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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.
| 301. |
If each pair of the three equations `x^(2)+ax+b=0, x^(2)+cx+d=0` and `x^(2)+ex+f=0` has exactly one root in common then show that `(a+c+e)^(2)=4(ac)+ce+ea-b-d-f` |
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Answer» Given equations are `x^(2)+ax+b=0`……….i `x^(2)+cx+d=0`…………..ii `x^(2)+ex+f=0`………….iii Let `alpha, beta` be the roots of Eq. (i) `beta, gamma` be the roots of Eq. ii and `gamma, delta` be the roots of Eq. iii then `alpha+beta=-a, alpha beta=b`.........iv ltbr `beta+gamma-c, beta gamma=d`.........v `gamma+alpha=-e,gamma alpha=f`...........vi `:.LHS=(a+c+e)^(2)=(-alpha-beta-beta-gamma-gamma-alpha)^(2)` [from eqs iv , v and vi] `=4(alpha+beta+gamma)^(2)`........vii `RHS=4(ac+ce+ea-b-d-f)` `=4{(alpha+beta)(beta+gamma)+beta+gamma)(gamma+alpha)+(gamma+alpha)` `(alpha+beta)-alpha beta-beta gamma-gamma alpha)}` [from eqs iv, v and vi] `=4(alpha^(2)+beta^(2)+gamma^(2)+2alpha beta+2beta tamma+2 gamma alpha)` `=4(alpha+beta+gamma)^(2)` ...........iii From Eq vii and viii then we get `(a+c+e)^(2)=4(ac+ce+cea-b-d-f)` |
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| 302. |
If `x`is real and `(x^2+2x+c)//(x^2+4x+3c)`can take all real values, of then show that `0lt=clt=1.` |
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Answer» Correct Answer - ` a in ( - (11)/(3), - 2 sqrt(2)]` Let `y = (x^(2) + 2x + c ) /(x^(2) + 4x + 3c)` or ` (y - 1 )x^(2) + (4y - 2) x + 3cy - c = 0` Now , x is real . Hence, `D = (4y - 2)^(2) - 4 (y - 1) (3cy - c) ge 0 AA y in ` R or ` (2y - 1 )^(2) - (y - 1)(3cy - c) ge 0 , AA y in ` R or ` (4 - 3c)y^(2) + (-4 + c + 3c) y + 1 - c ge 0, AA y in ` R. ` or 4-3c gt 0 and (4c - 4)^(2) - 4 (4 - 3c)(1 - c) le 0` or ` c lt (4)/(3) and 4 (c - 1)^(2) - (4-3c)(1 -c) le 0 ` or ` c lt (4)/(3) and (c - 1)xx (4c - 4 + 4 - 3c)le 0` or ` c lt (4)/(3) and (c - 1)(c)le 0` or ` c lt (4)/(3) and - le cle 1` or ` 0 le c le 1` . |
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| 303. |
Solve `(x^2-2x-3)/(x+1)=0.` |
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Answer» Equation `(x^(2)-2x-3)/(x + 1) = 0` is solveble over R - {-1} Now, `(x^(2)-2x-3)/(x + 1) = 0` or `x^(2)-2x -3 = 0 or (x - 3)(x+1)= = 0` or `x = 3 only (as x `in` R - {-1})` |
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| 304. |
Consider the following figure. Answer the following questions (i) What are the roots of the f(x) = 0? (ii) What are the roots of the f(x) = 4? (iii)What are the roots of the f(x) = g (x)? |
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Answer» (i) For the roots of the equation f(x) = 0, we check the points of intersection of y = f(x) with x -axis at x = - 1 and x = 2, which are roots of the equation f(x) = 0 . (ii) For the roors of the equation f(x) = 4, we check the points of intersection of the graphs y = f(x) and y = 4 ( a line parallel to x-axis at height 4 units above it) The x -values of the poinst of intersection of graphs is x = -2 and x = 3 , which are roots of the equation f(x) = 4. (iii) For the roots of the equation f(x) = g(x), we check the points of intersection of the graphs y = f(x) and y = g(x). The x -velues of the points of intersection of graphs are x = -2 and x = 4, which are roots of the equation f(x) = g(x) |
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| 305. |
Let `alpha,beta` be two real numbers satisfying the following relations `alpha^2+beta^2=5, 3(alpha^5+beta^5)=11(alpha^3+beta^3)1.` Possible value of `alpha beta` isA. `2`B. `-(10)/(3)`C. `-2`D. `(10)/(3)` |
