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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 value of P for which both the roots of the equation `4x^2-20Px + (25P^2 + 15P-66) = 0` are less than 2, lies inA. `(4//5, 2)`B. `(-1, -4//5)`C. `(2, oo)`D. `(-oo, -1)` |
| Answer» Correct Answer - D | |
| 152. |
If `ax^2 + bx+ c = 0`, `a!=0`, `a, b, c in R` has distinct real roots in ( 1,2) then a and 5a + 2b + c have (A) same sign (B) opposite sign (C) not determined (D) none of theseA. of same typeB. of opposite typeC. undeterminedD. none of these |
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Answer» Correct Answer - A Let `alpha , beta` be the roots of `ax^(2)+bx+c = 0`. Then, `alpha + beta = -(b)/(a) and alpha beta = (c)/(a)` Now, `a(5a+2b+c) = a^(2)(5+2(b)/(a)+(c)/(a))` `rArr" "a(5a+2b+c)=a^(2){5-2(alpha+beta)+alpha beta}` `rArr" "a(5a+2b+c)=a^(2){alpha beta - 2(alpha + beta)+4 + 1}` `rArr" "a(5a+2b+c)=a^(2){(alpha-2)(beta-2)+1} gt 0` `rArr" "a(5a + 2b + c) gt 0" "[because alpha, beta lt 2 therefore (alpha-2)(beta-2) gt 0]` Signs of and 5a + 2b + c are same. |
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| 153. |
If `7^(log 7(x^(2)-4x + 5))=x - 1`, x may have valuesA. 2, 3B. 7C. `-2, -3`D. 2, -3 |
| Answer» Correct Answer - A | |
| 154. |
For the equation `|x^2|+|x|-6=0`, the sum of the real roots is`1`(b) `0`(c) `2`(d) none of theseA. there is only one rootB. there are only two distinct rootsC. there are only three distinct rootsD. there are four distinct roots |
| Answer» Correct Answer - B | |
| 155. |
The expression `y = ax^2+ bx + c` has always the same sign as of a if `(A) 4ac < b^2 (B) 4ac > b^2 (C) 4ac= b2 (D) ac < b^2`A. `4 ac lt b^(2)`B. `4 ac gt b^(2)`C. `ac lt b^(2)`D. `ac gt b^(2)` |
| Answer» Correct Answer - B | |
| 156. |
If `alpha` and `beta` are the roots of `x^2 +px + q=0` and `alpha^4, beta^4` are the roots of `x^2-rx +s=0`, then the equation `x^2 -4qx+ 2q^2 -r=0` has alwaysA. two real rootsB. two negative rootsC. two positive rootsD. one positive and one negative roots |
| Answer» Correct Answer - A | |
| 157. |
If the roots of `x^(5) - 40 x^(4) + Px^(3) + Qx^(2) + Rx + S = 0` are in G.P. and sum of their reciprocals is 10, then |S| is equal toA. 4B. -4C. 8D. none of these |
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Answer» Correct Answer - D Let the roots be `(a)/(r^(2)), (a)/(r), a, ar, ar^(2)` Then, `a((1)/(r^(2))+(1)/(r)+1+r+r^(2))=40" "...(i)` It is given that `(1)/(a)(r^(2)+r+1+(1)/(r)+(1)/(r^(2)))=10" "...(i)` From (i) and (ii), we get `a^(2) = 4 rArr a = +- 2` Now, Product of the roots = -S `rArr" "a^(5) = - S rArr S = +- 32 rArr |S| = 32` |
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| 158. |
If `alpha, beta` are the roots of `ax^2 + bx + c = 0,` the equation whose roots are `2 + alpha, 2 + beta` isA. `ax^(2) + x(4a -b) + 4a - 2 b + c = 0`B. `ax^(2) + x(4 a - b) + 4a + 2b + c = 0`C. `ax^(2) + x(b-4a) + 4a + 2b + c = 0`D. `ax^(2) + x(b - 4a) + 4a - 2b + c = 0` |
