JEE Main Practice Test- 16 - JEE MCQ

The correct option is C.
Since.a, band c are in HP. Then 1/a, 1/b and 1/c are in AP. 2/b = 1/a + 1/c ⇒ 1/a - 2/b + 1/c = 0 Hence, straight line x/a + y/b + 1/c = 0 always passes through a fixed point (1, -2).

D is the correct option.sin⁻1(4/21)

D is the correct option.a.(b-c)=a.b-a.c is correct if a,b,c are any three vectors.

(z₁-z₂)/(z₂-z₃)= cos(-60)+sin(-60)=e-i60

taking mode on both sides

Absolute value of(z₁-z₂/z₂-z₃)=1

Therefore |z₁-z₂|=|z₂-z₃|

Also angle between z₁-z₂ & z₂-z₃ is 60

So given triangle is equilateral triangle.

Given y = cot^-1 (1+x)/(1-x)

Put x = tan θ

θ = tan^{-1 x}

Then y = cot^-1 (1 + tan θ)/(1- tan θ)

y = cot^-1 (tan π/4 + tan θ)/(1- tan π/4 tan θ)

= cot^-1 tan (π/4 +θ)

= cot^-1 cot (π/2 – (π/4 +θ))

= π/4 – θ

= π/4 – tan^-1 x

dy/dx = -1/(1+x²)

A statement is logically equivalent to its contrapositive.

To form the contrapositive, switch the "If" and  "then" sections

of the statement AND insert "NOTs" into each section.

Notice in this situation, that inserting

"NOTs" turns the thoughts positive (you are negating negative thoughts).

The correct option is Option C.

When f(x) = g(x)

Then the range includes 1. 

But it is already given that 1 is excluded.

Hence, the assertion is correct but the reason behind it is false.

Since A+B is defined, A and B are matrices of the same type,
say m×n; Also, AB is defined.
So, the number of columns in A must be equal to the number of rowsin B ie. n=m.
Hence, A and B are square matrices of the same order.

f ′ (x)   = 4 So, for 0 ≤ x < 1, we get f ′ x < 0, i . e . , f x is m.d. and for 1 < x ≤ 2 we get f ′ (x)   > 0, i . e . , f (x)    is m.i .

∴ min f (x)   =

It is given that the parabola passes through the point (2,-6)
Length of its latus rectum is 4a
y² = 4ax
i.e. 

(-6)² = 4a(2)
⇒a = 9/2
∴ Length of the latus rectum is 4 x 9/2 = 18

We know that ,
1) ⁿP_r. = n!/( n - r )!
2) ⁿC_r = n!/ [ ( n - r )! r ! ]
3) r! = ( ⁿP_r )/ ⁿC_r 

It is given that ,

nPr = 840 ---( 1 )

nCr = 35 ---( 2 )

r! = ( nPr )/ ( nCr )

= 840/35

= 24

r! = 4 × 3 × 2 × 1

r ! = 4!
Therefore,

r = 4

Substitute r value in equation ( 1 ),

We get

n!/ ( n - 4 ) ! = 840

n( n - 1 )( n - 2 )( n - 3 ) = 840

n( n - 1 )( n - 2 )( n - 3 ) = 7× 6 × 5 × 4

Therefore ,

n = 7

Let n be successive independent trials and p the probability of getting a success.

The the mean and variance of a binomial distribution is given by :

Mean = np

Variance = np(1 - p)

In this case we have :

p = 1/4

Variance = (standard deviation) ²

Variance = 3² = 9

From this we can get the value of n as follows :

Variance = np(1 - p)

Doing the substitution we have

9 = 1/4n( 1 - 1/4)

9 = 1/4n(3/4)

9 = 3/16n

n = 16/3 × 9

n = 48

Mean = np

Mean = 48 × 1/4 = 12

The mean of the binomial distribution is equal to 12.

A-B=π/4

Which implies A=B+π/4

Taking tan on both sides

tanA=tan(B+π/4)

tanA=(tanB + tan(π\4))/(1- tanB.tan(π/4))

tanA=(tanB+1)/(1-tanB)

Substituting the value of tanA in

(1+tanA).(1-tanB) we get

(1+(tanB+1)/(1-tanB)).(1-tanB)

= (1-tanB+tanB+1)/(1-tanB).(1-tanB)

=2