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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.
| 1. |
If roots of the equation `f(x)=x^6-12 x^5+bx^4+cx^3+dx^2+ex+64=0`are positive, then Which has the greatest absolute value ?A. bB. cC. dD. e |
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Answer» Correct Answer - A Let roots of equation `x^(6) - 12x^(5) + bx^(4) + cx^(3) + dx^(2) + ex + 64 = 0` be `x_(i), I = 1,2…..6` Now, `x_(1) + x_(2) + x_(3) + x_(4) + x_(5) + x_(6) = 12` and `x_(1) x_(2) x_(4) x_(5) x_(6) = 64` Thus, `(x_(1) + x_(2) …. + x_(6))/(6) = 2` and `(x_(1) x_(2) x_(3) x_(5) x_(6))^(1//6) = 2` `implies a.M = G.M` `implies x_(1) = x_(2) = x_(3) - x_(4) = x_(5) = x_(6) = 2` Hence, the given equation is equivalent to `(x - 2)^(6) = 0` or `x^(6) - 12 x^(5) + 60x^(4) - 160 x^(3) + 240x^(2) - 192 x - 64 = 0` `:. f(1) = 1 - 12 + 60 - 160 + 240 - 192 + 64 = 1` |
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| 2. |
If roots of the equation `f(x)=x^6-12 x^5+bx^4+cx^3+dx^2+ex+64=0`are positive, then remainder when `f(x)` is divided by `x-1` isA. 2B. 1C. 3D. 10 |
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Answer» Correct Answer - B Let roots of equation `x^(6) - 12x^(5) + bx^(4) + cx^(3) + dx^(2) + ex + 64 = 0` be `x_(i), I = 1,2…..6` Now, `x_(1) + x_(2) + x_(3) + x_(4) + x_(5) + x_(6) = 12` and `x_(1) x_(2) x_(4) x_(5) x_(6) = 64` Thus, `(x_(1) + x_(2) …. + x_(6))/(6) = 2` and `(x_(1) x_(2) x_(3) x_(5) x_(6))^(1//6) = 2` `implies a.M = G.M` `implies x_(1) = x_(2) = x_(3) - x_(4) = x_(5) = x_(6) = 2` Hence, the given equation is equivalent to `(x - 2)^(6) = 0` or `x^(6) - 12 x^(5) + 60x^(4) - 160 x^(3) + 240x^(2) - 192 x - 64 = 0` `:. f(1) = 1 - 12 + 60 - 160 + 240 - 192 + 64 = 1` |
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| 3. |
Find the greatest value of `x^2y^3z^4`if `x^2+y^2+z^2=1,w h e r ex ,y ,z`are positive. |
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Answer» We have `(2(x^2)/(2)+3(y^2)/(3)+4(z^2)/(4))/(9) ge(((x^2)/(2))^2((y^2)/(3))^3((z^2)/(4))^4)^((1)/(9))` `rArr (1)/(9^9) ge(x^4y^6x^8)/(2^23^34^4)` `rArr x^4y^6z^8 le (2^23^34^4)/(9^9) ` `rArr x^2y^3z^4 le 2^5xx3^(-(15)/(2))` |
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| 4. |
If a + b =1, a `gt` 0,b `gt` 0, prove that `(a + (1)/(a))^(2) + (b + (1)/(b))^(2) ge (25)/(2)` |
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Answer» We know that A.M. of mth power `gt` mth power of A.M. ` therefore (((a+1)/(a))^2+(b+(1)/(b))^2)/(2)gt [((a+(1)/(b))^2+(b+(1)/(b)))/(2)]^2," here "m=2` or ` (a+(1)/(a))^2+(b+(1)/(b))^2gt (1)/(2)[(a+b)+((1)/(a)+(1)/(b))]^2` Also, `(a^-1+b^-1)/(2)gt ((a+b)/(2))^2` or ` (1)/(2)((1)/(a)+(1)/(b))gt (2)/(a+b)` or ` (1)/(a)+(1)/(b)gt (4)/(a+b)` or ` (1)/(a)+(1)/(b)gt4`. Hence, from (1) From (i), ` (a+(1)/(a))^2+(b+(1)/(c))^2gt (1)/(2)(1+4)^2=(25)/(2)` |
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| 5. |
If x,y,z `gt` 0 and x + y + z = 1, the prove that `(2x)/(1 - x) + (2y)/(1 - y) + (2z)/(1 - z) ge 3`. |
