JEE Main Maths Mock Test- 1 - JEE MCQ

We have

S=2(x²+y²)+7x−7y+5=0

=x²+y²+7/2x−7/2y+5/2=0

Length of tangent from (0,0) =√S1

=√5/2

Hence, the required answer is √5/2.

y = e^cx
dy/dx = c. e^cx
y' = cy

The system of equation has a non-zero solution

Clearly, m₁m₂ = - 1. 
Hence, the two tangents arc mutually perpendicular. 
Statement 1 is true. 
Now, the locus of the point of intersection of two mutually perpendicular tangents to the circle x² + y² = r² is the director circle i.e. the circle x² +y² = 2r². 
For the given circle r = 13. .. 
Its director circle is x² + y² = 338. 
Hence, statement 2 is true and  a cogect explanation of statement as the point (17, 7) lies on the director circle of the circle (i). 

• ⁿC₁₂=ⁿC₈ 
• USing symmetry property, we know that
• ⁿC₁₂ = ⁿC_n-12 
• Therefore, ⁿC_n-12 = ⁿC₈ 
• Hence, n-12 = 8
• n = 20

Total outcomes = 36
Doublet are 6 (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)

Probability of getting doublet = 6/36
= 1/6

Covariance is a quantitative measure of the extent to which the deviation of one variable from its mean matches the deviation of the other from its mean. It is a mathematical relationship that is defined as:

  Cov(X,Y) = E[(X − E[X])(Y − E[Y])]

 Correlation between two random variables, ρ(X,Y) is the covariance of the two variables normalized by the variance of each variable. This normalization cancels the units out and normalizes the measure so that it is always in the range [0, 1]:

Given: Here,AB is the diagonal of square.
The vertices of a square A 
Let the mid−point of AB be EThen coordinates of E are 
Therefore equation of other diagonal is

Given y = f(x)

differentiating w.r.t x 

y' = f'(x) which is the slope of the tangent

Hence the slope of the normal is - 1/f'(x) = 3pi/4 = -1

therefore f'(x) = 1

Hence f'(3) = 1

f (x) = a cos (bx + c) + d ..(i)

As we know, -1 ≤ cos θ ≤ 1

For minimum, cos (bx + c) = −1

From (i), f (x) = −a + d = (d − a)

For maximum, cos (bx + c) = 1

From (i), f (x) = a + d = (d + a)

Range of f (x) = [d − a, d + a]

Alternatively,

-1 ≤ cos(bx + c) ≤1

-a ≤ a cos(bx + c) ≤ a

-a + d ≤ a cos(bx + c) + d ≤ a + d

Range of f(x) = [d – a, a + d]

OP² = x² + y²
y = e^x, y' = e^x,

ƒ(x) = ƒ(6 – x)
⇒ ƒ'(x) = –ƒ'(6 – x) .... (1)
put x = 0, 2, 5
ƒ'(0) = ƒ'(6) = ƒ'(2) = ƒ'(4) = ƒ'(5) = ƒ'(1) = 0
and from equation (1) we get ƒ'(3) = –ƒ'(3)
⇒ ƒ'(3) = 0
So ƒ'(x) = 0 has minimum 7 roots in
x ∈ [0, 6] ⇒ ƒ"(x) has min 6 roots in x ∈ [0,6]
h(x) = ƒ'(x).ƒ"(x)
h'(x) = (ƒ"(x))² + ƒ'(x) ƒ"'(x)
h(x) = 0 has 13 roots in x ∈ [0, 6]
h'(x) = 0 has 12 roots in x ∈ [0, 6]

x²(y + z)y²(z + x)z²(x + y) = a³b³c³ = x³y³z³
⇒ (x + y) (y + z) (z + x) = xyz
⇒ x²(y + z) + y²(z + y) + z²(x + y) + xyz = 0
⇒ a³ + b³ + c³ + abc = 0

x = t² – t + 1    .... (1)
y = t² + t + 1    .... (2)
y – x = 2t  &  x + y = 2(t² + 1)
________on elminating 't' we get
⇒ (x + y – 2) = 2(y - x)/2²
(x – y)² = 2(x + y – 2)
Axis : x – y = 0
Tangent at vertex : x + y – 2 = 0
Vertex : (1, 1) = (x, y)