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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.
| 201. |
If the slope of one of the pairs of lines represented by equation `a^(3) x^(2) + 2hxy + b^(3) y^(2) = 0` is square of the other, then prove that `ab(a+ b) = - 2h. ` |
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Answer» `a^(3) x^(2) + 2hxy + b^(3) y^(2)= 0` `rArr b^(3) ((y)/(x))^(2) + 2h ((y)/(x)) + x^(3) = 0` Lines are `y = mx, where m = m_(1) and m_(2).` `rArr b^(3) m^(2) + 2hm + a^(3) = 0` (1) Given that slope of one line is square of the other line. So, roots are `alpha and alpha^(2)`. `rArr alpha a//b,` which satisfies the equation (1) So, `b^(3) (a^(2))/(b^(2)) + 2h (a)/(b) + a^(3) = 0` `rArr ab^(2) + a^(2) b+ 2h = 0` `rArr ab (a+b) = - 2h` |
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| 202. |
Consider the equation `x^(2) + 2x - n = 0`m where `n in N` and `n in [5, 100]`. The total number of different values of n so that the given equation has integral roots isA. 8B. 3C. 6D. 4 |
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Answer» Correct Answer - 1 `x^(2) + 2x - n = 0 rArr (x + 1)^(2) = n + 1` `rArr x = -1 pm sqrt(n + 1)` Thus, n + 1 should be a perfect square. Now, `n in [5, 100] rArr n + 1 in [6, 101]` Perfect square values of n + 1 are 9, 16, 25, 36, 49, 64, 81, 100. Hence, number of values is 8. |
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| 203. |
If `a+b+c=0`then check the nature of roots of the equation `4a x^2+3b x+2c=0w h e r ea ,b ,c in Rdot` |
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Answer» Correct Answer - Roots are real and distinct. For the given equation `4ax^(2) + 3bx + 2c = 0`, we have `D = (3b)^(2) - 4 (4a) (2c)` = `9b^(2) - 32 ac` = `9 (-a -c)^(2) - 32 ax ` `= 9a^(2) - 14 ac + 9c^(2)` ` = 9c^(2) (((a)/(c))^(2) - (14)/(9) (a)/(c) + 1)` ` = 9 (((a)/(c)-(7)/(9))^(2) - (48)/(81) + 1)` Which is always positive. Hence, the roots are real and distinct |
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| 204. |
Let `f(x)=a x^2+b x+a ,b ,c in Rdot`If `f(x)`takes real values for real values of `x`and non-real values for non-real values of `x`, then`a=0`b. `b=0`c. `c=0`d. nothing can be said about `a ,b ,cdot`A. `a = 0`B. `b = 0`C. `c = 0`D. nothing can be said about a, b, c. |
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Answer» Correct Answer - 1 Suppose `a ne 0`. We rewrite f(x) as follows : `f(x) = a{x^(2) + b/a x + c/a}` `= {(x + (b)/(2a))^(2) + (4ac - b^(2))/(4a^(2))}` `f(-(b)/(2a) + i) = a{(-(b)/(2a) + i + (b)/(2a))^(2) + (4ac - b^(2))/(4a^(2))}` `= a{-1 + (4ac - b^(2))/(4a^(2))}`, which is a real number This is against the hypothesis. Therefore, a = 0. |
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| 205. |
Let `P(x0=x^3-8x^2+c x-d`be a polynomial with real coefficients and with all it roots beingdistinct positive integers. Then number of possible value of `c`is___________. |
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Answer» Correct Answer - 36 We have `x_(1)+x_(2)+x_(3)=8` `x_(1)+x_(2)+x_(3)=d` `x_(1)x_(2)+x_(2)x_(3)+x_(3)x_(1)=c` Possible roots `1,2,5or1,3,4` `:.d=10ord=12` `impliesc=2+10+5=17or3+12+4=19` Hence, `d=10andc=17ord=12andc=19` |
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| 206. |
The polynomial `P(x)=x^(3)+ax^(2)+bx+c` has the property that the mean of its roots, the product of its roots, and the sum of its coefficients are all equal. If the `y`-intercept of the graph of `y=P(x)` is `2`, The value of `b` isA. `-11`B. `-9`C. `-7`D. `5` |
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Answer» Correct Answer - A The `y`-intercept is at `x=0`, so we have `c=2`, meaning that the product of the roots is `-2`. We know that `-a` is the sum of the roots. The average of the roots is equal to the product, so the sum of the roots is `-6`, and `a=6`. Finally, `1+a+b+c=2` as well, so we have `1+6+b+2=-2` `impliesb=-11` `:.P(x)=x^(3)+6x^(2)-11x+2` `:.P(1)=1+6-11+2=-2` |
