InterviewSolution
Saved Bookmarks
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.
| 201. |
Find the value of `k`, so that the equation `2x^2+kx-5=0` and `x^2-3x-4=0` may have one root in common.A. `-3, (27)/(4)`B. `3, (-27)/(4)`C. `-3, (-27)/(4)`D. `3, (27)/(4)` |
|
Answer» Correct Answer - C Let `alpha` be the common root of the two equations. Then, `2alpha^(2) + k alpha - 5 = 0 and, alpha^(2) - 3alpha -4 = 0` Solving these two equations, we get `(alpha^(2))/(4k -15k)=(alpha)/(-5+8)=(1)/(-6-k)` `rArr" "alpha^(2) = (4k + 15)/(k+6) and alpha=(-3)/(k+6)` `rArr" "((-3)/(k+6))^(2) = (4k + 15)/(k+6)" "[therefore alpha^(2) = (alpha)^(2)]` `rArr" "(4k + 15) (k + 6) = 9` `rArr" "4k^(2) + 39k + 81 = 0 rArr k = - 3 or, k = -(27)/(4)` |
|
| 202. |
If the product of the roots of the equation `x^(2) - 2sqrt(2) kx + 2e^(2 log k) -1 = 0` is 31, then the roots of the equation are real for k equal toA. 1B. 2C. 3D. 4 |
|
Answer» Correct Answer - D It is given that the product of roots is 31 `therefore" "2e^(2 log k) -1 = 31` `rArr" "2k^(2) - 1 = 31 rArr 2k^(2) = 32 rArr k^(2) = 16 rArr k = +- 4` But, log k is defined for `k gt 0`. Therefore, k = 4 We have, Disc `= 8k^(2) - 8 e^(2 log k) + 4 = 8k^(2) - 8k^(2) + 4 = 4 gt 0` for all k. Hence, k = 4. |
|
| 203. |
The polynomiaal `(a x^2+b x+c)(a x^2-dx-c), a c!=0` hasA. our real rootsB. at least two real rootsC. at most two real rootsD. No real roots |
|
Answer» Correct Answer - B Ler `D_(1) and D_(2)` be the discriminants of `ax^(2) + bx + c = 0 and ax^(2) - dx - c = 0` respectively. Then, `D_(1) = b^(2) - 4ac and D_(2) = d^(2) + 4ac` `rArr" "D_(1) + D_(2) = b^(2) + d^(2) gt 0` `rArr" ""At least one of" D_(1) and D_(2)` is positive. Hence, at least one of the equations `ax^(2)+ bx + c = 0 and ax^(2) - dx - c = 0` has real roots. Thus, the given polynomial has at least two real roots. |
|
| 204. |
If p, q, r are real and `p ne q`, then the roots of the equation `(p-q)x^(2) +5(p+q) x-2(p-q) r` areA. real and equalB. unequal and rationalC. unequal and irrationalD. nothing can be said |
|
Answer» Correct Answer - D The discriminant D of the given equation is `D = 25 (p+q)^(2) + 8(p-q)^(2)+4r(p-q)` Clearly, its sign cannot be determined. So, nothing can be said about the nature of roots. |
|
| 205. |
The roots of the equtation `(x-3)/(x-2) + (2x)/(x+3) = 1`, where `x ne 2, -3` are |
| Answer» Correct Answer - `3,(-1)/(2)` | |
| 206. |
In a quadratic equation with leading coefficient 1, a student read the coefficient 16 of x wrong as 19 and obtain the roots as -15 and -4. The correct roots areA. 6, 10B. `-6, -10`C. `-7, -9`D. none of these |
|
