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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.
| 151. |
The sum of all the real roots of the equation `|x-2|^2+|x-2|-2=0` is |
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Answer» Correct Answer - 4 Given, `|x-2|^(2)+|x-2|-2=0` Case I When `c ge 2` `implies(x-2)^(2)+(x-2)-2=0` `impliesx^(2)+4-4x+x-2-2=0` `impliesx^(2)-3x=0` `impliesx(x-3)=0` `impliesx=0,3" "[0 "is rejected"]` `impliesx=3" "...(i)` Case II When `x gt 2` `implies{-(x-2))^(2)-(x-2)-3=0` `(x-2)^(2)-x+2-2=0` `impliesx^(2)+4-4x-x=0` `impliesx^(2)--4x-(x-4)=0` `impliesx(x-4)-1(x-4)=0` `implies(x-1)(x-4)=0` `impliesx=1,4" "["4 is rejected"]` `impliesx=1" "...(ii)` Hence, the sum of the roots is `3+1 =4.` Alternate Solution Given, `|x-2|^(2)+|x2|2=0` `implies(|x-2|+2)(|x-2|-1)=0` `therefore|x-2|=-2,1" "["neglacting"-2]` `implies|x-2|=1impliesx=3,1` `implies` Sum of the roots `=4` |
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| 152. |
The set of values of a for which ` (a - 1) x^(2) - (a + 1) x + a - 1` `ge 0 ` ture for all ` x ge 2` isA. `(-infty, 1)`B. `(1, (7)/(3))`C. `((7)/(3), infty)`D. none of these |
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Answer» Correct Answer - 3 Given, `(a - 1) x^(2) - (a + 1) x + a - 1 ge 0 ` or ` a(x^(2) - x + 1) - (x^(2) + x + 1) ge 0 ` ` rArr ge (x^(2) + x + 1)/(x^(2) - x + 1) = (2x)/(x^(2) - x + 1)` ` = 1 + (2)/(x + (1)/(x) - 1)` Let y = x + 1/x , Now , y is increasing in ` [ 2, infty)` . Hence, ` = 1 + (2)/(x + (1)/(x) - 1)in (1 , (7)/(3))` For all ` x ge 2,` Eq (1). should be true . Hence , ` a ge 7//3)` |
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| 153. |
If `a x^2+(b-c)x+a-b-c=0`has unequal real roots for all `c in R ,t h e n``b |
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Answer» Correct Answer - 3,4 We have, ` D = (b - c)^(2) - 4a(a - b - c) gt 0 ` or ` b^(2) + c^(2) - 2bc - 4a^(2) + 4ab + 4ac gt 0 ` ` or c^(2) + (4a - 2b) c - 4a^(2) + 4ab + b^(2) gt 0 ` for all ` c in ` R Discriminant of the above expression in c must be negative Hence, ` 1(4a = 2b)^(2) - 4 (-4a^(2) + 4ab + b^(2)) lt 0 ` or ` 4a^(2) - 4ab + b ^(2) + 4a^(2) - 4ab - b^(2) lt 0 ` or ` a (a - b) lt 0 ` `rArr a lt 0 and a - b gt 0 or a gt 0 and a - b lt 0 ` `rArr b lt a lt 0 or b gt a gt 0 ` . |
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| 154. |
Graph of ` y = ax^(2) + bx + c ` is as shown in the figure . If ` PQ= 9,` OR = 5 and OB = 2.5, the which of the following is /are ture? A. AB = 3B. ` y (-1) lt 0 `C. ` y (ge 7 for all x ge 3 `D. `ax^(2) + bx + c = ` mx has real roots for all real m |
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Answer» Correct Answer - 1,3,4 OR = 5 ` therefore f(0) = - 5= e` PQ = 9 . ` therefore - (D)/(4a) = - 9` `rArr (b^(2) + 20a)/(4a) = 9 ` `rArr b^(2) = 16 a ` …(1) OB - 2.5 So, one root of equation is 2.5 ` therefore 25a + 10 b - 20 = 0` ` or 5a + 2b - 4 = 0` ...(2) From (1) and (2), ` 5b^(2) + 32 b - 64 = 0 ` `rArr b = -8 ,(8)/(5) ` (Not possible . Since a ` gt ` 0, we must have ` b lt 0 `) ` therefore a = 4 ` `rArr y = 4x^(2) - 8x- 5 = 4x^(2) - 10x + 2x - 5 = (2x - 5) (2 + 1)` So, x = 2.5, - 0.5 are the roots of y(x) = 0 . Clealy, ` y(-1) g t0 ` ` y ge 7 ` `rArr 4x^(2) - 8x - 5 ge 7 ` `rArr x^(2) - 2x - 3 ge 0 ` `rArr (x - 3) (x +1) ge 0 ` `rArr x le - 1 or x ge 3` `ax^(2) + bx + c = mx ` `rArr 4x^(2) - (8 + m) x - 5 = 0 ` Clearly, above equation has two distinct real roots for any real values of m. |
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| 155. |
The set of values of `a`for which `a x^2+(a-2)x-2`is negative for exactly two integral `x ,`is`(0,2)`b. `[1,2)`c. `(1,2]`d. `(0,2]`A. (0,2)B. [1,2)C. (1, 2]D. (0,2] |
