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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.
| 101. |
If `a in (-1,1),`then roots of the quadratic equation `(a-1)x^2+a x+sqrt(1-a^2)=0`area. realb. imaginaryc. both equald. none of theseA. realB. imaginaryC. both equalD. none of these |
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Answer» Correct Answer - 1 `(a - 1) x^(2) + ax + sqrt(1 - a^(2)) = 0` `therefore D = a^(2) - 4(a - 1) sqrt(1 - a^(2))` `= a^(2) - 4a sqrt(1 - a^(2)) + 4 sqrt(1 - a^(2))` `= (a - 2 sqrt(1 - a^(2)))^(2) + 4sqrt(1 - a^(2))(1 - sqrt(1 - a^(2)))` `ge 0` for `a in (-1, 1)` Hence roots are real but not equal. |
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| 102. |
If `x`is real and the roots of the equation `a x^2+b x+c=0`are imaginary, then prove tat `a^2x^2+a b x+a c`is always positive. |
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Answer» Correct Answer - ` a in ( infty), -2)` Given, roots of `ax^(2) + bx + c = 0` are imaginary . Hence, ` b^(2) - 4ac lt 0 ` (1) Let us consider `f(x) = a^(2) x^(2) + abx + ac `. Her, coefficient of f(x) is `a^(2)` which is + ve, which makes graph concave upward. Also, ` D = (ab)^(2) - 4a^(2) (ac) = a^(2) (b^(2) - 4ac) lt 0` Hence, `f(x) gt 0 , AA x in ` R . |
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| 103. |
If `alpha`is a root of the equation `x^2+2x-1=0,`then prove that `4alpha^2-3alpha`is the other root. |
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Answer» We have `4x^(2) + 2x - 1 = 0` Let the other root be `beta` . `therefore alpha + beta = - (1)/(2), alpha beta = - (1)/(4)` Also, `4alpha^(2) + 2alpha - 1 = 0 as alpha` is a root, and we have to prove that `beta = 4alpha^(3) - 3 alpha.` Now `4alpha^(3) - 3alpha = 4 alpha ^(2) alpha - 3 alpha` = ` alpha (1 - 2 alpha) - 3 alpha ` `= - (1)/(2)[4alpha^(2) - 2alpha]` =`-(1)/(2)[1-2alpha + 4 alpha]` `= - (1)/(2) (1 + 2 alpha ) = - (1)/(2) - alpha = beta` [From (1)] Hence, the other root `beta is 4alpha^(3) - 3 alpha`. |
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| 104. |
For a `alt=0,`determine all real roots of theequation `x^2-2a|x-a|-3a^2=0.` |
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Answer» Correct Answer - `x={a(1-sqrt2),a (sqrt6-1)}` Here, `a ge 0` Given, `x^(2)-2a |x-a|-3a^(2)=0` Case I When `x ge a` `impliesx^(2)-2a(x-a)-3a^(2)=0` `impliesx^(2)-2ax -a^(2)=0` `impliesx=a +-sqrt2a` `" "[as a(1+sqrt2)ltaand a (1-sqrt2)gta]` `therefore"Neglecting" x=a(1+sqrt2)as x gea` `impliesx=a(1-sqrt2)" "...(i)` Case II When `x lt a impliesx^(2)+2a(x-a)-3a^(2)=0` `impliesx^(2)+2ax-5a^(2)=0impliesx=-aa+-sqrt6a` `" "[as a (sqrt6-1)lta (sqrt6-1)gta]` `therefore"Neglecting"x=a (-1-sqrt6)impliesx=a(sqrt6-1)" "...(ii)` From Eqs. (i) and (ii), we get `x={a(1-sqrt2),a (sqrt6-1)}` |
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| 105. |
If `x^2+p x+q=0`is the quadratic equation whose roots are `a-2a n d b-2`where `aa n db`are the roots of `x^2-3x+1=0,`then`p-1,q=5`b. `p=1,1=-5`c. `p=-1,q=1`d. `p=1,q=-1`A. `p=1,q=5`B. `p=1,q=-5`C. `p=-1,q=1`D. none of these |
| Answer» Correct Answer - D | |
| 106. |
If equation `(lamda^(2)-5lamda+6)x^(2)+(lamda^(2)-3lamda+2)x+(lamda^(2)-4)=0` is satisfied by more than two values of `x`, find the parameter `lamda`. |
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Answer» If an equation of degree to is satisfied by more than two values of unknown, then it must be an identity. Then, we must have `lamda^(2)-5lamda+6=0,lamda^(2)-3lamda+2=0lamda^(2)-4=0` `implieslamda=2,3` and `lamda=2,1` and `lamda=2,-2` Common value of `lamda` which satisfies each condition is `lamda=2`. |
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| 107. |
If `(-2,7)` is the highest point on the graph of` y =-2x^2-4ax +lambda`, then `lambda` equalsA. 31B. 11C. `-1`D. `-1/3` |
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Answer» Correct Answer - C We have `-((-4a)/(2(-2)))=-2` `impliesa=2` `:.y=-2x^(2)-8x+lamda`…..i Since Eq. (i) passes through points `(-2,7)` `:.7=-2(-2)^(2)-8(-2)+lamda` `implies7=-8+16+lamda` `:.lamda=-1` |
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| 108. |
Let `f(x) =4x^2-4ax+a^2-2a+2` be a quadratic polynomial in x,a be any real number.If x-coordinate ofd vertex of parabola y =f(x) is less thna 0 and f(x) has minimum value 3 for `x in [0,2]` then value of a isA. 1B. 2C. 3D. 0 |