Answer» Correct Answer - B Let the quadratic equation be `x^(2) + bs + c = 0`. It is given that b = 16. When b = 19 the roots are -15 and -4. `therefore" "`Product of roots `= c rArr -15 xx - 4 = c rArr c = 60`. Thus, the quadratic equation is `x^(2) + 16x + 60 = 0 rArr x = - 10, -6`. Hence, root are -10, and -6. |
|
| 207. |
Which one of the following is not true? The quadratic equation `x^(2) - 2x - a = 0, a ne 0`,A. cannot have a real root if `a lt -1`B. may not have a rational root even if a is a perfect squareC. cannot have an integral root if `n^(2)-1 lt a lt n^(2) + 2n`, where n = 0, 1, 2,......D. none of these |
|
Answer» Correct Answer - D Let D be the discriminant of `x^(2) - 2x - a = 0`. Then, D = 4 + 4a If `a lt -1, "then" D = 4(1+a) lt 0`. So, the equation cannot have real roots. Thus, option (a) is true. If a is a perfect square square, say `a = lambda^(2), lambda in Z`, then `D = 4 + 4a = 4(1+lambda^(2))`, which is not a perfect square. Thus, roots cannot be rational. So, option (b) is not true. Let `alpha` be an integral root of the given equation. Then, `alpha^(2) - 2 alpha - a = 0 rArr a = alpha^(2) - 2 alpha rArr a = (alpha-1)^(2) -1` `therefore" "n^(2) - 1 lt a lt n^(2) + 2n` `rArr" "n^(2) - 1 lt (alpha-1)^(2) -1 lt n^(2) + 2n` `rArr" "n^(2) lt (alpha - 1)^(2) lt (n + 1)^(2)`, which is not possible. So, the equation cannot have integral roots. Thus, option (c) is true. |
|
| 208. |
If `x^(2) -ax - 6 = 0` and `x^(2) + ax -2 = 0` have one common root, then a can be `"_________"`.A. -1B. 2C. -3D. 0 |
|
Answer» Correct Answer - A Find x in terms of a and substitute `x = a` in either of the equation. |
|
| 209. |
If b is the harmonic mean of a and c and `alpha, beta` are the roots of the equation `a(b-c)x^(2) + b(c-a) x+ c(a-b)=0`, thenA. `alpha + beta = 3`B. `alpha+beta=(1)/(2)`C. `alpha beta = 2`D. `alpha = 1, beta = 1` |
|
Answer» Correct Answer - D We observe that the sum of the coefficients of the given equation is zero. Therefore, 1 is a root of the given equation. Let `alpha = 1 and beta` be the other root. Then, `beta xx 1 =(c(a-b))/(a(b-c))rArr beta = (c)/(a) xx {(a-(2ac)/(a+c))/((2ac)/(a+c)-c)}=1" "[because b = (2ac)/(a+c)]` Hence, `alpha = beta = 1` |
|
| 210. |
Let `f(x) =ax^2 + bx+ c AA a, b, c in R, a != 0` satisfying `f (1) + f(2)=0`. Then, the quadratic equation `f(x)=0` must have :A. no real rootB. 1 and 2 as real rootsC. two equal rootsD. two distinct real roots |
|
Answer» Correct Answer - D We have, `f(x) = ax^(2) + bx + c` `therefore" "f(1) + f(2) = 0 rArr 5a + 3b + 2c - 0` Let D be the discriminant of f(x) = 0. Then, `D = b^(2) - 4ac` `rArr" "D=((5a+2c)/(-3))^(2)-4ac" "["Using (i)"]` `rArr" "D=((5a+2c)^(2)-36ac)/(9)` `rArr" "D=(25a^(2)+4c^(2)-16ac)/(9)=(1)/(9){(4a-2c)^(2)+9a^(2)} gt 0` Hence, f(x) = 0 has two distinct real roots. |
|
| 211. |
