Sample Questions of ISI ME I (Mathematics) 2010 Discussion

[Amit Goyal]

Solutions are the following:
The sum C(n+0, 0) + C(n+1, 1) +............C(n+m, m) equals
(d) C(n+m+1, n+1).

Consider the following 2-variable linear regression where the error e(i)’s are independently and identically distributed with mean 0 and variance 1; y(i) = a + b(x(i) − Mean(x)) + e(i), i = 1, 2, . . . , n. where Mean(x) = (x(1) + x(2) + ....x(n))/n
Let a^ and b^ be ordinary least squares estimates of a and b respectively.
Then the correlation coefficient between a^ and b^ is
(b) 0

[Amit Goyal]

Next two now:
Let f be a real valued continuous function on [0, 3]. Suppose that f(x) takes only rational values and f(1) = 1. Then f(2) equals
(a) 2,
(b) 4,
(c) 8.
(d) None of these.

Consider the function f(x, y) = ∫[0, √(sq(x)+sq(y))] exp(-sq(w)/(sq(x)+sq(y)))dw with the property that f(0, 0) = 0. Then the function f(x, y) is
(a) homogeneous of degree −1,
(b) homogeneous of degree 0.5,
(c) homogeneous of degree 1.
(d) None of these.

∫[a, b] g(w)dw can be read as definite integral of g(w) from a to b, sq(t) stands for square of t.

[Amit Goyal]

Solutions are the following:
Let f be a real valued continuous function on [0, 3]. Suppose that f(x) takes only rational values and f(1) = 1. Then f(2) equals
(d) None of these.
(f(2) = 1)

Consider the function f(x, y) = ∫[0, √(sq(x)+sq(y))] exp(-sq(w)/(sq(x)+sq(y)))dw with the property that f(0, 0) = 0. Then the function f(x, y) is
(c) homogeneous of degree 1.
(Hint: Its not necessary to compute integral to check for homogeneity)

∫[a, b] g(w)dw can be read as definite integral of g(w) from a to b, sq(t) stands for square of t.

[Amit Goyal]

Next four:
If f(1) = 0, f'(x) > f(x) for all x > 1, then f(x) is
(a) positive valued for all x > 1,
(b) negative valued for all x > 1,
(c) positive valued on (1, 2) but negative valued on [2,∞).
(d) None of these.

Consider the constrained optimization problem
max (ax + by) subject to (cx + dy) ≤ 100, x ≥ 0, y ≥ 0
where a, b, c, d are positive real numbers such that d/b > (c + d)/(a + b)
The unique solution (x*, y*) to this constrained optimization problem is
(a) (x*=100/a, y*=0),
(b) (x*=100/c, y*=0),
(c) (x*=0, y*=100/b),
(d) (x*=0, y*=100/d).

For any real number x, let [x] be the largest integer not exceeding x. The
domain of definition of the function f(x) = 1/ √(|[|x|-2]|-3) is
(a) [−6, 6],
(b) (−∞,−6) ∪ (+6,∞),
(c) (−∞,−6] ∪ [+6,∞).
(d) None of these.

Let f: R → R and g: R → R be defined as
f(x) = −1, x < −0.5
      = −0.5, −0.5 ≤ x < 0
      = 0, x = 0
      = 1, x > 0
and g(x) = 1 + x − [x], where [x] is the largest integer not exceeding x.
Then f(g(x)) equals
(a) −1,
(b) −0.5,
(c) 0,
(d) 1.

[priyanka p]

thank you sir,  2nd answer shud be option b,   1st ans i think can be option c

On Wed, Apr 21, 2010 at 11:56 AM, Amit Goyal [via Discussion forum] <[hidden email]> wrote:

If f(1) = 0, f'(x) > f(x) for all x > 1, then f(x) is
(a) positive valued for all x > 1,
(b) negative valued for all x > 1,
(c) positive valued on (1, 2) but negative valued on [2,∞).
(d) None of these.

Consider the constrained optimization problem
max (ax + by) subject to (cx + dy) ≤ 100, x ≥ 0, y ≥ 0
where a, b, c, d are positive real numbers such that d/b > (c + d)/(a + b)
The unique solution (x*, y*) to this constrained optimization problem is
(a) (x*=100/a, y*=0)
(b) (x*=100/c, y*=0)
(c) (x*=0, y*=100/b)
(c) (x*=0, y*=100/d)

---

View message @ http://n2.nabble.com/Sample-Questions-of-ISI-ME-I-Mathematics-2010-Discussion-tp4850019p4935346.html

To unsubscribe from Discussion forum, click here.

[priyanka p]

ans 1 is option (a) positive valued    ans 3 is option (c)  and ans 4 is option (d) as for ans 2 i think its option (b) but not very sure sir pls explain this question

[sonal]

sir can u explain hw the correlation coefficient b/w a and b is 0 in Q22 ISI ppr.shudn't it be 1.

