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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 `x^(2//3) -7 x^(1//3) + 10 = 0`, then the set of values of x, isA. {12, 5}B. {8}C. `phi`D. {8, 125} |
| Answer» Correct Answer - D | |
| 102. |
The value of k for which the equation `(k-2) x^(2) + 8x + k + 4 = 0` has both roots real, distinct and negative, is |
| Answer» Correct Answer - C | |
| 103. |
Let `(sin a) x^(2) + (sin a) x + 1 - cos a = 0`. The set of values of a for which roots of this equation are real and distinct, isA. `(0, 2 tan^(-1)(1)/(4))`B. `(o, (2pi)/(3))`C. `(0, pi)`D. `(0, 2pi)` |
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Answer» Correct Answer - A The roots of the given equation will be real and distinct, iff `sin^(2) a - 4 sin a(1-cos a) gt 0` `rArr" "(1-cos a){1+cos a - 4 sin a} gt 0` `rArr" "2 cos^(2)(a)/(2)-8 sin (a)/(2) cos (a)/(2) gt 0" "[because 1-cos a gt 0]` `rArr" "2 cos^(2)(a)/(2)(1-4"tan"(a)/(2))gt 0` `rArr" "4 "tan"(a)/(2) lt 1 rArr -(pi)/(2) lt (a)/(2) lt "tan"^(-1)(1)/(4) rArr -pi lt a lt 2 "tan"^(-1)(1)/(4)` Hence, option (a) is correct. |
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| 104. |
The smallest value of k, for which both the roots of the equation, `x^2-8kx + 16(k^2-k + 1)=0` are real, distinct and have values at least 4, isA. 2B. 3C. 4D. none of these |
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Answer» Correct Answer - A Let `f(x) = x^(2) - 8kx + 16 (k^(2)-k+1)`. Both the roots of the equation f(x) = 0 will be real, distinct and have values at least 4, if (i) `D gt 0` (ii) `f(4) ge 0` (iii) Vertex of u = f(x) lies on the right side of (4, 0) Now, (i) `D gt 0 rArr 64k^(2) - 64(k^(2) - k + 1) gt 0 rArr k - 1 gt 0 rArr k gt 1` (ii) `f(4) ge 0` `rarr" "16 - 32 k + 16 (k^(2)-k+1) ge 0` `rArr" "k^(2) - 3k + 2 ge 0 rArr k le 1 or k ge 2` (iii) x-coordinate of vertex `gt` 0 i.e.`" "4k gt 4 rArr k gt 1" "[because "x coordinate" = -(b)/(2a)]` Taking intersection of these, we have `k in [2, oo)`. Hence, the least value of k is 2. |
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| 105. |
All the values of `m`for whilch both the roots of the equation `x^2-2m x+m^2-1=0`are greater than -2 but less than 4 lie in the interval`-23`c. `-1A. `(-2, 0)`B. `(3, oo)`C. `(-1, 3)`D. (1, 4) |
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Answer» Correct Answer - C We have, `x^(2) - 2mx + m^(2) -1 = 0` `rArr" "(x-m)^(2) = 1 rArr x - m = +- 1 rArr x = m +-1` We have, `-2 lt m - 1 lt 4 and -2 lt m + 1 lt 4` `rArr" "-1 lt m lt 5 and -3 lt m lt 3 rArr -1 lt m lt 3` |
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| 106. |
If a, b are the real roots of `x^(2) + px + 1 = 0` and c, d are the real roots of `x^(2) + qx + 1 = 0`, then `(a-c)(b-c)(a+d)(b+d)` is divisible byA. `a - b - c - d`B. `a + b + c - d`C. `a + b + c + d`D. `a - b - c - d` |
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Answer» Correct Answer - C We have, `a+b = -p, ab = 1, c+d = -q and cd = 1`. Now, `(a-c)(b-c)(a+d)(b+d)` `=(a-c)(b+d)(b-c)(a+d)` `=(ab+ad-bc-cd)(ba+bd-ca-cd)` `=(ad-bc)(bd-ca)` `=abd^(2)+abc^(2)-cda^(2)-b^(2)cd` `=d^(2)+c^(2)-a^(2)-b^(2)` `=(c+d)^(2) -(a+b)^(2)" "[because ab = cd = 1]` `=(c+d+a+b)(c+d-a-b)` `=-(a+b+c+d)(a+b-c-d)` Hence, (a-c)(b-c)(a+d)(b+d) is divisible by a + b + c + d and a + b - c - d. |
