IIT JEE Main 2016 Maths Question 32

IIT JEE Main 2016 Maths Question 32

If f(x) + 2f(1/x) = 3x, x<> 0 and S = {x ∈ R : f(x)=f(-x)}; then S :

(1) contains exactly two elements (2) contains more than two elements (3) is an empty set (4) contains exactly one element

Solution

Solution to the problem requiring to find out the set S as per the relation specified is as follows:
f(x) + 2f(1/x) = 3x     …………. (1)
Replace x->(1/x)
=>f(1/x) + 2f(x) = 3/x
=> 2f(1/x) + 4f(x) =6/x      ……(2)
=> from (1) and (2)
3f(x) = 6/x – 3x
=> f(x) = (2/x) – x
=> f(-x) = (-2/x) + x
=> f(x) = f(-x)
=> (2/x) – x = (-2/x) + x
=> 4/x = 2x
=>x2 = 2
=>x= +/- √2

Hence set S contains two elements, i.e. +√2 , -√2.

Answer

Correct answer to given problem No 32 of IIT JEE Main Maths paper of 2016 is option number (1) that is set S contains exactly two elements.

IIT JEE Main 2016 Maths Question 31

IIT JEE Main 2016 Maths Question 31

The area (in sq. units) of the region {(x, y) : y2 >=2x and x2 + y2 <= 4x, x>= 0, y >= 0} is:

(1) Π – {(4√2)/3} (2) (Π/2) – {(2√2)/3} (3) Π – (4/3) (4) Π – (8/3)

Solution

Solution to the problem requiring to find out the area under a region bounded by given curves, is obtained by integration as follows:

y2 = 2x,  x2 + y2 = 4x
=> x2 + 2x = 4x
=> x2 = 2x
=> x = 0, 2
Required area = ∫02 [√(4x – x2) – √(2x)] dx
= ∫02 [√(4x – x2) – √2√x] dx
=[{(x-2)/2}√(4x – x2) + 2 sin-1{(x-2)/2} -{(2/3)√2}x3/2]]20
= 0 + 0 – (8/3) – [0 + 2(-Π/2)]
=Π – (8/3)

Answer

Correct answer to given problem No 31 of IIT JEE Main Maths paper of 2016 is option number (4).