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Answer» Correct Answer - A `alpha^(2)+beta^(2)=5` `3(alpha^(5)+beta^(5))=11(alpha^(3)+beta^(3))` `(alpha^(5)+beta^(5))/(alpha^(3)+beta^(3))=(11)/(3)` ` :. ((alpha^(3)+beta^(3))(alpha^(2)+beta^(2))-(alpha^(2)beta^(2)(alpha+beta)))/(alpha^(3)+beta^(3))=(11)/(3)` `:. alpha^(2)+beta^(2)-(alpha^(2)beta^(2)(alpha+beta))/((alpha+beta)(alpha^(2)+beta^(2)-alphabeta))=(11)/(3)` ` :. 5-(alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)` ltbrlt `:. (25-5alphabeta-alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)` Let `alphabeta=t` `(25-5t-t^(2))/(5-t)=(11)/(3)` `75-15t-3t^(2)=55-11 t` `75-15t-3t^(2)-55+11t=0` `-3t^(2)-4t+20=0` `(t-2)(3t+10)=0` ` :. t=2` or `(-10)/(3)` So `alpha beta=2`, `alphabeta=(-10)/(3)` If `alphabeta=2`, `alpha^(2)+beta^(2)=(alpha+beta)^(2)-2alphabeta` `:.5=(alpha+beta)^(2)-2xx2` `(alpha+beta)^(2)=9` `alpha+beta=+-3` for `alphabeta=(-10)/(3)`, `(alpha+beta)^(2) lt 0` `implies x^(2) +- 3x+2=0` |
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| 306. |
Let `alpha`, `beta` be two real numbers satisfying the following relations `alpha^(2)+beta^(2)=5`, `3(alpha^(5)+beta^(5))=11(alpha^(3)+beta^(3))` Possible value of `alpha + beta` isA. `+-2`B. `+-3`C. `+-1`D. `+-sqrt(3)` |
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Answer» Correct Answer - B `alpha^(2)+beta^(2)=5` `3(alpha^(5)+beta^(5))=11(alpha^(3)+beta^(3))` `(alpha^(5)+beta^(5))/(alpha^(3)+beta^(3))=(11)/(3)` ` :. ((alpha^(3)+beta^(3))(alpha^(2)+beta^(2))-(alpha^(2)beta^(2)(alpha+beta)))/(alpha^(3)+beta^(3))=(11)/(3)` `:. alpha^(2)+beta^(2)-(alpha^(2)beta^(2)(alpha+beta))/((alpha+beta)(alpha^(2)+beta^(2)-alphabeta))=(11)/(3)` ` :. 5-(alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)` ltbrlt `:. (25-5alphabeta-alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)` Let `alphabeta=t` `(25-5t-t^(2))/(5-t)=(11)/(3)` `75-15t-3t^(2)=55-11 t` `75-15t-3t^(2)-55+11t=0` `-3t^(2)-4t+20=0` `(t-2)(3t+10)=0` ` :. t=2` or `(-10)/(3)` So `alpha beta=2`, `alphabeta=(-10)/(3)` If `alphabeta=2`, `alpha^(2)+beta^(2)=(alpha+beta)^(2)-2alphabeta` `:.5=(alpha+beta)^(2)-2xx2` `(alpha+beta)^(2)=9` `alpha+beta=+-3` for `alphabeta=(-10)/(3)`, `(alpha+beta)^(2) lt 0` `implies x^(2) +- 3x+2=0` |
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| 307. |
If `a ,b ,c`be the sides of ` A B C`and equations `a x 62+b x+c=0a n d5x^2+12+13=0`have a common root, then find `/_Cdot` |
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Answer» Correct Answer - `angle C = 90^(@)` Roots of equation `5x^(2) + 12 x + 13 = 0` are imaginary Hence, both equation have same roots `therefore a : b:c = 5 :12: 13` `therefore a = 5k , b = 12k , c = 13k ` Here `c^(2) = a^(2) + b^(2) ` `rArr angle C = 90^(@)` |
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| 308. |
If m is chosen in the quadratic equation `(m^(2)+1)x^(2)-3x+(m^(2)+1)^(2)=0` such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is(A) `10sqrt5`(B) `8sqrt5`(C) `8sqrt3`(D) `4sqrt3`A. `10sqrt5`B. `5sqrt5`C. `8sqrt3`D. `4sqrt3` |
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Answer» Correct Answer - B Given quadratic equation is `(m^(2)+1)x^(2)-3x(m^(2)+1)^(2)=0" "…(i)` Let the roots of quadratic Eq. (1)mn are `alpha and beta.` so `alpha+beta(3)/(m^(2)+1)and alphabeta=m^(2)+1` According to the question, the sum of roots is greatest and it is possible only when `(m^(2)+1)` is minimu and minimum value of `m^(2)+1=1,` when `m=0"` `thereforealpha +beta=3` and `alpha beta=1, as m =0` Now, the absolute difference of the cubes of roots `=|alpha^(2)-beta^(3)|` `=|alpha-beta||alpha^(2)+alphabeta|` `=sqrt((alpha+beta)^(2)-4alphabeta)|(alpha+beta)^(2)-alpha beta|` `=sqrt(9-4)|9-1|=8sqrt5` |
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| 309. |
If `a + b + c = 0 and a^(2) + b^(2) + c^(3) = 4,` them find the value of `a^(4) + b^(4) +c^(4)`. |
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Answer» Correct Answer - `a^(4) + b^(4) +c^(4) = 8` `(a + b + c)^(2) = 0` `rArr a^(2) + b^(2) + c^(2) + 2(ab + bc + ca) = 0` `rArr ab + bc + ca = 2` On squaring, we get `a^(2) b^(2) + b^(2)c^(2)+ c^(2)a^(2)+ 2(ab^(2)c + 2a^(2) bc + 2bac^(2) = 4` `rArr a^(2) b^(2) + b^(2)c^(2)+ c^(2)a^(2)+ 2abc (a + b + c) = 4` `rArr a^(2) b^(2) + b^(2)c^(2)+ c^(2)a^(2) = 4` Now `a^(2) + b^(2) + c^(2) = 4` On squaring, we get `a^(4)+ b^(4)+ c^(4)+2(a^(2) b^(2) + b^(2)c^(2)+ c^(2)a^(2) = 16` `rArr a^(4)+ b^(4)+ c^(4)= 8` |