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Answer» Using `A.M. ge H.M.` we have `((x+y)+(y+z)+(z+x))/(3) ge (3)/((1)/(x+y)+(1)/(y+z)(1)/(z+x))` ` rArr 2(x+y+z)((1)/(x+y)+(1)/(y+z)+(1)/(z+x))ge 9` `rArr 2((x+y+z)/(x+y)+(x+y+z)/(y+z)+(x+y+z)/(z+x))ge 9` `rArr 2[(1+(z)/(x+y))+(1+(x)/(y+z))+(1+(y)/(x+z))]ge 9` ` rArr 2((x)/(y+z)+(y)/(z+x)+(z)/(x+y))ge3` `rArr 2((x)/(1-x)+(y)/(1-y)+(z)/(1-z)) ge3` |
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| 6. |
If `a,b,x,y` are real number and `x,y gt 0`, then `(a^(2))/(x)+(b^(2))/(y) ge ((a+b)^(2))/(x+y)` so on solving it we have `(ay-bx)^(2) ge 0`. Similarly, we can extend the inequality to three pairs of numbers, i.e, `(a^(2))/(x)+(b^(2))/(y)+(c^(2))/(z) ge ((a+b+c)^(2))/(x+y+z)` Now use this result to solve the following questions. The value of `(a^(2)+b^(2))/(a+b)+(b^(2)+c^(2))/(b+c)+(a^(2)+c^(2))/(a+c)` isA. ` ge (a+b+c)`B. `ge (1)/(2)(a+b+c)`C. `(3)/(2) le (a+b+c)`D. None of these |
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Answer» Correct Answer - A `(a)` `(a^(2)+b^(2))/(a+b)+(b^(2)+c^(2))/(b+c)+(a^(2)+c^(2))/(a+c)` `=(a^(2))/(a+b)+(b^(2))/(b+c)+(c^(2))/(a+c)+(b^(2))/(a+b)+(c^(2))/(b+c)+(a^(2))/(a+c)` `ge ((2a+2b+2c)^(2))/(4(a+b+c)) ge (a+b+c)` `implies(a^(2))/(a+b)+(b^(2))/(b+c)+(c^(2))/(a+c)+(b^(2))/(a+b)+(c^(2))/(b+c)+(a^(2))/(a+c) ge a+b+c` |
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| 7. |
If `a_1, a_2, ,a_n >0,`then prove that`(a_1)/(a_2)+(a_2)/(a_3)+(a_3)/(a_4)++(a_(n-1))/(a_n)+(a_n)/(a_1)> n` |
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Answer» Since `A.M gt G.M`., we have `(1)/(n) ((a_1)/(a_2)+(a_2)/(a_3)+(a_3)/(a_4)+.....+(a_n-1)/(a_n)+(a_n)/(a_1))((a_1)/(a_2)(a_2)/(a_3)(a_3)/(a_4)....(a_(n-1)a_n)/(a_na_1))^(1//n)` or `((a_1)/(a_2)+(a_2)/(a_3)+(a_3)/(a_4)+.....(a_(n-1))/(a_n)+(a_n)/(a_1)) gt n(1)^n` or `((a_1)/(a_2)+(a_2)/(a_3)+(a_3)/(a_4)+....+(a_(n-1))/(a_n))gt n` |
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| 8. |
If `a ,b ,c`are positive, then prove that `a//(b+c)+b//(c+a)+c//(a+b)geq3//2.` |
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Answer» we have `(a)/(b + c) + (b)/(c + a) + (c )/(a + b) ge (3)/(2)` `implies (a)/(b + c) + 1 + (b)/(c + a) + 1 + (c )/(a + b) + 1 ge (3)/(2) + 3` `implies (a + b + c)/(b + c) + (a + b + c)/(c + a) + (a + b+ c)/(a + b) ge (9)/(2)` `(1)/(b + c) + (1)/(c + a) + (1)/(a + b) ge (9)/(2(a + b + c))` Now, using A.M `ge` H.M we have `((1)/(b + c) + (1)/(c + a) + (1)/(a + b))/(3) ge (3) ((a + b) + (b + c) + (c + a))` `implies (1)/(b + c) + (1)/(c + a) + (1)/(a + b) ge (9)/(2(a + b + c))` |
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| 9. |
Prove that `(a b+x y)(a x+b y)>4a b x y(a , b ,x ,y >0)dot` |
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Answer» Using A.M `ge` G.M we have `(ab + xy)/(2) gt sqrt(abxy)` `implies ax + xy gt 2 sqrt(abxy)` similarly, ax + by `gt 2 sqrt(abxy)` Multiplying (1) and (2) we get `(ab + xy) (ax + by) gt 4abxy` |
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| 10. |
Prove that `[(x^2+y^2+z^2)/(x+y+z)]^(x+y+z)gtx^xy^yz^zgt[(x+y+z)/(3)]^(x+y+z)(x,y,zgt0)` |
