IIT JEE Main 2016 Maths Question 38

IIT JEE Main 2016 Maths Question 38: 

The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is:

(1) 2/√3
(2) √3
(3) 4/3
(4) 4/√3

Solution

Given in problem statement that for the hyperbola whose length of the latus rectum is equal to 8, i.e.
2b2/a=8
& the length of its conjugate axis is equal to half of the distance between its foci, hence, 2b =ae
=> 4b2 = a2e2
=> 4a2 (e2 -1) = a2e2
=> e = 2/√3

Answer to IIT JEE Main 2016 Maths Question 38

Correct answer to given problem No 38 of IIT JEE Main Maths paper of 2016 is option number (1) that is the eccentricity of hyperbola is 2/√3.

IIT JEE Main 2016 Maths Question 37

IIT JEE Main 2016 Maths Question 37: 

If the 2 , 5 and 9 terms of a non-constant AP (Arithematic Progression), are in GP (Geometric Progression), then the common ratio of GP (Geometric Progression) is:

(1) 1
(2) 7/4
(3) 8/5
(4) 4/3

Solution

Given terms of AP (Arithematic Progression) are
T2 = a  + d
T5 = a +  4d
T9 = a +  8d
If T2, T5 & T9 are in GP (Geometric Progression), then
 (a  + 4d)2 = (a + d)(a + 8d)
a2 + 16d2 + 8ad = a2 + ad + 8ad + 8d2
=> 8d2 = ad
=> d(8d – a) = 0
=> 8d = a (because d ≠ 0)
Now common ratio is (a+4d)/(a+d) = 12d/9d =4/3
.

Answer

Correct answer to given problem No 37 of IIT JEE Main Maths paper of 2016 is option number (4) that is the common ratio  for GP (Geometric Progression) is 4/3.

IIT JEE Main 2016 Maths Question 36

IIT JEE Main 2016 Maths Question 36: 

The sum of all real values of x satisfying the equation (x2 – 5x + 5)(x2 + 4x -60) = 1 is:

(1) 6
(2) 5
(3) 3
(4) -4

Solution

For ab = 1, three cases arise (a) when a is 1, then b ∈ R,
(b) If a is not zero, b = 0 and
(c) if a is -1, b ∈ even integer.

Based upon above three cases, we have
For (a) x2 – 5x + 5 = 1
=> x = 1, 4
For (b) x2 + 4x – 60 = 0
=> x = -10 , 6
For (c) x2 – 5x + 5 = -1 , x2 + 4x – 60 = even integer
x=2, 3
at x=2,  x2 + 4x – 60 = -48 (even integer)
at x = 3 (odd integer)
thus possible solutions are
x=1, 4, -10, 6, 2
and sum f solutions is 3.

Answer

Correct answer to given problem No 36 of IIT JEE Main Maths paper of 2016 is option number (3).

IIT JEE Main 2016 Maths Question 35

IIT JEE Main 2016 Maths Question 35: 

The centre(s) of those circles which touch the circle, x2 + y2 -8x -8y -4 = 0, externally and also touch the x-axis, lie on:

(1) a hyperbola
 (2) a parabola
(3) a circle
 (4) an ellipse which is not a circle:

Solution

Circles touch externally and required circle touches x-axis. Let the Centre be at point (h, k) and radius = | k |
C1C2 = r1 + r2
(h-4)2 + (k-4)2 = (6 + |k|)2
h2 -2.4h + 16 + k2 -8k + 16 = 36 + k2 + 12 |k|
h2 -8h -8k -12|k| =4
h2 -8x -8y -12|y|=4
It represents two-semi parabola.

Answer

Correct answer to given problem No 35 of IIT JEE Main Maths paper of 2016 is option number (2).

IIT JEE Main 2016 Maths Question 34

IIT JEE Main 2016 Maths Question 34: For x ∈ R, f
(x) = |log 2 – sin x| and g(x) = f(f(x)), then

(1) g'(0) =-cos(log 2)
(2) g is differentiable at x = 0 and g'(0)=-sin(log 2)
(3) g is not differentiable at x = 0
(4) g'(0) = cos(log 2)

Solution

f (x) = |log 2 – sin x|
g(x)=f(f(x))
g(x) =  |log 2 – sin (log 2 -sin x)|
g(x) = log 2 – sin (log 2 – sin x) (in neighbourhoodof x = 0)
g’ (0) =-cos(log 2 -sin 0) .(-cos 0)
g’ (0) = cos (log 2)

Answer

Correct answer to given problem No 34 of IIT JEE Main Maths paper of 2016 is option number (4).

IIT JEE Main 2016 Maths Question 33

IIT JEE Main 2016 Maths Question 33 requires to find if the integral
∫ [(2x12 + 5x9)/(x5 + x3 +1)3] dx is equal to:

(1) [(x5)/2(x5 + x3 + 1)2] + C (2) [(-x10)/2(x5 + x3 + 1)2] + C (3) [(-x5)/(x5 + x3 + 1)2] + C (4) [(x10)/2(x5 + x3 + 1)2] + C

where C is an arbitrary constant

Solution

Solution to the problem requiring to find out the required integral is obtained by using the substitution method as follows:
I = ∫ [(2x12 + 5x9)/(x5 + x3 +1)3] dx
=∫ [(2/x3 + 5x6)dx/(1/x5 + 1/x2 +1)3]
Let (1/x5 + 1/x2 +1) = t
=> (-2/x3 – 5x6)dx = dt
I = ∫dt/t3 = 1/2t2 + C
=> I = [1/2(1/x5 + 1/x2 +1)2] + C
=> I = [(x10)/2(x5 + x3 +1)2] + C

Answer

Correct answer to given problem No 33 of IIT JEE Main Maths paper of 2016 is option number (4).

IIT JEE Main 2016 Physics Question No 12

IIT JEE Main 2016 Physics Question No 12

‘n’ moles of an ideal gas undergoes a process A->B as shown in the figure. The maximum temperature of the gas during the process will be:

IIT JEE Main 2016 Physics Question No 12

(1) 9P0V0/(2nR)
(2) 9P0V0/(nR)
(3) 9P0V0/(4nR)
(4) 3P0V0/(2nR)

Solution

The equation of the process is y = mx + c
P = -(P0/V0)V + 3P0
Now PV = nRT
therefore P = nRT/V
=> (nRT/V)=-(P0/V0)V + 3P0
=> nRT=-(P0/V0)V2 + 3P0V
=> T=-(P0/nRV0)V2 + (3P0/nR)V                  … (1)
Now for maximum T,
dT/dV = 0                                                       … (2)
here dT/dV = -(2P0/nRV0)V + (3P0/nR)       … (3)
from (2) & (3) we get,
V=(3/2)V0
putting in (1) we get
T=-(P0/nRV0)(9/4V02) + (3P0/nR)(3/2)V0
=> T=(9/2)(P0V0/nR)-(9/4)(P0V0/nR)
=> Tmax = (9/4)(P0V0/nR)

Answer

Correct answer to given problem No 12 of IIT JEE Main Physics paper of 2016 is Tmax = (9/4)(P0V0/nR) i.e. option number 3.