InterviewSolution
Saved Bookmarks
InterviewSolution
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.
| 1. |
Let `alpha`, `beta (a lt b)` be the roots of the equation `ax^(2)+bx+c=0`. If `lim_(xtom) (|ax^(2)+bx+c|)/(ax^(2)+bx+c)=1` thenA. `(|a|)/(a)=-1`, `m lt alpha`B. `a gt 0`, `alpha lt m lt beta`C. `(|a|)/(a)=1`, `m gt beta`D. `a lt 0`, `m gt beta` |
|
Answer» Correct Answer - C `(c )` According to the given condition, we have `|am^(2)+bm+c|=am^(2)+bm+c` i.e., `am^(2)+bm+c gt 0` `implies` if `a lt 0` , then `m` lies in `(alpha,beta)` and if `a gt 0`, then `m` does not lie in `(alpha,beta)` Hence, option `(c )` is correct, since `(|a|)/(a)=1impliesa gt 0` and in that case `m` does not lie in `(alpha,beta)`. |
|
| 2. |
If `alpha, beta` are the roots of `x^(2)-3x+1=0`, then the equation whose roots are `(1/(alpha-2),1/(beta-2))` isA. `x^(2)+x-1=0`B. `x^(2)+x+1=0`C. `x^(2)-x-1=0`D. none of these |
| Answer» Correct Answer - C | |
| 3. |
let `alpha(a)` and `beta(a)` be the roots of the equation `((1+a)^(1/3)-1)x^2 +((1+a)^(1/2)-1)x+((1+a)^(1/6)-1)=0` where `agt-1` then, `lim_(a->0^+)alpha(a)` and `lim_(a->0^+)beta(a)`A. `(-5/2)` and 1B. `(-1/2)` and (1)C. `(-7/2)` and 2D. `(-9/2)` and 3 |
|
Answer» Correct Answer - D Let `a+1=h^(6)` `:.(h^(2)-1)x^(2)+(h^(3)-1)x+(h-1)=0` `implies((h^(2)-1)/(h-1))x^(2)+((h^(3)-1)/(h-1))x+1=0` As `ato0` then `hto1` `lim_(hto1)((h^(2)-1)/(h-1))x^(2)+lim_(hto1)((h^(3)-1)/(h-1))x+1=0` `implies2x^(2)+3x+1=0` `implies2x^(2)+2x+x+1=0` `implies(2x+1)(x+1)=0` `:.x=-1` and `x=-1/2` |
|
| 4. |
8. Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots (4,3). Rahul made a mistake in writing down coefficient of x to get roots (3, 2). The correct roots of equation are:A. `-4,-3`B. `6,1`C. `4,3`D. `-6,-1` |
|
Answer» Correct Answer - B Let the quadratic equation be `ax^(2)+bx+c=0` Sachin made a mistake in writing down constant term. `:.` Sum of the roots is correct. i.e. `alpha +beta=7` Rahul made a mistake in writing down coefficient of `x`. `:.` Product of the roots is correct. i.e. `alpha beta=6` `implies` Correct quadratic equation is `x^(2)-(alpha+beta)x+alpha beta=0` `impliesx^(2)-7x+6=0` `implies(x-6)(x-1)=0impliesx=6,1` Hence correct roots are 1 and 6. |
|
| 5. |
The number of real roots of the equation `5+|2^(x)-1|=2^(x)(2^(x)-2)is`A. 1B. 3C. 4D. 2 |
|
Answer» Correct Answer - A Given equation `5+|2^(x)-1|=2^(x)(2^(x)-2)` Case I If `2^(x)-1 ge0impliesx ge0,then 5+2^(x)-1=2^(x)(2^(x)-2)` Put `2^(x)=t,`then `5+t-1=t^(2)-2timpliest^(2)-3t-4=0` `impliest^(2)-4t+t-4=0impliest(t-4)+1(t-4)=0` `impliest=4or-1impliest=4" "(becauset=2^(x)gt0)` `implies2^(x)=4impliesx=2gt0` `impliesx=2` is the solution. m Case II `If 2^(x)-1lt0impliesxlt0,` then `5+1-2^(x)=2^(x)(2^(x)-2)` Put `2^(x)=y, then 6-y=y^(2)-2y` `impliesy^(2)-y-6=0impliesy^(2)-3y+2y-6=0` `implies(y+2)(t-3)=0impliesy=3or -2` `impliesy=3(as y=2^(x)gt0)implies2^(x)=3` `impliesx=log_(2)3gt0` So, `x=log_(2)3` is not a solution. Therefore number of real roots is one. |
|
| 6. |
The smallest value of k, for which both the roots of the equation, `x^2-8kx + 16(k^2-k + 1)=0` are real, distinct and have values at least 4, is |
|
Answer» Correct Answer - `k=2` (i) Given, `x^(2)-8kx+1(k^(2)-k+1)=0` `Now, D=64{ k^(2)-k+1}=64 (k-1)gt0 k gt 1` `(ii)-(b)/(2a)gt4implies(8k)/(2)gt4impliesk gt1` `(iii) f(4)ge0` `implies16-32k+16(k^(2)-k+1)ge0` `impliesk^(2)-3k+2ge0` `implies(k-2)(k-1)ge0` `impliesk le1 ork ge2` Hence, `k=3` |
|
| 7. |
Find the values of `a`for whilch the equation `sin^4x+asin^2x+1=0`will have ea solution. |