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Answer» Let `tan^(2) alpha = x` and `tan^(2) beta = y` `:. (sec^(4) alpha)/(tan^(2) beta) + (sec^(4) beta)/(tan^(2) alpha) = ((x + 1)^(2))/(y) + ((y + 1)^(2))/(x)` `= (x^(2))/(y) + (y^(2))/(x) + (1)/(x) + (1)/(y) + 2 (x)/(y) + 2 (y)/(x)` Now, using A.M `ge` G.M we get ltbegt `((x^(2))/(y) + (y^(2))/(x) + (1)/(x) + (1)/(y) + (x)/(y) + (x)/(y) + (y)/(x) + (y)/(x))/(8) ge ((x^(2))/(y). (y^(2))/(x).(1)/(x).(1)/(y).(1)/(y).(x)/(y).(x)/(y).(y)/(x).(y)/(x))^((1)/(8))` `implies (x^(2))/(y) + (y^(2))/(x) + (1)/(x) + (1)/(y) + (X)/(y) + (x)/(y) + (y)/(x) + (y)/(x) ge 8` `:. (sec^(4) alpha)/(tan^(2) beta) + (sec(4) beta)/(tan^(2) alpha) ge 8` |
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| 11. |
Prove that `b^2c^2+c^2a^2+a^2b^2gtabc(a+b+c)`, where `a,b,c gt 0` . |
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Answer» Using A.M. `gt` G.M., we have `(b^(2) c^(2) + c^(2) a^(2))/(2) gt sqrt((b^(2) c^(2))(c^(2) a^(2)))` `implies b^(2) c^(2) + c^(2) a^(2) gt 2 abc^(2)` similarly, `c^(2) a^(2) + a^(2) b^(2) gt 2 a^(2) bc` `a^(2) b^(2) + b^(2) c^(2) gt 2 ab^(2)c` Adding (1),(2 and (3), we get `2b^(2) c^(2) + 2 c^(2) a^(2) + 2 a^(2) b^(2) gt 2 abc^(2) + 2 bca^(2) + 2cab^(2)` `implies b^(2) c^(2) + c^(2) a^(2) + a^(2) b^(2) gt abc (a + b + c)` |
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| 12. |
If `a,b,c,in R^(+)`, such that `a+b+c=18`, then the maximum value of `a^2,b^3,c^4` is equal toA. `2^(18)xx3^(2)`B. `2^(18)xx3^3`C. `2^19 xx3^2`D. `2^19xx3^3` |
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Answer» Correct Answer - D `a + b + c = 18` `implies 2 xx (a)/(2) + 3 xx (b)/(3) + 4 xx (c )/(4) = 18` Using weighted A.M and G.M inequality, we get `(2 xx (a)/(2) + 3 xx (b)/(3) + 4 xx (c )/(4))/(9) ge (((a)/(2))^(2) ((b)/(3))^(3) ((c )/(4))^(4))^(1//9)` or `2^(9) ge (a^(2))/(2^(2)) xx (b^(2))/(3^(3)) xx (c^(4))/(4^(4))` or ` a^(2) b^(3) c^(4) le 3^(3) xx 2^(19)` |
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| 13. |
If `a gt 0,bgt 0,cgt0 and 2a +b+3c=1`, thenA. `a^4b^2c^2` is greatest then `a=(1)/(4)`B. `a^4b^2c^2` is greatest then `b=(1)/(4)`C. `a^4b^2c^2` is greatest then `c=(1)/(12)`D. greatest value of `a^4b^2c^2 is (1)/(9.4^8)` |
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Answer» Correct Answer - A::B::C::D A.M. `ge` G.M `implies ((2a)/(4) + (2a)/(4) + (2a)/(4) + (2a)/(4) + (b)/(2) + (b)/(2) + (3c)/(2) + (3c)/(2))/(8)` `ge .^(8)sqrt((2a)/(4).(2a)/(4).(2a)/(4).(2a)/(4).(b)/(2).(b)/(2).(3c)/(2).(3c)/(2))` `implies (2a + b + 3c)/(8) ge ((3^(2))/(2^(8)) a^(4) b^(2) c^(2))^(1/8)` The greatest value takes place when A.M =G.M ad `(2a)/(4) = (b)/(2) = (3c)/(2)` `implies a = b = 3c = K` Now, `2a + b + 3c = 1` `implies 2K + K + K = 1` `implies K = 1//4` `implies a = b = 1//4` and `c = 1//12` |
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| 14. |
Prove that `2/(b+c)+2/(c+a)+2/(a+b)0.` |
