BITSAT Mathematics Test - 2 - JEE MCQ

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

Let 
⇒
⇒

(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

a = 0 or a = 1/2, the equation becomes linear So 
Hence, only answer is, 

Number of ways = (10 + 1) (9 + 1) (7 + 1) -1
= 879

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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 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.

Given:

Again, consider 

From both the above results,
P = Q
Hence, this is the required solution.

Here,

Hence, this is the required solution.

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.

Given, vertex = (0, 0), point = P(2, 3) and axis is along X axis.
Since point (2, 3) lies in the first quadrant,
Therefore, equation of the parabola will be of the form y2 = 4ax , which passes through P(2, 3).
Therefore,
Therefore, required equation is

Hence, this is the 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.

Given:

Let 
So,

And,

Thus,

Hence, this is required solution.

Here,

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

Or,

Hence, this is the required solution.

Since, A & B are independent events.

New mean

∴ Given expression is 1 + 1 + 1 = 3

Since, fourth term of

Taking logarithm on both side

Given equation is ln(x + y) = 2xy …(1)

ln x + y = 2xy …1

For x = 0

ln(0 + y) = 2.0.y = 0

⇒ lny = 0

⇒ y = 1

Now, differentiating (1), we get,

At point (0,1), we get,

Given limit can be written as,

Using L'Hospital' rule,

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)

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

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.