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| 207. |
Let `f(x), g(x)`, and `h(x)` be the quadratic polynomials having positive leading coefficients and real and distinct roots. If each pair of them has a common root, then find the roots of `f(x) + g(x) + h(x) = 0`. |
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Answer» Let `f(x) = a_(1) (x - alpha) (x - beta),` ` g(x) = a_(2) )x - beta) (x - gamma),` and `h (x) = a_(3)(x - gamma) (x - alpha),` where `a_(1), a_(2), a_(3)` are positive. Let `f(x) + g(x) + h(x) = F(x)` `rArr F(alpha) = a_(2)(alpha - beta)(alpha - gamma)` `F(beta) = a_(3) (beta - gamma)(beta - alpha)` `f(gamma) = a_(1) (gamma - alpha)(gamma - beta)` `rArr F(alpha) F (beta) F(gamma) = - a_(1) a_(2) a _(3) (alpha - beta)^(2) (beta - gamma)^(2) (gamma - alpha )^(2) lt 0 ` So, roots of `F(x) = 0 are and distinct. |
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| 208. |
The polynomial `P(x)=x^(3)+ax^(2)+bx+c` has the property that the mean of its roots, the product of its roots, and the sum of its coefficients are all equal. If the `y`-intercept of the graph of `y=P(x)` is `2`, The value of `P(1)` isA. `0`B. `-1`C. `2`D. `-2` |
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Answer» Correct Answer - D The `y`-intercept is at `x=0`, so we have `c=2`, meaning that the product of the roots is `-2`. We know that `-a` is the sum of the roots. The average of the roots is equal to the product, so the sum of the roots is `-6`, and `a=6`. Finally, `1+a+b+c=2` as well, so we have `1+6+b+2=-2` `impliesb=-11` `:.P(x)=x^(3)+6x^(2)-11x+2` `:.P(1)=1+6-11+2=-2` |
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| 209. |
The polynomial `f(x)=x^4+a x^3+b x^3+c x+d`has real coefficients and `f(2i)=f(2+i)=0.`Find the value of `(a+b+c+d)dot` |
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Answer» Correct Answer - 9 If a polynomial has real coefficients, then roots occur in complex conjugate and roots are `pm 2i, 2 i.` Hence. ` f (x) = (x + 2i) (x - 2i) (x -2-i)(x -2-i)` `therefore f(1) = (1 + 2i) (1 + 2i) (1 - 2-i)(1 - 2 + i)` ` 5xx2 = 10 ` Also, ` f(1) = 1 + a + b + c + d` `therefore 1 + a + b + c + d = 10` `rArr a + b + c + d = 9` . |
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| 210. |
If `alpha,beta`are the roots of `x^2+p x+q=0a d nx^(2n)+p^n x^n+q^n=0a n di lf(alpha//beta),(beta//alpha)`are the roots of `x^n+1+(x+1)^n=0,t h e nn( in N)`a. must be an odd integerb. may be any integerc. must be an even integer d. cannot say anythingA. must be an odd integerB. may be any integerC. must be an even integerD. cannot say anything |
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Answer» Correct Answer - 3 We have, `alpha + beta = -p and alphabeta = q" "(1)` Also, since `alpha, beta` are the roots of `x^(2n) + p^(n)x^(n) + q^(n) = 0`, we have `alpha^(2n) + p^(n)alpha^(n) + q^(n) = 0 and beta^(2n) + p^(n)beta^(n) + q^(n) = 0` Subtracting the above relations, we get `(alpha^(2n) - beta^(2n)) + p^(n) (alpha^(n) - beta^(n)) = 0` `therefore alpha^(n) + beta^(n) = -p^(n)" "(2)` Given, `alpha//beta` or `beta//alpha` is a root of `x^(n) + 1 + (x + 1)^(n) = 0`. So, `(alpha//beta)^(n) + 1 + [(alpha//beta) + 1]^(n) = 0` `rArr (alpha^(n) + beta^(n)) + (alpha + beta)^(n) = 0` `rArr -p^(n) + (-p)^(n) = 0 " "` [Using (1) and (2)] It is possible only when n is even. |
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| 211. |
Let `f(x)=x^2+b x+c ,w h e r eb ,c in Rdot`If `f(x)`is a factor of both `x^4+6x^2+25a n d3x^4+4x^4+28 x+5`, then the least value of `f(x)`is`2`b. `3`c. `5//2`d. `4`A. 2B. 3C. `5//2`D. 4 |
