VITEEE Maths Test - 5 - JEE MCQ

Any point on the first line is (4r+k, 2r+1,r−1), and any point on the second line is (r′+k+1 ,−r′,2r′+1) for some values of r and r'. The lines are intersecting if these two points coincide i.e

4r + k = r′ + k + 1, 2r + 1 = −r′, r − 1 = 2r′ + 1 for some r and r'

⇒ 4r − r′ = 1, 2r + r′ = −1, r − 2r′ = 2

Now, 4r − r′ = 1, 2r + r′ = −1 ⇒ r = 0, r′ = −1 which satisfy r − 2 r′ = 2.

⇒ The given lines are intersecting for all real values of k.

We know that the word joining two simple statements to form a compound statement is called the connective.
Now, the given statement is
′′2+7>9 or  2+7<9′′
Two mathematical inequality cannot occur together. Either 2 + 7 can be greater than 9 or less than 9.
Hence, the connective word is 'or'.

Since, fourth term of

Taking logarithm on both side

Given:

Let 
So,

And,

Thus,

Hence, this is required solution.

Here,

Let V be the volume of the milk, then

Putting the value of r,

Differentiate on both sides with respect to t,

Given, the value of h = 6m and 
Put these values.

Thus,

Hence, this is required solution.

sinC=sin(A+B)=sinAcosB+cosAsinB

Here,

It means X and Y are independent events, so X' and Y' are also independent. Therefore,

Or,

Hence, this is the required solution.

Given mean:

And the median will remain same, i.e. 104.
Hence, this is the required solution.

Here,


Therefore, 3 such values of x are possible.
Hence, this is the required solution.

Total number of bijection from set of n elements to itself = n!

(x - a) (x - 10) + 1 = 0
∴ (x - a) (x - 10) = -1
∴ x - a = 1
and x - 1 0 = - 1 ,
or x - a = 1 and x - 10 = 1
∴ a = 8 ora = 12

Let the third vertex be A(α,β)

Using the diagram, OA⊥BC

⇒ Slope of OA × Slope BC = −1

Solving Equations(i)i and (ii)ii, we get

α = −1, β = 6

∴ The third vertex is (−1, 6)

Degree of the determinant is
n + (n + 2) + (n + 3) = 3n + 5 
and on R .H.S., degree = 2
3n + 5 = 2
⇒ n = -1

Given limit can be written as,

Using L'Hospital' rule,

Here,

Hence, this is the required solution.

The centre and radius of a circle x² + y² + 2gx + 2fy + c = 0 are (−g, −f) and

Hence, the centre of x² + y² − 4x −2y − 11 = 0 is A(2, 1) and the radius

If P(α, β) be the centre of C₂ of radius r₂ = 1

We know that, if two circles with centres A and P and radii r₁ and r₂ touches each other externally, then distance between their centres AP is equal to the sum of their radii i.e. AP = r₁ + r₂

The locus is obtained by replacing (α, β) by (x, y)

Hence, the locus is x² + y² − 4x − 2y − 20 = 0

As we know that maximum value of

According to the question,

Thus,
Maximum value of the given expression is 17.
Hence, this is required solution.

∵L.H.S.=(b + c)cos A + (c + a)cos B + (a + b)cos C
=bcos A +c cos A+ c cos B +a cos B +a cos C +b cos C
=(b cos A + a cosB ) + (c cos A + a cosC) + (c cos B + cos C)
=a + b + c = R.H.S
Hence L.H.S.=R.H.S.

New mean

​​​​

Let 
⇒
⇒

If a=4 cm, b=6 cm and c=8 cm,
then we have
2s = a + b + c 
⇒s=9 cm
By Heron's formula, we have

Now, we know that

Since, A & B are independent events.

Given equation is,

Semi-perimeter 

Given, sides of the triangle ABC are in AP.
Let sides are a, b, & c in which a is the smallest side.
Since, sides are in AP, So we have
2b = a+c⇒a=2b−c  ...(1)
Now, using cosine rule, we have

Using equations (1) & (2), we have

In ΔABC, circumcentre O and orthocentre  H
Given  OA=HA
⇒R = 2 R cos A
⇒ cos A = 1/2
⇒ A = π/3