Sample Questions of ISI ME I (Mathematics) 2010 Discussion

[duck]

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}

[Shweta Singhal]

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??

[priyanka p]

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?                                                          

[duck]

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

[N Uday Kumar]

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)] = '∞'

[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.

[Amit Goyal]

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.

[duck]

Q2>> its [0,1)

Q3>> its 2b-a-c

[N Uday Kumar]

Q1>> No. of solutions is 4(viz. 4,-4,1,-1)

[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

[Amit Goyal]

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]

[duck]

Q3>>(c) 2log(e)3 − 3log(e)2

[anisha]

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

[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]

[Amit Goyal]

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).

[priyanka p]

ans is option (c) (1-x)/x

On Sat, Apr 10, 2010 at 12:33 PM, Amit Goyal [via Discussion forum] <[hidden email]> 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).

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[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.

[sonal]

the answer is (a) 0

[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

[Amit Goyal]

Do the next two then,

Let X be a Normally distributed random variable with mean 0 and variance 1. Let F(.) be the cumulative distribution function of the variable X. Then the expectation of F(X) is
(a) −1/2 ,
(b) 0,
(c) 1/2 ,
(d) 1.

Consider any finite integer r ≥ 2. Then lim(x→0) f(r, x)/a(x) equals,
(where f(r, x) = log(e)(Σ(0, r) (x^k)); and a(x) = Σ(1, ∞) ((x^k)/k!))
(a) 0,
(b) 1,
(c) e,
(d) log(e)2.

Note: Σ(a, b) g(k) is summation of g(k) over values of k from a to b
x^k is x to the power k