WBJEE Maths Test - 5 - JEE MCQ

Since ω is a cube root of unity, it follows that ω³ = 1.

Therefore, (1 + ω³) = 2. For the expression (1 + ω²)³, we use the fact that ω² + ω + 1 = 0,

which simplifies to 1 + ω² = -ω. Consequently, (1 + ω²)³ = (-ω)³ = -ω³ = -1.

Therefore, the expression (1 + ω³) - (1 + ω²)³ simplifies to 2 - (-1) = 3.

We are given:

∫₀¹ f(x)·g(x) dx

Where:

1. f′(x) = f(x), with f(0) = 1

2. f(x) + g(x) = x²

Step 1: Solve for f(x)

From f′(x) = f(x), we separate variables:

f′(x) / f(x) = 1
∫ f′(x)/f(x) dx = ∫ 1 dx
ln|f(x)| = x + C
f(x) = e^(x+C) = k·eˣ

Using f(0) = 1:
1 = k·e⁰ → k = 1

So,
f(x) = eˣ

Step 2: Solve for g(x)

From f(x) + g(x) = x²:
g(x) = x² − eˣ

Step 3: Setup the integral

∫₀¹ f(x)·g(x) dx
= ∫₀¹ eˣ(x² − eˣ) dx
= ∫₀¹ x²·eˣ dx − ∫₀¹ e^(2x) dx

Step 4: Solve ∫₀¹ x²·eˣ dx using integration by parts

Let u = x², dv = eˣ dx
Then du = 2x dx, v = eˣ

∫ x²·eˣ dx = x²·eˣ − ∫ 2x·eˣ dx

Now, ∫ 2x·eˣ dx:
Let u = 2x, dv = eˣ dx
Then du = 2 dx, v = eˣ

∫ 2x·eˣ dx = 2x·eˣ − ∫ 2·eˣ dx
= 2x·eˣ − 2·eˣ

So,
∫ x²·eˣ dx = x²·eˣ − (2x·eˣ − 2·eˣ)
= x²·eˣ − 2x·eˣ + 2·eˣ
= eˣ(x² − 2x + 2)

Evaluate from 0 to 1:

At x = 1: e¹(1 − 2 + 2) = e
At x = 0: e⁰(0 − 0 + 2) = 2

So,
∫₀¹ x²·eˣ dx = e − 2

Step 5: Solve ∫₀¹ e^(2x) dx

= (1/2)·e^(2x) from 0 to 1
= (1/2)(e² − 1)

Step 6: Final calculation

∫₀¹ f(x)·g(x) dx
= (e − 2) − (1/2)(e² − 1)
= e − 2 − (e²/2 − 1/2)
= e − 2 − e²/2 + 1/2
= e − e²/2 − 3/2

Final Answer:
(C) e − e²/2 − 3/2

Given:
y = sin((1 + x²) / (1 − x²))

We are to find dy/dx.

Step 1: Identify inner and outer functions

Let:
u = (1 + x²) / (1 − x²) ← inner function
y = sin(u) ← outer function

Step 2: Differentiate using chain rule

Using the chain rule:
dy/dx = dy/du × du/dx

We know:
dy/du = cos(u)

Step 3: Differentiate u

To find du/dx for u = (1 + x²) / (1 − x²), use the quotient rule:

du/dx = [(1 − x²)(d/dx(1 + x²)) − (1 + x²)(d/dx(1 − x²))] / (1 − x²)²

Now compute derivatives:
d/dx(1 + x²) = 2x
d/dx(1 − x²) = −2x

So:
du/dx = [(1 − x²)(2x) − (1 + x²)(−2x)] / (1 − x²)²

Simplify numerator:
= (1 − x²)(2x) + (1 + x²)(2x)
= 2x(1 − x² + 1 + x²)
= 2x(2)
= 4x

So:
du/dx = 4x / (1 − x²)²

Step 4: Combine the results

dy/dx = cos((1 + x²) / (1 − x²)) × 4x / (1 − x²)²

Final Answer:
dy/dx = 4x × cos((1 + x²) / (1 − x²)) / (1 − x²)²

(A - B) x (A - B)
(A - B) x A - (A - B) x B
= A² - AB - BA + B²

We have a + b = 4 ⇒ b = 4 − a and b − a = 4 − 2 a = t (say)

Thus, f(a) is an increasing function of t. Hence, the given expression increases with increase in (b-a).