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| 310. |
Let `a,binRandabne1."If "6a^(2)+20a+15=0and15b^(2)+20b+6=0` then the value of `(4030b^(3))/(ab^(2)-9(ab+1)^(3))` is _____. |
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Answer» Correct Answer - 12 From the given relations, we can say that and `(1)/(b)` are the roots of the equation `6x^(2)+20x+15=0`. `:.a+(1)/(b)=(10)/(3)and(a)/(b)=(5)/(2)` `(b^(3))/(ab^(2)-9(ab+1)^(3))=(1)/(a.(1)/(b)-9(a+(1)/(b))^(3))` `=(1)/((5)/(2)+9.(1000)/(27))=(6)/(2015)` |
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| 311. |
Which of the following pair of graphs intersect ? (i) y = `x^(2) - x ` and y = 1 (ii) y = `x^(2) - 2x + 3` and y = sin x (iii) = `x^(2) - x + 1` and y = x - 4 |
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Answer» Graphs `y = x^(2) - x` and y = 1 intersect when `x^(2) `- x and y = 1 intersect when `x^(2)- x = 1 or x^(2) = x -1 = 0 ` Clearly, this equation has real solution. So graphs intersect. Graphs `y = x^(2) - 2x + 3` and y = sin x intersect when `x^(2)-2x + 3 = sin x(x-1)^(2)+ 2 = sin` x, which is not possible as L.H.S. has least value 2, whereas R.H.S. has maximun value 1. So, Graphs do not intersect. (iii) Graphs `y = x^(2)-x + 1` and y = x -4` intersect if `x^(2)-x + 1 = x - 4 or x^(2) = 2x + 5 = 0 or (x -1)^(2)+ 4 = 0`. Clearly. this equation has non-real roots . So, graphs do not intersect |
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| 312. |
Solve `(x^2+3x+2)/(x^2-6x-7)=0.` |
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Answer» Correct Answer - `x = -2` `(x^(2) _ 3x + 2)/(x^(2) - 6x - 7) = 0 or((x+ 1)(x+2))/((x-7)(x+ 1)) = 0` is solvable over R - {7, -1} Hence, from given equation, x = -2, which is the only roots of the equation. |
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| 313. |
Let `a`is a real number satisfying `a^3+1/(a^3)=18`. Then the value of `a^4+1/(a^4)-39`is ____. |
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Answer» Correct Answer - `a^(4) + (1)/(a^(4)) = 47` Let ` a + (1)/(a) = t` `therefore t^(3) - 3t - 18 = 0` t - 3 satisfies this equation `therefore (t - 3 ) (t^(2) _ 3t + 6 ) = 0` Clearly, t = 3 is the only soultion `therefore a ^(2) + (1)/(a) = 3` `rArr a^(2) + (1)/(a^(2)) = 7` `rArr a^(4) + (1)/(a^(4)) = 47` . |
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| 314. |
If `alpha and beta` are roots of the equation `x^(2)-2x+2=0,` then the least value of n for which `((alpha)/beta)^(n)=1` isA. 2B. 5C. 4D. 3 |
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Answer» Correct Answer - C Given, `alpha and beta` are the roots of the quadratic equation `x^(2)-2x+2=0` `implies(x-1)^(2)+1=0` `implies(x-1)^(2)=-1` `implies x-a=+-i" "["where i"=sqrt-1]` `impliesx=(1+i)or (1-i)` Clearly, if `alpha =1 +i, then beta=1-i` According to the equation `((alpha)/(beta))^(n)=1` `implies((1+i)/(1-i))^(n)=1` `implies(((1+i)(1+i))/((1-i)(1-i)))^(n)=1" "["by rationalization"]` `implies((1+i^(2)+2i)/(1-i^(2)))=1implies((2i)/(2))^(n)=1impliesi^(n)=1` So, minimum value of n is `4." "[becausei^(4)=1]` |
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| 315. |
Graph of `y = f(x)` is as shown in the following figure. Find the roots of the following equations `f(x) = 0` `f(x) = 4` ` f(x) = x + 2` |
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Answer» Correct Answer - (a) `x = -2 and x = 1` (b) `x = - 3 and x = 2` (c) `x = 2 and x = -2` Graphs of `y = f(x) ` intersect x-axis at x = -2 and x = 1, Which are roots of the equation `(x) = 0` (b) Graph of y = f(x) intersect the line y = 4 intersect at points whose x-value are - 3 and 2, which are the roots of the equation f(x) = 4 (c) For the roots of the equation f(x) = x + 2, we draw the graph of `y = x = 2`, which passes through the points (2, 4) and (-2, 0) which lies on the graph of y = f(x) also. so, roots of the equation f(x) = x + 2 are x = 2 and x = -2` |
