The Total solution for NCERT class 6-12
Find the approximate value of┬аf┬а(2.01),where┬аf┬а(x) = 4x2┬а+5x┬а+ 2
Let┬аx┬а= 2 and ╬Фx┬а=0.01. Then, we have:
f(2.01)=┬аf(x┬а+ ╬Фx) = 4(x┬а+ ╬Фx)2┬а+5(x┬а+ ╬Фx) + 2
Now, ╬Фy┬а=┬аf(x┬а+╬Фx) тИТ┬аf(x)
тИ┤┬аf(x┬а+ ╬Фx) =┬аf(x)+ ╬Фy
Hence, the approximate value of┬аf┬а(2.01)is 28.21.