| Answer» Correct Answer - D | |
| 159. |
If the roots of the equation `ax^2-bx-c=0` are changed by same quantity then the expression in a,b,c that does not change isA. `(b^(2)-4ac)/(a^(2))`B. `(b - 4c)/(a)`C. `(b^(2) + 4ac)/(a^(2))`D. `(b^(2) - 4ac)/(a)` |
| Answer» Correct Answer - C | |
| 160. |
p, q, r and s are integers. If the A.M. of the roots of `x^(2) - px + q^(2) = 0` and G.M. of the roots of `x^(2) - rx + s^(2) = 0` are equal, thenA. q is an odd integerB. r is an even integerC. p is an even integerD. s is an odd integer |
| Answer» Correct Answer - C | |
| 161. |
Let `alpha and beta` be the roots of equation `px^2 + qx + r = 0 , p != 0`.If `p,q,r` are in A.P. and `1/alpha+1/beta=4`, then the value of `|alpha-beta|` is :A. `(sqrt(34))/(9)`B. `(2sqrt(13))/(9)`C. `(sqrt(61))/(9)`D. `(2sqrt(17))/(9)` |
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Answer» Correct Answer - B Since `alpha and beta` are roots of the the equation `px^(2)+qx+r=0`. Therefore, `alpha+beta = - (q)/(p) and alpha beta = (r)/(p)` Now, `(1)/(alpha)+(1)/(beta)=4 rArr (alpha+beta)/(alpha beta)=4 rArr -(q)/(r)=4 rArr q =-4r` It is given that p,q, r are in AP. Therefore, `2q = p+r rArr -8r = p+r rArr p = -9r` `therefore" "alpha + beta = -(q)/(p) = -(4)/(9) and alpha beta =(r)/(p) = -(1)/(9)` Now, `(alpha - beta)^(2) = (alpha+beta)^(2) - 4 alpha beta` `rArr" "(alpha-beta)^(2)=(16)/(81)+(4)/(9)=(52)/(81) rArr |alpha-beta|=(2 sqrt(13))/(9)` |
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| 162. |
If `alpha,beta` are the roots of `ax^2 + c = bx`, then the equation `(a + cy)^2 =b^2y` in y has the rootsA. `alpha^(-1), beta^(-1)`B. `alpha^(2), beta^(2)`C. `alpha beta^(-1), alpha^(-1) beta`D. `root(alpha), root(beta)` |
| Answer» Correct Answer - B | |
| 163. |
If the difference between the roots of `x^2+ax+b=0` is same as that of `x^2+bx+a= 0` `a!=b`, then:A. `a + b + 4 = 0`B. `a + b - 4 = 0`C. `a - b - 4 = 0`D. `a - b + 4 = 0` |
| Answer» Correct Answer - A | |
| 164. |
If `secalpha, tanalpha` are roots of `ax^2 + bx + c = 0`, thenA. `a^(2) - b^(2) + 2ac = 0`B. `a^(3) + b^(3) + c^(3) - 2abc = 0`C. `a^(4) + 4 ab^(2) c = b^(4)`D. none of these |
| Answer» Correct Answer - C | |
| 165. |
If `alpha, beta, gamma` be the roots of `x^(3) + a^(3) = 0 (a in R)`, then the number of equation(s) whose roots are `((alpha)/(beta))^(2) and ((alpha)/(gamma))^(2)`, isA. 1B. 2C. 3D. 6 |
| Answer» Correct Answer - A | |
| 166. |
Let p, q, r `in` R and `r gt p gt 0`. If the quadratic equation `px^(2) + qx + r = 0` has two complex roots `alpha and beta`, then `|alpha|+|beta|`, isA. less than 2 but not equat to 1B. equal to 2C. equal to 1D. greater than 2 |