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| 107. |
The equation `x^(3) - 3x + 1 = 0` hasA. no rational but three irrational rootsB. one rational and two irrational rootsC. no real rootsD. three rational roots |
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Answer» Correct Answer - A If possible, let `(p)/(q)` be a rational roots of `x^(3) - 3x + 1 = 0`. Then, p is a factor of the constant term i.e. 1 and q is a factor of coefficient of `x^(3)` i.e. 1. `therefore" "p = +- 1 and q = +- 1 rArr (p)/(q) = 1, -1` But, 1 and -1 do not satisfy the given equation. Hence, the given equation has no rational roots. |
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| 108. |
The equation `|x+1||x-1|=a^(2) - 2a - 3` can have real solutions for x, if a belongs toA. `(-oo, -1]uu[3, oo)`B. `[1 - sqrt(5), 1 + sqrt(5)]`C. `[1-sqrt(5), 1] uu [3, 1 + sqrt(5)]`D. none of these |
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Answer» Correct Answer - C We have, `|x+1||x-1|=a^(2)-2a - 3" "...(i)` `rArr" "|x^(2)-1|=a^(2) - 2a -3` `rArr" "x^(2) -1 = +-(a^(2) -2a - 3)` `rArr" "x^(2) = a^(2) - 2a - 2, -a^(2) + 2a + 4` Thus, for equation (i) to have real solutiions, we must have `(a^(2)-2a - 3 ge 0 and a^(2) - 2a - 2 ge 0)` `or," "(a^(2)-3a-3 ge 0 and -a^(2) + 2a + 4 ge 0)` `or," "((a-3)(a+1) ge 0 and (a-1-sqrt(3))(a-1+sqrt(3)) ge 0)` `or," "((a-1+sqrt(5))(a-1-sqrt(5)) le 0 and (a-3)(a+1) ge 0)` `rArr" "a in (-oo, -1] uu [3, oo) or, a in [1-sqrt(5,)1 ] uu [3, 1 + sqrt(5)]` |
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| 109. |
If the equation `(1)/(x) + (1)/(x+a)=(1)/(lambda)+(1)/(lambda+a)` has real roots that are equal in magnitude and opposite in sign, thenA. `lambda^(2) = 3a^(2)`B. `lambda^(2) = 2a^(2)`C. `lambda^(2) = a^(2)`D. `a^(2) = 2 lambda^(2)` |
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Answer» Correct Answer - D We have, `(1)/(x)+(1)/(x+a)=(1)/(lambda)+(1)/(lambda+a)" "...(i)` Clearly, `x = lambda` is a root of this equation. It is given that equation (i) has real roots that are equal in magnitude and opposite in sign. Therefore, `x = - lambda` is a root of equation (i). `therefore" "-(1)/(lambda)+(1)/(a-lambda)=(1)/(lambda)+(1)/(lambda + a)` `rArr" "(2)/(lambda)=(1)/(a-lambda)-(1)/(a+lambda)` `rArr" "(2)/(lambda)=(2 lambda)/(a^(2)-lambda^(2)) rArr a^(2) - lambda^(2) = lambda^(2) rArr a^(2) = 2 lambda^(2)` |
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| 110. |
If two roots of the equation `x^3-px^2+qx-r=0` are equal in magnitude but opposite in sign, then:A. r = pqB. `r = 2p^(3) + pq`C. `r = p^(2)q`D. none of these |
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Answer» Correct Answer - A Let `alpha, beta, gamma` be the roots of the given equation such that `alpha = - beta`. Then, `alpha + beta + gamma = p rArr gamma = p`. Since `gamma` is a root of the given equation, so it satisfies the equation i.e. `gamma^(3) - p gamma^(2) + q gamma - r = 0 rArr p^(3) - p^(3) + pq - r = 0 rArr r = pq` |
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| 111. |
The value of m for which the equation `x^3-mx^2+3x-2=0` has two roots equal rea magnitude but opposite in sign, isA. `4//5`B. `3//4`C. `2//3`D. `1//2` |