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Answer» Correct Answer - 4 `f(x) = 4x^(2) - 4ax + a^(2)-2a + 2` represents a parabola whose vertex is at `((a)/(2), 2 - 2a)` Case-I : `0 lt(a)/(2) lt 2` In this case , `f(x)` attains the global minimum value at `x = (a)/(2)` . Thus , `f((a)/(2)) = 3` `rArr 3 = -2a + 2` `rArr a = - (1)/(2)` (Rejected) Case II : `(a)/(2)le2` In this cases, f(x) attains the global minimun value at x = 0 . Thus, `f(0) = 3` `rArr 3 = a^(2) - 2a + 2` `rArr a = 5 pm sqrt(10)` So, `a = 5 + sqrt(10)` Case III: `(a)/(2) le 0` In case,f(x) attains the global minimum value at x = 0. `rArr 3=a^(2) - 2a + 2 ` `rArr a = 1 p, sqrt(2)`. So, `a= 1 - sqrt(2)`. Hence, premissible values of a are `1-sqrt(2)` and `5+ sqrt(10)`. `f(x) = 4x^(2)-4ax + a^(2) -2a+ 2` is monotonic in `[0,2]` Hence, point of minima of funciton should not lie in `[0,2]`. Now, `f(x) = 0` `rArr 8x - 4a =0` If `(a)/(2)in [0,2]`, then `a in [0,4]`. For f(x) to be montomic in `[0,2], a notin [0,4]`. So , `a le 0 or a ge 4`. |
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| 109. |
Let `f(x) =4x^2-4ax+a^2-2a+2` be a quadratic polynomial in x,a be any real number.If x-coordinate ofd vertex of parabola y =f(x) is less thna 0 and f(x) has minimum value 3 for `x in [0,2]` then value of a isA. `a le 0 or a ge 4`B. `0 le a le 4`C. `a ge 0`D. none of these |
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Answer» Correct Answer - 1 `f(x) = 4x^(2) - 4ax + a^(2)-2a + 2` represents a parabola whose vertex is at `((a)/(2), 2 - 2a)` Case-I : `0 lt(a)/(2) lt 2` In this case , `f(x)` attains the global minimum value at `x = (a)/(2)` . Thus , `f((a)/(2)) = 3` `rArr 3 = -2a + 2` `rArr a = - (1)/(2)` (Rejected) Case II : `(a)/(2)le2` In this cases, f(x) attains the global minimun value at x = 0 . Thus, `f(0) = 3` `rArr 3 = a^(2) - 2a + 2` `rArr a = 5 pm sqrt(10)` So, `a = 5 + sqrt(10)` Case III: `(a)/(2) le 0` In case,f(x) attains the global minimum value at x = 0. `rArr 3=a^(2) - 2a + 2 ` `rArr a = 1 p, sqrt(2)`. So, `a= 1 - sqrt(2)`. Hence, premissible values of a are `1-sqrt(2)` and `5+ sqrt(10)`. `f(x) = 4x^(2)-4ax + a^(2) -2a+ 2` is monotonic in `[0,2]` Hence, point of minima of funciton should not lie in `[0,2]`. Now, `f(x) = 0` `rArr 8x - 4a =0` If `(a)/(2)in [0,2]`, then `a in [0,4]`. For f(x) to be montomic in `[0,2], a notin [0,4]`. So , `a le 0 or a ge 4`. |
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| 110. |
Consider equation `x^(4)-6x^(3)+8x^(2)+4ax-4a^(2)=0,ainR`. Then match the following lists: A. `{:(,a,b,c,d),((1),q,s,s,r):}`B. `{:(,a,b,c,d),((2),r,s,q,p):}`C. `{:(,a,b,c,d),((3),q,s,r,p):}`D. `{:(,a,b,c,d),((4),q,r,p,p):}` |
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Answer» Correct Answer - 4 `(x^(2)-2x-2a)(x^(2)-4x+2a)=0` `impliesx^(2)-2x-2a=0" "....(1)` or `x^(2)-4x+2a=0" "....(2)` Discrtiminant of eq. (1) is: `D_(1)=4+8a` Discriminant of eq. (2) is: `D_(2)=16-8a` (a) If equation has exactly two distinct roots then (1) `D_(1)gt0implies4+8agt0impliesagt-1//2` and `D_(2)implies16-8agt0impliesalt2` `:.ain(-1//2,2)` (b) If equation has exactly two distinct roots then (i) `D_(1)gt0andD_(2)lt0` `impliesain(2,oo)` (ii) `D_(1)lt0andD_(2)gt0` `impliesain(-oo,-1//2)` (c) If equation has no real roots then `D_(1)lt0andD_(2)lt0` `impliesainphi` (d) Clearly, rquation cannot have four roots positive. |
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| 111. |
Solve the equation `x^(2)+(x/(x-1))^(2)=8` |
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Answer» Correct Answer - `x_(1)=2,x_(2)=-1+sqrt(3)` and `x_(3)=-1-sqrt(3)` We have `x^(2)+(x/(x-1))^(2)=8` `implies(x+x/(x-1))^(2)-2.x.x/((x-1))=8` `implies(x^(2))/(x-1))^(2)-2((x^(2))/(x-1))-8=0`.........i Let `y=(x^(2))/(x-1)`. Then Eq. (i) reduces to `y^(2)-2y-8=0` `implies(y-4)(y+2)=0` `:.y=4,-2` If `y=4` then `4=(x^(2))/(x-1)` or `x^(2)-4x+4=0` or `(x-2)^(2)=0` or `x=2` `:.x_(1)=2` and if `y=-2`, then `-2=(x^(2))/(x-1)` or `x^(2)+2x-2=0` `:.x=(-2+-sqrt((4+8)))/2` `impliesx=-1+-sqrt(3)` `:.x_(2)=-1+sqrt(3),x_(3)=-1-sqrt(3)` |
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| 112. |
Solve the equation `x((3-x)/(x+1))+(x+(3-x)/(x+1))=2` |
