JEE Main Maths Test- 5 - JEE MCQ

To find the integrating factor for the equation (x log x) dy/dx + y = 2 log x, we first rewrite it in standard linear form:

dy/dx + (1/(x log x)) y = 2/x.

The integrating factor μ(x) is calculated as follows:

• Let P(x) = 1/(x log x).
• Then, μ(x) = e raised to the integral of P(x) dx.
• Let u = log x, hence du = (1/x)dx.

Transforming the integral:

• We have ∫(1/u)du = log|u| + C.
• Substituting back gives log(log x).

Thus, the integrating factor is:

μ(x) = e^{log(log x)} = log x.

Therefore, the integrating factor is log x.

The given differential equation x \\, dy - y \\, dx = 0 can be rewritten as \\( \\frac{dy}{dx} = \\frac{y}{x} \\). This is a homogeneous equation.

• Substituting v = \\( \\frac{y}{x} \\), we get:
• y = vx
• \\( \\frac{dy}{dx} = v + x \\frac{dv}{dx} \\)
• Substituting into the differential equation gives:
v + x \\frac{dv}{dx} = v
• This leads to:
x \\frac{dv}{dx} = 0

This implies v = C (a constant), so:

• y = Cx

This represents a straight line passing through the origin.

The given problem involves a first-order linear differential equation:

y' = 1 + y.

However, solving this equation leads to a contradiction when applying the boundary conditions, making it impossible for such a function to exist under the given constraints.

Therefore, the correct answer is D.

To find the primitive of sin x cos x, we use the double-angle identity:

• sin(2x) = 2 sin x cos x
• Thus, sin x cos x = (1/2) sin(2x)

Consequently, the integral is:

• ∫sin x cos x dx = ∫(1/2) sin(2x) dx
• = (1/2)(-1/2 cos(2x)) + C
• = -1/4 cos(2x) + C

Thus, the correct answer is C.

ƒ(x) is continuous  A + B = A + 3 – B
⇒ B = 3/2
ƒ(x) is differentiable 2B = 6 + A 
⇒ A = –3

ƒ'(x) + x²ƒ(x) = –2x, ƒ(1) = 1
⇒    ƒ'(1) + 1 = –2     ⇒    ƒ'(1) = –3
ƒ''(x) + 2xƒ(x) + x²ƒ'(x) = –2
ƒ''(1) + 2ƒ(1) + ƒ'(1) = –2
ƒ''(1) = 3 – 4 = –1  ⇒ |ƒ''(1)| = 1