| Answer» Correct Answer - C | |
| 112. |
If the equation `a/(x-a)+b/(x-b)=1`has two roots equal in magnitude and opposite in sign then the value of `a+b` isA. `-1`B. 0C. 1D. none of these |
| Answer» Correct Answer - B | |
| 113. |
The roots of `ax^(2) - bx + 2x = 0` are in the ratio of `2 : 3`, then `"____"`.A. `a^(2) = bc`B. `3b^(2) = 25 ac`C. `2b^(2) = 75 c`D. `5b^(2) = ac` |
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Answer» Correct Answer - B Let `alpha, beta` be the roots of `ax^(2 - bx + 2c = 0` Given `= alpha/beta = 2/3` `rArr = (2beta)/(3)` Product of the roots `(2beta)/(3) x beta = (2c)/(a)` `beta^(2) = (3c)/(a) " "(1)` Sum of the roots `= alpha + beta` `rArr (2beta)/(3) + beta = (b)/(a)` `rArr (5beta)/(3) = b/a` `beta = (3b)/(5a)` `beta^(2) = (9b^(2))/(25a^(2))" "(2)` From Eqs. (1) and (2) `(3c)/(a) = (9b^(2))/(25a^(2)) 3b^(2) = 25 ac`. |
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| 114. |
If the sum of the roots of a quadratic equation, is positive and product of the roots is negative, the numerically greater root has `"_____"` sign. [positive/negative] |
| Answer» Correct Answer - positive | |
| 115. |
If roots of `ax^(2) + bx + c = 0`are 2 , more than the roots of `px^(2) + qx + r = 0`, then the value of c in terms of p,q and r isA. `p+q+r`B. `4p-2q+r`C. `3p-q+2r`D. `2p+q-r` |
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Answer» Correct Answer - B Let `f(x) -= px^(2) + qx + r = 0` Given the quadratic equation whose roots are 2 more than the root of `f(x)` as `ax^(2) +bx + c = 0` `rArr f(x-2) = ax^(2) + bx + c` `rArr px^(2) - 4px + 4p + qx - 2q + r = ax^(2) + bx + c` `rArr px^(2) + (q-4p) x + (4p-2q+r) = ax^(2) + bx + c` `rArr c= 4p - 2q + r.` |
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| 116. |
If `(1-p)`is a root of quadratic equation `x^2+p x+(1-p)=0,`then find its roots.A. `-1, 2`B. `-1, 1`C. `0, -1`D. 0, 1 |
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Answer» Correct Answer - C Since (1-p) is a root of the given equation. `therefore" "(1-p)^(2) + p(1-p) + (1-p) = 0` `rArr" "(1-p) {(1-p)+p+1} = 0 rArr 2 (1-p) = 0 rArr p = 1` Substituting p = 1 in the given equation, we get `x^(2) + x = 0 rArr x = 0, -1` Hence, 0 and -1 are the roots of the given equation. |
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| 117. |
If `x+3` is the common factor of the expreassions `ax^(2) + bx + 1` and `px^(2) + qx - 3`, then `- ((9a+3p))/(3b+q) = "______"`.A. -2B. 2C. 3D. -1 |
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Answer» Correct Answer - D (i) If `x + k` is the common rot |
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| 118. |
If one root is square of the other root of the equation `x^2+p x+q=0`, then the relation between `pa n dq`is (2004, 1M)`p^3-(3p-1)q+q^2=0``p^3-q(3p+1)+q^2=0``p^3+q(3p-1)+q^2=0``p^3+q(3p+1)+q^2=0`A. `p^(3) - (3p - 1) q + q^(2) = 0`B. `p^(3) - (3p + 1) q + q^(2) = 0`C. `p^(3) + (3p - 1) q + q^(2) = 0`D. `p^(3) + (3p + 1)q + q^(2) = 0` |
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Answer» Correct Answer - A Let `alpha, alpha^(2)` be the roots of the equation `x^(2) + px + q = 0`. Then, `alpha + alpha^(2) = - p and alpha xx alpha^(2) = q` `rArr" "alpha + alpha^(2) = -p and alpha = q^(1//3)` `rArr" "q^(1//3) + q^(2//3) = -p` `q + q^(2) - 3pq = - p^(3) rArr p^(3) - q (3p -1) q^(2) = 0` |