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Answer» Correct Answer - 2 `f(x) = ax^(2) + (a - 2) x - 2 = (ax - 2 )(x + 1)` `f(0) = - 2 and f(-1) = 0` If a is negative then expression becomes negative for intinite vales of x, therefore it must be positive Expression to be negative for exactly two integral values of x So `(2)/(a) le 2 or a ge 1` and ` (2)/(a) gt 1 rArr a lt 2` `therefore a in [ 1, 2)` |
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| 156. |
If `a x^2+b x=6=0`does not have distinct real roots, then find the least value of `3a+bdot` |
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Answer» Given, equation `ax^(2) + bx + 6 le 0, AA x in` R, if a`lt` 0 or `f(x) = ax^(2)+ bx + 6 ge 0, AA x in` R, if `gt` 0 But `f(x) = 6 gt`0 `rArr f(x) = ax^(2) + bx + 6 ge 0, AA x in` R `rArr f(3) = 9a + 3b + 6 ge ` 0 `rArr 3a + b ge - 2` Therefore, the least value of 3a + b`is -2. |
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| 157. |
Solve the equation `(a-3)x+5=a+2`. |
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Answer» Case I for `a!=3`, this equatiion is linear, then `(a-3)x=(a-3)` `:.x+((a-3))/((a-3))=1` Case II For `a=3`, `0.x+5=3+2` `implies5=5` Therefore any real number is its solution. |
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| 158. |
Solve the equation `x/2+((3x-1))/6=1-x/2` |
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Answer» We have `x/2+((3x-1))/6=1-x/2` or `x/2+x/2+x/2=1+1/6` or `(3x)/2=7/6` or `x=7/9` |
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| 159. |
Find the value of `m` for which the expressiion `12x^(2)-10xy+2y^(2)+11x-5y+m` can be resolved into two rational linear factors. |
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Answer» Comparing the given expression with `ax^(2)+2hxy+by^(2)_2gx+2fy+c`, we have `a=12, h=-5, b=2, g=11/2,f=(-5/2),c=m` The given expression will have two linear factors if and only if `abc+2fgh-af^(2)-bg^(2)-ch^(2)=0` or `(12)(2)(m)(-5/2)(11/2)(-5)-(12)(-5/2)^(2)` `-(2)(11/2)^(2)-(m)(-5)^(2)=0` `implies24m+275/2-75-121/2-25m=0` or `m-2` |
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| 160. |
Find the linear factors of `x^(2)-5xy+4y^(2)+x+2y-2` |
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Answer» Given expression is `x^(2)-5xy+4y^(2)x+2y-2`………………I Its corresponding equation is `x^(2)-5xy+4y^(2)+x+2y-2=0` or `x^(2)-x(5y-1)+4y^(2)+2y-2=0` `:.x=((5y-1)+-sqrt((5y-1)^(2)-4.1.(4y^(2)+2y-2)))/2` `=((5y-1)+-sqrt((9y^(2)-18y+9))/2` `=((5y-1)+-sqrt((3y-3)^(2))/2` `=((5y-1)+-(3y-3))/2=4y-2,y+1` `:.` The required linear factors are `(x-4y+z)` and `(x-y-1)`. |
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| 161. |
Form a quadratic equation with real coefficients whose one root is `3-2idot` |
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Answer» Since the complex roots always occur in conjugate pairs, so the other root is 3 + 2i. The sum of the roots is (3+ 2i) + 3 - 2i) = 6 . The product of the roots is `(3 + 2i) (3 - 2i) = 9-4i^(2) = 9 + 4 = 13` Hence, the equation is `x^(2) -Sx + P = 0` `rArr x^(2) - 6x + 13 = 0` |
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| 162. |
If `(a^(2)-14a+13)x^(2)+(a+2)x-2=0` does not have two distinct real roots, then the maximum value of `a^(2)-15a` is _________. |
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Answer» Correct Answer - -9 Let `f(x)=(a^(2)-14a+13)x^(2)+(a+2)x-2` Given that the above equation does not have two distinct real roots. Therefore, either `f(x)geO or f(x)le O AA x in R`. But `f(0)=-2lt0` `therefore f(0)le0AAx in R` So, `f(-1)le0` `rArr(a^(2)-14a+13)-(a+2)-2le0` `rArr a^(2)-15a+9le0` `rArr a^(2)-15a+9le0` So, the maximum value of `a^(2)-15a " is "-9` |
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| 163. |
Form a quadratic equation whose roots are `-4a n d6.` |
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Answer» We have sum of the roots, S = -4 + 6 = 2, and product of the roots, `P = - 4xx6 = - 24`. Hence, the required equation is `x^(2) - Sx + P = 0` `rArr x^(2) - 2x - 24 = 0` |