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Answer» Hence `x+1!=0` and let `x((3-x)/(x+1))=u` and `x+(3-x)/(x+1)=v` `:.uv=2`……….i and `u+v=x((3-x)/(x+1))+x+((3-x)/(x+1))` `=(x+1)((3-x)/(x+1))+x=3-x+x=3` `:.u+v=3` and `uv=2` Then `u=2,v=1` or `u=1,v=2` Givnen equation is equivalent to the collection `:.{(x((3-x)/(x+1))=2),(x+(3-x)/(x+1)=1):}` or `{(x((3-x)/(x+1)))=1),(x+(3-x)/(x+1)=2):}` `implies{(x^(2)-x+2=0),(x^(2)-x+2=0):}` or `{(x^(2)-2x+1=0),(x^(2)-2x+1=0):}` `implies{(x^(2)-x+2=0),(x^(2)-2x+1=0):}implies{((x-1/2)^(2)+7/4!=0),((x-1)^(2)=0):}` `:.(x-1)^(2)=0` `impliesx=1` is a unique solution of the original equation. |
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| 113. |
Solve the equatioin `|x-|4-x||-2x=4` |
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Answer» The equation is equivalent to the collection of systems `{(|x-(4-x)|-2x=4, "if"4-xge0),(|x+(4-x)|-2x=4, "if"4-xlt0):}` `implies {(|2x-4|-2x=4, "if"xle4),(4-2x=4,"if"xgt4):}`……I The second system of this collection gives `x=0` but `xgt4` Hence, second system has no solution. The first system of collection Eq. (i) is equivalent to the system of collection `{(2x-4-2x=4, "if"2xge4),(-2x+4-2x=4,"if"2xlt4):}` `implies{(-4=4,"if"xge2),(-4x=0,"if"xlt2):}` The first system is failed and second system gives `x=0`. Hence `x=0` is unique solution of the give equation. |
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| 114. |
Solve `6x^(3)-11x^(2)+6x-1=0`, roots of the equatioin are in HP. |
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Answer» Put `x=1/y` in the given equation then `6/(y^(3))-11/(y^(2))+6/y-1=0` `impliesy^(3)-6y^(2)+11y-6=0`……..i Now, roots of Eq. (i) are in AP. Let the roots be `alpha-beta, alpha, alpha +beta` The, sum of roots `=alpha-beta+alpha+beta=6` `implies3 alpha=6` `:.alpha=2` Product of roots `=(alpha-beta).alpha.(alpha+beta)=6` `implies(2-beta)2(2+beta)=6implies4-beta^(2)=3` `:.beta=+-1` `:.` Roots of Eqs (i) are 1,2,3 or 3,2,1. Hence roots of the given equation are `1,1/2,1/3` or `1/3, 1/2, 1`. |
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| 115. |
Given that the expression `2x^3+3p x^2-4x+p`hs a remainder of 5 when divided by `x+2`, find the value of `pdot` |
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Answer» Correct Answer - `p = 1` Let `f(x) = 2x^(3) + 3px^(2) - 4x + p` Given `f(-2) = 2 (-2)^(3) + 3 (-2)^(2)p - 4 (-2)+ p = 5` or ` 13 p - 8 = 5 ` or ` p = 1` |
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| 116. |
If `(a^2-1)x^2+(a-1)x+a^2-4a+3=0`is identity in `x`, then find the value of `a`.A. `-1`B. `1`C. `3`D. `-1,1,3` |
| Answer» Correct Answer - B | |
| 117. |
If `a^(3)-3a^(2)+5a-17=0` and `b^(3)-3b^(2)+5b+11=0` are such that `a+b` is a real number, then the value of `a+b` isA. `-1`B. `1`C. `2`D. `-2` |
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Answer» Correct Answer - C `(c )` Let `a+b=lambdaimpliesb=lambda-a` So `(lambda-a)^(3)-3(lambda-a)^(2)+5(lambda-a)+11=0` `lambda^(3)-a^(3)-3lambda^(2)a+3lambdaa^(2)-3lambda^(2)+6lambda a-3a^(2)+5lambda-5a+11=0` ,brgt `implies a^(3)+(3-3lambda)a^(2)+(3lambda^(2)-6lambda+5)a-(lambda^(3)-3lambda^(2)+5lambda+11)=0` Comparing it with the equation `a^(3)-3a^(2)+5a-17=0`, we get `lambda=2` |
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| 118. |
If `a ,b ,c`are non-zero real numbers, then the minimum value of the expression`(((a^4 3a^2+1)(b^4+5b^2+1)(c^4+7c^2+1))/(a^2b^2c^2))`is not divisible by prime number. |
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Answer» Correct Answer - 315 We have `((a^(4)+3a^(2)+1)/a^(2))((b^(2)+5b^(2)+1)/(b^(2)))((c^(4)+7c^(2)+1)/(c^(2)))` `=(a^(2)+(1)/(a^(2))+3)(b^(2)+(1)/(b^(2))+5)(c^(2)+(1)/(c^(2))+7)` `=((a-(1)/(a))^(2)+5)((b-(1)/(b))^(2)+7)((c-(1)/(c))^(2)+9)` So, minimum value is `5xx7xx9=315`. |
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| 119. |
If roots of an equation `x^n-1=0` are `1,a_1,a_2,........,a_(n-1)` then the value of `(1-a_1)(1-a_2)(1-a_3)........(1-a_(n-1))` will be (a) `n` (b) `n^2` (c) `n^n` (d) `0`A. nB. `n^(2)`C. `n^(n)`D. 0 |
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Answer» Correct Answer - 1 Clearly, `x^(n) - 1= (x - 1)(x - a_(1)) (x - a_(2)) .. (x - a_(n - 1))` `rArr (x^(n) - 1)/(x - 1) = (x - a_(1))(x - a_(2)) .. (x - a_(n - 1)` `rArr 1 + x + x^(2) + .. + x^(n - 1) = (x - a_(1))(x - a_(2))..(x - a_(n - 1))` `rArr n = (1 - a_(1))(1 - a_(2))..(1 - a_(n - 1))` [putting x = 1] |