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| 119. |
Ilf `b_1, b_2=2(c_1+c_2,)`then at least one of the equation `x^2+b_1x+c_1=0a n dx^2+b_(2x)+c_2=0`hasa. imaginary roots b.real rootsc. purely imaginary roots d. none of theseA. real rootsB. purely imaginary rootsC. imaginary rootsD. none of these |
| Answer» Correct Answer - A | |
| 120. |
If `f(x)=a x^2+b x+c ,g(x)=-a x^2+b x+c ,w h e r ea c!=0,`then prove that `f(x)g(x)=0`has at least two real roots.A. at least three real rootsB. no real rootsC. at least two real rootsD. two real roots and two imaginary roots |
| Answer» Correct Answer - C | |
| 121. |
`alpha,beta` be the roots of the equation `x^2-px+r=0` and `alpha/2 , 2beta` be the roots of the equation `x^2-qx+r=0` then value of `r` isA. `(2)/(9)(p-q)(2q-p)`B. `(2)/(9)(q-p)(2p-q)`C. `(2)/(9)(q-2p)(2q-p)`D. `(2)/(9)(2p-q)(2q-p)` |
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Answer» Correct Answer - D Since `alpha, beta` are the roots of `x^(2) - px + r = 0` `therefore" "(alpha)/(2)+2beta = q and (alpha)/(2) xx 2 beta r` Solving `alpha + beta = p and (alpha)/(2)+2 beta = q`, we get `alpha =(2)/(3)(2p-q) and beta = (1)/(3)(2q-p)` `therefore" "alpha beta = r rArr r = (2)/(9) (2p-q) (2q-p)` |
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| 122. |
The square of one-sixth of the number of students in a class are studying in the library and the remaining eight students are playing in the ground. What is the total number of students of the class ? |
| Answer» Correct Answer - 12 or 24 | |
| 123. |
If `alpha` and `beta` are the roots of the equation `x^2-p(x+1)-q=0` then the value of `(alpha^2+2alpha+1)/(alpha^2+2alpha+q)` + `(beta^2+2beta+1)/(beta^2+2beta+q)` is(A)1 (B) 2 (C) 3 (D) 0A. 1B. 2C. 3D. 0 |
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Answer» Correct Answer - A The given equation is `x^(2) - px - (p+q) = 0` `therefore" "alpha+beta = p and alpha beta = -(p+q)` `rArr" "alpha beta + alpha + beta = - q` `rArr" "alpha beta + alpha + beta + 1 = - q+1` `rArr" "(alpha + 1)(beta + 1) = 1 - q" "....(i)` Now, `(alpha^(2)+2 alpha + 1)/(alpha^(2)+2 alpha + q)+(beta^(2)+2 beta+1)/(beta^(2)+2 beta + q)` `=((alpha+1)^(2))/((alpha+1)^(2)+q-1)+((beta+1)^(2))/((beta+1)^(2)+(q-1))` `=((alpha+1)^(2))/((alpha+1)^(2)-(alpha+1)(beta+1))+((beta+1)^(2))/((beta+1)^(2)-(alpha+1)(beta+1))" "["using (i)"]` `(alpha+1)/(alpha+beta)+(beta+1)/(beta-alpha)=(alph-beta)/(alpha-beta) = 1`. |
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| 124. |
If `alpha` and `beta` are the roots of `2x^(2) - x - 2 = 0`, then `(alpha^(3) + beta^(-3) + 2alpha^(-1) beta^(-1))` is equal toA. `-(17)/(8)`B. `(23)/(6)`C. `(37)/(9)`D. `-(29)/(8)` |
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Answer» Correct Answer - D () simplify the required expression and find `alpha + beta ` and `alphabeta`. (ii) Find the sum of the roots and product of the roots. (iii) Use relation, `a^(3) + b^(3) = (a+b)^(3) - 3ab(a+b)`. |
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| 125. |
If `alpha, beta` are the roots of the quadratic equation `lx^(2) + mx + n = 0`,then evaluate the following expressions. (a) `alpha^(2) + beta^(2)` (b) `alpha/beta + beta/alpha` (c ) `1/(alpha^(3)) + 1/(beta^(3))` |
