SRMJEEE Subject Wise & Full Length Maths Mock Test - 9 Free Online Test

Here,

Hence, this is the required solution.

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.

We know from the standard result of vector triple product that for any 3 vectors 

Now note that 

|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

t = T/|a|
  = 2π/π/2
  = 4

Given equation is

We know that,
Y = f(x)
So,

So, this is linear differential equation.
Where,

So,
I.F

So, complete equation is

Given that

Putting the value of c in equation (i), we
get,

Thus,

Hence, this is required solution.

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

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

Correct Answer :- D

Explanation : mm₁ = -1

(y₁/x₁)*m₁ = -1

m₁ = - x₁/y₁

So, equation of line passing through (x₁, y₁) 

y = m₁x

y = (-x₁/y₁)*x

yy₁+ xx₁ = 0

We need to evaluate the expression:

• The expression involves a series of binomial coefficients: 3ⁿC₀, 8ⁿC₁, 13ⁿC₂, and so on.
• The pattern alternates in sign, beginning with a positive term.
• Each term is derived from the binomial coefficient multiplied by a coefficient that follows the pattern of increasing integers.
• Upon analysis, the series can be represented as a polynomial of degree n.
• Using the identity for binomial coefficients, the sum evaluates to zero when summed over the complete set of coefficients.

Conclusion: The entire sum results in 0.

To find the equation of the circle:

• The given lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of the circle.
• First, find the intersection point of these lines, which is the centre of the circle.
• Solve the equations:
• 2x - 3y = 5
• 3x - 4y = 7
• Using simultaneous equations, solve to find the intersection point.
• The centre is (1, -1).

The area of the circle is 154 square units.

• Area formula: πr² = 154
• Calculate the radius: r² = 154/π ≈ 49
• The radius r is 7.

Thus, the equation of the circle with centre (1, -1) and radius 7 is:

• (x - 1)² + (y + 1)² = 49

Convert to standard form:

• Expand: x² - 2x + 1 + y² + 2y + 1 = 49
• Simplify: x² + y² - 2x + 2y = 47

The correct option is C.

R₃ cannot remain empty.
Total ways = ⁸C₆×6!=20160
When R₁ is empty =6!=720
When R₂ is empty =6!=720
∴ When no is empty =20160−720×2=18,720

The series given is an alternating series:

• 1/2! - 1/3! + 1/4! - 1/5! + ...

This resembles the expansion of e^x when x = -1, which is:

• e^-1 = 1 - 1/1! + 1/2! - 1/3! + 1/4! - ...

By comparing both, the given series is part of the expansion for e^-1:

• It starts from the 1/2! term, skipping the initial terms.

Thus, the series converges to e^-1.

Since, telephone number start with 67, so two digits is already fixed
Now, we have to arrangement of three digits from remaining eight digits (i.e., 0,1,2,3,4,5,8,9)

= 8 × 7 × 6
= 336 ways

Equation of AB is 
and that of AC is 
Hence 

Let A (−2,−6), B(–2,4) and C(1,3) are the vertices of the △ABC
Now using the distance formula, we get 

(AB)² = (BC)² + (CA)²
so the △ABC is a right angle triangle, right angle at C and we know that in right-angle triangle orthocenter is the point where right angle formed,
Therefore orthocenter is C(1, 3)

Given,

Using by parts rule,

Then, we get,

Thus,

Hence, this is required solution.

Given that,
 that are parallel to each other.
If and if these vectors are parallel to each other, then they must follow, 

⇒ sinθ = cosθ
⇒ θ = 45°

Let


Thus,

Hence, this is required solution

 Parallel vector =(2+b)i+6j−2k

Unit vector = (2+b)i+6j−2k / (b²+4b+44)^1/2

According to the question, 

1 = (2+b)+6−2/b²+4b+44

⇒b²+4b+44=b²+12b+36

⇒8b=8⇒b=1

Total number of outcomes when four normal dice are rolled
= 6 × 6 × 6 × 6 = 6⁴ = 1296
Total number of ways in which no dice shows up 2 i.e.
Each of the four dice shows up 1,3,4,5 or  6 as outcomes
=5 × 5 × 5 × 5 = 5⁴ = 625
Hence total number of possible outcomes when no dice shows up 2
=1296 − 625 = 671