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| 316. |
Prove that graphs of `y=x^2+2a n dy=3x-4`never intersect. |
| Answer» Graphs of `y = x^(3) + 2 and y = 3x + 4` intersect when `x^(2) + 2 = 3x - 4` or `x^(2) - 3x + 6 = 0` . But this equation has no real roots, hence graphs never intersect. | |
| 317. |
Let `x ,y ,z in R`such that `x_y+z=6a n xx y+y z+z x=7.`Then find the range of values of `x ,y ,a n dzdot` |
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Answer» Correct Answer - ` a in (- infty, - 1//2 )sup [ 4, nfty)` x,y,z,`in` R ` x + y + z = 6 and xy + yz + zx = 7` `rArr y (6 - y - z) + yz + z (6 - y - z) = 7` `rArr - y^(2) + (6 - z + z - z) y + z (6 -z) - 7 = 0` `rArr y^(2) + (z - 6 )y + 7 + z (z - 6) = 0` Now, y is real. Therefore , `(z - 6)^(2) - 4 [7 + z (z - 6)] ge 0 ` or ` 3z^(2) - 12z - 8 le 0` or `(12 - sqrt(144 + 96))/(6) le zle (12 + sqrt(144 + 96))/(6)` or ` (6 - 2 sqrt(15))/(3) le z le (6 + 2 sqrt(15))/(3)`. From symmetry, x and y have same range. |
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| 318. |
If two roots of `x^3-a x^2+b x-c=0`are equal inn magnitude but opposite in signs, then prove that `a b=cdot` |
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Answer» Correct Answer - `x^(3) - 64 = 0` Let the roots be ` x _(1), - x_(1), x_(2)`. Then ` x_(1) - x_(1) + x_(2) =a ` Hence, x =a is a root of the given equation `therefore a^(3) - a^(3) + ab - c = 0` `rArr ab = c ` |
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| 319. |
Let `f(x)=(1+m)x^2-2(1+3m)x+4(1+2m).` Numbe of interval values of `m` for which given qudratic expression is always positive is (A) `8` (B) `7` (C) `8` (D) `9`A. 6B. 8C. 7D. 3 |
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Answer» Correct Answer - C The quadratic expression `ax^(2)+bx+c,x in R` is always positive, if `a gt 0 and Dlt0.` So, the quadratic expression `(1+2m)x^(2)-2(1+3m)^(2)-4(2m+1)(1+m),x inR` will be always positive, if `1+2n gt 0 " "…(i)` and `D=4(1+3m)^(2)-4(2m+m)lt0" "(ii)` Form inequality Eq. (1), we get `m gt-1/2" "...(iii)` From inequatity Eq. (ii), we get `1+9m^(2)+6m-(2m^(2)+3m+1)lt0` `impliesm^(2)-6m-3lt0` `implies[m-(3+sqrt12)][m-(3sqrt12)]lt0` `"["becausem^(2)-6m-3=0impliesm=(6+-sqrt(36+12))/(2)=3+-sqrt12"]"` `implies3-sqrt12klt3+sqrt12" "...(iv)` From inequalities Eqs. (iii) and (iv), the integral values of m are `0,1,2,3,4,5,6` Hence, the number of integral values of m is 7. |
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| 320. |
In how many points the line `y + 14 = 0` cuts the curve whose equation is `x(x^(2) + x + 1) + y = 0 ? ` |
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Answer» Correct Answer - one When graphs of `y - 14 and y = -x(x^(2) + x + 1) = 0` intersect, we have `- (x^(3) + x^(2) + x ) = - 14 or x^(3) + x^(3) + x - 14 = 0` Now x = 2 satisfies the equation,then one root is x = 2. Dividing `x^(2) + x^(2) + x + 7) = 0` we have `(x - 2) (x^(2)+ 3x + 7 ) = 0` `rArr x = 2 or x^(2) + 3x + 7 = 0` Now `x ^(2) + 3x + 7 = 0` has non-real roots, Hence, graphs cut in only one real point. |
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| 321. |
What is the minimum height of any point on the curve `y=x^2+6x-5`above the x-axisdv? |
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Answer» `y = -x^(2) + 6x - 5` `=4- (x -3)^(2)` Now `(x - 3)^(2) ge 0` for all real x, Hence, the maximum value of `y(or - x^(2) + 6x - 5) ` is 4, which is the maximum height of the graph above x-axis. |
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| 322. |
If `x-c`is a factor of order `m`of the polynomial `f(x)`of degree `n(1 |