| Answer» Correct Answer - (a) `(m^(2)-2ln)/(l^(2))`, (b) `(m^(2) - 2ln)/(ln)`, (c ) `(3lmn - m^(3))/(n^(3))` | |
| 126. |
If the roots of the quadratic equation `x^2+p x+q=0`are `tan30^0a n dtan15^0,`respectively, then find the value of`2+q-pdot`A. 2B. 3C. 0D. 1 |
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Answer» Correct Answer - B It is given that `tan 30^(@) and tan 15^(@)` are roots of the given equation. `therefore" "tan 30^(@) + tan 15^(@) = - p and tan 30^(@) tan 15^(@) = q` Now, `tan 45^(@) = (tan 30^(@) + tan 15^(@))/(1-tan 30^(@) tan 15^(@))` `rArr" "1 = (-p)/(1-q) rArr 1-q = - p rArr q-p = 1 rArr q-p + 2 = 3` |
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| 127. |
Let `a ,b ,c`be real. If `a x^2+b x+c=0`has two real roots `alphaa n dbeta,w h e r ealpha1`, then show that `1+c/a+|b/a| |
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Answer» Correct Answer - A Let `f(x) = ax^(2) + bx + c`. Clearly, -1 and 1 lie between the roots of f(x) = 0. `therefore" "a f(1) lt 0 and a f(-1) lt 0` `rArr" "a(a+b+c) lt 0 and a(a-b-c) lt 0` `rArr" "a^(2)(1+(b)/(a)+(c)/(a)) lt 0 and a^(2) (1-(b)/(a)+(c)/(a)) lt 0` `rArr" "1+(b)/(a)+(c)/(a) lt 0 and 1 -(b)/(a)+(c)/(a) lt 0 rArr 1+|(b)/(a)|+(c)/(a) lt 0` |
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| 128. |
If `2alpha` and `3beta` are the roots of the equation `x^(2) + az +b = 0`, then find the equation whose roots are `a,b`. |
| Answer» Correct Answer - `x^(2) - (6alphabeta-2alpha-3beta) x - (2alpha+3beta)(6alphabeta) = 0` | |
| 129. |
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 ________.A. 1B. 0C. -1D. none of these |
| Answer» Correct Answer - C | |
| 130. |
Let `alphaa n dbeta`be the roots of `x^2-x+p=0a n dgammaa n ddelta`be the root of `x^2-4x+q=0.`If `alpha,beta,a n dgamma,delta`are in G.P., then the integral values of `pa n dq`, respectively, are`-2,-32`b. `-2,3`c. `-6,3`d. `-6,-32`A. `-2, -32`B. `-2, 3`C. `-6, 3`D. `-6, -32` |
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Answer» Correct Answer - A Since `alpha, beta` are the roots of the equation `x^(2) - x + p = 0 and gamma, delta` are the roots of the equation `x^(2) - 4x + q = 0`. Therefore, `alpha + beta = 1, alpha beta = p, gamma + delta =4, gamma delta = q` Let r be the common ratio of the G.P. `alpha, beta, gamma, delta`. Then, `beta = alpha r, gamma = alpha r^(2) and delta = alpha r^(3)`. Now, `alpha + beta = 1 rArr alpha + alpha r = 1` `gamma + delta = 4 rArr alpha r^(2) + alpha r^(3) = 4` `therefore" "(alpha r^(2) + alpha r^(3))/(alpha + alpha r) = 4 rArr r = +- 2`. CASE I When r = 2 In this case, we have `alpha + beta = 1` `rArr" "alpha + alpha r = 1 rArr 3 alpha = 1 rArr alpha = (1)/(3)" "[because r = 2]` `therefore" "p = alpha beta = alpha^(2) r = (2)/(9)` But, p is an integer. Therefore, r = 2 is not possible. CASE II When r = - 2 In this case, we have `alpha + beta = 1` `rArr" "alpha + alpha r = 1 rArr -alpha = 1 rArr alpha = -1" "[because r = -2]` `therefore" "alpha + beta = 1 rArr beta = 2` Now, `p = alpha beta rArr p = - 2` and, `q = gamma delta = (alpha r^(2))(alpha r^(3)) = - 32` Hence, `p = -2 and q = -32`. |
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| 131. |