[Amit Goyal]

If f(1) = 0, f'(x) > f(x) for all x > 1, then f(x) is
(a) positive valued for all x > 1

Consider the constrained optimization problem
max (ax + by) subject to (cx + dy) ≤ 100, x ≥ 0, y ≥ 0
where a, b, c, d are positive real numbers such that d/b > (c + d)/(a + b)
The unique solution (x*, y*) to this constrained optimization problem is
(b) (x*=100/c, y*=0)

For any real number x, let [x] be the largest integer not exceeding x. The
domain of definition of the function f(x) = 1/ √(|[|x|-2]|-3) is
(c) (−∞,−6] ∪ [+6,∞).

Let f: R → R and g: R → R be defined as
f(x) = −1, x < −0.5
      = −0.5, −0.5 ≤ x < 0
      = 0, x = 0
      = 1, x > 0
and g(x) = 1 + x − [x], where [x] is the largest integer not exceeding x.
Then f(g(x)) equals
(d) 1.

[Amit Goyal]

Last two now:
If f is a real valued function and af(x) + cf(−x) = b − dx for all x with a ≠ c and d ≠ 0. Then f(b/d) equals
(a) 0
(b) - (2bc/(sq(a) - sq(c)))
(c) (2bc/(sq(a) - sq(c)))
(d) More information is required to find the exact value of f(b/d).

For all x, y ∈ (0, ∞), a function f: (0, ∞) --> R satisfies the inequality
|f(x) - f(y)| ≤ cube(|x - y|). Then f is
(a) an increasing function
(b) a decreasing function
(c) a constant function
(d) None of these.

[priyanka p]

ans 1 is option (b) which can be calculated  by solving two equations which v get by putting f(b1/b2) and f(-b1/b2)                                                                        as for ans 2 it is option (c) a constant function

[Amit Goyal]

Well done Priyanka. So we have:
If f is a real valued function and af(x) + cf(−x) = b − dx for all x with a ≠ c and d ≠ 0. Then f(b/d) equals
(b) - (2bc/(sq(a) - sq(c)))

For all x, y ∈ (0, ∞), a function f: (0, ∞) --> R satisfies the inequality
|f(x) - f(y)| ≤ cube(|x - y|). Then f is
(c) a constant function

[Benhur]

i am having prob in understanding dese ques..can ny1 plz help??

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 sum C(n+0, 0) + C(n+1, 1) +............C(n+m, m) equals
(d) C(n+m+1, n+1).


Consider the function f(x, y) = ∫[0, √(sq(x)+sq(y))] exp(-sq(w)/(sq(x)+sq(y)))dw with the property that f(0, 0) = 0. Then the function f(x, y) is
(c) homogeneous of degree 1.
(Hint: Its not necessary to compute integral to check for homogeneity)


If f(1) = 0, f'(x) > f(x) for all x > 1, then f(x) is
(a) positive valued for all x > 1


Let f: R → R and g: R → R be defined as
f(x) = −1, x < −0.5
      = −0.5, −0.5 ≤ x < 0
      = 0, x = 0
      = 1, x > 0
and g(x) = 1 + x − [x], where [x] is the largest integer not exceeding x.
Then f(g(x)) equals
(d) 1.


For all x, y ∈ (0, ∞), a function f: (0, ∞) --> R satisfies the inequality
|f(x) - f(y)| ≤ cube(|x - y|). Then f is
(c) a constant function

[Viren]

Dear nidhi, the answer to this question (msqe 2010 Q2) shud be (D) because RHD is not equal to derivative at 0. PLease confirm.
viren

[duck]

Dear Viren,

The answer to this que is option "a" it is continuously differentiable
{LHD=RHD=3}


:)

[Viren]

please explain the answer to the following question:

Let f be a real valued continuous function on [0, 3]. Suppose that f(x) takes only rational values and f(1) = 1. Then f(2) equals
(a) 2,
(b) 4,
(c) 8.
(d) None of these

Consider the function f(x, y) = ∫[0, √(sq(x)+sq(y))] exp(-sq(w)/(sq(x)+sq(y)))dw with the property that f(0, 0) = 0. Then the function f(x, y) is
(c) homogeneous of degree 1.

and Q 25

[Viren]

Dear sir, can you explain the answer to Q23 and the last one Q30
thanks

[Viren]

can you explain the answer to Q23 and Q30.
thanks

[duck]

Hello.. :)

q 23> See, it says that the fuction takes only rational values and f(1)= 1
=> f(x) = 1
=>f(2) will also be equal to 1.
Therefor, none of the above.

q25> take f(x) = x log x
It satisfies all the conditions given in the que.. and we know it is positive valued as well.
So, option "a"

q30>
take any increasing or decreasing function, you will find tht for some (x,y) , the following inequality does not hold. while for constant function, it does.
So, option "c"

:)

[Viren]

hi,
In ques23.  how does the fact that it takes rational value imply that f(x)=1, how cn it be generalized?

[duck]

hello.. :)

take any other function, u will find tht either it is taking irrational values or f(1) is not equal to 1.
just check it.. if you still have some doubt then i will tell you.