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Answer» From the given information we have `f(x) = (x - c).^(m)g(x)` where g(x) is a polynomial of degree n - m . Then x = c is a common root for equation `f(x) = 0, f^(1) (x) = 0,f^(2) = 0, …, f^(m-1) (x) = 0` represents rth derivative of `f(x) w.r.t.x.` |
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| 323. |
If `a x^2+b x+c=0`has imaginary roots and `a+b+c |
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Answer» Correct Answer - No such a exists . `ax(2) + bx + c = 0` has imaginary roots . Hence . ` ax^(2) + bx + c lt AA x in R, if a gt 0` But, given a + c `lt ` b or ` a - b + c 0 or f (-1) lt 0` `rArr f(x) = ax^(2) + bx + c lt 0 AA x in R` `rArr f(-2) = 4a - 2b + c lt 0` `rArr 4a + c lt 2b` . |
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| 324. |
If the sum of the roots of an equation is 2 and the sum of their cubesis 98, then find the equation. |
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Answer» Correct Answer - `x^(2) - 2x - 15 = 0` Let the roots be ` alpha and beta ` . Then , ` alpha + beta = 2 and alpha^(3) + beta^(3) = 98` Now, ` alpha^(3) + beta^(3) = (alpha + beta ) (alpha^(2) - alpha beta + beta^(2))` `rArr 98 = 2 [(alpha + beta)^(2) - 3alpha beta]` or ` 49 = (4 - 3 alpha beta)` or ` alpha beta = - 15` Thus, the equation is `x^(2) - 2x - 15 = 0` |
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| 325. |
If `tanthetaa n dsectheta`are the roots of `a x^2+b x+c=0,`then prove that `a^4=b k^2(4a c-b^2)dot` |
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Answer» `tan theta + sec theta = - (b)/(a)` (1) ` tan theta sec theta = (c)/(a)` (2) Now, `sec^(2) theta - tan^(2) theta = 1` `rArr sec theta - tan theta = - (a)/(b) ` (3) From (1) and (3) . `rArr sec theta = - ((a^(2) + b^(2)))/(2ab) and tan theta = ((a^(2) - b^(2)))/(2ab)` substituting these values in Eq. (2), we have `((a^(2) + b^(2))(b^(2) - a^(2)))/(4a^(2)b^(2)) = (c)/(a)` or `b^(4) - a^(4) = 4acb ^(2)` or `a^(2) = b^(2)(b^(2) - 4ac)` |
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| 326. |
If the roots of the equation `a x^2+b x+c=0`are of the form `(k+1)//ka n d(k+2)//(k+1),t h e n(a+b+c)^2`is equal to`2b^2-a c`b. `a 62`c. `b^2-4a c`d. `b^2-2a c`A. `2b^(2) - ac`B. `a^(2)`C. `b^(2) - 4ac`D. `b^(2) - 2ac` |
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Answer» Correct Answer - 3 We have `(k + 1)/(k) + (k + 2)/(k + 1) = - b/a" "(1)` and `(k + 1)/(k) (k + 2)/(k + 1) = c/a` `rArr (k + 2)/(k) = c/a` or `2/k = c/a - 1 = (c - a)/(a)` or `k = (2a)/(c - a)" "(2)` Now, eliminate k. Putting the value of k in Eq. (1), we get `(c + a)/(2a) + (2c)/(c + a) = -b/a` or `(c + a)^(2) + 4ac = -2b (a + c)` or `(a + c)^(2) + 2b (a + c) = -4ac` Adding `b^(2)` to both sides, we have `(a + b + c)^(2) = b^(2) - 4ac` |
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| 327. |
If the roots of the equation `a(b-c)x^2+b(c-a)x+c(a-b)=0`are equal, show that `2//b=1//a+1//cdot` |
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Answer» Since the roots of the given equations are equal, its discriminant is zero, i. e., `b^(2)(c - a)^(2) - 4a(b - c) c(a - b) = 0` or `b^(2) (c^(2) + a^(2) - 2ac) - 4ac (ba - ca -b^(2) + bc) = 0` or `a^(2)b^(2) +b^(2)c^(2) + 4a^(2) c^(2) + 2b^(2)ac - 4a^(2) bc- 4abc^(2) = 0` `(ab + bc- 2ac)^(2) = 0 ` or `ab + bc -2ac = 0` or `ab + bc = 2ac` or `(1)/(c)+(1)/(a)=(2)/(b)` [dividing both sides byu abc] or `(2)/(b)=(1)/(a)+(1)/(c)` |
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| 328. |
If x is real, the expression ` (x+2)/(2x^2 + 3x+6)` takes all value in the intervalA. `(1/13,1/3)`B. `[-1/13,1/3]`C. `(-1/3,1/13)`D. none of these |
| Answer» Correct Answer - B | |
| 329. |
Let `alpha, beta ` be the roots of `x^(2) + bx + 1 = 0` . Them find the equation whose roots are `-(alpha + 1//beta) and -(beta + 1//alpha).` |