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| 120. |
If `alpha,beta,gamma,sigma`are the roots of the equation `x^4+4x^3-6x^3+7x-9=0,`then he value of `(1+alpha^2)(1+beta^2)(1+gamma^2)(1+sigma^2)`is`9`b. `11`c. `13`d. 5A. 9B. 11C. 13D. 5 |
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Answer» Correct Answer - 3 Since `alpha, beta, gamma, sigma` are the roots of the given equation, we have `x^(4) + 4x^(3) - 6x^(2) + 7x - 9 = (x - alpha)(x - beta)(x - gamma)(x - sigma)` Putting x = I and then x = -I, we get `1 - 4i + 6 + 7i - 9 = (I - alpha) (I - beta)(i - gamma) (i - sigma)` and `1 + 4i + 6 - 7i - 9 = (-i - alpha)(-i -beta)(-i -gamma)(-i - sigma)` Multiplying these two equations, we get `(-2 + 3i)(-2 - 3i) = (1 + alpha^(2))(1 + beta^(2))(1 + gamma^(2))(1 + sigma^(2))` or `13 = (1 + alpha^(2))(1 + beta^(2))(1 + gamma^(2))(1 + sigma^(2))` |
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| 121. |
Find the condition that the expressions `a x^2-b x y+c y^2a n da_1x^2+b_1x y+c_1y^2`may have factors `y-m xa n dm y-x ,`respectively. |
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Answer» y = mx is a factor of ` ax^(2) + bxy + c^(2)` Hence, ` ax^(2) + bxy + cy^(2)` will be zero when y - mx = 0 or y = mx `rArr ax^(2) + bx *mx + cm^(2) x^(2) = 0` or `cm^(2) + bm + a = 0` (1) Since my - x is a factor of ` a_(1)x^(2) + b_(1)xy + C_(1) y^(2) = 0` when my - x = 0 `rArr a_(1) m^(2) y^(2) + b_(1)*my* + c_(1)y^(2) = 0` [Putting x = my] `rArr a_(1) m^(2) + b_(1) m + c_(1) = 0` (2) Eliminating m from (1) and (2) , we get `(bc_(1) - ab_(1)) (cb_(1) - ba_(1)) = (aa_(1) - c c_(1))^(2)` . |
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| 122. |
If `(m_r ,1//m_r),r=1,2,3,4,`are four pairs of values of `xa n dy`that satisfy the equation `x^2+y^2+2gx+2fy+c=0`, then the value of `m_1, m_2, m_3, m_4`is`0`b. `1`c. `-1`d. none of these |
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Answer» Correct Answer - 2 If `[m_(r), (1//m_(r))]` satisfy the given equation `x^(2) + y^(2) + 2gx + 2fy + c = 0`, then `m_(r)^(2) + 1/m_(r)^(2) + 2gm_(r) + (2f)/(m_(r)) + c = 0` `rArr m_(r)^(4) + 2gm_(r)^(3) + cm_(r)^(2) + 2fm_(r) + 1 = 0` Now, roots of given equation are `m_(1), m_(2), m_(3), m_(4)`. The product of roots `m_(1)m_(2)m_(3)m_(4) = ("Constant term")/("Coefficient of " m_(r)^(4)) = 1/1 = 1` |
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| 123. |
Prove that if the equation `x^2+9y^2-4x+3=0`is satisfied for real values of `xa n dy ,t h e nx`must lie between 1 and 3 and`y`must lie between-1/3 and 1/3. |
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Answer» Given equation is `x^(2) + 9y^(2) - 4x + 3 =0` (1) or `x^(2) - 4x + 9y^(2) + 3 = 0` Since x is ral, we have `(-4)^(2) - 4(9y^(2)+ 3) ge 0` or `16 - 4 (9y^(2) + 3) ge 0` or ` 4 - 9y^(2) - 3 ge 0` or `9y^(2) - 1 le 0` or ` 9y^(2) le 1` or `y^(2)le (1)/(9)` `rArr -(1)/(3) le y le (1)/(3)` (2) Equation (1) can also be written as `9y^(2) + 0y + x^(2) - 4x + 3 = 0` (3) Since y is real, so or `0^(2) - 4x + 3 le 0` (4) or (x - 3) (x - 1) le 0` or ` 1 le x le 3` |
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| 124. |
If `a+b+c=24`, `a^(2)+b^(2)+c^(2)=210`, `abc=440`. Then the least value of `a-b-c` isA. `-2`B. `2`C. `8`D. `-14` |
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Answer» Correct Answer - D `(d)` `ab+bc+ca=((a+b+c)^(2)-(a^(2)+b^(2)+c^(2)))/(2)=183` Hence `a`, `b`, `c` are the roots of the equation. `t^(3)-24t^(2)+183t-440=0` `implies(t-5)(t-8)(t-11)=0` Thus `{a,b,c}={5,8,11}` |
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| 125. |
Find the condition on `a , b ,c ,d`such that equations `2a x^2+b^2+c x+d=0a n d2a x 62+3b x+4x=0`have a common root. |
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Answer» Let `alpha` be a common root of the given two equations. Thus `2aalpha^(3) + balpha^(2) + calpha + d = 0` (1) `2aalpha^(2) + 3balpha + 4c = 0` (2) Multiplying (2) with `alpha` and then subtracting (1) from it, we get `2balpha^(2) + 3calpha -d = 0` (3) Now, Eqs. (2) and (3) re quadratic having a common root `alpha `, so `(-2ad - 8bc)^(2) = (-3bd - 12c^(2)) (6ac - 6b^(2))` `rArr (ad + 4 bc)^(2) = (9)/(2) (bd + 4c^(2) (b^(2) - ac)` |
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| 126. |
If `alpha,beta,gamma` are the roots of the equation `x^3 + 4x +1=0` then `(alpha+beta)^(-1)+(beta+gamma)^(-1)+(gamma+alpha)^(-1)=` |