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Answer» Using A.M.`gt` H.M we get `((1)/(a) + (1)/(b))/(2) gt (2)/(a + b), ((1)/(b) + (1)/(c ))/(2) gt (2)/(b + c)` and `((1)/(c c ) + (1)/(a))/(2) gt (2)/(c + a)` Adding we get `(1)/(a) + (1)/(b) + (1)/(c ) gt (2)/(a + b) + (2)/(b + c) + (2)/(a + c)` |
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| 15. |
Prove that `1^1xx2^2xx3^3xx...xn^nlt ((2n+1)/(3))^((n(+1))/(2)` |
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Answer» `1^(2) = 1` `2^(2) = 2 + 2` `3^(2) = 3 + 3 + 3` `:.` Using weighted means `(1 + (2 + 2) + (3 + 3 + 3) + ….. + (n + n + ….n "times"))/(1 + 2 + 3 + …..+ n)` `ge (1^(1). 2^(2)……. N^(n))^((1)/(1 + 2 + 3 + … + n))` `implies (1 + 2^(2) + 3^(2) + .... + n^(2))/((n(n + 1))/(2)) ge (1^(1) 2^(2)...... n^(n))^((2)/(n(n + 1)))` `implies (n(n + 1)(2n + 1)/(6))/((n(n + 1))/(2)) ge (1^(1). 2^(2).... n^(n))^(n(n + 1))` `implies (2n + 1)/(3) ge (1^(1) 2^(2) .... n^(2))^((2)/(n(n + 1))` `implies 1^(1). 2^(2). 3^(3) .... n^(n) le ((2n + 1)/(3))^((n(n + 1))/(2))` |
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| 16. |
If first and `(2n-1)^(th)` terms of A.P., G.P. and H.P. are equal and their nth terms are a,b,c respectively, thenA. `a=b=c`B. `a+c=b`C. `agt bgt c`D. `ac-b^2=0` |
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Answer» Correct Answer - C::D Cosider the A.P since a is equidistant from the first term `alpha` and the last term `beta` of the A.P., therefore `alpha, a beta` are in A.P. Hence a is the A.M. of `alpha` and `beta`. So, `a = (alpha + beta)/(2)` Similarly, b and c are the geometric and hormoic means, i.e., `b = sqrt(alpha beta)` and `c = (2 alpha beta)/(alpha + beta)` Since A.M., G.M., and H.M. are in G.P and A.M `ge` G.M `ge` H.M., therefore a,b,c, are in G.P and `a ge b ge c` |
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| 17. |
`f(x)=((x-2)(x-1))/(x-3), forall xgt3`. The minimum value of `f(x)` is equal toA. `3+2sqrt(2)`B. `3+2sqrt(3)`C. `3sqrt(2)+2`D. `3sqrt(2)-2` |
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Answer» Correct Answer - A Let `x - 3 = t` `implies x - 2 (t + 1)` and `x - 1 = t + 2` `implies f(x) = ((x - 2) (x - 1))/(t)` `= ((t + 1) (t + 2))/(t)` `= ((t^(2) + 3t 2))/(t)` `t + (2)/(t) + 3` `le 3 + 2 sqrt(2)` (using A.M `ge` G.M as `t gt 0`) |
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| 18. |
Let A;G;H be the arithmetic; geometric and harmonic means between three given no. a;b;c then the equation having a;b;c as its root is `x^3-3Ax^2+3G^3/H x-G^3=0` |
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Answer» We have `A = (a + b + c)/(3) implies a + b + c = 34` `G (abc)^(1//3) implies abc = G^(3)` `(1)/(H) = ((1)/(a) + (1)/(b) + (1)/(c ))/(3) = (ab + bc + ca)/(3abc)` `implies (3G^(3))/(H) = ab + bc + ca` The equation having a, b and c as its roots is `x^(3) - (a + b+ c) x^(2) + (ab + bc + ca) x - abc = 0` or `x^(3) - 3Ax^(2) + (3G^(3))/(H) x - G^(3) = 0` |
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| 19. |
Let `x^2-3x+p=0` has two positive roots a and b, the minimum value of `((4)/(a)+(1)/(b))` is ____________. |