|
Answer» Let ` t = sin ^(2) x rArr t in [ 0 , 1] ` Hence , `t^(2) + at + 1 =0 ` should have at least one solution in [0,1]. Since product of roots is positive and equal to one , equation must have exactly one root in [0,1] . Hence , ` f(1) lt 0 ` `rArr 2 + a lt 0 ` `rArr a in (-infty, - 2)` . |
|
| 8. |
Solve the equation `2log_(3)x+log_(3)(x^(2)-3)=log_(3)0.5+5^(log_(5)(log_(3)8)` |
|
Answer» This equation can be written as `log_(3)x^(2)+log_(3)(x^(2)-3)=log_(3)0.5+log_(3)8` `log_(3)x^(2)+log_(3)(x^(2)-3)=log_(3)(4)` This is equivalent to the system `{(x^(2)gt0),(x^(2)-3gt0),(x^(2)(x^(2)-3)=4):}implies{(xlt0 "and" xgt0),(xlt-sqrt(3) "and" xgtsqrt(3)),((x^(2)-4)(x^(2)=1)=0):}` `impliesx^(2)-4=0:.x=+-2`, but `xgt0` Consequently `x_(1)=2` is only root of the original equation. |
|
| 9. |
All the values of `m`for whilch both the roots of the equation `x^2-2m x+m^2-1=0`are greater than -2 but less than 4 lie in the interval`-23`c. `-1A. `-2 lt m lt 0 `B. `m gt 3`C. `- 1 lt m lt 3`D. `1 lt m lt 4` |
|
Answer» Correct Answer - 3 The given equation is `x^(3) - 2mx +m^(2) - 1 = 0 ` or `(x - m + 1) (x - m - 1) = 0 ` or ` x = m - 1 , m + 1` From given condition, ` m - 1 gt - 2 and m+ 1 lt 4` `rArr m gt - 1 and m lt 3` Hence, ` - 1 lt m lt 3` . |
|
| 10. |
The values of x, satisfying the equation for `AA a > 0, 2log_(x) a + log_(ax) a +3log_(a^2 x)a=0` are |
|
Answer» Correct Answer - `x=a^(-1//2)or a^(-4//3)` The given equation can be rewritten as `(12)/(log_(a)x)+(1)/(log_(a)ax)+(3)/(log_(a)x)=0` `implies1/t+(1)/(1+t)+(3)/(2+t)=0, where t =log_(a)x` `implies2(1+t)(2+t)+3t(1+t)+t(2+t)=0` `implies6t^(2)+11t+4=0` `implies(2t+1)(3t+4)=0` `impliest=-1/2or log_(a)x=-4/3` `impliesx=a^(-1//2)` `or x=a^(-4//3)` |
|
| 11. |
Let `p ,q`be integers and let `alpha,beta`be the roots of the equation, `x^2-x-1=0,`where `alpha!=beta`. For `n=0,1,2, ,l e ta_n=palpha^n+qbeta^ndot`FACT : If `aa n db`are rational number and `a+bsqrt(5)=0,t h e na=0=bdot`If `a_4=28 ,t h e np+2q=`7 (b) 21(c) 14 (d) 12A. `a_(11)+2alpha_(10)`B. `2a_(11)+2a_(10)`C. `a_(11)-a_(10)`D. `a_(11)+a_(10)` |
|
Answer» Correct Answer - D `alpha^(2)=alpha+1` `beta^(2)=beta+1` `a_(n)=palpha^(n)=qbeta^(n)` `=p(alpha^(n-1)+alpha^(n-2))+q(beta^(n-1)+beta^(n-2))` `=a_(n-1)+a_(n-2)` `thereforea_(12)=a_(11)+a_(10)` |
|
| 12. |
The value of m for which one of the roots of `x^(2) - 3x + 2m = 0` is double of one of the roots of `x^(2) - x + m = 0` isA. -2B. 1C. 2D. none of these |
|
Answer» Correct Answer - 1 Let `alpha` be the root of `x^(2) - x + m = 0 and 2alpha` be the root of `x^(2) - 3x + 2m = 0`. Then, `alpha^(2) - alpha + m = 0 and 4alpha^(2) - 6alpha + 2m = 0` Eliminating `alpha, m^(2) = -2m rArr m = 0, m = -2` |
|
| 13. |
If `alpha and beta` be the roots of the equation `x^(2) + px - 1//(2p^(2)) = 0`, where `p in R`. Then the minimum value of `alpha^(4) + beta^(4)` isA. `2sqrt2`B. `2 - sqrt2`C. `2`D. `2 + sqrt2` |
|
Answer» Correct Answer - 4 Here, `alpha^(4) + beta^(4) = (alpha^(2) + beta^(2))^(2) - 2alpha^(2)beta^(2)` `= {(alpha + beta)^(2) - 2alphabeta}^(2) - 2(alphabeta)^(2)` `= (p^(2) + 1/p^(2))^(2) - (1)/(2p^(4))` `= p^(4) + (1)/(2p^(4)) + 2` `= (p^(2) - (1)/(sqrt2 p^(2)))^(2) + 2 + sqrt2 ge 2 + sqrt2` Thus, the minimum value of `alpha^(4) + beta^(4)` is `2 + sqrt2`. |
|
| 14. |
If `alpha,beta`are roots of `x^2+-p x+1=0a n dgamma,delta`are the roots of `x^2+q x+1=0`, then prove that `q^2-p^2=(alpha-gamma)(beta-gamma)(alpha+delta)(beta+delta)`. |
|