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Answer» Correct Answer - D Since `alpha and beta` are complex conjugate of each other. `therefore" "alpha = bar(beta) and , bar(alpha) = beta rArr |alpha| = |beta|` Now, `alpha = bar(beta) rArr alpha beta = bar(beta)beta rArr alpha beta = |beta|^(2)` Now, `alpha beta = (r)/(p) rArr |alpha|^(2) = |beta|^(2) = (r)/(p) gt 1 rArr |alpha| = |beta| gt 1` Hence, `|alpha|+|beta| gt 2`. |
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| 167. |
If `alpha, beta , gamma, delta` are the roots of the equation `x^4+x^2+1=0` then the equation whose roots are `alpha^2, beta^2, gamma^2, delta^2` isA. `(x^(2) - x + 1)^(2) = 0`B. `(x^(2) + x + 1)^(2) = 0`C. `x^(4) - x^(2) + 1 = 0`D. `x^(2) + x + 1 = 0` |
| Answer» Correct Answer - B | |
| 168. |
If `alpha` and `beta` are the roots of quadratic equation`x^2+ px + q=0` and `gamma` and `delta` are the roots of `x^2 + p x-r=0` then `(alpha-gamma)(alpha-delta)`A. p + qB. q - rC. r - qD. q + r |
| Answer» Correct Answer - D | |
| 169. |
Statement I: `x^2-5x+6 |
| Answer» Correct Answer - A | |
| 170. |
If the roots of the equation `ax^(2) - bx + 5c = 0` are in the ratio of `4 : 5` , thenA. `ab = 18c^(2)`B. `81b^(2) = 4ac`C. `bc = a^(2)`D. `4b^(2) = 81ac` |
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Answer» Correct Answer - D (i) Use the concept of sum and product of the roots of quadratic equation. (ii) Take the roots as `4alpha, 5alpha`. (iii) Using the sum of the roots and product of the roots eliminate `alpha`. |
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| 171. |
If `alpha, beta` are roots of the equation `2x^2 + 6x + b = 0 (b < 0),` then `alpha/beta+beta/alpha` is less thanA. 2B. -2C. 18D. none of these |
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Answer» Correct Answer - B We have, `alpha+beta = -3 and alpha beta = (b)/(2)` Since b `lt` 0, therefore discriminant `D = 36 - 4b gt 0`. So, `alpha and beta` are real. Now, `(alpha)/(beta)+(beta)/(alpha)=(alpha^(2)+beta^(2))/(alpha beta)=((alpha+beta)^(2)-2 alpha beta)/(alpha beta)=((alpha+beta)^(2))/(alpha beta)-2=(18)/(b)-2` `rArr" "(alpha)/(beta) + (beta)/(alpha) lt -2" "[because b lt 0]` ALITER We have, `alpha beta = (b)/(2) lt 0` `therefore" "alpha and beta` are of opposite signs `rArr" "(alpha)/(beta) lt 0` `rArr" "(alpha)/(beta)+(beta)/(alpha) lt - 2" "[because x + (1)/(x) lt - 2 "for all x" lt 0, x ne -1]` |
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| 172. |
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. 5B. 9C. 11D. 13 |
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Answer» Correct Answer - D Since `alpha, beta, gamma, sigma` are the roots of the given equation. `therefore" "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, we get `(-2+3i)(-2-3i)=(1+alpha^(2))(1+beta^(2))(1+gamma^(2))(1+sigma^(2))` `rArr" "13 = (1+alpha^(2))(1+beta^(2))(1+gamma^(2))(1+sigma^(2))` |