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Answer» For the given equation `alpha + beta + gamma = 0` `alpha beta + beta gamma + alpha gamma = 4, alpha beta gamma = -1` Now, `(alpha + beta)^(-1)+(beta + gamma)^(-1) + (gamma + alpha)^(-1).=(-gamma)^(-1)+(-alpha)^(-1) +(-beta)^(-1)` `=(alpha beta + beta gamma + alpha gamma)/(alpha beta gamma)` = `-(4)/(-1)` = 4. |
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| 127. |
If the cubic `2x^3=9x^2+12 x+k=0`has two equal roots then minimum value of `|k|`is______. |
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Answer» Correct Answer - 5 We have `2x^(3)-9x^(2)+12x +k =0` Let the roots are `alpha,alpha, beta` `2alpha+beta=(9)/(2)" "(1)` `alpha + 2alphabet=(12)/(2)=6" "(2)` are `alpha^(2)beta=-(k)/(2)" "(3)` putting `beta = ((9)/(2)-2alpha)` from (1) in (2), we have `alpha^(2)+2alpha((9)/(2)-2alpha)=6` or `alpha^(2)+9alpha-4alpha^(2)=6` or `3alpha^(2)-9alpha+6=0` or `alpha^(2)-3alpha+2=0` or `(alpha-2)(alpha-1)=0rArralpha=2 or1` `if alpha=2, "then "alpha=(1)/(2),` `if alpha=1, " then " beta=(5)/(2)` `therefore k=-2(2^(2))(1)/(2)=-4` `or k=-2 (1^(2))((5)/(2))=-5` |
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| 128. |
If the equation `x^(2)+2x+3=0 and ax^(2)+bx+c=0, a,b,c in R` have a common root, then `a:b:c` isA. `1:2:3`B. `3:2:1`C. `1:3:2`D. `3:1:2` |
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Answer» Correct Answer - A Given equations are `x^(2)+2x+3=0" "…(i)` and `ax^(2)+bx+c=0" "…(ii)` Since, Eq. (i) has imaginary roots, so Eq. (ii) will alos have both same as Eq. (i). Thus, `a/1=b/2=c/3` Hence,` a:b:c: is 1:2:3.` |
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| 129. |
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. `-sqrt2`B. `-isqrt3`C. `I sqrt5`D. `sqrt2` |
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Answer» Correct Answer - B If `a_(1)x^(2)+b_(1)x+c_(2)=0` have a common real roots, then `implies(a_(1)c_(2)-a_(2)c_(1))^(2)=(b_(1)c_(2)-b_(2)c_(1))(a_(1)b_(2)-a_(2)b_(1))` `therefore{:(x^(2)+bx-1=0),(x^(2)+x+b=0):}` have a common root. `implies(a+b)^(2)=(b^(2)+1)(1-b)` `implies b^(2)+2b+1=b^(2)-b^(3)+1-b` `impliesb^(3)+3b=0` `thereforeb(b^(2)+3)=0` `impliesb=0, pmsqrt3i` |
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| 130. |
The sum of the solutions of the equation `|sqrtx-2|+sqrtx(sqrtx-4)+2=(x gt0)` is equal toA. 9B. 12C. 4D. 10 |
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Answer» Correct Answer - D Key Idea Reduce the given equation into quadratic equation. Given equation is `|sqrtx-2|+sqrtx(sqrtx-4)+2=0 implies|sqrtx-2|+x-4sqrt+4=2` `implies|sqrtx-2|+(sqrtx-2)^(2)=2` `implies(|sqrtx-2|)^(2)+|sqrtx-2|-2=0` Let `|sqrtx-2|=y,` then above equation reduced to `y^(2)+y-2=0impliesy^(2)+2y-y-2=0` `impliesy(y+2)-1(y+2)=0implies(y+2)(y-1)=0` `impliesy=1,-2` `impliesthereforey=1" "[becausey="sqrtx-2|ge0]` `implies|sqrtx-2"=1` `impliessqrtx=pm1` `impliessqrtx=3or 1` `impliesx=9or1` `therefore" Sum of roots"=9+1=10` |
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| 131. |
Fill in the blanksIf the quadratic equations `x^2+a x+b=0a n dx^2+b x+a=0(a!=b)`have a common root, then the numerical value of `a+b`is ________. |
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Answer» Correct Answer - A Given equations are `x^(2)+ax+b=0` and `x^(2)+bx+a=0` have common root On subtracting above equations, we get `(a-b)x+(b-a)=0` `implies x=1` `thereforex=1` is the common root. `impliesa+a+b=0` `impliesa+b=-1` |
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| 132. |
If `alpha`is a real root of the quadratic equation `a x^2+b x+c=0a n dbeta`ils a real root of ` a x^2+b x+c=0,`then show that there is a root `gamma`of equation `(a//2)x^2+b x+c=0`whilch lies between `aa n dbetadot` |
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Answer» Let `f(x) = (a)/(2) x^(2) + bx + c` `rArr f(alpha) = (a)/(2) alpha^(2) + bx + c` =` aalpha^(2) + balpha + c (a )/(2) alpha ^(2)` ` = -(a)/(2) alpha^(2) (becouse alpha` is root of `ax^(2) + bx + c = 0` ) and `f(beta) = (a)/(2) beta^(2) + b beta + c` = - ` a beta^(2) + b beta + c + (3)/(2) a beta^(2)` `= (3)/(2) a beta^(2) (becouse beta` is a root of `-ax^(2) + bx + c = 0 ) ` Now, `f(alpha)f(beta) = (-3)/(4) a^(2) alpha^(2) beta^(2) lt 0` Hence, `f(x) = 0` has one real root between` alpha and beta` . |