Answer» Correct Answer - `q^(2)-p^(2)` Since, `alpha+beta=-, alphabeta=1and gamma+delta=-q,gamma delta=1` Now, `(alpha-gamma)(beta-gamma)(alpha+delta)(beta+delta)` `{alpha beta-gamma(alpha+beta)+gamma^(2)}{alphabeta+delta(alpha+beta)+delta^(2)}` `={1-gamma(-p)+gamma^(2)}{1+delta(-p)+delta^(2)}` `={1-gamma(-p)+gamma^(2)}{1+delta(-p)+delta^(2)}` `=(1+y^(2)+gammap)(1-deltap+delta^(2))=(-qgamma+gammap)(-deltap-deltap)` `[becausegamma^(2)+qgamma+1=0and delta^(2)+qdelta+1=0]` `=(q^(2)-p^(2))(gammadelta)=q^(2)-p^(2)" "[becausegamma delta=1]` |
|
| 15. |
If `alpha,beta`are the roots of `x^2+p x+q=0a n dgamma,delta`are the roots of `x^2+p x+r=0,`then `((alpha-gamma)(alpha-delta))/((beta-gamma)(beta-delta))=``1`b. `q`c. `r`d. `q+r`A. 1B. qC. rD. q + r |
|
Answer» Correct Answer - 1 `alpha, beta` be the roots of `x^(2) + px + q = 0` `gamma, delta` be the roots of `x^(2) + px + r = 0` `alpha + beta = -p` `alphabeta = q` `gamma + delta = -p` `gamma delta = r` Now, `(alpha - gamma)(alpha - delta) = alpha^(2) - alpha(gamma + delta) + gammadelta` `= alpha^(2) - alpha(alpha + beta) = r` `= - alphabeta + r = -q + r` By symmetry `(beta - gamma) (beta - delta) = -q + r` Hence, `((alpha - gamma)(alpha - delta))/((beta - gamma)(beta - delta)) = 1` |
|
| 16. |
Given that `alpha,gamma`are roots of the equation `A x^2-4x+1=0,a n dbeta,delta`the roots of the equation of `B x^2-6x+1=0,`such that `alpha,beta,gamma,a n ddelta`are in H.P., thena.`A=3`b. `A=4``B=2`d. `B=8`A. `A=3`B. `A=4`C. `B=2`D. `B=8` |
|
Answer» Since `alpha,beta, gamma` and `delta` are in HP hence `1/(alpha),1/(beta),1/(gamma)` and `1/(delta)` are in AP and they may be take as a `a-3d, a-d, a+d` and `a+3d`. Replaceing `x` by `1/x` we get the equation whose roots are `a-3d, a+d` is `x^(2)-4x+A=0` and equation whose roots are `a-d,a+3d` is `x^(2)-6x+B=0` then `(a-3d)+(a+d)=4implies2(a-d)=4` and `(a-d)+(a+3d)=6implies2(a+d)=6` `:.a=5/2` and `d=1/2` Now `A=(a-3d)(a+d)=(5/2-3/2)(5/2+1/2)=3` and `B=(a-d)(a+3d)(5/2-1/2)(5/2+3/2)=8` |
|
| 17. |
Let `p(x) =x^6-x^5-x^3-x^2-x` and `alpha, beta, gamma, delta` are the roots of the equation `x^4-x^3-x^2-1=0` then `P(alpha)+P(beta)+P(gamma)+P(delta)=`A. `4`B. `6`C. `8`D. `12` |
|
Answer» Correct Answer - B `(b)` `P(x)=x^(6)-x^(5)-x^(3)-x^(2)-x` and `Q(x)=x^(4)-x^(3)-x^(2)-1` Here` P(x)=Q(x)(x^(2)+1)+x^(2)-x+1` `:. sumP(alpha)=sumalpha^(2)-sumalpha+4` `=(sumalpha)^(2)-2sum(alphabeta)-sumalpha+4` `=1-2(-1)-(1)+4` `=6` |
|
| 18. |
If `x + y + z = 12, x^(2) + Y^(2) + z^(2) = 96` and `(1)/(x)+(1)/(x)+(1)/(z)= 36` . Then find the value `x^(3) + y^(3)+z^(3).` |
|
Answer» To get the value of `x^(3)+Y^(3) + z^(3) - 3xyz` `= (x + y + z ) (x^(2)+Y^(2)+x^(2) - xy- yz - zx)` …(1) We need the value of xyz and xy + yz + zx. We have `(x + Y + z)^(2) = 144` `therefore x^(2)+y^(2)+z^(2)+2xy+2yz+2xz=144` `rArr `96 + 2(xy + yz + xz) =144 `rArr `xy+ yz+ zx = 24 Given that `(1)/(x)+(1)/(x)+(1)/(z)= 36` `therefore (xy+yz+zx)/(xyz)=36` `rArr xyz = (24)/(36)=(2)/(3)` From (1), `x^(3)+y^(3)+z^(3) - 2 = 12xx(96-24)` =864 So, `x^(3)+y^(3)+z^(3) = 866` |
|
| 19. |
Polynomial `P(x)`is divided by `(x-3)`, the remainder if 6.If `P(x)`is divided by `(x^2-9)`, then the remainder is `g(x)`. Then the value of `g(2)`is ___________. |
|
Answer» Correct Answer - 4 As P(x) is an odd function Hence, `P(-x)=-P(x)impliesP(-3)=-P(3)=-6` Let `P(x)=Q(x^(2)-9)+ax+b` (where Q is quotient and (ax+b)=g(x)=remainder) Now `P(3)=3a+b=6" "(2)` Hence, b=0 and a=2 Hence, `g(x)=2ximpliesg(2)=4` |
|
| 20. |