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Answer» Correct Answer - 4 `f(x) = x^(2) + bx + c` f(x) is factor of both `x^(4) + 6x^(2) + 25 and 3x^(4) + 4x^(2) + 28x + 5` `therefore` f(x) will also be factor of `3(x^(4) + 6x^(2) + 25) - (3x^(4) + 4x^(2) + 28x + 5) = 14(x^(2) - 2x + 5)` `therefore x^(2) - 2x + 5` and f(x) have both roots common hence least value of `f(x) = (x - 1)^(2) + 4 ge 4` |
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| 212. |
If P(x) is a polynomial with integer coefficients such that for 4 distinct integers `a, b, c, d,P(a) = P(b) = P(c) = P(d) = 3`, if `P(e) = 5`, (e is an integer) thenA. e = 1B. e = 3C. e = 4D. no real value of e |
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Answer» Correct Answer - 4 `P(a) = P(b) = P(c) = P(d) = 3` `rArr P(x) = 3` has a, b, c, d as its roots `rArr P(x) - 3 = (x - a)(x - b)(x - c)(x - d) Q(x)` [`because Q(x)` has integral coefficient] Given P(e) = 5, then `(e - a) (e - b) (e - c) (e - d) Q(e) = 5` This is possible only when at least three of the five integers (e-a),(e-b),(e-c),(e-d), Q(e) re equal to 1 or -1. Hence, two of them will be equal, which is not possible. Since a, b, c, d are distinct integers, P(e) = 5 is not possible. |
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| 213. |
If both the roots of `a x^2+a x+1=0`are less than 1, then find the exhaustive range of values of `adot` |
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Answer» Both roots of `ax^(2) + ax + 1 = 0` are less then 1 . Compare this equation with `Ax^(2) + Bx + C = 0` . Then, the required conditions are (i) ` D = a^(2) - 4a ge 0 rArr a in (-infty, 0 ]cap [4, infty)` (1) (ii) `- (B)/(2A) gt 1 rArr - (a)/(2a)lt 1 ` (which is always true as a `ne`0) (iii) ` Af(1) gt 0 ` `rArr a(2a + 1) gt 0` `rArr ain (-infty,-(1)/(2)) cap (0, infty).` (2) From (1) and (2) , `a in (-infty, - 1//2) cap [4, infty)` . |
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| 214. |
If both the roots of `x^2+a x+2=0`lies in the interval (0, 3), then find the exhaustive range of value of`adot` |
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Answer» Both roots of `x^(2) + ax + 2 = 0 ` lies in the interval (0,3). Compare this equation with `Ax^(2) x + Bx + C = 0` . Then , the required conditions are (i) `D = a^(2) - 8 ge 0 rArr a in (-infty, - 2 sqrt(2)] cap [2sqrt(2), infty)` (1) (ii) `Af (0) gt 0 and Af (3) gt 0 ` `rArr 2 gt 0, 9 + 3a + 2 gt 0 ` `rArr a in (- (11)/(3), infty)` (2) `(iii) 0 lt - (B)/(2A) lt 3` `rArr 0 lt - (a)/(2) lt3` `rArr - 6 lt a lt 0 ` (3) From (1), (2) and (3) , ` a in (-11/ 3 , - 2 sqrt(2)]` . |
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| 215. |
If the equation `ax^(2) + bx + c = 0, a,b, c, in` R have non -real roots, thenA. `c(a - b +c)gt 0`B. `c (a + b + c) gt 0 `C. `c( 4a - 2b + c) gt0 `D. none of these |
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Answer» Correct Answer - 1,2,3 Since the roots if ` ax^(2) + bx + c = 0 ` are nonreal , so `f(x) = ax^(2) + bx + c ` Will have same sign for every value of x. Hence, ` f(0) = c. f(1) = a + b + c, f (-1) = a - b +c` ` f(-2) = 4 a - 2b + c ` `rArr c (a +b + c) gt 0, c(a - b + c) gt 0 , c (4a - 2b + c ) gt 0 ` . |
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| 216. |
Consider the inequation `x^(2) + x + a - 9 lt 0` The value of the parameter a so that the given inequaiton is ture `AA x in (-1,3)`A. `(-oo,-3]`B. `(-3,oo)`C. `[9,oo)`D. `(-oo, 34//4)` |
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Answer» Correct Answer - 1 If `x^(2)+ x + a-9 lt 0` is ture `AA x in (-1,3)`, then `f(-1) lt 0 and f(3)lt 0` `therefore 1-1 + a -9 lt 0 and 9 + 3 + a-9 lt 0` `rArr a lt 9 and a le -3` `rArr a le - 3` |
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| 217. |
For a `in` R, if `|x - a + 3| + |x-3a|=2x - 4a +3 |` is ture `AA x in`R. Then find the value of a. |