Statement-1: If a, b, c are distinct real numbers, then `a((x-b)(x-c))/((a-b)(a-c))+b((x-c)(x-a))/((b-c)(b-a))+c((x-a)(x-b))/((c-a)(c-b))=x` for each real x. Statement-2: `If a, b, c in R` such that `ax^(2) + bx + c = 0` for three distinct real values of x, then `a = b = c = 0` i.e. `ax^(2) + bx + c = 0` for all `x in R`.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. |
|
Answer» Correct Answer - A Clearly, statement-2 is true (see Theory). Let `f(x) = a((x-b)(x-c))/((a-b)(a-c))+b((x-c)(x-a))/((b-c)(b-a))+c((x-a)(x-b))/((c-a)(c-b))=x` Clearly, f(x) is a quadratic polynomial such that `f(a) = f(b) = f(c) = 0`. `therefore" "f(x) = 0 "for all" x in R" "["Using statement-2"]` So, statement-1 is true. |
|
| 212. |
The value of a for which exactly one root of the equation `e^ax^2 - e^(2a)x + e^a -1` lies between 1 and 2 are given byA. In `((5-sqrt(17))/(4))lt a lt "In"((5+sqrt(17))/(4))`B. `0 lt a lt 100`C. In `(5)/(4) lt a lt "In"(10)/(3)`D. none of these |
|
Answer» Correct Answer - A Let `f(x) = e^(a) x^(2) - e^(2a) x + e^(a)-1`. Clearly, y = f(x) represents a parabola opening upward. Also, Disc of f(x) = 0 is given by `D = e^(4a)-4e^(a)(e^(a) -10 = (e^(2a)-1) lt 0` So, the roots are real and distinct for all a `in` R. Thus, exactly one root will lie between 1 and 2, if f(1) f(2) `lt` 0 `rArr" "(e^(a)-e^(2a)+e^(a)-1)(e^(2a)-1)^(2) gt 0` `rArr" "2e^(2a)-5e^(a) + 1 lt 0` `rArr" "(5-sqrt(17))/(4)lt e^(a) lt (5+sqrt(17))/(4) rArr "In"(5-sqrt(17))/(4) lt a lt "In"(5+sqrt(17))/(4)` |
|
| 213. |
Let A, G, and H are the A.M., G.M. and H.M. respectively of two unequal positive integers. Then, the equation `Ax^(2) - Gx - H = 0` hasA. both roots as fractionsB. one root which is a negative fraction and other positive rootC. at least one root which is an integerD. none of these |
|
Answer» Correct Answer - B Let `alpha and beta` be the roots of `Ax^(2) - Gx - H = 0`. Then, `alpha + beta = (G)/(A) and alpha beta = -(H)/(A)` We know, that `A gt G gt H`. Also, A, G, H are A.M., G.M. and H.M. respectively of two unequal positive integers. Therefore, `A gt G gt H gt 0`. `rArr" "alpha+beta` is a positive fraction and `alpha beta` is a negative fraction. Let D be the discriminant of the given equation. Then, `D = G^(2) + 4 AH gt 0`. `rArr" "`Roots are real. Thus, the given equation has real roots such that their sum is a positive fraction and product is a negative fraction. This means that the equation has one positive fraction and one negative fraction as its roots. |
|
| 214. |
If `2x^(2) + (2p - 13) x + 2 = 0` is exactly divisible by `x-3`, then the value ofA. `(-16)/(6)`B. `(19)/(6)`C. `(16)/(6)`D. `(-19)/(6)` |
|
Answer» Correct Answer - B Substitute `x = 3` in the given equation and simplify. |
|
| 215. |