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| 164. |
The function k`f(x)=a x^2+b x^2+c x+d`has three positive roots. If thesum of the roots of `f(x)`is 4, the larget possible inegal values of `c//a`is ____________. |
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Answer» Correct Answer - 5 Let `ax^(2)+bx^(2)+cx+d=0` has roots p,q,r `pq+qr+rp=(c)/(a)" "(1)` but `pq+qr+rplep^(2)+q^(2)+r^(2)` `=(p+q+r)^(2)-2sumpq` `therefore3(pq+qr+rp)le(p+q+r)^(2)=16` `therefore3(c)/(a)le16` `rArr(c)/(a)le(16)/(3)` `rArr` largest integral value of `(c)/(a)` is 5 |
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| 165. |
If `b^(2)-4acle0` ("where" `ane0 and a,b,c,x,y in R`) satisfies the system `ax^(2)+x(b-3)+c+y=0 and ay^(2)+y(b-1)+c+3x=0`, then value of `(y)/(x)` is ___________. |
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Answer» Correct Answer - 3 Given equations are `ax^(2)+bx+c=3x-y" "...(1)` `ay^(2)+by+c=y-3x" "...(2)` Now, `b^(2)-4acge 0` `therefore ax^(2)+bx+cle0anday^(2)+by+cle0` `therefore 3x-yge0andy-3xge0` `rArr3x=y` or `therefore ax^(2)+bx+cle0anday^(2)+by+cle0`. `therefore3x-yle0 and y-3xle0` `rArr 3x=y` So, `(y)/(x)=3` |
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| 166. |
`ax^2+ by^2+ cz^2+2ayz+ 2bzx +2cxy` can be resolved into linear factors if a, b, c are such that |
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Answer» Given expression is `ax^(2)+by^(2)+cz^(2)+2ayz+2bzx+2cxy`……..i `=z^(2)[a(x/z)^(2)+b(y/z)^(2)+c+2a(y/z)+2b(x/z)+2c(x/z)(y/z)]` `=z^(2)[aX^(2)+bY^(2)+c+2aY+2bx+2cXY]` where `x/z=X` and `y/z=Y` Expression (i) will have two rational linear factors in x,y annd z if expression `aX^(2)+bY^(2)+2cXY+2bx+2aY+c` will have two linear factors if `abc+2abc-aa^(2)-bb^(2)-cc^(2)=0` or `a^(3)+b^(3)+c^(3)=3abc` |
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| 167. |
Show that `x^(2)-3|x|+2=0` is an equation. |
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Answer» Put `x=0` in `x^(2)-3|x|+2=0` `implies0^(2)-3|0|+2=2!=0` Since, the relation `x^(2)-3|x|+2=0` is not satisfied by `x=0`. Hence, it is an equation. |
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| 168. |
If `alpha,beta` are the roots of the equation `ax^2 + bx+c=0` then the roots of the equation `(a + b + c)x^2-(b + 2c)x+c=0` areA. cB. d - cC. 2cD. 0 |
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Answer» Correct Answer - 2,4 Let ` f(x) = x^(2) + ax + b `. Then ` x^(2) + (2c + a) x + c^(2) + ac + b = f(x + c)` Thus, the roots of f (x + c) = 0 will be 0, d - c . |
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| 169. |
If `cos^4theta+alpha`are the roots of the equation `x^2+2b x+b=0a n dcos^2theta+beta,sin^2theta+betaa r e`the roots of the equation `x^2+4x+2=0,`then values of `b`are`2`b. `-1`c. `-2`d. `2`A. 2B. -1C. -2D. 1 |
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Answer» Correct Answer - 1,2 We have ` cos ^(2) theta - sin^(2() theta = cos 2 theta` `rArr cos ^(4) theta theta - sin ^(4) theta = cos 2 theta ` `rArr (-2b)^(2) - 4b = - (-4)^(2) - 4xx2` (since L.H.S. is differnece of roots of first equation and R.H.S. is diffenece of roots of second equation). or ` 4b^(2) - 4b = 16 - 8= 8 ` or ` 4b^(2) - 4b - 8 = 0` or ` b^(2) - b - 2 = 0` or ` (b + 1)(b - 2) = 0` or ` b = 2, - 1` |
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| 170. |
if `c!=0` and the equation `p/(2x)=a/(x+c)+b/(x-c)` has two equal roots, then p can beA. `(sqrt(a) - sqrt(b))^(2) `B. `(sqrt(a)+ sqrt(b))^(2) `C. a+ bD. a - b |
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Answer» Correct Answer - 1.2 We can write the given equation as ` (p)/(2x) = ((a + b) x + c(b - a))/(x^(2) - c^(2))` or ` p(x^(2) - c^(2)) = 2 (a + b) x^(2) - 2 c (a - b) x ` or `(2a + 2b - p )x^(2) - 2c (a - b) x + pc^(2) = 0` for this equation to have equal roots, `c^(2) (a - b)^(2) - pc^(2) (2a + 2b - p) = 0 ` or ` (a - b)^(2) - 2p (a + b) + p^(2) = 0 [ because c^(2) ne 0 ]` or `[ p - (a + b)]^(2) = (a + b)^(2) - (a - b)^(2) = 4ab` or ` p - (a + b) = pm 2 sqrt(ab)` or `p = a + b pm 2 sqrt(ab) = (sqrt(a ) pm sqrt(b))^(2)` . |