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Answer» `|x - a + 3| + |x-3a|=2x - 4a +3|` `rArr |x - a + 3| + |x-3a|=|(x - a +3) + (x - 3a)|` `rArr (x - a + 3 )(x - 3a)ge0 AA x in`R. `rArr x^(2) + x (3 - 4a) + 3a (a - 3) ge 0, AA x in n` R. For the above inequality, `D le 0` `rArr (3 - 4 a)^(2) - 12 a (a - 3) le 0` `rArr (2a + 3)^(2) le 0` `rArr a = - 3//2` |
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| 218. |
Let `S={x in R: x ge 0 and 2|(sqrt(x)-3|+sqrt(x)(sqrt(x)-6)+6=0}` then S (1) is an empty set (2) contains exactly one element (3) contains exact;y two elements (4) contains exactly four elementsA. is an empty setB. xcontains exactly one elementC. contains exactly two elementsD. contains exactly four elements |
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Answer» Correct Answer - C We have, `2|sqrtx-3|+sqrtx(sqrtx-6)+6=0` Let `sqrtx-3=y` `impliessqrtx=y+3` `therefore2|y|+(y+3)(y-3)+6=0` `implies2|y|+y^(2)-3=0` `implies|y|^(2)+2|y|-3=0` `(|y|+3)(|y-1)=0` `implies|y|ne-3implies|y|=1` `impliesy=+-1impliessqrtx-3=+-1` `impliessqrtx=4,2impliesx=16,4` |
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| 219. |
If `cos x - y^(2) - sqrt(y - x ^(2) - 1 )ge 0 ` , thenA. ` y ge 1 `B. `x in ` RC. `y = 1 `D. `x = 0` |
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Answer» Correct Answer - 3,4 ` cos x- y^(2) - sqrt(y - x^(2) - 1 ) ge 0 ` (1) Now , `sqrt(y - x^(2) - 1)` is defined when ` y - x^(2) - 1 ge 0 or ge x^(2) + 1 ` so minimun values of y is 1. From (1), ` cos x - y^(2) ge sqrt(y - x^(2) - 1)` where `cos x - y^(2) ge 0 ` [ as when cos x is maximum `(=1) and y^(2) ` is minimum (=1), so cos x - `Y^(2)` is maximum ] . Also, ` sqrt(y - x ^(2) - 1)` Hence, `cos x - y^(2) = sqrt(y - x ^(2) - 1)= 0 ` `rArr y = 1 cos x = 1 , y = x^(2) = 1 ` `rArr x = 0, y = 1 ` |
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| 220. |
If `b^(2)ge4ac` for the equation `ax^(4)+bx^(2)+c=0` then all the roots of the equation will be real ifA. `bgt0,alt0,cgt0`B. `blt0,agt0,cgt0`C. `bgt0,agt0,cgt0`D. `bgt0,alt0,clt0` |
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Answer» Correct Answer - B::D Put `x^(2)=y` Then the given equation can be writte as `f(y)=ay^(2)+by+c=0`……….i The givenn equation will have real roots i.e. Eq. (i) has two non-negative roots. Then `-b/age0` ltbr `af(0)ge0` and `b^(2)-4acge0`[given] `impliesb/ale0` `acge0` `impliesagt0,blt0,cgt0` `or alt0,bgt0,clt0` |
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| 221. |
Let `S={x in R: x ge 0 and 2|(sqrt(x)-3|+sqrt(x)(sqrt(x)-6)+6=0}` then S (1) is an empty set (2) contains exactly one element (3) contains exact;y two elements (4) contains exactly four elementsA. contains exactly four elementsB. is an empty setC. contains exactly one elementD. contains exactly two elements |
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Answer» Correct Answer - 4 Given equation is `2 |sqrt(x) - 3|+ x - 6 sqrt(x)+6= 0 ` or ` 2 |sqrt(x) - 3|+ (sqrt(x) - 3)^(2) - 3 = 0 ` Let `|sqrt(x) - 3| = y` ` therefore y^(2) + 2y - 3 = 0 ` `rArr (y - 1) (y + 3) = 0 ` ` rArr y = 1, - 3 ` `therefore |sqrt(x) - 3| =1` `rArr sqrt(x) - 3 = pm 1` `rArr sqrt(x) = 2, 4` `rArr x = 4, 16 ` So, equation has two solutions. |
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| 222. |
if `x^3+ax+1=0 `and `x^4+ax^2+1=0` have common root then the exhaustive set of value of `a` isA. `a=2`B. `a=-2`C. `a=0`D. none of these |
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Answer» Correct Answer - B We have `x^(3)+ax+1=0` or `x^(4)+ax^(2)+x=0`……………..i and `x^(4)+ax^(2)+1=0`……………ii From Eqs. (i) and (ii) we get `x-1=0` `impliesx=1` Which is a common root. `:.1+a+1=0` `impliesa=-2` |
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| 223. |
Solution set of `x-sqrt(1-|x|)lt0`, isA. `[-1,(-1+sqrt(5))/2)`B. `[-1,1]`C. `[-1,(-1+sqrt(5))/2]`D. `(-1,(-1+sqrt(5))/2)` |