The set of values of a for which the inequality, `x^2 + ax + a^2 + 6a < 0` is satisfied for all `x belongs (1, 2)` lies in the interval:A. (1,2)B. `[1,2]`C. `[-7,4]`D. none of these |
| Answer» Correct Answer - D | |
| 21. |
If `alpha`, `beta`, `gamma` are such that `alpha+beta+gamma=2`, `alpha^2+beta^2+gamma^2=6`, `alpha^3+beta^3+gamma^3=8`, then `alpha^4+beta^4+gamma^4` |
|
Answer» We have, `(alpha+beta+gamma)^(2)=alpha^(2)+beta^(2)+gamma^(2)+2(betagamma+gammaalpha+alphabeta)` `rArr 4 = 6 + 2 (betagamma+gammaalpha+alphabeta)` `rArr betagamma+gammaalpha+alphabeta = -1` Also, `alpha^(3) + beta^(3) + gamma^(3) - 3alphabeta gamma` `=(alpha+beta+gamma)(alpha^(2)+beta^(2)+gamma^(2))(betagamma+gammaalpha+alphabeta)` or " " 8 - 3 `alpha+beta+gamma = 2(6+1)` or `3alpha+beta+gamma = 8 - 14 = -6` or `alpha+beta+gamma = - 2` Now, `(alpha^(2)+beta^(2)+gamma^(2))^(2)=Sigmaalpha^(4)+2Sigmaalpha^(2)beta^(2)` `=Sigmaalpha^(4)+2[(Sigmaalphabeta)^(2)-2alphabetagamma(Sigmaalpha)]` or `=Sigmaalpha^(4)= 36 - 2 [(-1)^(2)-2(-2)(2)] = 18` |
|
| 22. |
The remainder obtained when the polynomial `x+x^(3)+x^(9)+x^(27)+x^(81)+x^(243)` is divided by `x^(2)-1` isA. `6x+1`B. `5x+1`C. `4x`D. `6x` |
|
Answer» Correct Answer - B `(b)` Put `x^(2)=1` in the given polynomial `x+x^(2)*x+(x^(2))^(4)*x+(x^(2))^(13)*x+(x^(2))^(40)*x+(x^(2))^(121)*x=5x+1` |
|
| 23. |
Consider an unknow polynomial which divided by `(x - 3)` and `(x-4)` leaves remainder 2 and 1, respectively. Let R(x) be the remainder when this polynomial is divided by `(x-3)(x-4)`. If equations `R(x) = x^(2)+ ax +1` has two distint real roots, then exhaustive values of a are.A. `(-2,2)`B. `(-oo,-2) uu(2,oo)`C. `(-2,oo)`D. all real numbers |
|
Answer» Correct Answer - 4 Let unknow polynomial be P(x). Let Q(x) and R(x) be the quatient and remainder, respectively, when it is divided by the `(x-3) (x-4)`. Then `P(x) = (x-3)(x-4) Q(x) + R(x)` Then we have `R(x) = ax + b` ` rArr P(x) = (x-3) (x-4)Q(x) + ax+ b` Given that `P(3) =2 ` and `P(4)=1` . Hence, `3a + b = 2 and 4a + b = 1` `rArr a = -1 and b= 5` `rArr R(x) = 5 -x` `5-x = x^(2) + ax + 1 ` `rArr x^(2) + (a + 1) x - 4 =0` Given that roots and reall distinct. Therefore, `D lt 0 rArr (a+1)^(2) + 16 lt 0` Which is true for all real a. |
|
| 24. |
A certain polynomial `P(x)x in R`when divided by k`x-a ,x-ba n dx-c`leaves remainders`a , b ,a n dc`, resepectively. Then find remainder when `P(x)`is divided by `(x-a)(x-b)(x-c)w h e r eab, c`are distinct. |
|
Answer» By Remainder theorem `f(a)=a,f(b)=b` and `f(c)=c` Let the quotient be `Q(x)` and remainder is `R(x)`. `:.f(x)=(x-a)(x-b)(x-c)Q(x)+R(x)` `:.f(a)=0+R(a)impliesR(a)=a` `f(b)=0+R(b)impliesR(b)=b` and `f(c)=0+R(c)` `impliesR(c)=c` ltbr. So the equation `R(x)-x=0` has there roots a,b and c. But its degree is atmost two. So `R(x)-x` must be zero polynomial (or identity) Hence `R(x)=x`. |
|
| 25. |
Consider an unknow polynomial which divided by `(x - 3)` and `(x-4)` leaves remainder 2 and 1, respectively. Let R(x) be the remainder when this polynomial is divided by `(x-3)(x-4)`. If `R(x) = px^(2) + (q-1) x + 6` has no distinct real roots and `p gt 0`, then the least value of `3p + q` isA. `-2`B. `2//3`C. `-1//3`D. none of these |
|