The number of values of `a`for which equations `x^3+a x+1=0a n dx 64+a x^2+1=-`have a common root is`0`b. `1`c. `2`d. InfiniteA. 2B. -2C. 0D. none of these |
| Answer» Correct Answer - B | |
| 132. |
If the roots of the quadratic equation `x^(2) - 3x - 304 = 0` are `alpha` and `beta`, then the quadratic equation with roots `3alpha` and `3beta` isA. `x^(2) + 9x - 2736 = 0`B. `x^(2) - 9x - 2736 = 0`C. `x^(2) - 9x+ 2736 = 0`D. `x^(2) + 9x +2736 = 0` |
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Answer» Correct Answer - B (i) The product equation with thrice the roots of `f(x) = 0` as roots is `f(1/x) = 0`. (iii) The quadratic equation whose roots are m times of the roots of the equation `f(x) = 0` is `(-1)/(12), (55)/(18) = 0` |
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| 133. |
If `x^2=a x+10=0a n dx^2+b x-10=0`have common root,then `a^2-b^2`is equal to`10`b. `20`c. `30`d. `40`A. 10B. 20C. 30D. 40 |
| Answer» Correct Answer - D | |
| 134. |
If `alpha`is a root of the equation `x^2+2x-1=0,`then prove that `4alpha^2-3alpha`is the other root.A. `3 alpha^(3) - 4 alpha`B. `-2 alpha(alpha+1)`C. `4 alpha^(3) - 3 alpha`D. none of these |
| Answer» Correct Answer - C | |
| 135. |
If one root of the equation `ax^2+bx+c=0` is double the other, then the relation between `a,b,c` isA. `b^(2) = 9ac`B. `2b^(2) = 9ac`C. `2b^(2) = ac`D. `b^(2) = ac` |
| Answer» Correct Answer - B | |
| 136. |
If the quadratic equation `ax^2 + 2cx + b = 0 and ax^2 + 2bx + c = 0 (b!=c)` have a common root, then `a + 4b + 4c =`A. -2B. -1C. 0D. 1 |
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Answer» Correct Answer - C Let `alpha` be the common root of the given equations. Then, `a alpha^(2) + 2 c alpha + b = 0 and, a alpha^(2) + 2 b alpha + c = 0` `2 alpha(c-b) + (b-c) = 0" "["On subtracting"]` `alpha =(1)/(2)" "[because b ne c]` Putting `alpha = 1//2 "in" a alpha^(2) + 2 c alpha + b = 0`, we get `a + 4b + 4c = 0`. |
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| 137. |
Let a, b, c be real numbers such that `ax^(2) + bx + c = 0 and x^(2) + x + 1 = 0` have a common root. Statement-1: a = b = c Staement-2: Two quadratic equations with real coefficients cannot have only one imainary root common.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - A The equation `x^(2) + x + 1 = 0` has imaginary roots `omega and omega^(2). If ax^(2) + bx + c = 0 and x^(2) + x + 1 = 0` has one imaginary root common, then other imaginary root must also be common as imaginary roots occur in pairs. So, statement-2 is true. Given equations have both roots common. `therefore" "(a)/(1)=(b)/(1)=(c)/(1) rArr a = b = c` So, statement-1 is also true and statement-2 is a correct explanation for statement-1. |
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| 138. |
If one root of the quadratic equation `(a-b)x^2+ax+1=0` is double the other root where `a in R`, then the greatest value of b isA. `9//8`B. `7//8`C. `8//9`D. `8//7` |
| Answer» Correct Answer - A | |
| 139. |
If the equations `x^2+bx-1=0` and `x^2+x+b=0` have a common root different from `-1` then `|b|` is equal toA. `sqrt(2)`B. 2C. `sqrt(3)`D. 3 |
