IIT JAM Mathematics Practice Test- 3 - Mathematics MCQ

Now, apply leibnitz rule,

we get

Given

    ..(1)

Now apply leibnitz rule,

we get

f(x) = 2x cos x + x2 sin x + 1 At x = 0,

Let h(x) = {f(x) + f(-x)Kg(x) - g(-x)} h(-x)

= {f(-x) + f(x)Kg(-x) - g{x)}

= (f(—x) + f(x)Kg(x) - g(-x)}

= -h(x)

therefore,

we have

⇒ 

Now

On differentiating both sides, we get - sin2x f(sin x) cos x = - cos x

(By L-Hospital's rule)

since given that 

Here g is real valued continuous function on R. then

Now solve this integral by applying leibnitz rule, we have

since given that

Now differential both sides w.r.to. x , we have

Using Leibnitz'e rule, we have

Now put x = 1, both sides we have,

 = 2x + xcos2x.2+sin2x-sin2x

∴ f(x) - f(0) = 2x + 2x cos 2x 

Again differentiating w.r.to. x, we get 

Log I = 

Differentiating the equation of the curve

Therefore the required length

since

Hence

Put n =1

then

and we know th at Sum of the series is always greater than or equal to the calcu­lated value.

Given series is

or

Hence

[By Variable Seperation method]

log f(x) = logx +logc

=> f( X ) = C X
given f(1) = 1 then C = 1

So, f(x) = x 

now apply leibnitz rule, we get

f(x) = 2x sinx +x²cosx+3x²

Hence

Let

(by Leibnitz's rule)

So from F'(x) = 0, we get x = 0 or

Hence 

then apply L'Hospital rule, we get

=-f(0)

= -1 [by the definition of f(t)]

since

thus to find f(x)

On differentiating both sides using Newton Leibnitz formula

i.e

for  is obtained when 

∵ g(x) is odd function 

The line x = 1 meets the curves in (1,e) and B(1, 1/e). The second curve is lower cuive as ordinate of B is less than ordinate of A.
Both the curves pass through origin.

 
= 0.7357

The figure shows 1/8 of the desired surface. 

The equation of the surface has the form

so that

The domain of integration is a quarter circle  on the xz- plane, Thus, we have

The curve C consist of the arc OA, (y³ = x²) and the chord AO, (y = x)

∴