Answer» Correct Answer - 3 Let unknow polynomial be P(x). Let Q(x) and R(x) be the quatient and remainder, respectively, when it is divided by the `(x-3) (x-4)`. Then `P(x) = (x-3)(x-4) Q(x) + R(x)` Then we have `R(x) = ax + b` ` rArr P(x) = (x-3) (x-4)Q(x) + ax+ b` Given that `P(3) =2 ` and `P(4)=1` . Hence, `3a + b = 2 and 4a + b = 1` `rArr a = -1 and b= 5` `rArr R(x) = 5 -x` `- x + 5 = px^(2) + (q- 1)x + 6` `rArr px^(2) + qx + 1 = 0` Now, `p gt 0 ` and equation has no distinct real roots or equation has real and equal or imaginary roots. Then, `px^(2) + qx + 1 ge 0, AA x in R` `rArr f(3) ge rArr 9p + 3q + 1 ge 0 rArr 3p + q ge - 1//3` Hence, the least value of `3p + q` is `-1//3`. |
|
| 26. |
If a polynomial f (x) is divided by ` (x - 3) and (x - 4)` it leaves remainders as 7 and 12 respectively, then find the remainder when f (x) is divided by `(x-3)(x-4)`.A. `[-2,2]`B. `(-oo,-2- sqrt(3)] uu [-2+ sqrt(3),oo)`C. `(-oo,-2-3sqrt(3)] uu [-2+sqrt(3),oo)`D. none of these |
|
Answer» Correct Answer - 3 Let unknow polynomial be P(x). Let Q(x) and R(x) be the quatient and remainder, respectively, when it is divided by the `(x-3) (x-4)`. Then `P(x) = (x-3)(x-4) Q(x) + R(x)` Then we have `R(x) = ax + b` ` rArr P(x) = (x-3) (x-4)Q(x) + ax+ b` Given that `P(3) =2 ` and `P(4)=1` . Hence, `3a + b = 2 and 4a + b = 1` `rArr a = -1 and b= 5` `rArr R(x) = 5 -x` `f(x)= y = (-x + 5)/(x^(2) - 3x + 2)` `rArr yx^(2) + (1-3y) x + 2y - 5 = 0` Now, x is real, then `D ge 0` `rArr (1- 3y) ^(2) - 4y(2y - 5) ge 0` ` or y^(2) + 14y + 1 ge 0` ` y in (-oo, (-14- sqrt(192))/(2)]uu[(-14+sqrt(192))/(2),oo)` `(-oo, -7-4sqrt(3)] uu [ - 7 + 4sqrt(3),oo)` |
|
| 27. |
Let a `ne` 0 and P(x) be a polynomial of degree greater then 2.If P(x) leaves remianders a and a- when divided, respectively, by x + a and x - a, then find the remainder when P(x) is divided by `x^(2) - a^(2)`. |
|
Answer» According to the question P(-a) = and P(a) = - a. Let the remainder, when P(x) is divided by `x^(2)-a^(2)` be Ax + B. Then `P(x) = Q (x) (x^(2)-a^(2))` + Ax + B, where Q (x) is the quotient Putting x = a, we get P(a) = 0 + Aa + B or Aa + B = - a ...(1) Putting x = - a, we get `-Aa + B = a` ...(2) Solving (1) and (2) , we get B = 0 and A = - 1 Hence, the requried remainder = Ax + B = - x |
|
| 28. |
Given that `x^2+x-6`is a factor of `2x^4+x^3+b x+a+b-1,`find the value of `aa n dbdot` |
|
Answer» We have, ` x^(2) + x - 6 = (x+3)(x-2)` Let `f(X)=2x^(4)+x^(3)-ax^(2)+bx + a + b -1` Now, `f(-3) = 2 (-3)^(4)+ (-3)^(3)-3b + a+b - 1 = 0 ` or 134 - 8a -2b = 0 or 4a + b = 67 (1) Now `f(2)=2(2)^(4) + 2^(3)-a(2)^(2)+2b+ a + b - 1 = -0` or 39 - 3a + 3b = 0 or a - b = 13 (2) From (1) and (2), a = 16, b= 3. |
|
| 29. |
Let `a!=0a n dp(x)`be a polynomial of degree greater than 2. If `p(x)`leaves reminders `aa n d a`when divided respectively, by `+aa n dx-a ,`the remainder when `p(x)`is divided by `x^2-a^2`is`2x`b. `-2x`c. `x`d. ` x`A. `2x`B. `-2x`C. `x`D. `-x` |
| Answer» Correct Answer - D | |
| 30. |
If `x^2+p x+q=0a n dx^2+q x+p=0,(p!=q)`have a common roots, show that `p+q=0`. Also, show that their other roots are the roots of the equation `x^2+x+p q=0.` |
|
Answer» Given equation are `x^(2) + px + q = 0` (1) and `x^(2) + qx + p = 0` (2) Clearly for x = 1, from both equations we get `1+ p + q = 0` . Also for (1), product of roots = q Therefore, (1) has roots x = 1 and x = q For (2), product of roots = p Therefoure, (2) has roots x = 1 and x = p Thus, the equation having roots p and q is `x^(2) - (p + q) x + pq = 0` `therefore x^(2) + x + pq = 0 (as 1 + p + q = 0 )` |
|
| 31. |
The equations `x^2 +3x+5=0 `and `ax^2+bx+c=0` have a common root. If `a,b,c in N` then the least possible values of a+b+c is equal to |
|