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| 171. |
Find all values of the parameter a for which the quadratic equation `(a+1)x^(2)+2(a+1)x+a-2=0` (i) has two distinct roots. (ii) has no roots. (iii) has to equal roots. |
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Answer» By the hypothesis, this equation is quadratic and therefore `a!=-1` and the discriminant of this equation `D=4(a+1)^(2)-4(a+1)(a-2)` `=4(a+1)(a+1-a+2)` `=12(a+1)` (i) For `agt(-1)` then `Dgt0`, this equation has two distinct roots. (ii) For `a lt(-1)`, then `Dlt0`, this equation has no roots. (iii) This equation cannot have two equal roots. Sicne, `D=0` only for `a=-1` and this contradicts the hypothesis. |
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| 172. |
Match the following for the equation `x^(2)+a|x|+1=0` where, a is a parameter. |
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Answer» Correct Answer - `a to s; b to r; c to q; d to p ` `atos,btor,ctoq,d""to""p` Obviously when `age0`, we have no roots a all the terms are followed by+ve sign. Also for a=-2, we have `x^(2)-2|x|+1=0` or `|x|-1=0impliesx=pm1` Hence, the equation has two roots. Also when `alt-2`, for given equation `|x|=(-apmsqrt(a^(2)-4))/(2)gt0` Hence, the equation has four roots as `|-a|gtsqrt(a^(2)-4)`. Obviously, the equation has no three roots for any value of a. |
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| 173. |
If `a^(2)-4a+1=4`, then the value of `(a^(3)-a^(2)+a-1)/(a^(2)-1)(a^(2)ne1)` |
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Answer» Correct Answer - 4 Given `a^(2)-4a+1=4impliesa^(2)+1=4(1+a)` `y=((a-1)(1+a^(2)))/(a^(2)-1)=(a^(2)+1)/(a+1)=(4(a+1))/(a+1)=4` |
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| 174. |
Match the following for lists: |
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Answer» Correct Answer - `a to r; b to r; c to q; d to p ` `ator,btor,ctoq,d""to""p` `(m-2)x^(2)-(8-2m)x-(8-3m)=0` has roots of opposite signs. The product of roots is `-(8-3m)/(m-2)lt0` or `(3m-8)/(m-2)lt0` or `2ltmlt8//3` (b) Exactly one root of equation `x^(2)-m(2x-8)-15=0` lies in interval (0,1). `f(0)f(1)lt0` `implies(0-m(-8)-14)(1-m(-6)-15)lt0` `implies(8m-15)(6m-14)lt0` `implies15//8ltmlt7//3` (c) `x^(2)+2(m+1)x+9m-5=0` has both roots negative. Hence, sum of roots is `-2(m+1)ltormgt-1" "(1)` Product of roots is `9m-5gt0impliesmgt5//9" "(2)` Discriminant, `Dge0implies4(m+1)^(2)-4(9m-5)ge0` `impliesm^(2)-7m+6ge0` `impliesmle1ormge0" "(3)` Hence, for (1), (2),and (3), we get `m""in((5)/(9),1]uu[6,oo)` (d) `f(x)=x^(2)+2(m-1)x+m+5=0` has one root less than 1 and the other root greater than 1. Hence, `f(1)lt0` `implies1+2(m-1)+m+5lt0` `impliesmlt-4//3` |
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| 175. |
If `p ,q in {1,2,3,4,5}`, then find the number of equations of form `p^2x^2+q^2x+1=0`having real roots. |
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Answer» Correct Answer - 16 equations. `because D = q^(4) - 4p^(2) gen 0` `=(q^(2) + 2p)(q^(2) - 2p) ge 0` `rArr q^(2) ge 2p` If p = 1, then q = 2, 3, 4, ,5 p = 2, then q = 2, 3, 4, ,5 p = 3, then q = 3, 4, ,5 p = 4, then q = 3, 4, ,5 p = 5, then q = 4, ,5 So, number of possible equation is 16 |
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| 176. |
If `a ,b ,c in R`such that `a+b+c=0a n da!=c`, then prove that the roots of `(b+c-a)x^2+(c+a-b)x+(a+b-c)=0`are real and distinct. |
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Answer» Given equation is `(b + c - a)^(2) + ( c + a - b) x + ( a + b - c) = 0` or `(-2a)x^(2) + (-2b)x + (-2c) = 0` or `ax^(2) + bx + c = 0` `rArr D = b^(2) - 4ac` `= (-c - a)^(2) - 4ac` `= (c - a)^(2) gt 0` Hence, roots are real and distinct |