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Answer» Correct Answer - `x^(2) - x (2b) + 4 = 0` Since ` alpha , beta ` roots of ` x^(2) + bx + 1 = 0` , we have ` alpha + beta = -b , alpha beta = 1` Now ,` (-alpha - (1)/(beta)) + (- beta - (1)/(alpha )) = - (alpha + beta) - ((1)/(beta + (1)/(alpha))` = ` b + b 2b` and `( - alpha - (1)/(beta) ( - beta - (1)/(alpha )) = alpha beta + 2 + (1)/(alpha beta ) = 1 + 2 + 1 = 4` Thus, the equation whose roots are ` - alpha - 1//beta and - beta - 1//alpha is x^(2) - x (2b) + 4 = 0` . |
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| 330. |
Prove that the roots of the equation `(a^4+b^4)x^2+4a b c dx+(c^4+d^4)=0`cannot be different, if real. |
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Answer» The discriminant of the given equation is `D= 16a^(2)b^(2)c^(2)d^(2) = (a^(4)+b^(4))(c^(4)+d^(4))` `=-4[(a^(4)+b^(4))(c^(4)+d^(4)) -4 = a^(2)b^(2)c^(3)d^(2)]` `=-4[a^(4)c^(4)+a^(4)d^(4)+ b^(4)c^(4)+b^(4)d^(4)-a^(2)b^(2)c^(2)d^(2)]` `=-4[(a^(4)c^(4)+b^(4)d^(4) -2 a^(2)b^(2)c^(2)d^(2)) + (a^(4)d^(4)+b^(4)c^(4) -2 a^(2)b^(2)c^(2)d^2)] ` `=-4[(a^(2)c^(2)-b^(2)d^(2))^(2) + (a^(2)d^(2)-b^(2)c^(2))^(2)]` (1) Since rootd of the givrn equation are real, we have `D ge 0` `rArr =-4[(a^(2)c^(2)-b^(2)d^(2))^(2) + (a^(2)d^(2)-b^(2)c^(2))^(2)] ge0` or `(a^(2)c^(2)-b^(2)d^(2))^(2) + (a^(2)d^(2)-b^(2)c^(2))^(2)le 0` or `(a^(2)c^(2)-b^(2)d^(2))^(2) + (a^(2)d^(2)-b^(2)c^(2))^(2)= 0 (2)` ( Since sum of two positive quantities connot be negative) From (1) and (2), we get D = 0. Hence, the roots of the given quadratic equation are not different, if real. |
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| 331. |
What is the minimum height of any point on the curve `y=x^2-4x+6`above the x-axis? |
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Answer» `y = x^(2) - 4x + 6` = `(x - 2)^(2) + 2` Now `(x - 2)^(2) ge 0` for all real x. Then `(x - 2)^(2) + 2 ge 2` for all real .x Hence, the maximum value of `y (or - x^(2) + 6x - 5)` is 4 , which is the maximum height of the graph of the graph above x -axis. |
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| 332. |
If `x=1+1/(3+1/(3+1/(2...oo)))` then the value of x isA. `sqrt(5/2)`B. `sqrt(3/2)`C. `sqrt(7/3)`D. `sqrt(5/3)` |
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Answer» Correct Answer - 4 `x = 1 + (1)/(3+(1)/(2+(1)/(3+1/(2...oo))))` `= 1 + (1)/(3+(1)/(1 + x))` `= 1 + (1 + x)/(4 + 3x)` or `3x^(2) = 5` or `x = sqrt(5/3)` (as x is positive) |
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| 333. |
If `x` is real, then the minimum value of the expression `x^2-8x+17` isA. `-1`B. `0`C. `1`D. `2` |
| Answer» Correct Answer - C | |
| 334. |
Consider the equation `x^4 + 2ax^3 + x^2 + 2ax + 1 = 0` where `a in R`. Also range of function `f(x)= x+1/x` is `(-oo,-2]uu[2,oo)` If equation has at least two distinct positive real roots then all possible values of a areA. 2B. 1C. 0D. 3 |
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Answer» Correct Answer - 3 If exctly two roots are positive, the other two roots are negative. The -2 and 2 must lie between the roots. So, `f(-2) lt 0 and f(2) lt 0` `rArr a gt 3//4 and a gt lt - 3//4` Hence, no such values of a exist. |
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| 335. |
If for all real values of a one root of the equation `x^2-3ax + f(a)=0` is double of the other, then f(x) is equal toA. `2x`B. `x^(2)`C. `2x^(2)`D. `2sqrt(x)` |
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Answer» Correct Answer - C Let `alpha` beone of `x^(2)-3ax+f(a)=0` `impliesalpha+2alpha=3alphaimplies3alpha=3a` `impliesalpha=a` …………i and `alpha.2alpha=f(a)` `impliesf(a)=2alpha^(2)=2alpha^(2)` [ using Eq. (i)] `impliesf(x)=2x^(2)` |
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| 336. |
If roots of equation `x^3-2c x+a b=0`are real and unequal, then prove that the roots of `x^2-2(a+b)x+a^2+b^2+2c^2=0`will be imaginary. |
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Answer» Roots pf eqution `x^(2) - 2cx + ab = 0` are real and unequal `therefore D_(1) gt 0` `therefore 4c^(2) - 4ab gt 0` or `c^(2) - ab gt 0` (1) For equation `x^(2) - 2 (a + b) x + (a^(2) + b^(2) + 2c^(2))=0` `D_(2) = 4 (a + b)^(2) - 4 (a^(2) + b^(2) + 2c^(2))` = `- 8 ((c^(2) - ab) lt 0 (from (1))` Therefoue , roots are imaginary. |