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Answer» Correct Answer - A We have `x-sqrt(1-|x|)lt0`…i which is defined only when `1-|x|gt0` `implies|x|le1` `implies x epsilon [-1,1]` Now, from Eq. (i) we get `x lt sqrt(1-|x|)` Case I If`x ge0` i.e. `0lexle1` `x-sqrt((1-|x|))lt0` `impliesx lt sqrt((1-x))` On squaring both sides we get `x^(2)+x-1lt0` `implies(-1-sqrt(5))/2lt xlt (-1+sqrt(5))/2` But `xge0` `:.x epsilon[0,(-1+sqrt(5))/2)` Case II If `x lt 0` i.e. `-1lexlt0` `x-sqrt((1+x))lt0` `impliesx lt sqrt(1+x)` [always true] ` x epsilon [-1,0)` Combining both cases we get `x epsilon [-1,(-1=sqrt(5))/2)` |
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| 224. |
If `p ,q , in {1,2,3,4},`then find the number of equations of the form `p x^2+q x+1=0`having real roots.A. 15B. 9C. 8D. 7 |
| Answer» Correct Answer - D | |
| 225. |
Solution of the equation `3^(2x^2)-2.3^(x^2+x+6)+3^(2(x+6))` =0 isA. `{-3,2}`B. `{6,-1}`C. `{-2,3}`D. `{1,-6}` |
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Answer» Correct Answer - C We have `3^(2x^(2))-2.3^(x^(2)+x+6)+3^(2(x+6))=0` `implies(3^(x^(2))-3^(x+6)^(2)=0` `implies3^(x^(2))-3x^(x+6)=0` `implies3^(x^(2))-3^(x+6)=0` `=3^(x^(2))=3^(x+6)impliesx^(2)-x+6` `impliesx^(2)-x-6=0` `implies(x-3)(x+2)=0` `:.x={-2,3}` |
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| 226. |
Number of real roots of the equation `sqrt(x)+sqrt(x-sqrt((1-x)))=1` is |
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Answer» Correct Answer - B We have `sqrt(dx)+sqrt(x-sqrt((-x)))=1` `impliessqrt(x-sqrt(1-x))=1-sqrt(x)` On squaring both sides we get `x-sqrt(1-x)=1+x-2sqrt(x)` `implies-sqrt(1-x)=1-2sqrt(x)` Again, squaring on both sides we get `1-x=1+4x-4sqrt(x)` `4sqrt(x)=5x` `impliessqrt(x)=4/5` [on squaring both sides] `impliesx=16/25` Hence the number of real solutions is 1. |
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| 227. |
If `x_(1)` and `x_(2)` are the arithmetic and harmonic means of the roots fo the equation `ax^(2)+bx+c=0`, the quadratic equation whose roots are `x_(1)` and `x_(2)` isA. `abx^(2)+(b^(2)+ac)x+bc=0`B. `2abx^(2)+(b^(2)+4ac)x+2bc=0`C. `2abx^(2)+(b^(2)+ac)x+bc=0`D. none of these |
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Answer» Correct Answer - B Let `alpha` and beta` be the roots of `ax^(2)+bx+c=0` `:.x_(1)=(alpha+beta)/2=-b/(2a)` and `x_(2)=(2alpha beta)/(alpha+betsa)=(2 . c/a)/(-b/a)=-(2c)/b` `:.` The required equation is `x^(2)-[(-b/(2a))+(-(2c)/b)]x+(2bc)/(2ab)=0` i.e. `2abx^(2)+(b^(2)+4ac)x+2bc=0` |
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| 228. |
If one root of the equation `x^(2)-lamdax+12=0` is even prime while `x^(2)+lamdax+mu=0` has equal roots, then `mu` isA. 8B. `16`C. `24`D. 32 |
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Answer» Correct Answer - B Since 2 is only even prime. Therefore we have `2^(2)+lamda.2+12=0` `implieslamda=8` `:.x^(2)+lamdax+mu=0` `impliesx^(2)+8x+mu=0`….i But Eq. (i) has equal roots. `:.D=0` `implies 8^(2)-4.1=mu=0` `impliesmu=16` |
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| 229. |
If the difference of the roots of `x^(2)-lamdax+8=0` be 2 the value4 of `lamda` isA. `+-2`B. `+-4`C. `+-6`D. `+-8` |
| Answer» Correct Answer - C | |
| 230. |
If `a gt 1` , then the roots of the equation `(1-a)x^(2)+3ax-1=0` areA. one positive and one negativeB. both negativeC. both positiveD. both non real complex |
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Answer» Correct Answer - C `(c )` Sum of roots `=(3a)/(a-1)implies +ve` Product of roots`=(1)/(a-1)implies+ve` Also `D=9a^(2)+4(1-a)=a(9a-4)+4 gt 0` Hence, both positive roots. |
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| 231. |
The equation `ax^(4)-2x^(2)-(a-1)=0` will have real and unequal roots ifA. `o lt a lt 1`B. `a gt 0`, `a ne 1`C. `a lt 0` , `a ne 1`D. none of these |