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| 133. |
Number of real solutions of `sqrt(2x-4)-sqrt(x+5)=1` isA. `0`B. `1`C. `2`D. infinite |
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Answer» Correct Answer - B `(b)` We have `sqrt(2x-4)=1+sqrt(x+5)` Squaring `2x-4=1+(x+5)+2sqrt(x+5)` `implies x-10=2sqrt(x+5)` `implies x^(2)+100-20x=4x+20` `impliesx^(2)-24x+80=0` `impliesx=4,20` Putting `x=4`, we get `sqrt(4)-sqrt(9)=1`, which is not possible Putting `x=20`, we get `sqrt(36)-sqrt(25)=1` Hence, `x=20` is the only solution. |
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| 134. |
The number of solutions of `sqrt(3x^(2)+x+5)=x-3` is(A) `0`(B) `1`(C) `2`(D) `4`A. `0`B. `1`C. `2`D. `4` |
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Answer» Correct Answer - A `(a)` We have `sqrt(3x^(2)+x+5)=x-3` We must have `3x^(2)+x+5 ge 0` and `x-3 ge 0` or `x ge 3` `sqrt(3x^(2)+x+3)=x-3` ……….`(i)` Squaring both sides of `(i)`, we get `3x^(2)+x+5=x^(2)-6x+9` `implies 2x^(2)+7x-4=0` `implies (2x-1)(x+4)=0` `implies x=1//2,-4` These values does not satisfy the inequality `x ge 3`. Thus, `(i)` has no solution. |
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| 135. |
The number of real or complex solutions of `x^(2)-6|x|+8=0` isA. `6`B. `7`C. `8`D. `9` |
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Answer» Correct Answer - A `(a)` If `x` is real, `x^(2)-6|x|+8=0` `implies |x|^(2)-6|x|+8=0` `implies |x|=2,4` `implies x=+-2,+-4` If `x` is non-real, say `x=alpha+I beta`, then `(alpha+I beta)^(2)-6sqrt(alpha^(2)+beta^(2))+8=0` `:.alpha^(2)-beta^(2)+8-6sqrt(alpha^(2)+beta^(2))+2i alpha beta=0` Comparing real and imaginary parts, `alpha beta=0 implies alpha=0` (if `beta=0`, then `x` is real) and `-beta^(2)+8-66sqrt(beta^(2))=0` `:. beta+-6beta-8=0` `implies beta=(overset(-)(+)6overset(-)(+)sqrt(68))/(2)` `implies beta=+-(3-sqrt(17))` Hence, `+-(3-sqrt(17))i` are non-real roots. |
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| 136. |
Number of real solutions of `sqrt(x)+sqrt(x-sqrt(1-x))=1` isA. `0`B. `1`C. `2`D. infinite |
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Answer» Correct Answer - B `(b)` We have `sqrt(x)+sqrt(x-sqrt(1-x))=1` `impliessqrt(x-sqrt(1-x))=1-sqrt(x)` Squaring `x-sqrt(1-x)=1+x-2sqrt(x)` `implies2sqrt(x)-sqrt(1-x)=1` ..........`(i)` `implies (2sqrt(x)-sqrt(1-x))(2sqrt(x)+sqrt(1-x))=(2sqrt(x)+sqrt(1-x))` `implies4x-(1-x)=2sqrt(x)+sqrt(1-x)` `implies 2sqrt(x)+sqrt(1-x)=5x-1` ................`(ii)` Adding `(i)` and `(ii)`, `4sqrt(x)=5x` `implies 16x=25x^(2)` `implies x=0,(16)/(25)` Clearly `x=0` does not satisfy the equation. Putting `x=(16)/(25)` in equation `L.H.S=(4)/(3)+sqrt((16)/(25)-(3)/(5))=(4)/(5)+(1)/(5)=1` So `x=(16)/(25)` is the only solution. |
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| 137. |
Let `a ,b ,c`be real numbers `a!=0.`If `alpha`is a root of `a^2x^2+b x+c=0.beta`is the root of `a^2x^2-b x-c=0a n d0A. `gamma=(alpha+beta)/(2)`B. `gamma=alpha+(beta)/(2)`C. `gamma=alpha`D. `alphaltgammalt beta` |
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Answer» Correct Answer - D Since, `alpha` is a root of `a^(2)x^(2)+bx+c=0` `impliesa^(1)alpha^(2)+balpha+c=0" "...(i)` `and beta"is a root of"a^(2)x^(2)-bx-c=0` `impliesa^(2)beta^(2)-b beta-c=0" "...(ii)` Let `f(x)=a^(2)alpha^(2)+2bx+2c` `therefore f(alpha)a^(2)alpha^(2)+2balpha+2c` `=a^(2)alpha^(2)-2a^(2)alpha^(2)=-a^(2)alpha^(2)" "["from Eq."(i)]` and `f(beta)=alpha^(2)beta^(2)+2b beta+2c` `= a^(2)beta^(2)+2a^(2)beta^(2)=3a^(2)beta^(2)" "["from Eq."(ii).]` `implies f(alpha) f(beta)lt0` f(x) must have a root laying in the open interval `(alpha, beta).` `therefore alpha lt gamma lt beta` |
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| 138. |
Let `a ,b ,c`be real numbers with `a!=0a n dl e talpha,beta`be the roots of the equation `a x^2+b x+c=0.`Express the roots of `a^3x^2+a b c x+c^3=0`in terms of `alpha,betadot` |
