SRMJEEE Maths Mock Test - 3 - JEE MCQ

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

The area A is given by the integral of the difference between the top and bottom functions from 0 to 3:

A = ∫ from 0 to 3 of [ (1/3) * sqrt(9 - x²) - (1 - x/3) ] dx

Now simplify the expression inside the integral:

This becomes:
A = ∫ from 0 to 3 of [ (1/3) * sqrt(9 - x²) - 1 + x/3 ] dx

Now break the integral into three simpler parts:

1. First part: (1/3) * ∫ sqrt(9 - x²) dx from 0 to 3

2. Second part: ∫ 1 dx from 0 to 3

3. Third part: (1/3) * ∫ x dx from 0 to 3

Now evaluate each part:

1. ∫ sqrt(9 - x²) dx from 0 to 3 = (9π) / 4

2. ∫ 1 dx from 0 to 3 = 3

3. ∫ x dx from 0 to 3 = (1/2) * (3²) = 9/2

Put them all together:

• First term: (1/3) * (9π / 4) = (3π) / 4

• Second term: -3

• Third term: (1/3) * (9/2) = 3/2

Now compute:

A = (3π) / 4 - 3 + 3/2

Convert everything to a common denominator:

• (3π)/4 is already in 4ths

• 3 = 6/2

• 3/2 stays the same

So:
A = (3π)/4 - 6/2 + 3/2 = (3π)/4 - 3/2

If a.b = 0
⇒ ab(cosθ) = 0
or cosθ = 0
⇒ θ = 90°
Therefore, a⊥b

We know that:

sin²x + cos²x = 1

So the left-hand side of the equation becomes:

81 × (sin²x + cos²x) = 81 × 1 = 81

This means the value of the left-hand side is always 81, no matter what x is.

But the right-hand side of the equation is 30, which is clearly not equal to 81.

Conclusion:

There is no real value of x in the interval [0, π] that can satisfy the equation.

No solution exists.

Two lines are perpendicular then product of slope be -1

m₁​m₂ ​= −1

(3 − λ)/2​ × 2 = −1

3 − λ = −1

3 + 1 = λ.

λ = 4.

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

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

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

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

Given, g(x) = 1 + √x​

And f(g(x)) = 3 + 2√x​ + x

Now, f(g(x)) = 3 + 2√x​ + x [ terms splitting]

f(g(x)) = 2 + (1 + 2√x​ + x)

Also, (1 + √x​)² = 1 + x + 2√x​

⇒ f(g(x)) = 2 + (1 + √x​)²

⇒ f(x) = 2 + x²

f(x) = x² + 2​

∵ 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​

Sum of the roots ⇒ a+b = −1                ...(1)
Product of the roots ⇒ ab = 1               ...(2)
We know that,
(a+b)² = a² + b² + 2ab
⇒ a² + b² = (a+b)² − 2ab
Substituting (1) and (2) in the above equation, we get
⇒ a² + b² = (−1)² − 2(1)
∴ a² + b² = −1 

Since, |B| = 0
AB and A⁽⁻¹⁾B is singular ie their det = 0

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.

Let 
We have
If A is a square matrix of order n then A(adjA) = |A|.I_n
Here, n = 2

Let λ:1 be the ratio in which the line joining the points (a,b,c) and (-a, -c, -b) is divided by xy- plane.

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.

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

Integrating both sides, we get

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

Third term of G.P. = ar² = 3

General term = arⁿ⁻¹

Product of first five terms = a⁵⋅r^0+1+2+3+4

= a⁵r¹⁰=(ar²)⁵

= (3)⁵

= 243.