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| 337. |
Find the range of `f(x)=sqrt(x-1)+sqrt(5-xdot)` |
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Answer» Correct Answer - `x^(2) - 4abx - (a^(2) - b^(2))^(2) = 0` ` (alpha + beta)^(2) = (alpha + beta)^(2)` and `(alpha - beta)^(2) = (alpha + beta) ^(2) - 4alpha beta ` ` (a + b )^(2) - 4 ((a^(2) + b^(2))/(2))` ` 2 ab - (a^(2) + b^(2))` 2 ab - (a^(2) + b^(2))` = - (a - b)^(2)` Now , the required equation whose roots are ` (alpha + beta)^(2) and (alpha - beta )^(2)` is `x^(2) - {(alpha + beta)^(2) + (alpha + beta)^(2)} x + (alpha + beta)^(2) (alpha - beta)^(2) = 0` or `x^(2) - {(a +b)^(2) - (a- b)^(2)} x - (a + b)^(2) (a-b)^(2) = 0` or ` x^(2) - 4 abx - (a^(2) - b^(2))^(2) = 0` . |
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| 338. |
Find the value of `2+1/(2+1/(2+1/(2+oo)))` |
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Answer» Correct Answer - 1 + sqrt(2)`, Let the given expression be equal to x . Them `x = 2+(1)/(x)` or `x^(2)-2x - 1 = 0` or `x= (2pmsqrt(4+ 4))/(2) = (2pm2 sqrt(2))/(2) = 1 pm sqrt(2)` But, the given expression is positive. Therefore, ` x = 1 + sqrt(2)` is the value of the given expression |
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| 339. |
The difference of maximum and minimum value of `(x^(2)+4x+9)/(x^(2)+9)` isA. `1//3`B. `2//3`C. `-2//3`D. `4//3` |
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Answer» Correct Answer - D `(d)` Let `(x^(2)+4x+9)/(x^(2)+9)=y` `implies(y-1)x^(2)-4x+9(y-1)=0` For real value of `x`, `D ge 0` `(5-3y)(3y-1) ge 0` `implies (1)/(3) le y le (5)/(3)` `:.` Difference `=(5)/(3)-(1)/(3)=(4)/(3)` |
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| 340. |
Solve `4^x+6^x=9^xdot` |
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Answer» Correct Answer - `x = log_(2//3)((sqrt(5)-1)/(2))` `4^(x) + 6^(x) = 9^(x)` or `((4)/(9))^(x)+((2)/(3))^(x) = 1` Putting `((2)/(3))^(x) = y ,` we have `y^(2) + y - 1 = 0` or ` y = (-1pmsqrt(5))/(2)` `rArr ((2)/(3))^(x) = (sqrt(5-1))/(2) (because((2)/(3))^(x) gt0)` `rArr x = log_(2//3)((sqrt(5-1))/(2)) ` |
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| 341. |
If `alpha,beta`are the roots of the quadratic equation `a x^2+b x=c=0`, then which of the following expression will be the symmetric functionof roots`|logalpha/beta|`b. `alpha^2beta^5+beta^2alpha^5`c. `t a n(alpha-beta)`d. `(log1/alpha)^2+(logbeta)^2`A. `|"log"(alpha)/(beta)|`B. `alpha^(2) beta^(5) + beta ^(2) alpha ^(5)`C. `tan (alpha - beta)`D. `("log"(1)/(alpha))^(2)+ (log beta)^(2)` |
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Answer» Correct Answer - 1,2,4 Symmetric functions are those which do not change by interc |
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| 342. |
Let `r ,s ,a n dt`be the roots of equation `8x^2+1001 x+2008=0.`Then find the value of.A. 751B. 752C. 753D. 754 |
| Answer» Correct Answer - C | |
| 343. |
Find a quadratic equation whoseproduct of roots `x_1a n dx_2`is equal to 4 an satisfying the relation `(x_1)/(x_1-1)+(x_2)/(x_2-1)2.` |
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Answer» Since `x_(1) x_(2) = 4` or `x_(2) = (4)/(x_(1))` `therefore (x_(1))/(x_(1)-1) + (x_(2))/(x_(2)-1) = 2` or `(x_(1))/(x_(1)-1) + ((4)/(x_(1)))/((4)/(x_(1))-1) = 2` or `(x_(1))/(x_(1)-1) + (4)/(4-x_(1)) =2` or `4x_(1) -x_(1)^(2) + 4x_(1)-4 =2(x_(1) -1) (4-x_(1))` or `x_(1)^(2) -2x_(1)+4=0` or `x^(2)-2x+ 4 =0`, Which is required equation. |
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| 344. |
The real numbers `x_1, x_2, x_3`satisfying the equation `x^3-x^2+b x+gamma=0`ar ein A.P. Find the intervals in which `betaa n dgamma`lie.A. ` (- (1)/(9), infty)`B. `(-(1)/(27) , + infty)`C. `((2)/(9), + infty)`D. none of these |
