Let f and g are two linear functions such that f (x – 1) = 2x – 3 + g(1) – f (1) and g(x – 1) = 4x + 5 – g (1)

Linked Question 1

Area enclosed by the curves y = f (x), y = g (x) and y = 0, is

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{8}$

No Solution

Linked Question 2

If $\int \frac{f (x) + 2}{(\frac{f (x) - 1}{2}) (\frac{f (x) + 1}{2}) (\frac{g (x) + 7}{4}) (\frac{g (x) + 11}{4}) + 1} dx = C - \frac{1}{h (x)}$ where h (x) is of the form of ax² + bx + c, then (a + b + c) equals

None

Put x = 2 in both the equation we get

f (1) = 3 and g (1) = 5

$∴$ f (x) = 2x + 1 and g (x) = 4x + 1

$\int \frac{2 x + 3}{(x^{2} + 3 x) (x^{2} + 3 x + 2) + 1} dx$

Put x² + 3x = t ⇒ (2x + 3)dx = dt

$\int \frac{dt}{t (t + 2) + 1}$ ;

$\int \frac{dt}{{(t + 1)}^{2}} = C - \frac{1}{t + 1} = C - \frac{1}{x^{2} + 3 x + 1}$

⇒ a = 1, b = 3, c = 1

⇒ a + b + c = 5

Linked Question 3

The value of $\underset{x \to 0}{Lim} {(\frac{g (x)}{f (x)})}^{cosecx}$ is equal to

e⁴

e³

e²

Put x = 2 in both the equation we get

f (1) = 3 and g (1) = 5

$∴$ f (x) = 2x + 1 and g (x) = 4x + 1

$\underset{x \to 0}{Lim} {(\frac{4 x + 1}{2 x + 1})}^{cosecx} = \frac{p}{q} = e^{2}$