2)((2x)/5)lt1):}` On solving the first system, we have `implies{(x^(2)-12x+20gt0),((2x)/5gt2):}` `hArr{((x-10)(x-2)gt0),(xgt5):}` `hArr{(xlt2 "and" xgt10),(xgt5):}` Therefore the system has solution `x gt10` On solving the second system, we have `implies{(0ltx^(2)-12x+30lt10),(1lt(2x)/5lt2):}` `hArr{(x^(2)-12x+30gt0 "and" x^(2)-12x=20lt),(5//2ltxlt5):}` `hArr {(xlt6-sqrt(6) "and "xgt6+sqrt(6) "and" 2ltxlt10),(0lt xlt5):}` Therefore the system has solution `2ltxlt6-sqrt(6)` combining both system, then solution of the original inequations is ` x epsilon (2,5,sqrt(6))uu(10,oo)` |
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| 287. |
If `alpha,beta and gamma` are roots of equation `x^(3)-3x^(2)+1=0`, then the value of `((alpha)/(1+alpha))^(3)+((beta)/(1+beta))^(3)+((gamma)/(1+gamma))^(3)` is __________. |
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Answer» Correct Answer - -1 Let `alpha_(1)=(alpha)/(1+alpha),beta_(1)=(beta)/(1+beta),andgamma_(1)=(gamma)/(1+gamma)`, Also, let `y=alpha_(1),beta_(1) and gamma_(1)` be roots of an equation is y. Then `y=(x)/(1+x)` `therefore x =(y)/(1-y)` Now,`x^(3)-3x^(2)+1=0` `rArr((y)/(1-y))^(3)-3((y)/(1-y))^(2)+1=0` `rArr3y^(3)-3y+1=0` `rArralpha_(1)+beta_(1)+gamma_(1)=0 and alpha_(1)beta_(1)gamma_(1)=-(1)/(3)` `therefore alpha_(1)^(3)+beta_(1)^(3)+gamma_(1)^(3)=3alpha_(1)beta_(1)gamma_(1)=3(-(1)/(3))=-1` |
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| 288. |
The sum of the non-real root of `(x^2+x-2)(x^2+x-3)=12`is`-1`b. `1`c. `-6`d. `6`A. -1B. 1C. -6D. 6 |
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Answer» Correct Answer - 1 Put `x^(2) + x = y`, so that Eq. (1) becomes `(y -2)(y - 3) = 12` or `y^(2) -5y -6 =0` or `(y - b)(y + 1) = 0` or y = 6, -1 When y = 6, we get `x^(2) + x - 6 = 0` `rArr (x + 3)(x - 2) = 0` or x = -3, 2 When y = -1, we get `x^(2) + x + 1 = 0` which has nonreal roots and sum of roots is -1. |
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| 289. |
Let `alphaa n dbeta`be the solutions of the quadratic equation `x 62-1154 x+1=0`, then the value of `alpha4+beta4`is equal to ______. |
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Answer» Correct Answer - 6 `alpha+beta=1154andalphabeta=1` `(sqrtalpha+sqrtbeta)^(2)=alpha+beta+2sqrt(alphabeta)=1154+2=1156=(34)^(2)` `impliessqrtalpha+sqrtbeta=34` Again `(alpha^(1//4)+beta^(1//4))^(2)=sqrtalpha+sqrtbeta+2(alphabeta)^(1//4)=34+2=36` `alpha^(1//4)+beta^(1//4)=6` |
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| 290. |
If `a ,b ,a n dc`are odd integers, then prove that roots of `a x^2+b x+c=0`cannot be rational. |
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Answer» Here, discriminant `D =b^(2) - 4ac.` Suppose the roots are rational. Thus , D will be a perfect square. Let `b^(2) - 4 ac = d^(2)` . Since a, b and c are odd integers, d will be odd .Now, `b^(2) - d^(2) = 4ac` Let b = 2k + 1 and d = 2m + 1 . Then `b^(2) - d^(2) = (b -d) (b + d)` `2(k -m) 2(k + m + 1)` Now, either (k - m) or (k + m + 1) is always even. Hence, `b^(2)-d^(2)` is always a multiple of 8. But , 4ac is a multiple of 4 (not of 8 ), Which is a contradiction. Hence, the roots of `ax^(2) + bx + c =0` connot be rational. |
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| 291. |
Integral part of the product of non-real roots of equation `x^(4)-4x^(3)+6x^(2)-4x=69` is _______. |