Statement-1: If a, b, c, A, B, C are real numbers such that `a lt b lt c`, then `f(x) = (x-a)(x-b)(x-c) -A^(2)(x-a)-B^(2)(x-b)-C^(2)(x-c)` has exactly one real root. Statement-2: If f(x) is a real polynomical and `x_(1), x_(2) in R` such that `f(x_(1)) f(x_(2)) lt 0`, then f(x) has at least one real root between `x_(1) and x_(2)`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. |
|
Answer» Correct Answer - D Clearly, statement-2 is true. (See Theory) We have, `f(a) = - B^(2)(a-b)-C^(2)(a-c) = - [B^(2)(a-b)+C^(2)(a-c)] gt 0 and, f(c) = - [A^(2)(c-a)+(c-b)B^(2)] lt 0` Also, `f(x) rarr - oo "as" x rarr - oo and f(x) rarr oo "as" x rarr oo`. Thus, f(x) changes its signs from positive to negative or negative to positive in the intervals `(-oo, a),(a,c) and (c, oo)`. Hence, f(x) has three real roots. |
|
| 216. |
Statement (1) : If a and b are integers and roots of `x^2 + ax + b = 0` are rational then they must be integers. Statement (2): If the coefficient of `x^2` in a quadratic equation is unity then its roots must be integersA. 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. |
|
Answer» Correct Answer - C Let `(p)/(q), (p,q in Z, , q ne 0 and HCF (p, q) = 1)` be a root of `x^(2) + ax + b = 0`. Then, `(p^(2))/(q^(2))+(ap)/(q) + b = 0` `rArr" "p^(2) + apq + bq^(2) = 0` `rArr" "p^(2) = -q (ap + bq)` `rArr" "q "divides" p^(2)" "[therefore q "divides-q"(ap + bq)]` `rArr" "q "divides p"" "[therefore HCF (p, q) = 1]` `rArr" "q = 1" "[therefore HCF (p, q) = 1]` Thus `x^(2) + ax + b = 0` has integer roots. So, statement-1 is true. Statement-2 is false as `x^(2) + x + 1 = 0` does not have integer roots. |
|
| 217. |
If the roots of the equation `ax^2 + bx + c = 0, a != 0 (a, b, c` are real numbers), are imaginary and `a + c < b,` thenA. 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. |
|
Answer» Correct Answer - B Clearly, statement-2 is true. To check the truth of statement-1, let us assume that each one of the given equation has real roots. Then, `b^(2) - 4ac gt 0, c^(2) - 4ab ge 0 and a^(2) - 4bc ge 0` `rArr" "b^(2) ge 4ac, c^(2) ge 4ab and a^(2) ge 4bc` `rArr" "a^(2) b^(2) c^(2) ge 64 a^(2) b^(2) c^(2)`, which is not possible. So, our supposition is wrong. Therefore, at least one of the given equations has imaginary roots. Hence, statement-1 is true. Clearly, statement-2 is not a correct explanation for statement-1. |
|
| 218. |
Given that `a x^2+b x+c=0`has no real roots and `a+b+c |
| Answer» Correct Answer - C | |
| 219. |
Let `alpha,beta`are the roots of `x^2+b x+1=0.`Then find the equation whoseroots are `(alpha+1//beta)a n d(beta+1//alpha)`.A. `x^(2) = 0`B. `x^(2) + 2b x + 4 = 0`C. `x^(2) - 2 bx + 4 = 0`D. `x^(2) - bx + 1 = 0` |
| Answer» Correct Answer - C | |
| 220. |
If `x in R`, then the expression `9^(x) - 3^(x) + 1` assumesA. all real valuesB. all real values greater than 0C. all real values greater than 3/4D. all real values greater than 1/4 |