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| 173. |
If `alpha` and `beta` are the roots of quadratic equation`x^2+ px + q=0` and `gamma` and `delta` are the roots of `x^2 + p x-r=0` then `(alpha-gamma)(alpha-delta)`A. p + qB. q - rC. r - qD. q + r |
| Answer» Correct Answer - D | |
| 174. |
`cosalpha`is a root of the equation `25 x^2+5x-12=0,-1A. `(12)/(25)`B. `(-12)/(25)`C. `(-24)/(25)`D. `(20)/(25)` |
| Answer» Correct Answer - C | |
| 175. |
Statement-1: `If a, b, c in Q and 2^(1//3)` is a root of `ax^(2) + bx + c = 0`, then a = b = c = 0. Statement-2: A polynomial equation with rational coefficients cannot have irrational roots.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 - C Multiplying both sides of `ax^(2) + bx + c = 0` by the LCM of denominators of rational numbers a, b and c, we obtain a quadratic equation with integer coefficients. So, let us assume that a, b, c are integers and they do not have any common factors. It is given that `2^(1//3)` satisfies `ax^(2) + bx + c = 0`. `therefore" "2^(2//3) a+2^(1//3)b+c = 0` `rArr" "c =-(2^(1//3)a+2^(1//3)b)` `rArr" "c^(3) =-{4a^(3) + 2b^(3)+6ab(2^(2//3) a+2^(1//3)b)}` `rArr" "c^(3) = - 4a^(3) - 2b^(3) + 6abc` `rArr" "c^(3) =-2(2a^(3) + b^(3) - 3abc)` `rArr" "2|c^(3)` `rArr" "2|c` `rArr" "c = 2 c_(1)` for some integer `c_(1)`. Puttion `c = 2c_(1)` in (i), we get `8c_(1)^(3) = -4a^(3) - 2b^(3) + 6 abc_(1)` `rArr" "4c_(1)^(3) = -(2a^(3)+b^(3) - 6 abc_(1))` `rArr" "b^(3) = -2(a^(3) + 2c_(1)^(3) - 3 abc_(1))` `rArr" "2|b^(3) rArr 2| b rArr b = 2b_(1)` for some integer `b_(1)`. Putting `b = 2b_(1) and c = 2c_(1)` in (i), we get `8c_(1)^(3) = - 4a^(3) - 16 b_(1)^(3) + 24 ab_(1) c_(1)` `rArr" "a^(3) = - 2(c_(1)^(3)+2b_(1)^(3) - 3ab_(1) c_(1)) rArr 2|a^(3) rArr 2|a` This is a contradiction. Therefore, a = b = c = 0. So, Statement-1 is true. Statement-2 is false, because `x^(3) - 2 = 0` is a polynomial equation with rational coefficients having `2^(1//3)` as its root. |
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| 176. |
Let a be a non-zero real number and `alpha, beta` be the roots of the equation `ax^(2) + 5x + 2 = 0`. Then the absolute value of the difference of the roots of the equation `a^(3)(x+5)^(2) - 25 a(x+5) + 50 = 0`, isA. `|alpha^(2) - beta^(2)|`B. `|alpha beta (alpha^(2) - beta^(2))|`C. `|(alpha^(2) - beta^(2))/(alpha beta)|`D. `|(alpha^(2) - beta^(2))/(alpha^(2) beta^(2))|` |
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Answer» Correct Answer - A Clearly, `alpha + beta = -(5)/(a) and alpha beta = (2)/(a)` Let p, q be the roots of the equation `a^(3) x^(2) + 5a (2a^(2)-5) x + 25(a^(3) - 5a + 2) = 0`. Then, `p+q = -(5)/(a^(2))(2a^(2) - 5) and, pq = (25)/(a^(3))(a^(3) - 5a + 2)` `therefore" "|p-q| = sqrt((p+q)^(2)-4pq)` `rArr" "|p-q|=sqrt((25)/(a^(4))(2a^(2)-5)^(2)-(100)/(a^(3))(a^(3)-5a + 2))` `rArr" "|p-q|=(5)/(a^(2))sqrt(4a^(2)-20a^(2)+25-4a^(4)+20a^(2)-8a)` `rArr" "|p-q|=(5)/(a^(2))sqrt(25-8a)=sqrt(((5)/(a))^(4)-4((5)/(a))^(2)xx(2)/(a))` `rArr" "|p-q|=sqrt((alpha+beta)^(4)-4(alpha+beta)^(2)alpha beta)` `rArr" "|p-q|=|alpha+beta|sqrt((alpha+beta)^(2)-4alpha beta)` `rArr" "|p-q|=|alpha+beta||alpha-beta|=|alpha^(2)-beta^(2)|` |