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Answer» Correct Answer - 3 a + b = 3 H.M `ge` A.M for numbe `(a)/(2), (a)/(2)`, b. We have `(3)/((2)/(a) + (2)/(a) + (1)/(b)) le ((a)/(2) + (a)/(2) + 6)/(3) = 1` `:. 1 ge (3)/((2)/(a) + (2)/(a) + (1)/(b))`or `(2)/(a) + (2)/(a) + (1)/(b) ge 3` `:. (4)/(a) + (1)/(b) ge 3` |
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| 20. |
Find the greatest value of `x^2y^3, w h e r exa n dy`lie in the first quadrant on the line `3x+4y=5.` |
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Answer» Given that 3x + 4y = 5 since we have expression `x^(2) y^(3)` we consider `2((3x)/(2)) + 3 ((4y)/(3)) = 3x + 4y = 5` Using A.M `ge` G.M for weighted means, we get `(2((3x)/(2)) + 3 ((4)/(3)))/(2 + 3) ge [((3x)/(2))^(2) ((4y)/(3))^(3)]^((1)/(5))` `implies ((3x + 4y)/(5))^(5) ge ((3)/(2))^(2) ((4)/(3))^(3) x^(2) y^(3)` `implies x^(2) y^(2) le ((2)/(3))^(2) ((3)/(4))^(3)` `implies x^(2) y^(3) le (3)/(16)` so, the greatest value of `x^(2) y^(3)` is `(3)/(16)` |
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| 21. |
If `a ,b , a n dc`are distinct positive real numbers such that `a+b+c=1,`then prove tha `((1+a)(1+b)(1+c))/((1-a)(1-b)(1-c))> 8.` |
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Answer» a + b + c = 1 Or a = 1 - b = c `implies 1 + a = (a - b) + (1 - c) gt 2 sqrt((1 - b)(1 - c))` Using A.M `ge` G.M Similarly, `1 + b gt 2 sqrt((1 - c)(1 - a))` and `1 + c gt 2 sqrt((1 - a)(1 - b))` Multiply, we get `(1 + a) (1 + b) (1 + c) gt 8 (1 - a) (1 - b) (1 - c)` or `((1 + a) (1 + b) (1 + c))/((1- a) (1 - b) (1 - c)) gt 8` |
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| 22. |
Find all positive real solutions to `4x+(18)/(y)=14,2y+(9)/(z)=15,9z+(16)/(x)=17`. |
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Answer» Adding all the given equation, we have `4x + (16)/(x) + 2y + (18)/(y) + 9z + (9)/(0) = 46` Now, using A.M `ge` G.M we have `4x + (16)/(x) ge 16, 2y + (18)/(y) 12` and `9z + (9)/(z) 18` `:. 4x + (16)/(x) + 2y + (18)/(y) + 9z + (9)/(z) ge 46` So, equation (1) holds only if `4x + (16)/(x) = 16, 2y, + (18)/(y) =12` and `9z + (9)/(z) = 18` `:. 4x = (16)/(x), 2y = (18)/(y)` and `9z = (9)/(z)` `:. x = 2, y = 3` and `z = 1` |
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| 23. |
For positive real numbers `a ,bc`such that `a+b+c=p ,`which one holds?`(p-a)(p-b)(p-c)lt=8/(27)p^3``(p-a)(p-b)(p-c)geq8a b c``(b c)/a+(c a)/b+(a b)/clt=p``non eoft h e s e`A. `(p-a)(p-b)(p-c)le(8)/(27)p^3`B. `(p-a)(p-b)(p-c)gt 8abc`C. `(bc)/(a)+(ca)/(b)+(ab)/(c)le p`D. none of these |
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Answer» Correct Answer - A::B Using A.M `ge` G.M one can show that `(b + c) (c + a) (a + b) ge 8abc` `implies (p - a) (p - b) (p -c ) ge 8abc` Therefore (2), hold. Also `((p -a) + (p - b) + (p -c))/(3) ge [(p - a)(p - b) (p - c)]^(1//3)` or `(3p - (a + b + c))/(3) ge [(p - a)(p - b)(p - c)]^(1//3)` or `(2p)/(3) ge [(p - a)(p - b)(p - c)]^(1//3)` or `(p -a) (p - b) (p - c) le (8p^(2))/(27)` Therefore (1) holds. Again `(1)/(2) ((bc)/(a) + (ca)/(a)) ge sqrt(((bc)/(a) (ca)/(a)))` and so on. Adding the inequalities, we get `(bc)/(a) + (ca)/(b) + (ab)/(c ) ge a + b + c = p` Therefore, (3) does not hold. |
