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Here we cannot solve for x easily. 2a \geqslant sinx" src="http://latex.codecogs.com/gif.latex?%3D%3E2a%20%5Cgeqslant%20sinx"> 2a \geqslant 1"...

D^n (x^n logx)

let, g’(0)=f’(1)e f(0) +f(1)e f(0) .f(0) =f’(1)+f’(0)=0+2=2

Buy Cengage’s Playing with Graphs, after reading it you will feel comfortable with this part.

Where is the figure?

This is not fair!!!!!

YES THANK YOU SO MUCH!!!!!!!!!!!

This is again of 0/0 form. Again use L-Hospital rule Now putting x=1.

for f(x) divide numerator denominaor by x^3.... then use expansion series... I couldn’t understand what g(x) is so pls rewrite that

e^(x^2)-cosx/x^2 using l hopital rule 2xe^2+sinx/2x 4x^2e^x+cosx/2 1/2

the line is y=5-x since the line is perpendicular to line y=x, the slope of this line is -1. and acc. to point slope formula i.e., (y-y’)=m(x-x’) hence (y-2)=(-1)(x-3) y-2=3-x y=5-x

0/0 form but neumerator has value exactly = 0 but denominator tends to 0 therefore its value is

Monotonocity part in Ml khanna is in chapters like derivatives,differentiations etc

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