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| 177. |
If `alpha` and `beta` are the zeros of the Quadratic Polynomial F(X) =` 6x^2 + x − 2`, Find the Value of`alpha/beta + beta/alpha`A. `1/2`B. `-1/2`C. `1/(sqrt(2))`D. 1 |
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Answer» Correct Answer - D Factorize LHS of the given to find `alpha` and `beta`. |
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| 178. |
If `x-c`is a factor of order `m`of the polynomial `f(x)`of degree `n(1A. `f^(m)(x)`B. `f^(m-1)(x)`C. `f'(x)`D. none of these |
| Answer» Correct Answer - B | |
| 179. |
If one root of the polynomial `f(x)=5x^2+13 x+k`is reciprocal of the other, thenthe value of `k`is(a) `0`(b) `5`(c) `1/6`(d) `6` |
| Answer» Correct Answer - B | |
| 180. |
The polynomial, `sqrt(3)x^(2) + 2x +1` is a `"______"` expression. |
| Answer» Correct Answer - quadratic | |
| 181. |
The set of values of a for which each on of the roots of `x^(2) - 4ax + 2a^(2) - 3a + 5 = 0` is greater than 2, isA. `a in (1, oo)`B. `a = 1`C. `a in (-oo, 1)`D. `a in (9//2, oo)` |
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Answer» Correct Answer - D Let `f(x) = x^(2) - 4 ax + 2a^(2) - 3a + 5`. Clearly, y = f(x) represents a parabola opening upward. So, its both roots will be greater than 2, if (i) Discriminat `ge 0` (ii) x-coordinate of vertex `gt 2` (iii) 2 lies outside the roots i.e. `f(2) gt 0` Now, (i) Discriminat `ge 0` `rArr" "16a^(2) - 4(2a^(2)-3a + 5)ge 0` `rArr" "2a^(2) + 3a - 5 ge 0` `rArr" "(2a+5)(a-1) ge 0 rArr a le -(5)/(2) or, a ge 1" "...(i)` (ii) x-coordinate of vertex `gt 2` `rArr" "2a gt 2 rArr a gt 1" "...(ii)` (iii) `f(2) gt 0` `rArr" "2a^(2) - 11 a + 9 gt 0` `rArr" "(a-1)(2a-9) gt 0 rArr a lt 1 or, a gt (9)/(2)" "...(iii)` From (i), (ii) and (iii), we get `a gt (9)/(2)`. |
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| 182. |
If the roots of the equation `x^2-2a x+a^2-a-3=0`are ra and less than 3, then`a |
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Answer» Correct Answer - A Let `f(x) = x^(2) - 2ax + a^(2) + a -3`. Clearly, y = f(x) represents a parabola opening upward. If roots of f(x) = 0 are less than 3, we must have (i) Discriminant `gt 0` (ii) x-coordinate of vertex of y = f(x) is less than 3 (iii) 3 lies outside the roots of `f(x) = 0 i.e. f(3) gt 0` Now, (i) Discriminant `ge 0` `rArr" "4a^(2) -4(a^(2) + a - 3) ge 0 rArr -4 (a-3) ge 0 rArr a - 3 le 0` `rArr" "a le 3" "...(i)` (ii) x-coordinate of vertex `lt 3` `rArr" "a lt 3" "...(ii)` (iii) `f(3) lt 0 rArr a^(2) - 5a + 6 gt 0 rArr (a-2) (a-3) gt 0` `rArr" "a lt 2 or a gt 3" "...(iii)` From (i), (ii) and (iii), we get `a lt 2`. |
