# Sample Questions of ISI ME I (Mathematics) 2010 Discussion Classic List Threaded 128 messages 12345 ... 7
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 Q1>> answer is 8{2}, {2, 3}, {2, 4}, {2, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {2, 3, 4, 5} are the 8 sets whose intersection with Q is {2} :)
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 In reply to this post by duck The function f(x) = x(square_root(x) + square_root(x + 9)) is (a) continuously differentiable at x = 0, (b) continuous but not differentiable at x = 0, (c) differentiable but the derivative is not continuous at x = 0, (d) not differentiable at x = 0. Note: square_root(g(x)) stands for square root of g(x) HEY Nidhi... Can u plz explain dis ques with full solution??
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## MSQE2006MACRO

 In reply to this post by Amit Goyal Consider a typical keynesian closed economy producing a single good and having a single household.There are 2 types of final expenditure--investment autonomously given 36 units and household consumption (c) equaling household's labor income wl, its given w=4. production fn is y=24 l^1/2. find the eql level of output and employment.is there any involuantry unemployment? how much?  sir , i am giving the solution, pls check.      Y=C+I =4L+36 ALSO Y=24L^1/2. SO using these, we get l=9  which gives Y=72 units. but how would be involuantry unemployment be calculated?
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 In reply to this post by Shweta Singhal hey hi shweta.. just follow the following steps..u will get the answer>> first check the differentiablity of f(x) at x=0 , you will find that it is differentiable at that point.. Now to see whether the function is continuosly differentiable at x=0 , we need to check whether the derivative of f(x) is continuous at x=0 or not.. if the derivative of f(x) is continuous at x=0 then, the function f(x) is continuosly differentiable at x=0 so, second step is to find the derivative of f(x) third step: check the continuity of the derivative at x=0 bus answer aa gya..  function f(x) is continuosly differentiable at x=0  :) ... i hope :)
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 In reply to this post by Amit Goyal Then Expectation of X is Ans. (b) " ∞ " E(x) = (-∞)INTEGRATION(+∞)[x*f(x)] = (-∞)INT(c)[x*f(x)] + (c)INT(+∞)[x*f(x)] = = (-∞)INT(c)[x*0] +(-∞)INT(c)[x*(c/sq(x))] = (c)INT(+∞)[c/x] = c*[log(∞)-log(c)] = '∞'
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 Administrator In reply to this post by Amit Goyal Well done Shweta, Nidhi and Uday. Let P = {1, 2, 3, 4, 5} and Q = {1, 2}. The total number of subsets X of P such that X ∩ Q = {2} is (c) 8 An unbiased coin is tossed until a head appears. The expected number of tosses required is (b) 2 Let X be a random variable with probability density function f(x) = c/sq(x) if x ≥ c, 0 if x < c. Then Expectation of X is (b) ∞ Note: sq(x) stands for square of x.
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 Administrator Ok, Next three now: The number of real solutions of the equation sq(x) − 5|x| + 4 = 0 is (a) two, (b) three, (c) four. (d) None of these. Range of the function f(x) = sq(x)/1+sq(x) is (a) [0, 1), (b) (0, 1), (c) [0, 1]. (d) (0, 1]. If a, b, c are in AP, then the value of the determinant |x + 2; x + 3; x + 2a| |x + 3; x + 4; x + 2b| |x + 4; x + 5; x + 2c| is (a) sq(b) − 4ac, (b) ab + bc + ca, (c) 2b − a − c, (d) 3b + a + c.
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 Q2>> its [0,1)Q3>> its 2b-a-c :)
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 Q1>> No. of solutions is 4(viz. 4,-4,1,-1)
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 Administrator In reply to this post by Amit Goyal Perfect answers Nidhi and Uday The number of real solutions of the equation sq(x) − 5|x| + 4 = 0 is (c) four Range of the function f(x) = sq(x)/(1+sq(x)) is (a) [0, 1) If a, b, c are in AP, then the value of the determinant |x + 2; x + 3; x + 2a| |x + 3; x + 4; x + 2b| |x + 4; x + 5; x + 2c| is (c) 2b − a − c
