VITEEE Maths Test - 9 - JEE MCQ

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

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.

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.

Equation of AB is 
and that of AC is 
Hence 

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

⇒ 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