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Answer» Correct Answer - 9 We have `x^(4)-4x^(3)+6x^(2)-4x=69` `rArrx^(4)-4x^(3)+6x^(2)-4x+1=70` `rArr (x-1)^(4)=70` `rArr (x-1)^(2)=pm sqrt70` For the product of non-real roots, consider `(x-1)^(2)=-sqrt70` `rArr x^(2)-2x+1+sqrt70=0` Therefore, product is `1+sqrt70`,whose intgral part is 9. |
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| 292. |
If `f(x)=x^2+b x^2+c x+da n df(0),f(-1)`are odd integers, prove that `f(x)=0`cannot have all integral roots. |
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Answer» `f(0) = d, f(-1) = -1 + b - c + d` `rArr d = odd and - 1 + b - c + d = odd` `rArr b - c = 1 + odd -d` `= (1+ odd) -(odd) = even - odd = odd` (1) Thus, both d and b - c are odd If possible let the three roots `alpha, beta, gamma` be all integers. Now, `alpha beta gamma = - (d)/(1) = -d =` negative odd integers (2) `rArr alpha, beta, gamma` are three integers whose product is odd `rArr alpha, beta, gamma `all are odd Again `alpha+ beta + gamma = - b and alpha beta + beta gamma + alpha gamma = c` (3) `rArr ` b and c both will be odd `rArr (b - c)` will be even which contradicts with (1) Hence, the three roots connot be all integers. |
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| 293. |
If `alpha,beta`are the roots of lthe equation `2x 62-3x-6=0,`find the equation whose roots are `alpha^2+2a n dbeta^2+2.` |
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Answer» Since `alpha, beta ` are roots of the equation `2x^(2) - 3x - 6 = 0, ` we have `alpha + beta = 3//2 and alpha beta = -3` `rArr alpha^(2) + beta^(2) = (alpha + beta)^(2) - 2alpha beta` = `(90/(4)+6 = (33)/(4)` Now , `(alpha^(2)+beta^(2))+(beta^(2)+2) = (alpha^(2)+ beta^(2)) + 4` `=(33)/(4) + 4 = (49)/(4)` and `(alpha ^(2) + 2 ) (beta^(2)+2)= alpha^(2) + beta^(2) + 2(alpha^(2)+ beta^(2))+4` `=(-3)^(2) + 2((33)/(4)) + 4` `= (59)/(2)` So , the equatioon whose roots are `alpha^(2) + 2 and beta^(2) + 2` is given by `x^(2) - x [(alpha^(2) + 2) + (beta^(2) + 2)] + (alpha^(2) + 2 ) (beta^(2) + 2 ) = 0` `rArr x^(2) - (49)/(4)x+(59)/(2) = 0` or `4x^(2) - 49x + 118 = 0` |
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| 294. |
Find the largest natural number a for which the maximum value of `f(x)=a-1+2x-x^2`is smaller thante ninimum value of `g(x)=x^2-2a x=10-2adot` |
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Answer» `f(x) = a - 1 + 2x - x^(2)` `= a - (x^(2)-2x + 1)` `= a (x - 1) ^(2)` Hence, the maximum value of `f(x)` is a when `(x - 1)^(2) = 0 or x = 1` `g(x) = x^(2) - 2ax + 10 -2a` `=(x - a)^(2) + 10 - 2a - a^(2)` Hence, the minimum value of `g(x) is 10 -2a - a^(2)` when `(x - a)^(2) = 0 or x= a`. Now given that maximum of `f(x)` is smaller than the minimum of g(x) . Thus, `a lt - a^(2)+ 10 - 2a` or `a^(2) + 3a - 10 lt 0` or ` (a + 5) (a - 2) lt 0` `therefore -5 lt a lt 2` So, The largest natural number a = 1. |
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| 295. |
If `x^2+ax+1=0` is a factor of `ax^3+ bx + c`, then which of the following conditions are not valid |
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Answer» Correct Answer - `b - a + a^(3) = 0, a^(2) + c = 0` `x^(2) + ax + 1` must divide `ax^(3) + bx + c .` Now `(ax^(3) +bx + c)/(x^(2) + ax + 1)= a (x - a) + ((b - a + a)^(3) x+c + a^(2))/(x^(2) + ax + 1)` Ther remainder must be zero . Hence, `(b - a + a^(3)) x + c + a^(2) = 0 AA x in` R So, `b - a + a^(3) = 0, a^(2) + c = 0` |
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| 296. |
