`f(x) = sqrt(1 + cos^(2) (x^(2)) `の場合、`f(sqrt(pi)/(2))`の値は

f(x)=√1+cos²(x²)のとき、f’(√π/2)の値は？ RD Sharma|Ncert|微分|CET|Objective

' limit x tends to 0 '1-cosx(sqrt(cos2x))/x^2' || '1-cosx(sqrt(cos2x))/x^2' ||

If f(x)=(1-cos⁡4x)/x^2,xless0a,x=0(√x)/(√(16+√x) −4),xgreater0)is continuous at x=0,find a|jee mains

find the domain and range of `f(x)=sqrt(cos(sin x))+sin^-1((1+x^2)/(2x))`

Increasing Decreasing RD SHARMA example 28 show that 2x + cot inverse x + log (sqrt 1+x^2 - x

Let x=2 be a root of equation x^2+px+q=0 &f(x)=1-cos(x2-4px+q2+8q+16)/(x-2p)^2,x#2p &0at x=2p then

The function `f(x)=sqrt(cos(sinx))+sin^-1((1+x^2)/(2x))` is defined for :

Revision session -MLF W 9 and 10

Find the domain `f(x)=sqrt(cos2x)+sqrt(16-x^2)`

JEE MAINS 2018 Show that `f(x) = 2x + cot^-1 x + log(sqrt(1+x^2)-x)` is increasing in `R`

Let x = 2 be a root of the equation x^4 +px +q = 0 and f(x) = { 1- cos(x^2 - 4px +q^2 +8q +16)

`int_(0)^(pi)(xsinx)/((1+cos^(2)x))dx`を評価します。

How to Find the Domain of a Function

Find the range of the following functions given by || f(x) = 3/(2 - x^2) || f(x)-1+3cos-2x |#range

Range of `f(x)=cos^-1(sqrt(1+2x^2)/(1+x^2))`

【全パターン網羅】数3の積分で無双して、ライバルに差をつけろ！

let `f(x)=sqrt(1+x^2)` then

関数 f(x)=cos x +cos (sqrt(2)x) が最大値に達する x の値の数は...

とします：f(x) ` {:(=(1-cos 4x)/(x^(2))", ... "x lt 0 ),(= a", ... " x = 0 ),(= (sqrt(x))