JEE Advanced (Single Correct MCQs): Definite Integrals and Applications

Now we know that if  then it means that
f (x) is + ve on some part of (α, β) and – ve on other part of (α, β).
But here 1 + cos⁸ x is always + ve,
∴ ax² + bx + c is + ve on some part of [1 , 2] and – ve on other part [1, 2]
∴ ax² + bx + c= 0 has at least one root in (1, 2).
⇒ ax² + bx + c = 0 has at least one root in (0,2).

Differentiating both sides w.r.t b, we get
f(b) = 3(b – 1) cos(3b + 4) + sin(3b + 4)
⇒ f(x) = 3(x – 1) cos(3x + 4) + sin(3x + 4)

We have,

Let F (x) = (f (x) + f (- x)) (g (x) - g (-x))
then F (-x) = (f (- x) + f (x))(g (- x)- g (x)) = - [f (x) + f (- x)][ g(x) - g (-x)]
=-F (x)
∴ F(x) is an odd function, ∴ we get I = 0

We have

Adding (1) and (2), we get

In the range   we have to find the value of

(applying line integral on inequality)

     … (1)

(applying line integral on inequality)

    … (2)

From (1) and (2), we get

[∵ e^{cos x} sin x is an odd function.]

We know that for    and hence 

and for 

               .....(1)
 
Adding (1) and (2),

 (even function)
             ...(3)

          ....(4) 
Adding (3) and (4) 
∴ I = π /2

The given lines are
y = x – 1; y = – x – 1;
y = x  + 1 and  y =  – x + 1
which are two pairs of parallel lines and distance between the lines of each pair is √2 Also non parallel lines are perpendicular. Thus lines represents a square of side √2 Hence, area = (√2)² = 2 sq. units.

Now the given equation x² - f ' (x)=0 becomes

Given that T > 0 is a fixed real number. f is continuous

⇒ f is a periodic function of period T

Intergrating by parts considering (1 + t)ⁿ as first function, we get

∴ f(x) increases when  x < 0

The curves given are

and x-axis y = 0 ...(3)
Eqn. (1), [y² = x] represents right handed parabola but with +ve values of y i.e., part of curve lying above x-axis.
Solving (1) and (2) we get,

Also (2) meets x-axis at (3, 0)

Shaded area is the required area given by

Differentiating both sides w.r.t. t  

[Using Leibnitz theorem]

y = ax² and x = ay²

Points of intersection are O (0, 0) and 

The given curves are

y = (x+ 1)²     ...(1)

upward parabola with vertex at (–1 ,0) meeting  y –axis at (0, 1)
y = (x -1)²                 ...(2)

a line parallel to x–axis meeting (1) at  

and meeting (2) at 

The graph is as shown 

The required area is the shaded portion given by ar (BPCQB)  = 2 Ar(PQCP) (by symmetry)

The given curves are 

∴ The area bounded by the above curves, by the lines

Also when  

Given that f is a non negative function defined on

Differentiating both sides with respect to x, we get

Integrating both sides with respect to x, we get

∵ Given that f (0) = 0   ⇒ C = 0

Hence f (x) = ± sin x

But as f (x) is a non negative function on  [0, 1]
∴ f (x) = sinx.