BITSAT Maths Test - 1 - JEE MCQ

The intersection points of x² + 9y² = 9  and x + 3y = 3 are (0,1) and (3,0).


∴ A = 6π − 3π = 3π.

The term x^-9 will be T(10). Therefore T(10) will be
⇒ (-2/x)⁹
⇒ -512/x⁹
⇒ -512x^-9

Let y = mx be the family of lines through origin. Therefore, dy/dx = m
Eliminating m, we get y = (dy/dx).x or x(dy/dx) – y = 0.

Given circle is  x² + y² + 6x - 4y - 3 = 0 .....(i)
Given point is (5, 1) Let P = (5,1)
Now length of the tangent from P(x, y) to circle (i) = √ x² + y² + 2gx + 2fy + c​.
Now length of the tangent from P(5, 1) to circle (i) = √ 5²+1²+ 6.5 − 4.1 − 3​ = 7

(siny+ y.cosy)dy = [x(2logx + 1)]dx

Integrating both sides, we get

We have π/4 ​< arg (z) < π/3​
Let z = x + iy
∴ arg (z) = tan^-1(y/x​)
The given inequality can be written as
π/4 ​< tan^-1(y/x​) < π/3​
⇒ tan(π/4) ​< y/x ​< tan(π/3)​
⇒ 1 < y/x ​< √3​
⇒ x < y < √3​x
This inequality represents the region between the lines
 y = x and y = √3​x

ydy = (c−x)dx
Integrating both sides,
∫ydy = ∫(c−x)dx
y²​/2 = cx − x²​/2 + k
x²/2 + y²/2 ​− cx = k
x² + y² − 2cx = 2k
(x−c)² + y² = 2k + c²
This is equation of a circle with center on x-axis.

Given, Perpendicular = 15 m
Base = 15 m
Angle of depression = ​tanθ = 15/15 = 1
θ = 45^°

t = T/|a|
  = 2π/π/2
  = 4

|A+B| = 0
⇒ A + B = O, where O is a zero matrix.
⇒ A = O − B = −B
Take determinant on both sides
⇒ |A| = −|B|
⇒ |A| + |B| = 0

(A - B)²
⇒ (A - B) x (A - B)
⇒ (A - B) x A - (A - B) x B
⇒  A² - AB - BA + B²

f(x) = (x+1)³ + 1      
∴ f'(x) = 3(x+1)²
f'(x) = 0  ⇒  x = -1
Now, f" (-1 - ∈) = 3(-∈)² > 0, f'(-1 + ∈)² = 3∈² > 0
∴ f(x) has neither a maximum nor a minimum at x = -1
Let f'(x) = φ ′ (x) = 3(x+1)²    
∴ φ′(x) = 6(x+1).
φ′(x) = 0  ⇒  x = -1
φ ′ (-1-∈) = 6(-∈) < 0, φ ′ (-1-∈) = 6∈ > 0
∴ φ (x) has a minimum at x = -1

f(n) = n³ + 2n
put n=1, to obtain f(1) = 1³ + 2.1 = 3
Therefore, f(1) is divisible by 3
Assume that for n=k, f(k) = k³ + 2k is divisible by 3
Now, f(k+1) = (k+1)³ + 2(k+1)
⇒ k³ + 2k + 3(k² + k + 1)
⇒ f(k) + 3(k² + k + 1)
Since, f(k) is divisible by 3
Therefore, f(k+1) is divisible by 3
and from the principle of mathematical induction f(n) is divisible by 3 for all n∈N

Let P(4t₁, 2t₁²) and Q(4t₂, 2t₂²) be the points on x² = 8y. Equation of the normals at P and Q are 
y - 2t₁² = -1/t₁(x - 4t₁²)  ....(1) 
y - 2t₂² = -1/t₂ (x - 4t₂²)   ....(2)
Since the normals are at right angles, we have
(-1/t₁)(-1/t₂) = -1
⇒ t₁t₂ = -1   ....(3)
Solving Eqs. (1) and (2) and from Eq. (3), we have

⇒ 2y = x² + 12  which is the required locus.

The boys are in majority, if groups formed are 4B 3G, 5B 2G, 6B 1G.
Total number of such combinations
⇒ ⁶C₄ x ⁴C₃ + ⁶C₅ x ⁴C₂ + ⁶C₆ x ⁴C₁
⇒ 15 x 4 + 6 x 6 + 1 x 4
⇒ 60 + 36 + 4
⇒ 100

For a number to be even its last digit should be divisible by 2.
Therefore, only 2,4,6 can be used as last digits.
Number of such combinations,
⇒ 4!×3 = 24×3 = 72

The total number of ways in which 5 person can be chosen out of 9 person is ⁹C₅ = 126
The couple serves the committee in ⁷C₃ x ²C₂ = 35 ways
The couple does not serve the committee in
⁷C₅ - ⁷C₂ = 21 ways
Since the couple will either be together or not at all
∴ favourable number of cases = 35 + 21 = 56
∴ required porbability = 56 126 = 4/9

∵ probability for odd = p

∴ probability for even = 2p

∵ p + 2p = 1

⇒ 3p = 1

⇒ p = 1/3​

∴ probability for odd = 1/3​ , probability for even = 2/3​

Sum of two no. is even means either both are odd or both are even

∴ required probability = (1/3​ × 1/3)​ + (2/3 ​× 2/3) ​= 1/9 ​+ 4/9 ​= 5/9​