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Answer» Correct Answer - 2 From the question, the real roots of `x^(3) - x^(2) + betax + y = 0` are `x_(1),x_(2),x_(3)` and the they are in A.P.As `x_(1),x_(2),x_(3)` are in A.P., let `x_(1) = a-d, x_(2) = a, x_(3) = a + d`. Now, `x_(1) + x_(2) + x_(3) = -(-1)/(1) = 1` `rArr a-d + a+d = 1` ` rArr a= (1)/(3)" "(1)` `x_(1),x_(2)+x_(x_(2)x_(3) + x_(3)+x_(1) = (beta)/(1) = beta` `rArr (a-d)a+a(a+d) + (a+d) (a-d) = beta" "(2)` `x_(1)x_(2)x_(3)= -(gamma)/(1) = gamma` `rArr (a-d)a(a+d) = - gamma` Form (1) and (2), we get `3a^(2) -d^(2) = beta ` `rArr 3 (1)/(9) - d^(2) = beta , so beta = (1)/(3) - d^(2) lt (1)/(3)` From (1) and (3) , we get `(1)/(3) ((1)/(9) - d^(2)) = - gamma` `rArr gamma = (1)/(3) ( d^(2)-(1)/(9)) gt (1)/(3) (-(1)/(9)) = - (1)/(27)` ` gamma in (-(1)/(27), + infty)` |
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| 345. |
The real numbers `x_1, x_2, x_3`satisfying the equation `x^3-x^2+b x+gamma=0`ar ein A.P. Find the intervals in which `betaa n dgamma`lie.A. ` (-infty, (1)/(3))`B. `(-infty, - (1)/(3))`C. `((1)/(3) , infty)`D. ` (-(1)/(3), infty)` |
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Answer» Correct Answer - 1 From the question, the real roots of `x^(3) - x^(2) + betax + y = 0` are `x_(1),x_(2),x_(3)` and the they are in A.P.As `x_(1),x_(2),x_(3)` are in A.P., let `x_(1) = a-d, x_(2) = a, x_(3) = a + d`. Now, `x_(1) + x_(2) + x_(3) = -(-1)/(1) = 1` `rArr a-d + a+d = 1` ` rArr a= (1)/(3)" "(1)` `x_(1),x_(2)+x_(x_(2)x_(3) + x_(3)+x_(1) = (beta)/(1) = beta` `rArr (a-d)a+a(a+d) + (a+d) (a-d) = beta" "(2)` `x_(1)x_(2)x_(3)= -(gamma)/(1) = gamma` `rArr (a-d)a(a+d) = - gamma` Form (1) and (2), we get `3a^(2) -d^(2) = beta ` `rArr 3 (1)/(9) - d^(2) = beta , so beta = (1)/(3) - d^(2) lt (1)/(3)` From (1) and (3) , we get `(1)/(3) ((1)/(9) - d^(2)) = - gamma` `rArr gamma = (1)/(3) ( d^(2)-(1)/(9)) gt (1)/(3) (-(1)/(9)) = - (1)/(27)` ` gamma in (-(1)/(27), + infty)` |
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| 346. |
Match the following lists: |
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Answer» Correct Answer - `a to s; b to p; c to q; d to r ` `atos,bto""p,ctoq,d""tor` (a) `x^(2)ax+b=0` has root `lapha`. Hence. `alpha^(2)+aalpha+ b=0" "(1)` `x^(2)+px+q=0` has root `-alpha`. Hence, `alpha^(2)-palpha+q=" "(2)` Eliminating `alpha` from (1) and (2), we get `(q-b)^(2)=(aq+bp)(-p-a)` or `(q-b)^(2)=-(aq+pb)(p+a)` (b) `x^(2)ax+b=0` has root `alpha`. Hence, `alpha^(2)+aalpha+b=0 " "(1)` `x^(2)+px+q=0` has root `1//alpha`. Hence, `qalpha^(2)+palpha+1=0" "(2)` Eliminating `alpha` from (1) and (2), we get `(1-bq)^(2)=(a-pb)(p-aq)` (c) `x^(2)+ax+b=0` has roots, `alpha,beta`. Hence, `alpha^(2)+aalpha+b=0" "(1)` `x^(2)+px+q=0` has roots `-2//alpha,gamma`. Hence, `qalpha^(2)-2palpha+4=0" "(2)` Eliminating `alpha` from (1) and (2), we et `(4-bq)^(2)=(4a+2pb)(-2p-aq)` (d) `x^(2)+ax+b=0` has roots `alpha,beta`. Hence, `alpha^(2)+aalpha+b=0" "(1)` `x^(2)+px+q=0` has roots `-1//2alpha,gamma`. Hence, `4qalpha^(2)-2palpha+1=0" "(2)` Elimintaing `alpha` from (1) and (2), we get `(1-4bq)^(2)=(a+2bp)(-2p-4aq)` |
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| 347. |
If `sqrt(sqrt(sqrt(x)))=x^4+4 4 4 4,`then the value of `x^4`is____. |
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Answer» Correct Answer - 4 `x^(1//8)=(3x^(4)+4)^(1//64)` `impliesx^(8)=3x^(4)+4` `impliesx^(4)=4` |
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| 348. |
Sum of the valus of x satisfying the equation `sqrt(2x+sqrt(2x+4))=4` is ______. |
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Answer» Correct Answer - 6 `sqrt(2x+sqrt(2x+4))=4` `impliessqrt(2x+4)=16-2x` `implies2x+4=4(8-x)^(2)` `implies2x^(2)-33x+126=0` `implies(2r-21)(x-6)=0` `impliesx=6,21//2` But `x=21//2` does not satisfy the given equation. |
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