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| 24. |
If `a gt 0`, the least value of `(a^3+a^2+a+1)^2` isA. `64a^2`B. `16a^4`C. `16a^3`D. none of these |
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Answer» Correct Answer - C We have `(a^(3) + 1)/(2) ge sqrt(a^(3) xx 1)` and `(a^(2) + a^(1))/(2) ge sqrt(a^(2) a^(1))` Adding, we get `(a^(3) + a^(2) + a + 1)/(2) ge 2 sqrt(a^(3))` |
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| 25. |
If `l ,m ,n`are the three positive roots of the equation `x^3-a x^2+b x-48=0,`then the minimum value of `(1//l)+(2//m)+(3//n)`equals`1``2``3//2``5//2`A. 1B. 2C. `3//2`D. `5//2` |
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Answer» Correct Answer - C Consider 1/l, 2/m, 3/n and use A.M `ge` G.M. Then `:. (1)/(3) ((1)/(l) + (2)/(m) + (2)/(n)) ge ((6)/(48))^(1//3) = (1)/(2)` `:. ((1)/(l) + (2)/(m) + (3)/(n))_("min") = (3)/(2)` |
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| 26. |
Find all real solutions to `2^x+x^2=2-(1)/(2^x)`. |
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Answer» Given equation is `2^(x) + (1)/(2^(x)) = 2 - x^(2)` Using A.M `ge` G.M., we have `2^(x) + (1)/(2^(x)) ge 2` But `2 - x^(2) le 2` So, given equation holds only if `2^(x) + (1)/(2^(x)) = 2 - x^(2) = 2` Therefore, x = 0 is the only solution. |
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| 27. |
Find the minimum value of `4^sin^(2x)+4^cos^(2x)`. |
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Answer» Using A.M `ge` G.M. we have `(4^("Sin"^(2)x) + 4^("cos"^(2)x))/(2) ge sqrt(4^("Sin"^(2)x), 4^("cos"^(2)x))` `:. 4^("sin"^(2) x) + 4^("cos"^(2) x) ge 2 sqrt(4^("sin"^(2)x + "cos"^(2)x))` `implies 4^("sin"^(2)x) + 4^("cos"^(2)x) ge 2 sqrt(4^(1))` Hence, the minimum value of `4^("sin"^(2)x) + 4^("cos"^(2)x)` is 4. |
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| 28. |
If A is the area and 2s the sum of the sides of a triangle,thenA. `Ale (s^2)/(4)`B. `Ale(s^2)/(3sqrt(3))`C. `Alt (s^2)/(sqrt(3)`D. none of these |
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Answer» Correct Answer - A::B We have `2s = a + b + c` `A^(2) = s (s - a) (s - b)(s - c)` Now, `A.M. ge G.M` `implies (s + (s - a) + (s - b) + (s - c))/(4) ge [s(s-a)(s-b)(s-c)]^(1//4)` `implies (4s - 2s)/(4) ge [A^(2)]^(1//4)` `implies s//2 ge A^(1//2)` or `A le s^(2)//4` Also, `((s - a) + (s - b) + (s - c))/(3) ge [(s - a) (s - b) (s - c)]^(1//3)` `implies (s)/(3) ge [(A^(2))/(s)]^(1//3)` or `(A^(2))/(s) le (s^(3))/(27)` or `A le (s^(2))/(3sqrt(3))` |
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| 29. |
For `x^2-(a+3)|x|=4=0`to have real solutions, the range of `a`is`(-oo,-7]uu[1,oo)``(-3,oo)``(-oo,-7]``[1,oo)`A. `(-oo,-7]cup [1,oo)`B. `(-3,oo)`C. `(-oo,-7]`D. `[1,oo)` |
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Answer» Correct Answer - D `a = (x^(2) + 4)/(|x|) - 3` `= |x| + (4)/(|x|) - 3` `ge 2 sqrt(|x| xx (4)/(|x|)) - 3` `implies a ge 1` |