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 Administrator Next 4 now. If a < b < c < d, then the equation (x − a)(x − b) + 2(x − c)(x − d) = 0 has (a) both the roots in the interval [a, b], (b) both the roots in the interval [c, d], (c) one root in the interval (a, b) and the other root in the interval (c, d), (d) one root in the interval [a, b] and the other root in the interval [c, d]. Let f and g be two differentiable functions on (0, 1) such that f(0) = 2, f(1) = 6, g(0) = 0 and g(1) = 2. Then there exists θ ∈ (0, 1) such that f'(θ) equals (a) (1/2)g'(θ), (b) 2g'(θ), (c) 6g'(θ), (d) (1/6)g'(θ). The minimum value of log(x)a + log(a)x, for 1 < a < x, is (a) less than 1, (b) greater than 2, (c) greater than 1 but less than 2. (d) None of these. The value of ∫(1/(2x(1+√x)))dx over [4, 9] equals (a) log(e)3 − log(e)2, (b) 2log(e)2 − log(e)3, (c) 2log(e)3 − 3log(e)2, (d) 3log(e)3 − 2log(e)2. Read log(x)a as log value of a when x is the base, And likewise others. Read ∫(1/(2x(1+√x)))dx over [4, 9] as the value of definite integral of the said function over the range [4, 9]
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 Q3>>(c) 2log(e)3 − 3log(e)2 :)
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 In reply to this post by Amit Goyal Q1- c, One root in the (a,b)and the other in the interval (c,d) Q2 - b, 2g'(@) Q3-b, >2 Q4 - c, 2log3 -3log2
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 Administrator In reply to this post by Amit Goyal Good work Nidhi and Anisha If a < b < c < d, then the equation (x − a)(x − b) + 2(x − c)(x − d) = 0 has (a) both the roots in the interval [a, b], (b) both the roots in the interval [c, d], (c) one root in the interval (a, b) and the other root in the interval (c, d), (d) one root in the interval [a, b] and the other root in the interval [c, d]. About this problem i am skeptical that there is some missing information because if we pick a = 0, b = 1, c = 100, d = 101, then there is no x for which (x)(x − 1) + 2(x − 100)(x − 101) = 0. The reason is if i pick x in the interval [0, 1] then x(x -1) will lie between -1/4 and 0 and 2(x − 100)(x − 101) will take the value larger than 2(99)(100) so the sum (x)(x − 1) + 2(x − 100)(x − 101) cannot be 0. And if i pick x in the interval [100, 101] then 2(x − 100)(x − 101)  will lie between -1/2 and 0 and x(x -1) will take value larger than (99)(100) so the sum (x)(x − 1) + 2(x − 100)(x − 101) cannot be 0. Thus none of the options. Let f and g be two differentiable functions on (0, 1) such that f(0) = 2, f(1) = 6, g(0) = 0 and g(1) = 2. Then there exists θ ∈ (0, 1) such that f'(θ) equals (b) 2g'(θ) The minimum value of log(x)a + log(a)x, for 1 < a < x, is (b) greater than 2 The value of ∫(1/(2x(1+√x)))dx over [4, 9] equals (c) 2log(e)3 − 3log(e)2 Read log(x)a as log value of a when x is the base, And likewise others. Read ∫(1/(2x(1+√x)))dx over [4, 9] as the value of definite integral of the said function over the range [4, 9]
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 Administrator Next one: The inverse of the function f(x) = 1/(1 + x) , x > 0, is (a) (1 + x), (b) (1 + x)/x, (c) (1 − x)/x, (d) x/(1 + x).
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 ans is option (c) (1-x)/x On Sat, Apr 10, 2010 at 12:33 PM, Amit Goyal [via Discussion forum] wrote: Next one: The inverse of the function f(x) = 1/(1 + x) , x > 0, is (a) (1 + x), (b) (1 + x)/x, (c) (1 − x)/x, (d) x/(1 + x). View message @ http://n2.nabble.com/Sample-Questions-of-ISI-ME-I-Mathematics-2010-Discussion-tp4850019p4880822.html To unsubscribe from Discussion forum, click here.
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 Administrator This post was updated on . In reply to this post by Amit Goyal Well answered Priyanka. Next one now: Let X(i), i = 1, 2, . . . , n be identically distributed with variance sq(s). Let cov(X(i),X(j)) = p for all i ≠ j. Define x(n) = (1/n)ΣX(i) = (1/n)(X(1)+.........+X(n)) and let a(n) = Var(x(n)). Then lim(n→∞) a(n) equals (a) 0, (b) p, (c) sq(s) + p, (d) sq(s) + sq(p). Note: x(n) is the mean and a(n) is the variance of mean. And you need to find limit of variance of mean as n goes to infinity.
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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

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## Re: Sample Questions of ISI ME I (Mathematics) 2010 Discussion

 Administrator In reply to this post by Amit Goyal Hi Sonal, a(n) will converge to 0 if the X(i)'s are independent. Here cov(X(i),X(j)) = p and Simple computation will give the following value of variance of a(n), a(n) = [n(sq(s)) + (sq(n) - n)p]/(sq(n)) = (sq(s)/n) + p - (p/n) Hence, lim(n→∞) a(n) = p