| Answer» Correct Answer - C | |
| 221. |
Find the value of `a`for which the sum of the squares of the roots of the equation `x^2-(a-2)x-a-1=0`assumes the least value. |
|
Answer» Correct Answer - B Let `alpha, beta` be the roots of the given equation. Then, `alpha + beta = a -2 and alpha beta = - (a+1)`. `therefore" "alpha^(2)+beta^(2)=(alpha+beta)^(2)-2 alpha beta = (a-2)^(2) + 2(a+1)` `rArr" "alpha^(2) + beta^(2) = a^(2) - 2a + 6 = (a-1)^(2) + 5` Clearly, `alpha^(2) + beta^(2) ge 5`. So, the minimum value of `alpha^(2)+beta^(2)` is 5 which it attains at = 1. |
|
| 222. |
If `alpha, beta` are the roots of ` a x^2 + bx + c = 0` and `k in R` then the condition so that `alpha < k < beta` is :A. `ak^(2) + bk + c lt 0`B. `a^(2) k^(2)+ abk + ac lt 0`C. `a^(2)k^(2) + abk + ac gt 0`D. none of these |
|
Answer» Correct Answer - B Let `f(x) = ax^(2) + bx + c`. It is given that `alpha, beta` are real roots of `f(x) = 0`. So, k lies between `alpha and beta`, if `a f(k) lt 0 rArr a (ak^(2) + bk + c) lt 0 rArr a^(2) k^(2) + abk + ac lt 0` |
|
| 223. |
All possible values of a, so that 6 lies between the roots of the equation `x^2 + 2(a-3)x +9 =0`A. `a in [-3//4, oo)`B. `a in (oo, -3//4)`C. `a in (-oo, 0) uu(6, oo)`D. `a in (-3//4, 6)` |
|
Answer» Correct Answer - B Let `f(x) = x^(2) + 2(a-3) x + 9`. If 6 lies between the roots fo f(x) = 0, then we must have (i) `"Disc" gt 0, and" "(ii) f(6) lt 0" "[therefore "Coeff of" x^(2) "is positive"]` Now, Disc `gt` 0 `rArr" "4(a-3)^(2) - 36 gt 0` `rArr" "(a-3)^(2) - 9 gt 0` `rArr" "a^(2) - 6a gt 0` `rArr" "a(a-6) gt 0 rArr a lt 0 or a gt 6" "...(i)` and, `f(6) lt 0` `rArr" "36 + 12(a-3) + 9 lt 0 rArr a lt -(3)/(4)" "....(ii)` From (i) and (ii), we get `a lt - 3//4 i.e. a in (-oo, -3//4)`. |
|
| 224. |
If `x` is real then the value of `(x^2 - 3x +4) / (x^2 + 3x +4)` lies in the intervalA. `(0, 1//7)`B. `7, oo)`C. `[1//7, 7]`D. `[-1//7, 7]` |
|
Answer» Correct Answer - C Let `y = (x^(2)-3x + 4)/(x^(2) + 3x + 4)`. Then, `x^(2) (y-1) + 3x(y+1)+4(y-1) = 0` This equation gives the values of x for given values of y. But, y is the value when x is real. So, the roots of this equaton are real. `therefore" "9(y+1)^(2) - 16(y-1)^(2) ge 0" "["Using Discriminant" ge 0]` `rArr" "7y^(2) - 50y + 7 le 0 rArr (7y-1) (y-7) le 0 rArr 1//7 le y le 7`. Hence, the given expression lies between 1/7 and 7. |
|
| 225. |
Given that, for all real x, the expression `(x^2+2x+4)/(x^2-2x+4)` lies between `1/3` and 3. The values between which the expression `(9.3^(2x)+6.3^x+4)/(9.3^(2x)-6.3^x+4)` lies areA. `3^(-1) and 3`B. `-2 and 0`C. `-1 and 1`D. 0 and 2 |
|