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Answer» Correct Answer - A `(a)` Putting `x^(2)=y`, the given equation in `x` reduces to `ay^(2)-2y-(a-1)=0` ……….`(i)` The given biquardratic equation will have four real and distinct roots, if the quadratic equation `(i)` has two distinct and positive roots. For that, we must have `D gt 0impliesa^(2)-a+1 gt 0`, which is true `AA a in R` Product of roots `gt 0implies0 lt a lt 1` Sum of roots `gt 0 implies a gt 0` Hence, the acceptable values of `a` are `0 lt a lt 1`. |
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| 232. |
Find the rang of`f(x)=(x^2+34 x-71)/(x^2+2x-7)``f(x)=(x^2-x+1)/(x^2+x+1)` |
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Answer» Correct Answer - `(a) (-infty, 5 )cup [9, infty)` `[1//3,3]` Let `(x^(2) 34 x - 71)/(x^(2) + 2x - 7) = y` or `( 1 - y) + 2(17 - y) x + (7y - 71) = 0` For real value of x `b^(2) - 4ac ge 0` `rArr y^(2) - 14 y + 45 ge 0` `rArr (y - 5) (y - 9) ge 0` `rArr y le 5 or y ge 9` Hence, the range is `(- infty, 5] cup [ 9, infty)` (b) Let `y= (x^(2) - x + 1)/(x^(2)+ x + 1)` `(1 -y )x^(2) - 4 ( 1 - y)^(2) ge 0` Now, if x is real, then `D ge 0` `rArr ( 1 + y)^(2) - 4 (1 - y )^(2) ge 0` or `(1 + y - 2 + 2y) ( 1 + y + 2 - 2y) ge 0` or `(3y - 1) ( 3 - y) ge 0` or ` 3(y - (1)/(2))^(2 )(y - 3) le 0` or `(1)/(3) le y le 3` Hence, the range is `[1//3, 3)].` |
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| 233. |
Solve the equation `5^(x)root(x)(8^(x-1))=500` |
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Answer» We have `5^(x)root(x)(8^(x-1))=5^(3).2^(2)` `implies5^(x).7^(((x-1)/3))=5^(3).2^(2)` `implies5^(x).2^((3x-3)/x)=5^(3).2^(2)` `implies5^(x_3).2(((x-3)/x))=1` `implies(5.2^(1//x)(x-3))=1` is equivalent to the equation `10^((x-3)log5.2^(1//x))=1` `implies(x-3)log(5.2^(1//x))=0` Thus, original equation is equivalent to the collection of equations `x-3=0,log(5.2^(1//x))=0` `:.x_(1)=3,5.2^(1//x)=1implies2^(1//x)=(1/5)` `:.x_(2)=-log_(5)2` Hence roots of the original equation are `x_(1)=3` and `x_(2)=-log_(5)2`. |
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| 234. |
Solve the inequation `log_(2x+3)x^(2)ltlog_(2x+3)(2x+3)` |
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Answer» This inequation is equivalent to the collection of the systems `[{(2x+3gt1),(x^(2)lt2x+3),(0lt2x+3lt1),(x^(2)gt2x+3):}implies[{(xgt-1),((x-3)(x+1)lt0):},{((-3/2ltxlt-1),((x-3)(x+1)gt0):}):}` `implies[{(xgt-1),(-1ltxlt3):}, {(-3/2lt x lt-1),(xlt-1 "and" xgt3)implies-3/2ltxlt-1):}` Hence the solution of the original inequation is `x epsilon(-3/2,-1)uu(-1,3)`. |
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| 235. |
Solve `sqrt(3x^2-7x-30)+sqrt(2x^2-7x-5)=x+5.` |
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Answer» Correct Answer - `x = 6, - 5//2` `sqrt(3x^(2) - 7x - 30) + sqrt(2x^(2) - 7x - 5) = x + 5` or `sqrt(3x^(2) - 7x - 30) =(x+5)- sqrt(2x^(2) - 7x - 5) = ` or On squaring, we get `3x^(2) - 7x - 30 =x^(2) + 10x + 25 - 2(x+5) xxsqrt(2x^(2) - 7x - 5)+ 2x^(2) - 7x - 5` or `10x + 50 = 2(x + 5) sqrt(2x^(2) - 7x - 5)` `rArr x = - 5 or sqrt(2x^(2) - 7x - 5) = 5` `rArr x = - 5 or 2x^(2) - 7x - 30 = 0` `rArr x = - 5 or x = 6 or x = - 5//2` But `x = - 5` does not satisfy the original equation. |
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| 236. |
Solve `3^2x^(2-7)^=9.` |
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Answer» Correct Answer - `x = 1, 5//2` `3^(2x^(2)-7x + 7) = 9.` or `3^(2x^(2)-7x + 7) = 3^(2)` or `2x^(2)-7x + 7 = 2` or `2x^(2)-7x + 5 = 0 or x = 1, 5 //2` |
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| 237. |
If the roots of equation `(a + 1)x^2-3ax + 4a = 0` (a is not equals to -1) are greater than unity, thenA. `[-(10)/(7),1]`B. `[-(12)/(7),0]`C. `[-(16)/(7),-1)`D. `(-(16)/(7),0)` |
