BITSAT Mathematics Test - 6 - JEE MCQ

Consider the expression:

Hence, this is the required solution.

Here,


Hence, this is the required solution.

Apart from the librarian's book, there are 49 other books. When he revisits the library, 29 other books are left in the shelf (excluding his book).
Now,
n(s) = Number of ways of selecting 29 books out of 49 books = ⁴⁹C₂₉
n(E) = Number of ways of selecting 29 books out of 47 books (leaving two neighbouring positions vacant) = ⁴⁷C₂₉
Therefore,
required probability
Hence, this is the required solution.

Here,

Let

We have to differentiate both sides with respect to x.
Then,

Substitute these values,

And,

Then, we have to find I.F.

So, the complete solution is

Thus,

Hence, this is the required solution.

Here,
y log x = x - y
Differentiate with respect to x.

Hence, this is the required solution.

We know that the equation of the chord will be

Here, ∝ is the angle subtended at the centre by the chord.
Therefore,

As chord is subtending 90^o at the centre, therefore

The general term in in the expression is:

For coefficient of a⁴, we put
10 - 3r = 4
r = 2
Therefore, coefficient of a⁴

Hence, this is the required solution.

Since P is orthogonal, each row is orthogonal to the other rows.
Therefore,

Hence, this is the required solution.

We have

Thus, this will give all the integral values which are not less than 0.
So, all the whole numbers are the range of the function.
Hence, this is the required solution.

Consider the expression:


Now,

Hence, this is the required solution.

The given equation is
y=ae^bx

2log (y + 3) = log x + log c
(y+3)²=xc, which is parabola.

4^tanx − 3.2^tanx + 2 ≤ 0
Let 2^tanx = t
⇒ t² − 3t + 2 ≤ 0
⇒ (t − 2) (t − 1) ≤ 0
⇒1 ≤ t ≤ 2
⇒20 ≤ 2^tanx ≤ 2¹
⇒0 ≤ tanx ≤1

Given that: |tan2x|=sinx in [0,π]
Drawing the graph of y=|tan2x| and y=sinx

Another Method

∴sin3θ = 3sinθ − 4sin³θ

We have,

Integrating, we get,

We have, (1 + tan y)(dx − dy)+2x dy = 0
⇒(1 + tan y) dx = (1 + tan y − 2x) dy

It is a linear differential equation.

=(cos y + sin y)e^y
So, the solution is
xe^y(sin y+cos y) = ∫e^y (sin y + cos y)dy + c
⇒ xe^y(sin y+cos y)=e^y sin y + c
⇒ x(sin y+cos y)=siny+ce^−y.

We have,
⇒ (ax + 3) dx = (2y + f) dy
On integrating, we obtain

This will represent a circle, if

Given, 

This is a linear equation, comparing with the equation

Equation (i) passing through (1,2)

Put in (i)

x (1 + y²) dy = (y + yx² – x²) dx

Solution is:

∴ y′y′′′=3(y′′)²

∴ x=A₁y²+A₂y+A₃

sin36° sin72° sin108° sin144°
=sin²36°sin²72°

We have,

Applying componendo and dividendo, we get

⇒tan²B=tanA tanC
⇒tanA, tanB & tanC are in G.P.

Thus, the greatest and least value of f (x) are 1/2 and respectively.
Hence, the correct  option is B

We have, (1+tany)(dx−dy)+2xdy=0
⇒(1+tany)dx=(1+tany−2x)dy

It is a linear differential equation.

=(cosy+siny)e^y
So, the solution is
xe^y (sin y + cos y) = ∫e^y (sin y + cos y)dy + c
⇒ xe^y(sin y+cos y)=e^ysin y + c
⇒ x(sin y + cos y) = sin y + ce^−y.

Given:
ϕ(x)=ϕ′(x)

Integrating,

⇒log ϕ (x) = x + k
⇒ϕ(x)=e^x+k = ex⋅c
⇒ϕ(x) = ce^x

∴ ϕ(x)=2e^x−1
⇒ ϕ (3) = 2e²

(Linear differential equation in x)
∴ General solution is 
Now as at y = 0, x=1; so c=1
Finally, 

Given, 2^{1+|cosx|+∣∣cos2x∣∣+∣∣cos3x∣∣+...} = 2²


Thus, total four solutions are possible for the given equation, they are