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| 30. |
Find the maximum value of `(7-x)^4(2+x)^5w h e nx`lies between `-2a n d7.` |
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Answer» Since `- 2 lt x lt 7` `:. X + 2 gt 0` and `7 - x gt 0` We have to find maximum value of `(7 - x)^(4) (2 + x)^(5)` or `p^(4) q^(5` where P + q = 9 `(4((p)/(4)) + ((q)/(5)))/(5 + 5) ge [((p)/(5))^(4) ((q)/(5))^(5)]^((1)/(9)` `implies [((p)/(4))^(4) ((q)/(5))^(5)]^((1)/(9)) le (p + q)/(9)` `implies ((p)/(4))^(4) ((q)/(5))^(5 le 1` `implies p^(4) q^(5) le 4^(4) 5^(5)` Therefore, maximum value of `(7 - x)^(4) (2 + x)^(5)` is `4^(4) 5^(5)` |
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| 31. |
If `(log)_2(a+b)+(log)_2(c+d)geq4.`Then find the minimum value of the expression `a+b+c+ddot` |
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Answer» `log_(2) (a + b) + log_(2) (c + d) ge 4` `:. Log_(2) {(a + b) (c + d)} ge 4` `:. (1 + b) (c + d) ge 2^(4)` Now `((a + b)+(c + d))/(2) ge sqrt((a + b)(c + d)) ge 2^(2)` or `a + b + c + d ge 8` Hence, the minimum value of a + b + c + d is 8 |
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| 32. |
If `a,b,c,in R^(+)`, and a,b,c, d are in H.P. thenA. `a+dgt b+c`B. `a+bgt c+d`C. `a+cgt b+d`D. none of these |
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Answer» Correct Answer - A a,b,c,d are in H.P. Hence, b is H.M. of a and c, c is H.M of b and d. Using A.M `ge` H.M., we get `(a + c)/(2) gt b` or `a + c gt 2b` and `(b + d)/(2) gt c` or `b + d 2c` or `a + c + d gt 2b + 2c` or `a + d gt b + c` |
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| 33. |
If `a ,b ,c`are the sides of a triangle, then the minimum value of `a/(b+c-a)+b/(c+a-b)+c/(a+b-c)`is equal to`3``6``9``12`A. 3B. 6C. 9D. 12 |
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Answer» Correct Answer - A `2E = (2a)/(b + c - a) + (2b)/(c + a - b) + (2c)/(a + b - c)` `= (2a)/(b + c + a) + 1 + (2b)/(c + a + b) + 1 + (2c)/(a + b -c) + 1 - 3` `= (a + b + c) ((1)/(b + c -a) + (1)/(c + a - b) + (1)/(a + b - c)) - 3` Using A.M `ge` H.M we have `((1)/(b + c - a) + (1)/(c + a - b) + (1)/(a + b -c))/(3) ge (3)/(a + b + c)` Or `(a + b+ c) ((1)/(b + c -a)+(1)/(c+a+b) + (1)/(a + b+ c)) ge 9` or `(a + b+ c) ((1)/(b + c -a) + (1)/(c + a - b) + (1)/(a + b - c)) - 3 ge 6` `implies E ge 3` |
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| 34. |
If `a ,b ,c in R^+`, then the minimum value of `a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)`is equal to`a b c``2a b c``3a b c``6a b c`A. abcB. 2abcC. 3abcD. 6abc |
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Answer» Correct Answer - D `a (b^(2) + c^(2)) + b (c^(2) + a^(2)) + c (a^(2) + b^(2))` `= ab^(2) + ac^(2) + bc^(2) + ba^(2) + ca^(2) + cb^(2)` Using A.M `ge` G.M we get `(ab^(2) + ac^(2) + bc^(2) + ba^(2) + ca^(2) + cb^(2))/(6) ge (a^(6) b^(6) c^(6))^(1//6)` Or `a (b^(2) + c^(2)) + b(c^(2) + a^(2)) + c (a^(2) + b^(2)) ge 6 abc` |
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| 35. |