Answer» Roots of `x^(2) + 3x + 5 = 0 ` are non- real. Thus, given equation will have two common roots. `rArr (a)/(1) = (b)/(3) = (c)/(5) = lambda` `rArr a + b + c = 9 lambda, a, b, c in` N Thus, minimum value of a + b + c is 9. |
|
| 32. |
If equations `x^2+a x+12=0. x^2+b x+15=0a n dx^2+(a+b)x+36=0,`have a common positive root, then find the values of `aa n dbdot` |
|
Answer» We have , `x^(2) + ax + 12 = 0` (1) `x^(2) + bx + 15 = 0` (2) `x^(2) + (a + b) x + 36 = 0` (3) Adding (1) and (2), we get `2x^(2)+ (a+ b) x + 27 = 0` Now subtracting it from the third equation, we get `x^(2) - 9 = 0 rArr x = 3, 3` Thus , the common positive root is 3, Hence, `9 + 12 + 3a = 0` `rArr a = - 7 and 9 + 3b + 15 = 0 ` `rArr b = -8` . |
|
| 33. |
If the equations `a x^2+b x+c=0a n dx^3+3x^2+3x+2=0`have two common roots, then`a=b=c`b. `a=b!=c`c. `a=-b=c`d. none of theseA. a = b = cB. `a = b ne c`C. `a = -b = c`D. none of these |
|
Answer» Correct Answer - 1 By observation x = -2 satisfies equation `x^(3) + 3x^(2) + 3x + 2 = 0` then we have `(x + 2)(x^(2) + x + 1) = 0` `x^(2) + x + 1 = 0` has nonreal roots. Since nonreal roots occur in conjugate pair, `x^(2) + x + 1 = 0` and `ax^(2) + bx + c = 0` are identical `rArr a = b = c` |
|
| 34. |
Find the value of `x` in `sqrt(x+2sqrt(x+2sqrt(x+2sqrt(3x))))=x.` |
|
Answer» Rewrite the given equation `sqrt(x+2sqrt(x+2sqrt(x+………….+2sqrt(x+2sqrt(x+2x)))))=x`……i On replacing the last letter x on the LHS of Eq i by the value of `x` expressed by Eq. i w get `x=ubrace(sqrt(x+2sqrt(x+2sqrt(x+……….+2sqrt(x+2sqrt(3x))))))_(2n"radical signs")` Further let us replace the last letter x by the same expression again and again yields. `:. `x=ubrace(sqrt(x+2sqrt(x+2sqrt(x+……….+2sqrt(x+2sqrt(3x))))))_(3n"radical signs")` `x=ubrace(sqrt(x+2sqrt(x+2sqrt(x+……….+2sqrt(x+2sqrt(3x))))))_(4n"radical signs")=......` We can write `x=sqrt(x+2sqrt(x+2sqrt(x+...)))` `=ubrace(lim_(Ntooo)sqrt(x+2sqrt(x+2sqrt(x+……….+2sqrt(x+2sqrt(3x))))))_(N"radical signs")` it follows that `x=sqrt(x+2sqrt(x+2sqrt(x+....)))` `=sqrt(x+2(sqrt(x+2sqrt(x+...))))=sqrt((x+2x))` Hence `x^(2)=x+2x` `impliesx^(2)-3x=0` `:.x=0,3` |
|
| 35. |
The number of values of `a`for which equations `x^3+a x+1=0 and x^4+a x^2+1=0`have a common root is |
|
Answer» Correct Answer - 2 Given equation are `x^(3) + ax + 1 = 0` or `x^(4) + ax^(2) + x = 0 " "(1)` and `x^(4) + ax^(2) + 1 = 0" "(2)` From (1) - (2), we get x = 1. Thus, x = 1 is the common root Hence, `1 + a + 1 = 0 rArr a = -2` |
|
| 36. |
If the equation `x^(2)-4x-(3k-1)|x-2|-2k+8=0,kinR`, has exaclty three distinct solutions, then k is eaual to _____. |
|
Answer» Correct Answer - 2 Given equation is `(x-2)^(2)-4-(3k-1)|x-2|-2k+8=0` or `r^(2)-(3k-1)t-2k+4=0`, where `t=|x-2|` For exactly 3 distinct solutions, one of the roots is zero and other root is positive. `:.k=2` |
|
| 37. |
Statement 1 Roots of `x^(2)-2sqrt(3)x-46=0` are rational. Statement 2 Discriminant of `x^(2)-2sqrt(3)x-46=0` isa perfect square.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» In `ax^(2)+bx+c=0,a,b,c epsilonq` [here Q is the set of rational number] If `Dgt0` and is a perfect square, then roots are real, distinct and rational. But here `2sqrt(3)!inQ` `:.` Roots are not rational. Here roots are `(2sqrt(3)+-sqrt((12+184)))/2` i.e. `sqrt(3)+-7` [irrational ] But `D=12+184=196=(14)^(2)` `:.` Statement -1 is false and statement -2 is true. |
|
| 38. |
If the expression `x^2+2(a+b+c)x+3(bc+ca+ab)` is a perfect square thenA. a = b = cB. `a = pm b = pm c`C. `a = b ne c`D. none of these |
|