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Answer» Correct Answer - `x=alpha^(2)beta,alphabeta^(2)` Since, `ax^(2)+bx+c=0` has roots `alpha and beta.` `impliesalpha+beta=-b//a` and ` alphabeta=c//a` `Now, a^(3)x^(2)+abcx+c^(3)=0" "...(i)` On dividing the equation by `c^(2),` we get `(a^(3))/(c^(2))x^(2)(abcx)/(c^(2))+(c^(3))/(c^(2))=0` `impliesa((ax)/(c))^(2)+b((ax)/(c))+c=0` `impliesx=c/aalpha,c/abeta` are the roots `implies x=alphabeta, alpha beta beta` are the roots `impliesx=alpha^(2)beta,alphabeta^(2)` are teh roots Divide the Eq. (i) by `a^(3),` we get `x^(2)+b/c.c/ax+((c)/(a))^(3)=0` `impliesx^(2)-(alpha+beta)(alphabeta)x+(alphabeta)^(3)=0` `impliesx^(2)-alpha^(2)betax-alpha beta ^(2)x+(alphabeta)^(2)=0` `impliesx(x-alpha^(2)beta)-alphabeta^(2)(x-alphabeta^(2))=0` `implies (x-alpha^(2)beta)(x-alphabeta^(2))=0` `impliesx=alpha^(2)beta,alpha beta^(2)` which is the required answer. Alternate Solution Since, `a^(3)x^(2)+abcx+c^(3)=0` `impliesx=(-abcpmsqrt((abc)^(2)-4.a^(3).c^(3)))/(2a^(3))` `impliesx=(-(b//a)(c//a)pmsqrt((b//c)^(2)(c//a)^(2)-4(c//a)^(3)))/(2)` `impliesx=((alpha+beta)(alphabeta)pmsqrt((alpha+beta)^(2)(alphabeta)^(2)-4(alphabeta)^(3)))/(2)` `impliesx=((alpha+beta)(alphabeta)pmalphabetasqrt((alpha+beta)^(2)-4alphabeta))/(2)` `impliesx=alphabeta[((alpha+beta)pmsqrt((alpha-beta)^(2)))/(2)]` `impliesx=alphabeta[((alpha+beta)pmsqrt((alpha-beta)))/(2)]` `impliesx=alphabeta[(alpha+beta+alpha-beta)/(2),(alpha+beta-alpha+beta)/(2)]` `impliesx=alpha beta[(2alpha)/(2),(2 beta)/(2)]` `impliesx=alpha^(2)beta, alpha beta^(2)` which is the required answer. |
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| 139. |
The largest interval for which`x^(12)+x^9+x^4-x+1>0``-4A. `-4ltx le0`B. `0ltx lt1`C. `-100 lt x lt100`D. `-ooltxltoo` |
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Answer» Correct Answer - D Given, `x^(12)-x^(9)+x^(4)-x+1gt0` here, three cases arises: Case I When `x le0impliesx^(12)gt0,-x^(9)gt0,x^(4)gt0,-xgt0` `thereforex^(12)-x^(9)+x^(4)-x+1 gt 0,AA x le0" "...(i)` Case II When `0 lt x le 1` `x^(9)lt x^(4)and x lt1implies-x^(9)+x^(4)gt0and a-xgt0` `thereforex^(12)-x^(9)+x^(4)-x+1gt0 ltx le1` Case III When `x gt 1 implies x^(12)gtx^(9)and x^(4)gtx` `thereforex^(12)-x^(9)+x^(4)-x+1gt0,AAxgt1" "...(iii)` From Eqs. (i), (ii), (iii), the above equation holds for all ` x in R.` |
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| 140. |
Consider the polynomial f`(x)= 1+2x+3x^2+4x^3`. Let s be the sum of all distinct real roots of `f(x)`and let `t= |s|`.A. `(-(1)/(4),0)`B. `(-11,-(3)/(4))`C. `(-(3)/(4),(1)/(2))`D. `(0,(1)/(4))` |
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Answer» Correct Answer - C Given, `f(x)4x^(3)+3x^(2)+2x+1` `f(x)=2(6x^(2)+3x+1)` `impliesD=9-24lt0` Hence, `f(x)=0` has only one real root. `f(-(1)/(2))=1-1+3/4-4/8gt0` `f(-(3)/(4))=1-6/4+27/16-108/64` `=(64-96+108-108)/(64)lt0` f(x) changes its sign in `(-(3)/(4),-(1)/(2))` Hence, f(x)=0 has a root in `(-(3)/(4),-(1)/(2)).` |
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| 141. |
If the equations `x^(2)+2lambdax+lambda^(2)+1=0`, `lambda in R` and `ax^(2)+bx+c=0` , where `a`, `b`, `c` are lengths of sides of triangle have a common root, then the possible range of values of `lambda` isA. `(0,2)`B. `(sqrt(3),3)`C. `(2sqrt(2),3sqrt(2))`D. `(0,oo)` |
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Answer» Correct Answer - A `(a)` `(x+lambda)^(2)+1=0` has clearly imaginery roots So, both roots of the equations are common `:. (a)/(1)=(b)/(2lambda)=(c )/(lambda^(2)+1)=k`(Say) Then `a=k`, `b=2lambdak`, `c=(lambda^(2)+1)k` As `a`, `b`, `c` are sides of triangle `a+b gt implies 2lambda+1 gt lambda^(2)+1implieslambda^(2)-2lambda lt 0` `implies lambda in (0,2)` The other conditions also imply same relation. |
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| 142. |
`4 x^2-2 x+a=0,` has two roots lies in`(-1,1)` then ? |
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Answer» Let `f(x)=4x^(2)-2x+a` as both roots of the equation `f(x)=0` are lie between `(-1,1)` we can take `D ge0` `af(-1)gt0,af(1)gt0` and `-1lt1/4lt1` (i) Consider `Dge0` `(-2)^(2)-4.4.age0impliesale1/4`…….i (ii) Consider `af(-1)gt0` `4(4+2+a)gt0` `impliesage-6impliesa epsilon(-6,oo)`.............ii (iii) Consider `af(1)gt0` `4(4-2+a)gt0impliesagt-2` `impliesa epsilon(-2,oo)`.........iii Hence the values of `a` satisfying Eqs. (i), (ii) and (iii) at the same time are `a epsilon(-2,1/4]` |
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| 143. |
The number of real solutions of `|x|+2sqrt(5-4x-x^2)=16`is/area. 6b. 1 c.0 d. 4A. 6B. 1C. 0D. 4 |