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Answer» Correct Answer - C `(c )` `(a+1)x^(2)-3ax+4a=0` `D=9a^(2)-16a(a+1) ge 0` ………`(i)` `x_(1) gt 1`, `x_(2) gt 1` `:. (a+1)f(1) gt 0` and `(3a)/(2(a+1)) gt 1` `implies(a+1)(2a+1) gt 0` and `(a-2)/(a+1) gt 0` ……..`(ii)` `(2a+1)/(a+1) gt 0`………`(iii)` Solving `(i)`, `(ii)` and `(iii)`, we get `-(16)/(7) le a lt -1`. |
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| 238. |
Solve the equation `3^(x-4)+5^(x-4)=34` |
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Answer» Here `3^(2)+5^(2)=34`, then given equation has a solution `x-4=2` `:.x_(1)=6` is a root of the original equation. |
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| 239. |
Solve `(8^x+27^x)/(12^x+18^x)=7/6` |
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Answer» Correct Answer - `x = -1 or 1` `(8^(x) + 27^(x))/(12^(x) + 18^(x) )=(7)/(6)` or `(8)^(x)/(12)^(x) ((1 +(27//8)^x))/((1 +(18//12)^x))= (7)/(6)` or `((2)/(3))^(x)((1 +(3//2)^(3x))/(1 +(3//2)^x))= (7)/(6)` Let `((3)/(2))^(x) = t` . Then `(1 +t^(3))/(t(1 + t)) = (7)/(6)` or `((1 +t)(t^(2) + 1 - t))/(t(1 + t)) = (7)/(6)` or `(t^(2)+ 1 - t)/(t) = (7)/(6) (because t + 1 ne` 0) or `6t^(2) - 13t + 6 = 0` or `(2t - 3)(3t-2) = 0` `rArr t = (2)/(3) or (3)/(2)` `rArr x = - 1 or 1` . |
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| 240. |
Solve the equation `15.2^(x+1)+15.2^(2-x)=135` |
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Answer» This equation rewrite in the form `30.2^x)+60/(2^(x))=135` Let `t=2^(x)` Then `30t^(2)-135t+60=0` `implies6t^(2)-27t+12=0` `implies6t^(2)-24t-3t+12=0` `implies(t-4)(6t-3)=0` Then `t_(1)=4` and `t_(2)=1/2` Thus, given equation is equivalen to `2^(x)=4` and `2^(x)=1/2` Then `x_(1)=2` and `x_(2)=-1` Hence roots of the original equation are `x_(1)=2` and `x_(2)=-1`. |
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| 241. |
If `(b^2-4a c)^2(1+4a^2) |
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Answer» Correct Answer - 2 `((b^(2) - 4ac)^(2))/(16a^(2)) lt (4)/(1 + 4a^(2))` Now , max `(ax^(2) + bx + c) = -((b^(2) - 4ac)/(4a)` Also, `(-2)/(sqrt(1 + 4 a^(2))) lt - (b^(2) - 4ac)/(4a) lt (2)/(sqrt(1 + 4a^(2)))` So, maximum value is alwasys less then 2 (when a `to` 0) . |
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| 242. |
Solve the equation `64.9^(x)-84(12^(x))+27(16^(x))=0` |
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Answer» Here `9xx16=(12)^(2)` Then, we divide its bothsides by`12^(x)` and obtain `implies64.(3/4)^(x)-84+27.(4/3)^(x)=0`…..i Let `(3/4)^(x)=t`, the eq. (i) reduce in te form `64t^(2)-84t+27=0` `:.t_(1)=3/4` and `t_(2)=9/16` then`(3/4)^(x)=(3/4)^(1)` and `(3/4)^(x)=(3/4)^(2)` `:.x_(1)=1` and `x_(2)=2` Hence, roots of the original equation are `x_(1)=1` and `x_(2)=2` |
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| 243. |
Solve the equation `5^(2x)+24.5^(x)-25=0` . |
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Answer» Let `5^(x)=t` then the given equation can reduce in the form `t^(2)-24t-25=0` `rarr(t-25)(t_1)=0impliest!=-1` `:.t=25` then `5^(x)=25=5^(2)` then `x=2` Hence `x_(1)=2` is only one root of the original equation. |
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| 244. |
Solve the equation `5^(x^(2)+3x+2)=1` |
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Answer» This equation is equivalent to `x^(2)+3x+2=0` `implies(x+1)(x+2)=0` `:.x_(1)=-1,x_(2)=-2` consequently, this equation has two roots `x_(1)=-1` and `x_(2)=-2`. |
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| 245. |
If `alpha` and `beta` be the roots of equation `x^(2) + 3x + 1 = 0` then the value of `((alpha)/(1 + beta))^(2) + ((beta)/(1 + alpha))^(2)` is equal toA. `18`B. `19`C. `20`D. `21` |