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| 177. |
Find the number of points of local extrema of `f(x)=3x^4-4x^3+6x^2+ax+b` where `a,b in R` |
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Answer» We have `f(x) = 3x^(4) - 4x^(3) + 6x^(2) + ax + b` `therefore (dy)/(dx) = 12x(x^(2) - x + 1) + a` `rArr (d^(2)y)/(dx^(2)) = 12(3x^(2) - 2x + 1) gt 0` So, `(dy)/(dx)` is an increasing function. But `(dy)/(dx)` is polynomial of degree 3. So, it has exactly one real root. Thus, derivative becomes zero at exactly one point. |
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| 178. |
If `x=1a n dx=2`are solutions of equations `x^3+a x^2+b x+c=0a n da+b=1,`then find the value of `bdot` |
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Answer» Correct Answer - ` b= 5` Since x = 1 is a root of the given equation, it satisfiles the equation. Hnece, putting x = 1 in the given equation, we get `a + b + c = 1 (1)` but given that `a + b = 1 ` `rArr c = -2` (2) Now put x = 2 in the given equation, we have ` 8 + 4a + 2b - 2 = 0` or `6 + 2a + 2 (a + b) = 0` or ` 6 + 2a + 2 = 0` or `a = -4` `rArr b = 5` |
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| 179. |
The number of real solutions of the equation `(9//10)^x=-3+x-x^2`isa. 2b. 0 c.1 d.none of theseA. 2B. 0C. 1D. none of these |
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Answer» Correct Answer - 2 Let `f(x) = -3 + x - x^(2)`. Then `f(x) lt 0` for all x, because coefficient of `x^(2)` is negative and `D lt 0`. Thus, L.H.S. of the given equation is always positive, whereas the R.H.S. is always less than zero. Hence, there is no solution. |
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| 180. |
The number of integral values of a for which the quadratic equation `(x+a)(x+1991)+1=0`has integral roots area. 3 b. 0c. 1 d. 2A. 3B. 0C. 1D. 2 |
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Answer» Correct Answer - 4 `(x + a)(x + 1991) + 1 = 0` or `(x + a)(x + 1991) = -1` `rArr (x + a) = 1 and x + 1991 = -1` `rArr a = 1993` or `x + a = -1 and x + 1991 = 1 rArr a = 1989` |
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| 181. |
the largest negative integer which satisfies `(x^2-1)/((x-2)(x-3))>0`A. `-4`B. `-3`C. `-2`D. `-1` |
| Answer» Correct Answer - C | |
| 182. |
Find the values of `m`for which the expression `2x^2+m x y+3y^2-5y-2`can be resolved into two rational linear factors. |
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Answer» We know that `ax^(2) + 2hxy + by^(2) + 2gx + 2fy + c` can be resolved into two linear factors if and only if ` abc + 2fgh - af^(2) - bg^(2) -ch^(2) = 0` Given expression is `2x^(2) + mxy + 3y^(2) - 5y - 2` Here, `a = 2, h = m//2` , `b = 3 ,g = 0` , `f = - 5//2,` `c = -2` Therefore, expression `2x^(2) + mxy + 3y^(2) - 5y - 2` will have two linear factors if and only if or `2xx3(-2) + 2 ((-5)/(2))(0) ((m)/(2))-2 ((-5)/(2))^(2) - 3 xx0^(2) - (-2)((m)/(2))^(2) =0` or `-12 - (25)/(2) + (m^2)/(2) =0` or ` m^(2) = 49` or ` m = pm 7` |
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| 183. |
Find he linear factors of `2x^2-y^2-x+x y+2y-1.` |
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Answer» Given expression is `2x^(2) - y^(2) - x + xy + 2y -1` (1) Its corresponding equation is `2x^(2) - y^(2) - x +xy + 2y - 1 = 0` or `2x^(2)-(1 -y)x - (y^(2) - 2y + 1) = 0` `therefore x = (1-y pm sqrt((1-y)^(2) + 4.2(y^(2)-2y + 1)))/(4)` `= (1-y pm sqrt((1-y)^(2) + 8(y-1)^(2)))/(4)` `= (1-y pm sqrt(9(1-y^(2))))/(4)` `= (1-y pm 3(1-y))/(4)` `1 - y, - (1 -y)/(2)` Hence, the required factors are (x + y -1) and (2x - y + 1). |
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| 184. |
Find the values of `m`for which the expression `2x^2+m x y+3y^2-5y-2`can be resolved into two rational linear factors.A. 3B. 5C. 7D. 9 |