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Answer» Correct Answer - 3 We have `2sqrt(5 - 4x - x^(2)) = 16 - |x|` `rArr 4(5 - 4x - x^(2)) = 256 - 32|x| + x^(2)` `rArr 5x^(2) - 32|x| + 16x + 236 = 0` `rArr 5x^(2) - 16x + 236 = 0`, when `x ge 0` and `5x^(2) + 48x + 236 = 0`, when `x lt 0`. In both cases, equation has non-real roots. |
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| 144. |
If `x^2+a x-3x-(a+2)=0`has real and distinct roots, then the minimum value of `(a^2+1)/(a^2+2)`isA. 1B. 0C. `1/2`D. `1/4` |
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Answer» Correct Answer - 3 `D gt 0` `rArr (a - 3)^(2) + 4(a + 2) gt 0` or `a^(2) - 6a + 9 + 4a + 8 gt 0` or `a^(2) - 2a + 17 gt 0` `rArr a in R` `therefore (a^(2) + 1)/(a^(2) + 2) = 1 - (1)/(a^(2) + 2) ge 1/2` |
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| 145. |
If `x`is real, then `x//(x^2-5x+9)`lies between`-1a n d-1//11`b. `1a n d-1//11`c. `1a n d1//11`d. none of theseA. `-1 and -1//11`B. `1 and -1//11`C. `1 and 1//11`D. none of these |
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Answer» Correct Answer - 2 Let `(x)/(x^(2) - 5x + 9) = y` or `yx^(2) - 5yx + 9y = x` or `yx^(2) - (5y + 1)x + 9y = 0` Now, x is real, so `D ge 0` `rArr (-(5y + 1))^(2) - 4 * y *(9y) ge 0` or `-11y^(2) + 10y + 1 ge 0` or `11y^(2) - 10y - 1 le 0` or `(11y + 1) (y - 1) le 0` or `-1/11 le y le 1` |
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| 146. |
Find the range of `f(x)=x^2-x-3.` |
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Answer» Correct Answer - `[-13//4, infty)` `f(x) = x^(2) - x - 3 (x - (1)/(2))^(2) - (1)/(4) - 3 = (x - (1)/(2))^(2)-(13)/(4)` ` (x - (1)/(2))^(2 )ge 0, AA x in` R ` (x - (1)/(2))^(2 )-(13)/(4)ge (-13)/(4), AA x in`R Hence, the range is `[13//4. infty)`. |
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| 147. |
In the given figure, vertices of `Delta ABC` lie on `y = f(x) = ax^2 + bx + c.` The `DeltaABC` is right angled isosceles triangle whose hypotenuse `AC = 4sqrt2` units. Number of integral values of `k` for which one root of `f(x) = 0` is more than `k` and other less than `k`A. 6B. 4C. 5D. 7 |
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Answer» Correct Answer - 3 `f(x) = 0` `rArr x^(2)//(2sqrt(2)) - 2sqrt(2) = 0` or `x = pm 2sqrt(2)` Therefore, number of integral values of K for which for which k lies in `(-2sqrt(2), 2sqrt(2))` is 5. |
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| 148. |
The interval of `a`for which the equation `t a n^2x-(a-4)tanx+4-2a=0`has at least one solution `AAx in [0,pi//4]``a in (2,3)`b. `a in [2,3]`c. `a in (1,4)`d. `a in [1,4]`A. ` a in (2,3)`B. `a in [2, 3]`C. ` a in (1, 4)`D. ` a in [1 ,4]` |
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Answer» Correct Answer - 2 `tan x = (a - 4 pm sqrt(a - 4)^(2) - 4 (4 - 2a))/(2)` ` = (a - 4 pm a)/(2) = a - 2 , - 2` `therefore tan x = a- 2 ( because tan x ne - 2)` ` because a in [ 0 , (pi)/(4)]` `because 0 le a - 2 le ` `rArr 2 le a le 3` |
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| 149. |
Find the least value of `n`such that `(n-2)x^2+x+n+4>0,AAx in R ,w h e r en in Ndot` |
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Answer» `(n-2)x^(2) + 8 x + n + 4 gt 0, AA x in `R `rArr D = 64 - 4 (n- 2) (n + 4) lt 0 and n - 2 gt 0` `rArr 16 - (n^(2) + 2n - 8) lt 0 and n gt 2` `rArr n^(2) + 2n - 24 gt 0 and n gt 2` `rArr (n + 6) (n - 4) gt 0 and n gt 2` `rArr n gt 4 as n in N and n gt 2` `rArr n ge 5` Hence, the least value of n is 5. |
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| 150. |
Find the values of `k`for which`|(x^2+k x+1)/(x^2+x+1)| |
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Answer» We have `|x| lt a `rArr - a lt x lt + a` Therefore, the given inequality implies `-2lt(x^(2) + kx + 1)/(x^(2) + x + 1) lt 2` (2) Now, `x^(2) + x +1 = (x+ 1//x)^(2) + (3//4)` is positive for all values of x. Multiplying (1) by `x^(2) + x + 1,` we get ` -2(x^(2) + x+1) lt x^(2) + kx + 1 lt 2(x^(2) + x + 1)` This yields two inequalities, viz., `3x^(2) + (2+k)x + 3) gt`0 and `x^(2) + (2-k) x + 1 lt 0` For above inequalities to be. ture for all values of x, their discrimainats must be negative . Hence, `(2+k)^(2) - 36 lt 0 and (2 - k)^(2) - 4 lt 0` (2) `rArr (k+ 8) (k - 4) lt 0 and k (k - 4) lt 0` (3) `rArr - 8 lt k lt 4 and 0 lt k lt 4` Therefore, `0lt k lt 4` |
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