If `a,b,c,d in R^(+)-{1}`, then the minimum value of `log_(d) a+ log_(c)b+log _(a)c+log_(b)d` isA. 4B. 2C. 1D. none of these |
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Answer» Correct Answer - A A.M `ge`G.M `implies (log_(4) a + log_(c ) b + log_(a) C + log_(b) d)/(4)` `ge .^(4)sqrt((log a)/(log d) xx (log_ b)/(log c) xx (log c)/(log a) xx (log d)/(log b))` or `log_(d) a + log_(b) d + log_(a) c + log_(c ) b ge 4` |
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| 36. |
If `a ,b ,c in R^+, t h e n(b c)/(b+c)+(a c)/(a+c)+(a b)/(a+b)`is always`lt=1/2(a+b+c)``geq1/3sqrt(a b c)``lt=1/3(a+b+c)``geq1/2sqrt(a b c)`A. `le (1)/(2)(a+b+c)`B. `ge (1)/(2)sqrt(abc)`C. `le (1)/(3)(a+b+c)`D. `ge (1)/(2)sqrt(abc)` |
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Answer» Correct Answer - A Using A.M and H.M inequalit, we get `(2 bc)/(b + c) le (b + c)/(2), (2 ac)/(a + c) le (a + c)/(2), (2ab)/(a + b) le (a + b)/(2)` Adding, `(bc)/(b + c) + (ac)/(a + c) + (ab)/(a + b) le (1)/(2) (a + b + c)` |
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| 37. |
Prove that `2^n >1+nsqrt(2^(n-1)),AAn >2`where `n`is a positive integer. |
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Answer» `2^(n) gt 1 + n sqrt(2^(n - 1))` `implies (2^(n) - 1)/(2 - 1) gt n x 2^((n - 1)//2)` Now, `(2^(n) - 1)//(2 - 1)` is the sum of a G.P whose first term is 1 and common ratio is 2. We have to prove that `1 + 2 + 2^(2) + 2^(3) + …. + 2^(n - 1) ge n xx 2^((n - 1)//2)`. Now, Using A.M. `ge` G.M., we get `(1 + 2 + 2^(2) + ..... 2^(n - 1))/(n) ge (1 xx 2 xx 2^(2) xx 2^(3) ... 2^(n - 1))^(1//n)` Now, `R.H.S = (2^(1 + 2 + 3 +... ( n - 1)))^(1//n)` `= [2^((n - 1) n//2)]^(1//n) = 2^((n - 1)//2)` Hence, `1 + 2 + 2^(2) + .... + 2^(n - 1) gt n xx 2^((n - 1)//2)` |
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| 38. |
If `a , b , c ,`are positive real numbers, then prove that (2004, 4M)`{(1+a)(1+b)(1+c)}^7>7^7a^4b^4c^4` |
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Answer» Using A.M `ge` G.M `((a_(1))/(2) + (a_(1))/(2) + a_(2) + a_(3) + a_(4) + …. + a_(n))/(n + 1) ge (((a_(1))/(2))^(2).a_(2).a_(3).a_(4)……..a_(n))^((1)/(n + 1))` `implies ((1)/(n + 1))^(n + 1) ge (a_(1)^(2) a_(2) a_(3) a_(4)….a_(n))/(4)` `implies (4)/((n + 1)^(n + 1)) ge a_(1)^(2) a_(2) a_(3) a_(4)...... a_(n)` Hence, required maximum value is `(4)/((n + 1)^(n + 1))` |
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| 39. |
Prove that `(sec^4alpha)/(tan^2beta)+(sec^4beta)/(tan^2alpha)ge8`. If each term in the expression is well defined. |
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Answer» Given that a,b,c are positive real numbers. To prove that `(a + 1)^(7) + (b + 1)^(7) (c + 1)^(7) gt 7^(7) a^(4) b^(4) c^(4)`. `L.H.S = (1 + a)^(7) (1 + b)^(7) (1 + c)^(7)` `= [(1 + a) (1 + b) (1 + c)]^(7)` (1) Now, A.M. `ge` G.M `implies (a + b + c + ab + bc + ca + abc)/(7) ge (a^(4) b^(4) c^(4))^(1//7)` or `(a + b + c +ab + bc + ca + abc)^(7) ge 7^(7) (a^(4) b^(4) c^(4))` (2) From Eqs. (1) and (2) we get `[(1 + a)(1 + b)(1 + c)]^(7) gt 7^(7) a^(4)b^(4)c^(4)` |
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