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Answer» Correct Answer - C Let `alpha` a common root different from -1. Then, `alpha^(2) + b alpha - 1 = 0 and alpha^(2) + alpha + b = 0` `rArr" "(alpha^(2) + b alpha -1)-(alpha^(2) + alpha + b)=0 rArr alpha(b-1) = b+1 rArr alpha = (b+1)/(b-1)` Putting this value of `alpha "in" alpha^(2) + b alpha - 1 = 0`, we obtain `(b+1)^(2) + b(b^(2) -1) - (b-1)^(2) = 0 rArr b^(3) + 3b - 0 rArr b = 0, b = +- i sqrt(3)` For b = 0, `alpha^(2) + b alpha - 1 = 0` gives `alpha = - 1`. So b `ne` 0. `therefore" "b = +-i sqrt(3) rArr |b| = sqrt(3)` |
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| 140. |
The values of the parameter a for which the quadratic equations `(1-2a) x^(2) - 6ax - 1 =0 and ax^(2) - x + 1 = 0` have at least one root in common, areA. `0, (1)/(2)`B. `(1)/(2), (2)/(9)`C. `(2)/(9)`D. `0, (1)/(2), (2)/(9)` |
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Answer» Correct Answer - C For a = 0 or a = 1/2, one of the quadratic equations becomes linear. So, `a ne 0, a ne (1)/(2)`. Hence, the only answer is a = 2/9. |
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| 141. |
If `a and b(!=0)` are the roots of the equation `x^2+ax+b=0` then the least value of `x^2+ax+b` isA. `(9)/(4)`B. `-(9)/(4)`C. `-(1)/(4)`D. `(1)/(4)` |
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Answer» Correct Answer - B Since a and b are the roots of the equation `x^(2) + ax + b = 0`. `therefore" "a+b =-a and ab = b` Now, `ab = b rArr (a-1)b=0 rArr a=1" "[therefore b ne 0]` Putting a = 1 in a + b = - a, we get b = - 2. Clearly, `y=x^(2)+ax + b` is a parabola opening upward. `therefore" "y_(min)=-(D)/(4)" "[therefore y_(min)=-(D)/(4a)"for" y=ax^(2)+bx+c]` `rArr" "y_(min)=-(a^(2)-4b)/(4)=-(9)/(4)` |
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| 142. |
25. The integer k for which the inequality `x^2 -2(4k 1)x 15k 2k-7 0 is valid for any real x is (2) 3 (3) 4 (4) 5A. 2B. 3C. 4D. none of these |
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Answer» Correct Answer - B Let `f(x) = x^(2) - 2(4k -1)x+15k^(2) - 2k -7`. Then, `f(x) gt 0` for all x `in` R `rArr" ""Disc" lt 0` `rArr" "4(4k-1)^(2)-4(15k^(2)-2k-7) lt 0` `rArr" "k^(2)-6k + 8 lt 0 rArr 2 lt k lt 4` |
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| 143. |
`a, b, c, in R, a ne 0` and the quadratic equation `ax^(2) + bx + c = 0` has no real roots, then which one of the following is not true?A. `a + b + c gt 0`B. `a(a+b+c)gt 0`C. `ac(a+b+c)gt 0`D. `c(a+b+c)gt 0` |
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Answer» Correct Answer - A Let `f(x) = ax^(2) + bx + c`. It is given that f(x) = 0 has no real roots. So, either `f(x) gt 0` for all x `in` R or f(x) `lt` 0 fir all x `in` R i.e. f(x) has same sign for all values of x. `therefore" "f(0) f(1) gt 0 rArr c(a+b+c) gt 0` and, `a f(1) gt 0 rArr a(a+b+c) gt 0`. Also, ac `gt` 0. `therefore" "ac f(1) gt 0 rArr ac (a+b+c) gt 0` |
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| 144. |
If the roots of `ax^(2) + bx + c = 0 (a gt 0)` be each greater than unity, thenA. a + b + c = 0B. a + b + c `gt` 0C. a + b + c `lt` 0D. none of these |
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Answer» Correct Answer - B Let `f(x) = ax^(2) + bx + c`. Since 1 lies outside the roots of `f(x) = 0`. So, `a f(1) gt 0` `rArr" "f(1) gt 0 rArr a + b + c gt 0" "[because a gt 0]` |