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Answer» Correct Answer - A `(a)` `alpha+beta=-3` , `alpha beta=1`. Also `alpha^(2)+3alpha+1=0` and `beta^(2)+3beta+1=0` `implies` where `alpha^(2)=-(3alpha+1)` and `beta^(2)=-(3beta+1)` `y=(alpha^(2))/((1+beta)^(2))+(beta^(2))/((alpha+1)^(2))` `= (alpha^(2))/(1+2beta+beta^(2))+(beta^(2))/(1+2alpha+alpha^(2))` `=((-(3alpha+1))/(-beta))+((-(1+3beta))/(-alpha))` `=(1+3alpha)/(beta)+(1+3beta)/(alpha)` `=(alpha(1+3alpha)+beta(1+3beta))/(alpha beta)` `=3(alpha^(2)+beta^(2))+(alpha+beta)` (as `alpha beta=1`) `=3[9-2]+(-3)=21-3=18` |
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| 246. |
Solve the inequation `sqrt((x+14))lt(x+2)` |
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Answer» We have `sqrt((x+14))lt(x+2)` This inequation is equivalent to the system `{(x+14ge0),(x+2ge0),(x+14lt(x+2)^(2)):}implies{(xge-14),(xgt-2),(x^(2)+3x-10gt0):}` `implies{(xge-14),(xgt-2),((x+5)(x-2)ge0):}implies{(xge-14),(xge-2),(xlt-5 "and" xgt2):}` On combining all three inquation of the system, we get `xgt2`, i.e. `x epsilon (2,oo)` |
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| 247. |
Let `f(x)=ax^(2)+bx+c`, `g(x)=ax^(2)+px+q`, where `a`, `b`, `c`, `q`, `p in R` and `b ne p`. If their discriminants are equal and `f(x)=g(x)` has a root `alpha`, thenA. `alpha` will be `A.M.` of the roots of `f(x)=0`, `g(x)=0`B. `alpha` will be `G.M.` of the roots of `f(x)=0`, `g(x)=0`C. `alpha` will be `A.M.` of the roots of `f(x)=0` or `g(x)=0`D. `alpha` will be `G.M.` of the roots of `f(x)=0` or `g(x)=0` |
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Answer» Correct Answer - A `(a)` `a alpha^(2)+b alpha+c=a alpha^(2)+p alpha+q=0` `implies alpha=(q-c)/(b-p)`…………`(i)` and `b^(2)-4ac=p^(2)-4aq` `impliesb^(2)-p^(2)=4a(c-q)` `implies b+q=(4a(c-q))/(b-p)=-4a alpha` (from `(i)`) `:.alpha=(-(b+p))/(4a)=((-b)/(a)-(p)/(a))/(4)` which is `AM` of all the roots of `f(x)=0` and `g(x)=0` |
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| 248. |
Solve the equation `sqrt((6-x))(3^(x^(2)-7.2x+3.9)-9sqrt(3))=0` |
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Answer» We have `sqrt((6-x))(3^(x^(2)-7.2x+3.9)-9sqrt(3))=0` This equation is defined for `6-xge0`. i.e. `xle6` …….i This equation is equivalent to the collection of equations `sqrt(6-x)=0` and `3^(x^(2)-7.2x+3.9)-9sqrt(3)=0` `:.x_(1)=6` and `3^(x^(2)-7.2x+3.9)=3^(2.5)` Then `x^(2)-7.2x+3.9=2.5` `x^(2)-7.2x+1.4=0` We find that `x_(2)=1/5` and `x_(3)=7` Hence solution of the original equation are [which satisfies Eq.(i) ] `x_(1)=6,x_(2)=1/5` |
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| 249. |
Solve the equation `sqrt(x)=x-2` |
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Answer» We have `sqrt(x)=x-2` On squaring both sides we obtain `x=(x-2)^(2)` `impliesx^(2)-5x+4=0implies(x-1)(x-4)=0` `:.x_(1)=1` and `x_(2)=4` Hence `x_(1)=4` satisfies the original equation, but `x_(2)=1` does not satisfy the original equation. `:.x_(2)=1` is the extraneous root. |
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| 250. |
Let `f(x)=ax^(2)+bx+c`, `a ne 0`, `a`, `b`, `c in I`. Suppose that `f(1)=0`, `50 lt f(7) lt 60 ` and `70 lt f(8) lt 80`. Number of integral values of `x` for which `f(x) lt 0` isA. `0`B. `1`C. `2`D. `3` |
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Answer» Correct Answer - B `f(x)=ax^(2)+bx+c`, `a ne 0`, `a,b,c in I` `f(1)=0` ……..`(i)` `implies a+b+c=0` `50 lt f(7) lt 60` `50 lt 49a+7b+c lt 60` `implies 50 lt 48a+6b lt 60` `implies (50)/(6) lt 8a+b lt 10` `implies 8a+b=9`……`(ii)` Also `70 lt f(8) lt 80` `implies 70 lt 64a+8b+c lt 80` `implies70 lt 63a+7b lt 80` `implies 10 lt 9a+b lt (80)/(7)` `implies 9a+b=11`.........`(iii)` From `(i)`, `(ii)` and `(iii)`, `a=2`, `b=-7`, `c=5` `implies f(x)=2x^(2)-7x+5=(2x-5)(x-1)` `=2(x^(2)-(7)/(2)x+(5)/(2))` `=2((x-(7)/(4))^(2)-(9)/(16))` `implies f(x)` has least value `-(9)/(8)` `f(x) lt 0 implies (2x-5)(x-1) lt 0implies1 lt x lt 5//2` |
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