| Answer» Correct Answer - C | |
| 185. |
Find all real value of a for which the equation `x^4 + (a - 1)x^3+ x^2 + (a - 1)x + 1 = 0` possesses at least two distinct positive roots |
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Answer» We have ltbegt `x^(4) - (a+ 1) x^(3)+ x^(2) + (a + 1) x- 2 = 0` Clearly, x = 1 and x = - 1 satisfy the above equation `because` Given equation is `(x^(2) - 1) (x^(2) - (a + 1) x + 2) = 2` So, `f(x) = x^(2) - (a + 1)x + 2 = 0` should have at least one positive real root . Since product of roots of above equation is 2, both the roots have same sign. If equation has two positive roots, then `Dge0` and sum of roots is positive (as product is already positive ). `therefore (a + 1)^(2) - 8 ge - and a + 1 gt 0` `rArr a le -1 -2 sqrt(2) - 1 and a gt - 1` `therefore a in [2sqrt(2) - 1 , infty)` . |
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| 186. |
If `(x^2+x=2)62=(a-3)(x^2+x+1)(x^2+x+2)+(a-4)(x^2+x+1)^2=0`has at least one root, then find the complete set of values of `adot` |
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Answer» Let `t = x^(2) + x + 1 rArr r in [(3)/(4), infty)` Hence, `(t+1)^(2) - (a -3)t(t+ 1) + (a - 4)t^(2)= 0` or `t^(2)+2t + 1 - (a -3)(t^(2)+ 1) + (a - 4)t^(2)= 0` or ` t(2 - a + 3) + 1 = ` or `t= (1) /(a -5)` `rArr (1)/(a -5) ge (3)/(4)` `rArr (19 - 3a)/((a - 5)) gt 0` `rArr a in(5, (19)/(3)]` |
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| 187. |
Let `a,b,c,d` be distinct real numbers and a and b are the roots of the quadratic equation `x^2-2cx-5d=0` . If c and d are the roots of the quadratic equation ` x^2-2ax-5b=0` then find the numerical value of `a+b+c+d` |
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Answer» Correct Answer - 3 `a+b=2c`.i `ab=-5x`..ii `c+d=2a`……iii `cd=5b`.(iv) From Eqs (i) and (iii) we get `a+b+c+d=2(a+c)` `:.a+c=b+d`………….v From Eqs (i) and (iii) we get `b-d=3(c-a)` …………vi Also `a` is a root of `x^(2)-2cx-5d=0` `:.a^(2)-2ac-5d=0`.......vii As c is a root of `c^(2)-2ac-5b=0`....viii From eqs vii and viii we get `a^(2)-c^(2)-5(d-b)=0` `implies(a+c)(a-c)+5(b-d)=0` `implies(a+c)(a-c)+15(c-a)=0` [ from Eq. (vi)] `implies(a-c)(a+c-15)=0` `:.a+c=15,a-c!=0` From Eq. (v) we get `b+d=15` `:.a+b+c+d=a+c+b+d=15+15=30` `implies` Sum of digits of `a+b|c+d=3+0=3` |
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| 188. |
The number of solutions of the equation `sqrt(x^(2))-sqrt((x-1)^(2))+sqrt((x-2)^(2))=sqrt(5)` is |
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Answer» Correct Answer - 2 We have `sqrt(x^(2))-sqrt((x-1)^(2))+sqrt((x-2)^(2))=sqrt(5)` `implies|x|-|x-1|+|x=2|=sqrt(5)` Case I If `xlt0` then `-x+(x-1)-(x-2)=sqrt(5)` `x=1-sqrt(5)` Case II I `0lexlt1` then `x+(x-1)-(x-2)=sqrt(5)` `impliesx=sqrt(5)-1` which is not possible. Case III If `1lexlt2`, then `x-(x-1)-(x-2)=sqrt(5)` `impliesx=3-sqrt(5)` which is not possible. Case IV If `xgt2`, then `x-(x-1)+(x-2)=sqrt(5)` ltbr `impliesx=1+sqrt(5)` Hence number of solutiions is 2. |
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| 189. |
The equation `(x+3-4(x-1)^(1//2))^(1//2)+(x+8-6(x-1)^(1//2))^(1//2)=1` has(A) no solution(B) only `1` solution(C) only `2` solutions(D) more than `2` solutionsA. no solutionB. only `1` solutionC. only `2` solutionsD. more than `2` solutions |
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Answer» Correct Answer - D `(d)` Put `(x-1)^(1//2)=t` or `x=t^(2)+1` Therefore, the given equation becomes `(t^(2)+4-4t)^(1//2)+(t^(2)+9-6t)^(1//2)=1` `implies [(t-2)^(2)]^(1//2)+[(t-3)^(2)]^(1//2)=1` `implies|t-2|+|t-3|=1` This equation is satisfied for all values of `t` lying between `2` and `3` i.e., `2 le t le 3` Thus, the given equation is satisfied for all values of `x` lying between `5` and `10`. |
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| 190. |