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| 183. |
If the sum of the roots of the equation `(a+1)x^2=(2a+3)x+(3a+4)=0`is -1, then find the product of the roots. |
| Answer» Correct Answer - C | |
| 184. |
If the absolute value of the difference of the roots of the equation `x^(2) + ax + 1 = 0 "exceeds" sqrt(3a)`, thenA. `a in (-oo, -1) uu (4, oo)`B. `a in (4, oo)`C. `a in (-1, 4)`D. `a in [ 0, 4)` |
| Answer» Correct Answer - A | |
| 185. |
C-1 If ` 2+ isqrt3` is a root of the equation `x^2 + px + q = 0`, Where p,q `in` R, then find the ordered pair ( p,q).A. `p = -4, q = 7`B. `p = 4, q = 7`C. `p = 4, q = - 7`D. `p = -4, q = -7` |
| Answer» Correct Answer - A | |
| 186. |
The zeroes of the quadratic polynomial `x^(2) - 24x + 143` are |
| Answer» Correct Answer - 11,13 | |
| 187. |
Find the value of m for which the quadratic equation , `3x^(2) + 10x + (m-3) = 0` has roots which are reciproal to each other. |
| Answer» Correct Answer - `6` | |
| 188. |
For a `alt=0,`determine all real roots of theequation `x^2-2a|x-a|-3a^2=0.`A. `a(1- sqrt(2)), a(-1 + sqrt(6))`B. `a(1+sqrt(2)), a(1-sqrt(6))`C. `a(1-sqrt(2)), a(1-sqrt(6))`D. none of these |
| Answer» Correct Answer - A | |
| 189. |
Determine the nature of the roots of the following equatins : (a) `x^(3) + 2x +4 = 0` (b) `3x^(2) - 10x + 3 = 0` (c ) `x^(2) - 24x + 144 = 0` |
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Answer» Correct Answer - (a) Complex conjugates (b) Real and distinct (c ) Real equal |
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| 190. |
If the roots of the equation `x^2-px+q=0` differ by unity thenA. `p^(2) = 4q`B. `p^(2) = 4q + 1`C. `p^(2) = 4q - 1`D. none of these |
| Answer» Correct Answer - B | |
| 191. |
If a, b `in` R, then the equation `x^(2) - abx - a^(2) = 0` hasA. one positive and one negative rootB. both positive rootsC. both negative rootsD. non-real roots |
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Answer» Correct Answer - A Let `alpha , beta` be the roots of the givn equation. Then, `alpha + beta = ab and alpha beta = - a^(2)` Also, Disc `= a^(2)b^(2) + 4a^(2) gt 0`. Thus, `alpha, beta in R` such that `alpha beta lt 0`. Therefore, one of `alpha and beta` is positive and other is negative. |
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| 192. |
If `tan A` and `tan B` are the roots of `x^2-px + q = 0` , then the value of` sin^2(A+ B)` isA. `(p^(2))/(p^(2)+(1+q)^(2))`B. `(p^(2))/(p^(2)+q^(2))`C. `(q^(2))/(p^(2)+(1-q)^(2))`D. `(p^(2))/((p+q)^(2))` |
