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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!!!!!
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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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