Q. Let p and q real number such that `p!= 0`,`p^2!=q` and `p^2!=-q`. if `alpha` and `beta` are non-zero complex number satisfying `alpha+beta=-p` and `alpha^3+beta^3=q`, then a quadratic equation having `alpha/beta` and `beta/alpha` as its roots isA. `(p^(3)+q)x^(2)-(p^(3)+2q)x+(p^(3)+q)=0`B. `(p^(3)+q)x^(2)-(p^(3)-2q)x+(p^(3)+q)=0`C. `(p^(3)-q)x^(2)-(5p^(3)-2q)x+(p^(3)-q)=0`D. `(p^(3)-q)x^(2)(5p^(3)+2q)x+(p^(3)-q)=0` |
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Answer» Correct Answer - C `(alpha)/(beta)+(beta)/(alpha)=(alpha^(2)+beta^(2))/(alpha beta)=((alpha +beta)^(2)-2 alpha beta)/(alpha beta)`……….i and given `alpha^(3)+beta^(3)=q, alpha + beta=-p` `implies(alpha+beta^(3)=q,alpha + beta=-p` `=(alpha +beta)^(3)-3alpha beta(alpha +beta)=q` `implies-p^(3)+3p alpha beta=q` ltbr or `alpha beta=(q+p^(3))/(3p)` `:.` From eq. (i) we get `(alpha)/(beta)+(beta)/(alpha)=(p^(2)-(2(q+p^(3)))/(3p)/(((q+p^(3)))/(3p))=(p^(3)-2q)/((q+p^(3)))` and product of the roots `=(alpha)/(beta). (beta)/(alpha)=1` `:.` Required equation is `x^(2)-((p^(3)-2q)/(q+p^(3)))x+q=0` or `(q+p^(3))x^(2)-(p^(3)-2q)x+(q+p^(3))=0` |
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| 297. |
Q. Let p and q real number such that `p!= 0`,`p^2!=q` and `p^2!=-q`. if `alpha` and `beta` are non-zero complex number satisfying `alpha+beta=-p` and `alpha^3+beta^3=q`, then a quadratic equation having `alpha/beta` and `beta/alpha` as its roots isA. `(p^(3)+q)x^(2)-(p^(3)+2q)x+(p^(3)+q)=0`B. `(p^(3)+q)x^(2)-(p^(3)+2q)x+(p^(3)+q)=0`C. `(p^(3)-q)x^(2)-(5p^(3)-2q)x+(p^(3)-q)=0`D. `(p^(3)-q)x^(2)-(5p^(3)+2q)x+(p^(3)-q)=0` |
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Answer» Correct Answer - B Sum of roots `=(alpha^(2)+beta^(2))/(alphabeta)` and product = 1 Given, `alpha+beta=-pand alpha^(3)+beta^(3)=q` `implies(alpha+beta)(alpha^(2)-alphabeta+beta^(2))=q` `thereforealpha^(2)+beta^(2)-alphabeta=(-q)/(p)" "....(i)` and `(alpha+beta)^(2)=p^(2)` `impliesalpha^(2)+beta^(2)+2alphabeta=p^(2)" "(ii)` From Eqs. (i) and (ii) we get `alpha^(2)+beta^(2)=(p^(a)-2q)/(3p)and alpha beta=(p^(3)+q)/(3p)` `therefore` Required equation is, `x^(2)((p^(3)-2q)x)/((p^(3)+q))+1=0` `implies(p^(3)+q)x^(2)-(p^(3)-2q)x+(p^(3)+q)=0` |
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| 298. |
Suppose `a ,b ,c in I`such that the greatest common divisor fo `x^2+a x+ba n dx^2b x+ci s(x+1)`and the least common ultiple of `x^2+a x+ba n dx^2+b x+c`is `(x^3-4x^2+x+6)dot`Then the value of `|a+b+c|`is equal to ___________. |
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Answer» Correct Answer - 6 `x^(2)+ax+b-=(x+1)(x+b)rArrb+1=a" "(1)` also `x^(2)+bx+c-=(x+1)(x+c)rArrc+1=b` or `b+1=c+2" "(2)` Hence `b+1=a=c+2` Also `(x+1)(x+b)(x+c)-=x^(3)-4x^(2)+x+6` `rArrx^(3)+(1+b+c)x^(2)+(b+bc+c)x+bc-=x^(3)-4x^(2)+x+6` `rArr1+b+c=-4` `rArr 2c+2=-4` `rArr c=-3,b=-2 and a =-1` `rArr a+b+c=-6` |
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| 299. |
If G and L are the greatest and least values of the expression `(x^(2)-x+1)/(x^(2)+x+1), x epsilon R` respectively then `G` and `L` are the roots of the equationA. `3x^(2)-10x+3=0`B. `4x^(2)-17x+4=0`C. `x^(2)-7x+10=0`D. `x^(2)-5x+6=0` |
| Answer» Correct Answer - Eq | |
| 300. |
If `a`is the root (having the least absolute value) or the equation `x^2-b x-1=0(b in R^+)`, then prove that `-1 |
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Answer» Let `f(x) = x^(2) - bx - 1 (b in R^(+))` `f( -1) = b + ve` `f(0) = - 1= - ve` `f(1) = - b = -ve ` Clearly, one root lies in `(-1,0)` and the other in `(1,infty)` . Now , sum of roots is b, which is positive . So .`alpha ` (having the least absolute value) `in (-1, 0)` . |
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