Answer» Correct Answer - A It is given that `(1)/(3) lt (x^(2)-2x+4)/(x^(2)+2x+4) lt 3` for all x `in` R. `rArr" "(1)/(3) lt (x^(2)+2x+4)/(x^(2)-2x+4) lt 3` for all x `in R" "...(i)` Let `3^(x+1) = y`. Then, `y in R` for all x `in` R. Also, `(9*3^(2x)+6*3^(x)+4)/(9*3^(2x)-6*3^(x)+4)=(3^(2x+2)+2*3^(x+1)+4)/(3^(2x+2)-2*3^(x+1)+4)=(y^(2)+2y+4)/(y^(2)-2y+4)` But, `(1)/(3) lt (y^(2)+2y+4)/(y^(2)-2y+4) lt 3" "["From (i)"]` `therefore" "(1)/(3) lt (9*3^(2x)+6*3^(x)+4)/(9*3^(2x)-6*3^(x)+4) lt 3` |
|
| 226. |
If x `in` R then `(x^(2)+2x+a)/(x^(2)+4x+3a)` can take all real values ifA. `a in (0, 2)`B. `a in [0, 1]`C. `a in [-1, 1]`D. none of these |
|
Answer» Correct Answer - B Let `y = (x^(2)+2x + a)/(x^(2)+4x+3a)`. Then, `x^(2)(y-1)+2(2y-1) x + a(3y-1)=0` `rArr" "4(2y-1)^(2) - 4(y-1) a (3y-1) ge 0` for all `y in R" "[because x in R]` `rArr" "(4-3a)y^(2)-4(1-a)y+1 - a ge 0` for all y `in` R `rArr" "4 - 3a gt 0` and Discriminant `le 0` `rArr" "a lt (4)/(3) and 16 (1-a)^(2) - 4(1-a)(4-3a) le 0` `rArr" "a lt (4)/(3) and a (a-1) le 0` `rArr" "a lt (4)/(3) and 0 le a le 1 rArr 0 le a le 1 rArr a rArr [0, 1]` |
|
| 227. |
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. `+- 7`B. `+-5`C. `+-4`D. `+-1` |
|
Answer» Correct Answer - A Comparing the given equation with `ax^(2) + 2hxy + by^(2) + 2gx + 2 fy + c = 0`, we have `a = 2, h = m//2, b = 3, c = - 2, f = - 5//2, g = 0` The given expression is resolvable into linear factors, if `abc + 2 fgh - af^(2)-bg^(2) - ch^(2) = 0` `rArr" "-12-(25)/(2) + 2((m^(2))/(4))= 0 rArr m^(2) = 49 rArr m = +- 7` |
|
| 228. |
If `alpha,beta` are the roots of the equation `ax^2+bx+c=0` and `S_n=alpha^n+beta^n` then evaluate `|[3, 1+s_1, +s_2] , [1+s_1, 1+s_2, 1+s_3] , [1+s_2, 1+s_3, 1+s_4]|`A. `(b^(2)-4nc)/(a^(4))`B. `((a+b+c)(b^(2)+4ac))/(a^(4))`C. `((a+b+c)(b^(2)-4ac))/(a^(4))`D. `((a+b+c)^(2)(b^(2)-4ac))/(a^(4))` |
| Answer» Correct Answer - D | |
| 229. |
If the roots of `a_1x^2 + b_1x+ c_1 = 0` are `alpha_1 ,beta_ 1` and those of `a_2x^2+b_2x+c_2=0` are `alpha_2,beta_2` such that `alpha_1alpha_2=beta_1beta_2=1` thenA. `(a_(1))/(a_(2))=(b_(1))/(b_(2))=(c_(1))/(c_(2))`B. `(a_(1))/(c_(2))=(b_(1))/(b_(2))=(c_(1))/(a_(2))`C. `a_(1) a_(2) = b_(1) b_(2) = c_(1) c_(2)`D. none of these |
| Answer» Correct Answer - B | |
| 230. |
Let `Delta^(2)` be the discriminant and `alpha, beta` be the roots of the equation `ax^(2) + bx + c = 0`. Then, `2a alpha + Delta and 2 a beta - Delta` can be the roots of the equationA. `x^(2) + 2bx + b^(2) = 0`B. `x^(2) - 2bx + b^(2) = 0`C. `x^(2) + 2bx - 3b^(2) - 16 ac = 0`D. `x^(2) - 2bx - 3b^(2) + 16 ac = 0` |
|
