VITEEE Maths Test - 3 - JEE MCQ

Since the sphere passes through origin, its equation is of the form
x² + y² + z² + 2ux + 2vy + 2wz = 0.  ...(1)
Substituting the co - ordinates of the other given points, we get
u = 1, v = 2, w = 4
Centre of sphere = (u/2, v/2, w/2)
⇒ (1/2, 1, 2)

The graph of the logarithmic function y = log_ax is the reflection about the line y = x of the graph of the exponential function y = a^x, as shown in Figure.

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.

The product of the perpendiculars drawn from any point on a hyperbola to its asymptotes is a constant value. This value can be calculated using the hyperbola's parameters.

• Consider a hyperbola defined by the equation: x²/a² - y²/b² = 1.
• The asymptotes of this hyperbola are given by the equations: y = (b/a)x and y = -(b/a)x.
• For any point (x₁, y₁) on the hyperbola, the perpendicular distances to these asymptotes are calculated.
• The product of these perpendicular distances is always a²b²/(a² + b²).

Thus, the correct answer is a²b²/(a² + b²), which corresponds to option C.

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

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

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

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

∵ 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.

(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.