If `m`, `n` are positive integers and `m+nsqrt(2)=sqrt(41+24sqrt(2))`, then `(m+n)` is equal toA. `5`B. `6`C. `7`D. `8` |
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Answer» Correct Answer - C `(c )` We have `m+nsqrt(2)=sqrt(41+24sqrt(2))` Squaring both sides, we get `m^(2)+2n^(2)+2sqrt(2)mn=4a+24sqrt(2)` `:.` On comparing we get `m^(2)+2n^(2)=41`…….`(i)` and `mn=12`…….`(ii)` `:.` On solving `(i)` and `(ii)`, we get `implies m^(2)+(2(144))/(m^(2))=41` `impliesm^(4)=41m^(2)+288=0` `implies (m^(2)-32)(m^(2)-9)=0` `implies m^(2) ne 32` `:. m^(2) =9 implies m =3` and `n=4` Hence, `(m+n)=7` |
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| 191. |
Number of positive integers `x`for which `f(x)=x^3-8x^2+20 x-13`is a prime number is______. |
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Answer» Clearly, `f(1) = 0` `therefore f(x) = (x - 1) (x^(2) - 7x + 13)` For f(x) to be prime, at least one of the factors must be prime. Hence, either x - 1 = 1 or `x^(2) - 7x + 13 = 1` If x - 1 = 1, then x = 2 If `x^(2) - 7x + 13 = 1 ,` then` x^(2) - 7x + 12 = 0` `rArr x = 3 or 4` So, x = 2, 3, 4 |
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| 192. |
If `alpha, beta and gamma` the roots of the equation `x^(3) + 3x^(2) - 4x - 2 = 0.` then find the values of the following expressions: (i) `alpha ^(2) + beta^(2) + gamma^(2)` (ii)`alpha ^(3) + beta^(3) + gamma^(3)` (iii) `(1)/(alpha)+(1)/(beta)+(1)/(gamma)` |
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Answer» Given that `alpha,beta and gamma` are the roots of the equation `x^(3) + 3x^(2) - 4x - 2 = 0` Therefore, `alpha + beta + gamma = -(3)/(1) = -3, alpha beta + beta gamma + gamma alpha = (-4)/(1) = -4`, and `alpha beta gamma = ((-2))/(1) = 2.` `(alpha + beta + gamma)^(2) = alpha^(2) + beta^(2) + gamma^(2) + 2alphabeta + 2beta gamma + 2gammaalpha` ltbgt `rArr alpha^(2) + beta^(2) + gamma^(2) = (alpha + beta + gamma)^(2) - 2(alphabeta + beta gamma + gammaalpha)` `= (-3)^(2) - 2(-4) = 9+ 8 = 17` (ii) `alpha^(3) + beta^(3) + gamma^(3) -3alpha beta gamma = (alpha + beta + gamma) (alpha^(2) + beta^(2) + gamma^(2)-alphabeta - beta gamma - gammaalpha)` `rArr alpha^(3) + beta^(3) + gamma^(3)-3(2) = (-3)(17-(-4))` `rArr alpha^(3) + beta^(3) + gamma^(3)-63 + 6 = -57` (iii) `(1)/(alpha)+(1)/(beta)+(1)/(gamma)=(alphabeta + beta gamma + gammaalpha)/(alphabetagamma)=(-4)/(2) = -2`. |
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| 193. |
If r is positive real number such that `4sqrt(r)-(1)/(4sqrt(r))= 4,` then find the value of ` 6sqrt(r)+(1)/(6sqrt(r)).` |
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Answer» We have `4sqrt(r)-(1)/(4sqrt(r))= 4` On squaring, we get `sqrt(r)-(1)/(sqrt(r))-2= 16` or `sqrt(r)-(1)/(sqrt(r))= 18` Now, let `6sqrt(r)+(1)/(6sqrt(r))= x` On cubing both sides, we get `sqrt(r)+(1)/(sqrt(r))+3(6sqrt(r) +(1)/(6sqrt(r))) = x^(3)` `rArr 18 + 3x = x^(3)` `rArr x^(3) - 3x - 18 = 0` `rArr (x - 3) (x^(2) + 3x + 6) = 0` `rArr x = 3 (as x^(2) + 3x + 6 = 0 has complex roots)` `therefore 6sqrt(r) +(1)/(6sqrt(r) = 3` |
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| 194. |
If the roots of `x^(4)+qx^(2)+kx+225=0` are in arthmetic progression, then the value of `q`, isA. `15`B. `25`C. `35`D. `-50` |
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Answer» Correct Answer - D `(d)` Let the `4` roots be `x-3y`, `x-y`, `x+y`, `x+3y` `"sum"=4x` and if `4x=0` then `4` roots are `-3y`, `-y`, `y`,`3y` and the product is `9y^(4)=225impliesy^(2)=5`. The product taken two at a time is `-10y^(2)` `implies q=-50` |
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| 195. |
The value of `b` for which the equation `x^2+bx-1=0 and x^2+x+b=0` have one root in common isA. ` - sqrt(2)`B. `-i sqrt(3)`C. `sqrt(2)`D. `sqrt(3)` |