Answer» Correct Answer - A We have, `alpha,beta =(-b +- Delta)/(2a)` Now, two case arise: CASE I When `alpha = (-b + Delta)/(2a) and, beta = (-b-Delta)/(2a)` `rArr" "2a alpha + Delta = -b + 2 Delta and, 2 alpha beta - Delta = -b - 2 Delta` So, the required equation is `x^(2) - x(-2b) + b^(2) - 4 Delta^(2) = 0` `rArr" "x^(2) + 2bx + b^(2) - 4(b^(2) - 4ac) = 0` `rArr" "x^(2) + 2bx - 3b^(2) + 16 ac = 0` CASE II When `alpha = (-b-Delta)/(2a) and, beta=(-b+Delta)/(2a)` `rArr" "2a alpha + Delta = -b and 2 a beta - Delta = - b` So, the required equation is `x^(2) - x(-2b)+b^(2) = 0 or, x^(2) + 2bx + b^(2) = 0`. |
|
| 231. |
If `a(p+q)^2+2b p q+c=0a b da(p+r)^2+2b p r+c=0(a!=0)`, then`q r=p^2`b. `q r=p^2+c/a`c. `q r=p^2`d. none of theseA. `p^(2)+(c)/(a)`B. `p^(2) + (a)/(c)`C. `p^(2) + (a)/(b)`D. `p^(2) + (b)/(a)` |
| Answer» Correct Answer - A | |
| 232. |
Let `a ,b ,a n dc`be real numbers such that `4a+2b+c=0a n da b > 0.`Then the equation `a x^2+b x+c=0.``com p l e xroot s`b. `e x a c t l yon eroot`c. `r e a lroot s`d. `non eoft h e s e`A. real rootsB. complex rootsC. exactly one rootD. none of these |
| Answer» Correct Answer - A | |
| 233. |
`e^(|sinx|)+e^(-|sinx|)+4a=0`will have exactly four different solutions in `[0,2pi]`if.`a in R`(b) `a in [-3/4,-1/4]``a in [(-1-e^2)/(4e),oo]`(d) none of theseA. `a in [-(e)/(4),-(1)/(4)]`B. `a in R`C. `a in [-(-1-e^(2))/(4e),oo)`D. none of these |
|
Answer» Correct Answer - D Let `t = e^(|sin x|) "Clearly", t in [1, e]`. `therefore" "e^(|sin x|)+e^(-|sin x|) + 4a = 0` `rArr" "t^(2) + 4at + 1 = 0 or f(t) = 0, "where" f(t) = t^(2) + 4at + 1` This will have two distinct real roots in [1, e], if `f(1) gt 0, f(e) gt 0, 1 lt -2a lt e and 16 a^(2) - 4 gt 0` `rArr" "4a+2 gt 0, e^(23)+4ae + 1 gt 0, -(e)/(2)lt a lt (-1)/(2) and a lt (-1)/(2) or a gt (1)/(2)` `rArr" "a gt (-1)/(2), a gt -(1+e^(2))/(4e), -(e)/(2) lt a lt (-1)/(2) and a lt (-1)/(2) or a gt (1)/(2)` Clearly, there is no value of a satisfying the above in equalities. |
|
| 234. |
The ratio of the roots of the equation `ax^2+ bx+c =0` is same equation `Ax^2+ Bx + C =0`. If `D_1 and D_2` are the discriminants of `ax^2+bx +C= 0 and Ax^2+Bx+C=0` respectively, then `D_1 : D_2`A. `(a^(2))/(p^(2))`B. `(b^(2))/(q^(2))`C. `(c^(2))/(r^(2))`D. none of these |
|
Answer» Correct Answer - B Let `alpha_(1), beta_(1)` be the roots of `ax^(2) + bx + c = 0 and alpha_(2), beta_(2)` be the roots of `px^(2) + qx + r = 0`. Then, `(alpha_(1))/(beta_(1))=(alpha_(2))/(beta_(2))" "["Given"]` `rArr" "(alpha_(1)+beta_(1))/(alpha_(1)-beta_(1))=(alpha_(2)+beta_(2))/(alpha_(2)-beta_(2))" "[{:("Applying componendo"),(" ""and dividendo"):}]` `(alpha_(1)+beta_(1))^(2)/((alpha_(1)-beta_(1))^(2))=((alpha_(2)+beta_(2))^(2))/((alpha_(2)-beta_(2))^(2))` `rArr" "((alpha_(1)+beta_(1))^(2))/((alpha_(1)+beta_(1))^(2)4 alpha_(1)beta_(1))=((a_(2)+beta_(2))^(2))/((a_(2)+beta_(2))^(2)-4 alpha_(2) beta_(2))` `rArr" "(b^(2)//a^(2))/((b^(2)-4ac)/(a^(2)))=(q^(2)//p^(2))/((q^(2)-4rp)/(p^(2))) rArr (b^(2))/(D_(1))=(q^(2))/(D_(2))rArr (D_(1))/(D_(2))=(b^(2))/(q^(2))` |
|