Answer» Correct Answer - 1 Given quadratic expression is `x^(2) + 2(a + b + c)x + 3(bc + ca + ab).` This quadratic expression will be a perfect square if the discriminant of its corresponding equation is zero. Hence, `4(a + b + c)^(2) - 4 xx 3(bc + ca + ab) = 0` or `(a + b + c)^(2) - 3(bc + ca + ab) = 0` or `a^(2) + b + c^(2) + 2ab + 2bc + 2ca - 3(bc + ca + ab) = 0` or `a^(2) + b^(2) + c^(2) - ab bc - ca = 0` or `1/2[2a^(2) + 2b^(2) + 2c^(2) - 2ab - 2bc - 2ca] = 0` or `1/2[(a^(2) + b^(2) - 2ab) + (b^(2) + c^(2) - 2bc) + (c^(2) + a^(2) - 2ca)] = 0` or `1/2[(a-b)^(2) + (b - c)^(2) + (c - a)^(2)] = 0` which is possible only when `(a - b)^(2) = 0, (b - c)^(2) = 0` and `(c - a)^(2) = 0,` i.e., a = b = c. |
|
| 39. |
The number of value of `k`for which `[x^2-(k-2)x+k^2]xx""[x^2+k x+(2k-1)]`is a perfect square is`2`b. `1`c. `0`d. none of theseA. 2B. 1C. 0D. none of these |
|
Answer» Correct Answer - 2 For given situation, `x^(2) - (k - 2)x + k^(2) = 0 and x^(2) + kx + 2k - 1 = 0` should have both roots common or each should have equal roots. If both roots are common, then `1/1 = (-(k - 1))/(k) = (k^(2))/(2k - 1)` `rArr k = -k + 2 and 2k - 1 = k^(2) rArr k = 1` If both the equations have equal roots, then `(k - 2)^(2) - 4k^(2) = 0 and k^(2) - 4(2k - 1) = 0` `rArr (3k - 2) (-k - 2) = 0 and k^(2) - 8k + 4 = 0` There is no common value of k. Therefore, k = 1 is the only possible value. |
|
| 40. |
Let `a gt 0, b gt 0 and c lt0.` Then, both the roots of the equation `ax^(2) +bx+c=0`A. are real and nagativeB. have negative real partsC. have positive real partsD. None of the above |
|
Answer» Correct Answer - B Since, `a, b, c gt 0 an ax^(2)+bx+c=0` `impliesx=(-b)/(2a)+-(sqrt(b^(2)-4ac))/(2a)` Case I When `b^(2)-4acgt0` `impliesx=(-b)/(2a)-(sqrt(b^(2)-4ac))/(2a)and (-b)/(2a)+(sqrt(b^(2)-4ac))/(2a)` both roots, are negative. Case II When `b^(2)-4ac=0` `impliesx(-b)/(2a),` i.e. both roots are equal and negative Case III When `b^(2)-4ac lt0` `impliesx=(-b)/(2a)+-i(sqrt(4ac-b^(2)))/(2a)` have negative real part. `therefore` Form above discussion, both roots have negative real parts |
|
| 41. |
For real solution of equation `3sqrt(x+3p+1)-3sqrt(x)=1`, we haveA. `p ge 1//4`B. `p ge -1//4`C. `p ge 1//3`D. `p ge -1//3` |
|
Answer» Correct Answer - B `(b)` `3sqrt(x+3p+1)=3sqrt(x)+1` Let `3sqrt(x)=h` `implies3sqrt(x+3p+1)=h+1` `impliesx+3p+1=h^(3)+3h^(2)+3h+1` `implies h^(3)+3p+1=h^(3)+3h^(2)+3h+1` `implies3h^(2)+3h-3p=0` `h^(2)+h-p=0` For real roots ` D ge 0` `implies 1+4p ge 0` `implies p ge -1//4` |
|
| 42. |
If the equation x^4 -λx^2 +9 has only two real roods, then the set of values of `λ` isA. `(-oo,-6)`B. `(-6,6)`C. `{6}`D. `phi` |
|
Answer» Correct Answer - D From above discussion we have for `lamdain(-oo,6)uu(6,oo)` equation has either four distinct real or no real roots. For `lamda=6`, both roots are `pmsqrt3` Alternate method: `x^(4)-lamdax^(2)+9=0` or `x^(2)+(9)/(x^(2))=lamda` Now `(x-(3)/(x))^(2)=lamda-6` For above equation (i) four distinct real roots if `lamda-6gt0orlamdagt6`. (ii) no real roots if `lamda-6lt0orlamdalt6` (iii) real and equal repeating roots if `lamda=6` |
|
| 43. |
The sum of all real values of X satisfying the equation `(x^2-5x+5)^(x^2 + 4x -60) = 1` is:A. 6B. 5C. 3D. `-4` |
|
Answer» Correct Answer - C `(x^(2)-5x+5)^(x^(2)+4x-60)=1` Case I `x^(2)-5x+5=1` and `x^(2)+4x-60` can be any real number `impliesx=1,4` Case II `x^(2)-5x+c=-1` and `x^(2)+4x-60` has to be an even number ltbr gt`impliesx=2,3` For `x=2,3` For `x=3, x^(2)+4x-60` is odd `:.!=3` Hence `x=2` Case III `x^(2)-5x+5` can be any real number and `x^(2)+4x-60=0` `impliesx=-10,6` `implies` Sum of all values o `x=1+4+2-10+6=3` |