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Answer» Correct Answer - A We have, `tan A + tan B = p and tan A tan B = q`. `therefore" "tan(A+B) = (tan A + tan B)/(1-tan A tan B)=(p)/(1-q)` Now, `sin^(2)(A+B) = (1)/(2){1 - cos 2 (A+B)}` `rArr" "sin^(2)(A+B)=(1)/(2){1-(2-tan^(2)(A+B))/(1+tan^(2)(A+B))}=(tan^(2)(A+B))/(1+tan^(2)(A+B))` `rArr" "sin^(2)(A+B)=(p^(2)//(1-q)^(2))/(1+(p^(2))/(1-q)^(2))=(p^(2))/(p^(2)+(1-q)^(2))` |
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| 193. |
A triangle `PQR , /_R=90^@` and `tan(P/2)` and `tan(Q/2)` roots of the `ax^2+bx+c=0` then prove that `a+b=c`A. a+b = cB. b+c = 0C. a+c = bD. b = c |
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Answer» Correct Answer - A It is given that `"tan"(P)/(2) and "tan"(Q)/(2)` are roots of `ax^(2) + bc + c = 0`. `therefore" ""tan"(P)/(2) + "tan"(Q)/(2) = - (b)/(a) and "tan"(P)/(2) "tan"(Q)/(2) = (c)/(a)` We have, `R = (pi)/(2) and P+Q+R = pi rArr P+Q = (pi)/(2) rArr (P)/(2)+(Q)/(2) = (pi)/(4)` `therefore" "tan((P)/(2)+(Q)/(2)) = "tan"(pi)/(4)` `rArr" "(tan P//2 + tan Q//2)/(1-tan P//2 tan Q//2) = 1` `rArr" "(-(b)/(a))/(1-(c)/(a))=1` `rArr" "1-(c)/(a) = - (b)/(a) rArr a-c = - b rArr a+b = c` |
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| 194. |
Let `p, q` be integers and let `alpha,beta` be the roots of the equation `x^2-2x+3=0` where `alpha != beta` For `n= 0, 1, 2,.......,` Let `alpha_n=palpha^n+qbeta^n` value `alpha_9=`A. `a_(n) + 1 = a_(n) + a_(n) -1`B. `a_(n) + 2 = a_(n) +1 + a_(n) -1`C. `a_(n)+1 = a_(n) + 1`D. `a_(n)+1 = a_(n) -1 + 1` |
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Answer» Correct Answer - A It is given that `alpha, beta` are roots of `x^(2) - x - 1 = 0`. `therefore" "alpha + beta = 1 and alpha beta = - 1`. We have, `a_(n) = p alpha^(n) + q beta^(n)` `therefore" "a_(n+1) = p alpha^(n+1) + q beta^(n+1)` `rArr" "a_(n+1) = (p alpha^(n) + q beta^(n)) (alpha+beta) - alpha beta (p alpha^(n-1) + q beta^(n-1))` `rArr" " a_(n+1) = a_(n) + a_(n-1)" "[therefore alpha+beta = 1, alpha beta = - 1]` Hence, option (a) is correct. |
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| 195. |
In illustration 13, if `a_(4) = 28, "then" p + 2q =`A. 21B. 11C. 7D. 12 |
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Answer» Correct Answer - D From illustration 13, we obtain `a_(n+1) = a_(n) + a_(n-1)" "...(i)` `rArr" "a_(4) = a_(3) + a_(2)" "["Putting n = 3"]` `rArr" "a_(4) = (a_(2)+a_(1)) + (a_(1) + a_(0))" ["Putting n = 2, 1 in (i)"]` `rArr" "a_(4) = a_(2) + 2a_(1) + a_(0)` `rArr" "a_(4) = (a_(1) + a_(0)) + (2a_(1) + a_(0))" "["Putting n = 1 in (i)", a_(2) = a_(1) + a_(0)]` `rArr" "a_(4) = 3a_(1) + 2a_(0)` ltbgt `rArr" "28 = 3 (p alpha+q beta) + 2(p+q)" "[therefore a_(n) = p alpha^(n) + q beta^(n)]` `rArr" "28 = 3p ((1+ sqrt(5))/(2))+3q((1- sqrt(5))/(2))+2(p+q)" "[therefore alpha = (1+sqrt(5))/(2), beta = (1-sqrt(5))/(2)]` `rArr" "28 = (7)/(2)(p+q)+(3 sqrt(5))/(2) (p-q)` `rArr" "(7)/(2)(p+q) = 28 and p - q = 0` `rArr" "p + q = 8 and p = q rArr p = q = 4` Hence, p + 2q = 4 + 8 = 12. |
