NDA I - Mathematics Question Paper 2016 - NDA MCQ

P and Q are points on complex plane. Angle between OP and OQ is

x and y are positive numbers.
x ≤ y²
x < x²   positive numbers.
Hence relation is reflexive.
Transitive -​

Thus relation is not transitive.
Symmetric
1 ≤ (2)² while 2  (I)²
Hence relation is not symmetric.
Thus x ≤ y²  positive numbers is reflexive, but not transitive and symmetric.

(a) x²-px + 4> 0  real values of x.
If b² - 4ac < 0
⇒p² - 4(1)(4)<0
⇒p² < 16 ⇒|p| < 4

z = x + iy




 

As we know that line PQ intersects x-axis andy-axis at Rand Q.
∵ M is the mid point of PQ

⇒ x = 6 and y = 10
Hence area of triangle OPQ
 

Distance from the centre to the point of line which touches circle is OM = radius

f(θ) = 4 (sin²θ + cos⁴ θ)
= 4 (sin² θ + cos² θ(1- sin² θ))
= 4 (sin² θ + cos² θ - sin² θ cos² θ)

For maximum value of f(θ), sin²2θ should be minimum.
i.e. sin²2θ = 0
f(θ)l_max=4(l1-0) = 4

f(θ) = 4 (sin²θ + cos⁴ θ)
= 4 (sin² θ + cos² θ(1- sin² θ))
= 4 (sin² θ + cos² θ - sin² θ cos² θ)

 For minimum value ot f(θ), sin²2θ should be maximum i.e. sin²2θ= 1.

f(θ) = 4 (sin²θ + cos⁴ θ)
= 4 (sin² θ + cos² θ(1- sin² θ))
= 4 (sin² θ + cos² θ - sin² θ cos² θ)



Since sin θ cannot have vlaue greater than 1 & less than -1.
Hence f(θ) = 2 has no solution.

Hence f(x ) and g(x) intersects at ( -1 , -2 ) and (2 ,3 ).
 

This is a cubic equation.
If we put y = then ( 3y + 1) = 0 is a factor o f cubic equation.

Similarly for (f) = -1 we will get 27y³ - 27y² - 4 = 0 and after solving it we will find that it has two solutions. 
y₀=1.1184,-0.05922.

As g(x) is a gretest integer function so value of g(x) in integral limit will be

The value of g(x) in value  will be 2 and in range

form(l)


 

f(x) = | x - l |  + x²  x  ∈R
f₁ (x ) = |x - l |, f₂(x) = x²
f₁ (x) and f₂{x) both are continuous.
Hence f(x) is continuous.
f(x) in differentiable at x = 0
f₁(x) is not differentiable at x = 1.
Hence(fx) is continuous but not differentiable at x= 1

As we know,

f(x) is in quadratic form (parabola). Hence f(x) is decreasing in  and increasing 

f(x) has local minimum at one point only in (-∞ ,∞ ).

Hence area required for given region is

Area required for given region is

Since it is a definite integral will have a definite value. The sequence {a_2n} is in AP with common difference. Statement (1) is correct.
The sequence {a_2n + 1} is also in AP with common difference.
Statement (2) is correct.

∵ given sequence a_n also AP with no difference.
Thus a_n-1 - a_n-4 = 0

 x+ | y |= 2y
x = 2y - |y |
2y-| y | = x



∵ by checking
y as a function of x is continuous at x = 0, but not differentiable at x = 0.
So all of the statements are not correct.

Option (d) is correct.

OAB is triangle

Coordinates o f O, A, B are (0, 0) respectively.


 

f(x) = (x-l)²(x + l ) (x-2)³
f'(x) = 2(x - l)(x + l)(x - 2)³+ ( x - l)²(x - 2)³+(x - 1)² (x + l)³(x - 2)²
= (x - l)(x -1)² [2(x+1)(x - 2 ) +( x - l) ( x - 2) + 3 ( x - l ) ( x + l)]
f '(x) = ( x - l)(x - 2 )²[2x² - 2x - 4 + x² - 3x + 2 + 3x² - 3]
= (x - l)(x - 2)² [6x² - 5x - 5]
For maxima and minima 
f'(x )= 0
(x - l)(x - 2)² [6x² - 5x - 5] = 0 

The change in signs of f(x) for dififrent values of x is shown:

∵ Local Minima are

 

Local Maxima is [x = 1 ]