|
| 44. |
(B) (2, 9/4) If two roots of the equation `(a-1) (x^2 + x + 1 )^2-(a + 1) (x^4 + x^2 + 1) = 0` are real and distinct, then a lies in the intervalA. `(- oo, 3]`B. `(-oo, -2) uu (2, oo)`C. `[-2, 2]`D. `[-3, oo)` |
|
Answer» Correct Answer - 2 `x^(4) + x^(2) + 1 = (x^(2) + 1)^(2) - x^(2)` `= (x^(2) + x + 1)(x^(2) - x + 1)` `x^(2) + x + 1 = (x + 1/2)^(2) + 3/4 ne 0 AA x` Therefore, we can cancel this factor and we get `(a - 1)(x^(2) - x + 1) = (a + 1)(x^(2) - x + 1)` or `x^(2) - ax + 1 = 0` It has real and distinct roots if `D = a^(2) - 4 gt 0`. |
|
| 45. |
If `ax^(2) + bx + c = 0 ` has imaginary roots and a - b + c ` gt` 0 . then the set of point (x, y) satisfying the equation `|a (x^(2) + (y)/(a)) + (b + 1) x + c| = |ax^(2) + bx + c|+ |x + y|` of the region in the xy- plane which isA. on or above the bisector of I and III quadrantB. on or above the bisector of II and IV quadentC. on or below the bisector of I and III quadrantD. on or below the bisector of II and IV quadrant . |
|
Answer» Correct Answer - 2 `|a(x^(2) + (y)/(a)) + (b + 1)x + c |= |ax^(2) + bx + c| + |x + y|` `rArr [ (ax^(2)= bx + c) + (x + y)| = |ax^(2) + bx + c |+ |x + y|` (1) Now ` f(x) = ax^(2) + bx + c = 0 ` has imaginary roots and ` a - b + c gt 0 or f(-1) gt 0 ` `rArr f(x) = ax^(2) + bx + c gt 0` for all real values of x `rArr x + y ge 0 ` `rArr (x , y)` lies on or above the bisector of II and Iv quadrants . |
|
| 46. |
If one root is square of the other root of the equation `x^2+p x+q=0`, then the relation between `pa n dq`is (2004, 1M)`p^3-(3p-1)q+q^2=0``p^3-q(3p+1)+q^2=0``p^3+q(3p-1)+q^2=0``p^3+q(3p+1)+q^2=0`A. `p^(3)-q(3p-1)+q^(2)=0`B. `p^(3)-q(3p+1)+q^(2)=0`C. `p^(3)+q(3p-1)+q^(2)=0`D. `p^(3)+q(3p+1)+q^(2)=0` |
|
Answer» Correct Answer - A Let the roots of `x^(2)+px+q=0` be `alphaand alpha^(2).` `impliesalpha+alpha^(2)=-pand alpha^(3)=q` `impliesalpha(alpha+1)=-p` `impliesalpha^(3){alpha^(3)+1+3alpha(alpha+1)}=-p^(3)["cubing both sides"]` `impliesq(q+1-3p)=-p^(3)` `impliesp^(3)-(3p-1)q+q^(2)=0` |
|
| 47. |
The sum of values of `x`satisfying the equation `(31+8sqrt(15))^x^(2-3)+1=(32+8sqrt(15))^x^(2-3)`is`3`b. `0`c. `2`d. none of theseA. 3B. 0C. 2D. none of these |
| Answer» Correct Answer - B | |
| 48. |
If the equation x^4-λx^2+9=0 has four real and distinct roots, then `lamda` lies in the intervalA. `(-oo,-6)uu(6,oo)`B. `(0,oo)`C. `(6,oo)`D. `(-oo,-6)` |
|
Answer» Correct Answer - C `x^(4)-lamdax^(2)+9=0` Let `x^(2)=tge0` `:.f(t)=t^(2)-lamdat+9=0" "(1)` If given equation has four real and distinct roods, then both roots of (1) must be positive `:.Dgt0implieslamda^(2)-36gt0" "(2)` `(-b)/(2a)gt0implies(lamda)/(2)gt0orlamdagt0" "(3)` `f(0)gtor9gt0" "(4)` `:.lambda in(-6,oo)" "["from(2),(3),(4)"]`. |
|
| 49. |
The set of all `x` satisfying `3^(2x)-3x^(x)-6gt0` is given byA. `0ltxlt1`B. `xgt1`C. `xgt3^(-2)`D. none of these |
| Answer» Correct Answer - B | |
| 50. |
Let `P(x)=5/4+6x-9x^2a n dQ(y)=-4y^2+4y+(13)/2dot`if there exists unique pair of real numbers `(x , y)`such that `P(x)Q(y)=20`, then the value of `(6x+10 y)`is _____. |
|
Answer» Correct Answer - 3 We have `P(x)=(5)/(3)-6 x-9x^(2)=-(3x+1)^(2)=-(3x+1)^(2)+(8)/(3)` `impliesP_("max")=(8)/(3)` Similarly, `Q(y)=-4y^(2)+4y+(13)/(2)=-(2y-1)^(2)+(15)/(2)` `Q_(max)=(15)/(2)` Now, `P_(max)xxQ_("max)=(8)/(3)xx(15)/(2)=20` So, `(x,y)-=(-(1)/(3),(1)/(2))` Hence, `6x+10y=6((-1)/(3))+10((1)/(2))=-2+5=3` |
|