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| 196. |
the roots of the equation `(a+sqrt(b))^(x^2-15)+(a-sqrt(b))^(x^2-15)=2a` where `a^2-b=1` areA. `+-2,+-sqrt(3)`B. `+-4,+-sqrt(14)`C. `+-3,+-sqrt(5)`D. `+-6, +- sqrt(20)` |
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Answer» Correct Answer - B We have, `a-sqrt(b)=((a-sqrt(b))(a + sqrt(b)))/(a + sqrt(b))=(a^(2)-b)/(a+sqrt(b))=(1)/(a+sqrt(b))[because a^(2)-b = 1]` So, by puttiong `(a + sqrt(b))^(x^(2)-15)=y`, the given equation becomes `y+(1)/(y) = 2a` `rArr" "y^(2) - 2 ay + 1 = 0` `rArr" "(y-a)^(2) = a^(2) - 1` `rArr" "y - a = +-sqrt(a^(2)-1)` `rArr" "y-a = +- sqrt(b)" "[because a^(2) - 1 = b]` `rArr" " y = a +- sqrt(b)` `rArr" "(a + sqrt(b))^(x^(2)-15) = a + sqrt(b) , a - sqrt(b)` `rArr" "x^(2) - 15 = 1 or, x^(2) - 15 = - 1 rArr x = +- 4, x = +- sqrt(14)` ALITER We have, `(a + sqrt(b))^(x^(2)-15)+(a-sqrt(b))^(x^(2)+15)=(a+sqrt(a))^(1)+(a-sqrt(b))^(1)` `rArr" "x^(2) - 15 = +- 1 rArr x = +- 4, x = +- sqrt(14)` |
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| 197. |
The number of values of the pair (a, b) for which `a(x+1)^2 + b(-x^2 – 3x - 2) + x + 1 = 0` isan identity in x, is |
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Answer» Correct Answer - B We have, `a(x + 1)^(2) + b(-x^(2) - 3x - 2) + x + 1 = 0` `rArr" "x^(2) (a-b) + x(2a - 3b + 1) + a - 2b + 1 = 0` For this to be an identity in x, we must have `a-b=0, 2a-3b + 1 = 0 and a - 2b + 1 = 0` `rArr" "a = b = 1` Hence, there is only one pair (1, 1). |
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| 198. |
If `m in Z` and the equation `m x^(2) + (2m - 1) x + (m - 2) = 0` has rational roots, then m is of the formA. `n(n+2), n in Z`B. `n (n+1), n in Z`C. `n (n-2), n in Z`D. none of these |
| Answer» Correct Answer - B | |
| 199. |
if `(1+k)tan^2x-4tanx-1+k=0` has real roots `tanx_1` and `tanx_2` thenA. `k^(2) le 5`B. `tan(x_(1) + x_(2)) = 2`C. for `k = 2, x_(1) = pi//4`D. all of these |
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Answer» Correct Answer - D We have, `tan x_(1) and tan x_(2)` as the roots of the equation `(1+k) tan^(2) x - 4 tan x + (k-1) = 0` `therefore" "tan x_(1)+tan x_(2)=(4)/(1+k)and tan x_(1) tan x_(2)=(k-1)/(k+1)` `rArr" "tan(x_(1) + x_(2))=(tan x_(1) + tan x_(2))/(1-tan x_(1) tan x_(2)) = 2` So, option (b) is correct. For k = 2, the equation reduces to `3 tan^(2) x -4 tan x+1 = 0` `rArr" "(tan x-1)(3 tan x-1)=0` `rArr" "tan x = 1 or tan x=(1)/(3)` `rArr" "x_(1)=pi//4 and x_(2) = tan^(-1) (1)/(3)` Since the equation has real roots. Therefore, Disc `ge 0 rArr 16 - 4 (k^(2) -1) ge 0 rArr k^(2) le 5` Hence, all options are correct. |
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| 200. |
if `(1+k)tan^2x-4tanx-1+k=0` has real roots `tanx_1` and `tanx_2` thenA. `k^(2) le 5`B. `k^(2) ge 6`C. `k = 3`D. none of these |
| Answer» Correct Answer - A | |