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| 145. |
If the roots of a quadratic equation `ax^(2) + bx + c` are complex, then `b^(2) lt "____"` |
| Answer» Correct Answer - 4ac | |
| 146. |
Let `f(x) =ax^(2) + bx + c and f(-1) lt 1, f(1) gt -1, f(3) lt -4 and a ne 0`, thenA. `a gt 0`B. `a lt 0`C. sign of a cannot be determinedD. none of these |
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Answer» Correct Answer - B We have, `f(-1) lt 1, f(1) gt -1 and f(3) lt -4` `rArr" "a-b+c lt 1" "...(i)` `rArr" "a+b+c gt -1" "...(ii)` `rArr" "9a + 3b + c lt -4" "...(iii)` From (ii), we have `-a -b -c lt 1" "...(iv)` Multiplying (i) by 3 and adding to (iii), we get `12a + 4c lt -1 rArr 3a + c lt -(1)/(4)" "...(v)` Multiplying (iv) by 3 and adding to (ii), we get `6a - 2c lt -1 rArr 3a - c lt -(1)/(2)" "...(vi)` Adding (v) and (vi), we get `6a lt -(3)/(4) rArr a lt -(1)/(8) rArr a lt 0`. |
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| 147. |
If the equation `ax^(2) + 2 bx - 3c = 0` has no real roots and `(3c)/(4) lt a + b`, thenA. `c lt 0`B. `c gt 0`C. `c ge 0`D. `c = 0` |
| Answer» Correct Answer - A | |
| 148. |
If a,b,c are positive real numbers, then the number of positive real roots of the equation `ax^(2)+bx+c=0` isA. are real and positiveB. real and negativeC. have negative real partD. have positive real part. |
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Answer» Correct Answer - C Roots of `ax^(2) + bc + c = 0` are given by `x = - (b)/(2a) +-(sqrt(b^(2)-4ac))/(2a)` We have, `a, b gt 0 rArr -b//2a lt 0`. If `b^(2) - 4ac lt 0`, then roots are imaginary of which real part is negative. If `b^(2) - 4ac gt 0`, then roots are real and negative since `sqrt(b^(2) - 4ac) lt b`. Hence, in either case, both the roots have negative real part. |
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| 149. |
If `ax^(2) + bx + 10 = 0` does not have two distinct real roots, then the least value of 5a + b, isA. -3B. -2C. 3D. none of these |
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Answer» Correct Answer - B It is given that `ax^(2) + bx + 10 = 0` does not have two distinct real roots. `therefore" "b^(2) - 40a le 0 rArr a ge (b^(2))/(40)` Let y = 5a + b. Then, `y = 5 xx (b^(2))/(40)+b=(b^(2)+8b)/(8)=(1)/(8) (b^(2) + 8b)` `rArr" "y=(1)/(8) (b+4)^(2) - 2 ge -2` Hence, the least value of 5a + b is -2. |
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| 150. |
If a, b, c are positive and a = 2b + 3c, then roots of the equation `ax^(2) + bx + c = 0` are real forA. `|(a)/(c)-11| ge 4 sqrt(7)`B. `|(c)/(a)-11| ge 4 sqrt(7)`C. `|(b)/(c)+4| ge 2 sqrt(7)`D. `|(c)/(b)-4| ge 2 sqrt(7)` |
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Answer» Correct Answer - A For real roots, we must have `b^(2) - 4ac ge 0` `rArr" "((a-3c)/(2))^(2)-4ac ge 0` `rArr" "a^(2)+9c^(2)-6ac-16ac ge 0` `rArr" "((a)/(c))^(2)-22((a)/(c))+9 ge 0 rArr ((a)/(c)-11)^(2) ge (4 sqrt(7))^(2)` `rArr" "|(a)/(c)-11| ge 4 sqrt(7)` Again, `b^(2) - 4ac ge 0 and a = 2b + 3c` `rArr" "b^(2)-4x(ab+3c)ge0` `rArr" "b^(2)-8bc-12c^(2) ge 0` `rArr" "((b)/(c))^(2)-8((b)/(c))-12 ge 0` `rArr" "((b)/(c)-4)^(2) ge (2 sqrt(7))^(2) rArr |(b)/(c)-4| ge 2 sqrt(7)` Hence, options (a) is true. |
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