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Answer» Correct Answer - 2 ` x^(2) + bx - 1 = 0 ` `x^(2) + x + b = 0 ` Subtracting, we get ` (b - 1) x - 1 - b = 0 ` (1) `rArr x = (b + 1)/(b- 1) ` This value of x satisfies equation (1) `rArr x = ((b + 1) ^(2))/((b- 1) ^(2)) + (b + 1)/(b- 1) + b = 0 ` `rArr b = sqrt(3i), - sqrt(3i), 0` . |
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| 196. |
Let `alpha` and `beta` be the roots of `x^2-6x-2=0` with `alpha>beta` if `a_n=alpha^n-beta^n` for ` n>=1` then the value of `(a_10 - 2a_8)/(2a_9)`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - 2 ` a_(n) = a^(n) - beta^(n)` Also `alpha^(8)` on both sides `rArr alpha ^(10) - 6alpha ^(9) - 2 alpha^(8) = 0` (1) Similarly ` beta ^(10) - 6 beta^(9) - 2 beta ^(8) = 0 ` (2) Subtracting (2) from (1) we have ` alpha ^(10) - beta^(10) - 6 (alpha ^(9) - beta^(9)) = 2 (alpha ^(8) - beta^(8))` `rArr a_(10) - 6 a_(9) = 2a_(8)` `rArr (a_(10) - 2 a_(8))/(2a_(9)) = 3 ` |
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| 197. |
The quadratic equation `p(x)=0`with real coefficients has purely imaginary roots. Then the equation `p(p(x))=0`hasonly purely imaginary rootsat real rootstwo real and purely imaginary rootsneither real nor purely imaginary rootsA. only purely imaginary rootsB. all real rootsC. two real and two purely imaginary rootsD. neither real nor purely imaginary roots |
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Answer» Correct Answer - D If quadratric equation has purely imaginary roots, then coefficient of x must be equal to zero. Let `p(x)=ax^(2)+b` with a, b of same sing and `a, b in R.` Then `p[p(x)=a(ax^(2)+b)^(2)+b` P(x) has imaginary roots say ix. Then , also `ax^(2)+b in Rand (ax^(2)+b)^(2)gt0` `a (ax^(2)+b)^(2)+b ne0, AAx` Thus, `p[p(x)] ne0, AAx` |
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| 198. |
A quadratic equation with integral coefficients has two different primenumbers as its roots. If the sum of the coefficients of the equation isprime, then the sum of the roots is`2`b. `5`c. `7`d. `11`A. 2B. 5C. 7D. 11 |
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Answer» Correct Answer - 2 Let equation is `a (x - alpha) (x - beta) = 0`, where `alpha, beta` are prime numbers Sum of coefficients `= a - a(alpha + beta) + aalphabeta` `= a(alpha - 1) (beta - 1)`, which is prime number `rArr a = 1` and one of `(alpha - 1) and (beta - 1)` will be 1 Let `alpha - 1 = 1` or `alpha = 2`. then `beta - 1` will be 2. So `beta = 3` Hence, sum of roots = 5 |
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| 199. |
The quadratic equation `p(x)=0`with real coefficients has purely imaginary roots. Then the equation `p(p(x))=0`hasonly purely imaginary rootsat real rootstwo real and purely imaginary rootsneither real nor purely imaginary rootsA. only purely imaginary rootsB. all real rootsC. two real and two purely imaginary rootsD. neither real nor purealy imaginary roots |
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Answer» Correct Answer - 3 Since ` p(x) = 0` has purelu imaginary roots, ` p(x) = ax^(2) + c `, where a and c have same sign. Also , `p(p(x)) = 0 ` `rArr p(x) ` is purely imaginary `rArr ax^(2) + c ` is purely imaginary Hence ,x cannot be either purely real or purely imaginary. |
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| 200. |
Show that the minimum value of `(x+a)(x+b)//(x+c)dotw h e r ea > c ,b > c ,`is `(sqrt(a-c)+sqrt(-c))^2`for real values of `x >-cdot` |
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Answer» Given expression is `(x + a) (x + b) //(x + c).` Let `x + c + y `. Then `((x + a) (x + a))/((x + c)) = ((y+ (a - c))(y+(b-c)))/(y)` ` = (y^(2)+ [(a - c)+(b -c)]y+(b-c)(b - c))/(y)` ` = y+((a - c)+(b -c))/y+(b-c)+(b - c)` `=[sqrt(y)-sqrt((a-c)(b-c))/(y)]^(2) + [sqrt(a-c)+sqrt(b-c)]^(2)` ` =ge[sqrt(a - c)+sqrt(b -c)]^(2)` Hence, the least value is `[sqrt(a - c)+sqrt(b -c)]^(2)` |
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