BITSAT Mathematics Test - 3 - JEE MCQ

If tan^-1 x = θ, then x = tanθ, 
⇒ sec² θ = 1 + tan² θ = 1 + x²
⇒ secθ = 
∴ 

⁼

(∵ x > a, thereforelx - al = x - a)

If the scalar triple product of three non-zero vectors is zero, then the vectors are coplanar.

=1/2 x √( 2² + 14² + 10² )
=1/2 x √( 4 + 196 + 100 )
=1/2 x √( 300)
=1/2 x 10√3
=5√3

Vertices of the triangle are (0, 0) (-2, -2), (-4, - 8).
∴ Required area

Centre of the circles is (1, 2) and therefore every normal passes through (1,2). Required line is x + y = 3.

cos 5° - sin 25° = sin 85° - sin 25° 
= 2 cos 55° sin30°.

Correct Answer is A.

The binomial expansion is

The general term

The term which is independent of x is

Thus,
r = 8
Therefore, independent term = (r + 1)th = 9th
Hence, this is the required solution.

Given: P, Q and R are the angles of a triangle.
P + Q + R = π
P + Q = π - R
Now,

Now, let

Hence, this is the required solution.

Here,

Hence, this is the required solution.

Given, mid-points of sides are S(p, 0, 0), T(0, q, 0) and U(0, 0, r). Consider the diagram shown below.

Also, by mid-point theorem,

Similarly,

Therefore,

Hence, this is the required solution.

Any point on L1 is  and on L₂ is 
The lines will intersect, when

From above results, we get

Therefore,

Hence, this is the required solution.

We find RHL at x = 1

Then, find LHL at x = 1

Then, we have to find

Since square root function is used in the expression, the function does not exist for x < 0.
Hence, this is the required solution.

Given,

Then, differentiate this equation on both sides.

Thus, c = 0
Hence, this is the required solution.

Given equation of two curves


Both equations make parabola,
So,
Area of curve,

Thus, Area 
Hence, this is the required solution.

The equation of tangent at (3, 1) is

The coordinates of point P are

Then, we have to find slope of the perpendicular line through P.

Thus, 
Hence, this is the required solution.

Let P be an event that drawn pen is blue and Q be an event that drawn pen is red.
Then, PQ and QP are two disjointed cases of the given event.
Therefore,

Hence, this is the required solution.

∴  3−2x>0 and 2x−1>0 for log to exist

Take antilog
3−2x>2x−1⇒4x<1⇒ x<1   ...(2)
From (1) and (2)x∈ 

Given in equation: 

By (i) and (ii), we get

Now, combining equations  (iii) and (iv), we get 

Here, it is given that

From equations, (1),(2) and (3) we get

If a>0, and ax² + bx + c > 0 ∀x∈R
Then, discriminant of the quadratic equation ax² +bx+c=0 will be negative,
i.e. b²−4ac<0
and the roots will be imaginary.