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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 2variable 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

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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.

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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.

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This post was updated on .
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/ √([x2]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.


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)


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


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

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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/ √([x2]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.

Administrator

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.


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

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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


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


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


Dear Viren,
The answer to this que is option "a" it is continuously differentiable
{LHD=RHD=3}
:)
:)


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


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


can you explain the answer to Q23 and Q30.
thanks


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"
:)
:)


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


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.
:)

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