Mathematics_Today_-_October_2016

and^11
n

f x nxdx

n

fx nxdx

a

b

a

b

∫ ′()cos ≤⋅∫|()||cos |′

≤^1 ∫ = − → 0

n

Mdx Mb a

a n

b ()

where |f ′(x)| ≤ some M

Hence, f x nxdx n
a

b

∫ ()sin →→∞0 as

11. (d) : tan−−−tan = ⇒ = +
    −

11

4

1

1

ba ba

a

π

So, length of interval lba a
a

= − = +

−

1

1

2

Using, dl
da

=0,at lmin, we have a=^12 −

12. (c) : When integrated once,
    fx′()=+∫δδ 12 ()x ()xdx⋅ = θ 1 (x) + θ 2 (x) + c
    f ′(0) = c = 0
    So, f ′(x) = θ 1 (x) + θ 2 (x)
    To find f (x), weintegrate f ′(x) or take the area covered
    by θ 1 (x) and θ 2 (x).



  

 

  



Now the graphs of θ 1 (x) and (θ 2 )x are

So, f(5) = area of the shaded region

= (4 × 1) + (3 × 1) = 7

13. (a) : Combine the graphs
    of θ 1 (x) and θ 2 (x) to get the
    graph of f ′(x).
    So, f ′(5) = 2


14. (c) : From the graph, it is clear that f ′(x) is not
    differetiable at two points x = 1 and x = 2.


15. (b) : The given curve equation is xy^2 = (x – 3y)^3 , a
    homogeneous equation.
    So,

dy

dx

y

x

==ie Q x.. ( ) x

So, y^2 + (Q(x))^2 = logπ represents a circle.

16. (c) : Q′(x) = 1 ⇒ Q′(1) = 1


17. (b) : limsin ( ) limsin
    xx

Qx

x

x

→→x

==

00

1

””









 

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