# differentiation-6

9) Find _{ }

Solution:

We have _{ }

Therefore by taking log on both sides we get,

logy = y log sinx.

Differentiating both sides w.r.t x we get,

_{ }

10) Is f(x) differentiable at x=0 if f(x) is defined as follow

_{ }

Solution:

_{ }

_{ }

Here f(x) is not differentiable at x=0.

11) Determine the values of x for which the following function fails to be continuous or differentiable.

_{ }

Justify your Answer.

Solution:

By given definition it is clear that the function f is continuous and differentiable at all points except possibility at x=1 and x=2.

Continuity at x=1,

L.H.L = _{ }

R.H.L = _{ }

Also f (1) = 0, Hence f(x) is continuous at x=1.

Now differentiability at x=1.

_{ }

Since _{ } so we get f(x) is differentiable at x=1.

Continuity at x=2.

_{ }

_{ }

Since R.H.L ¹ L.H.L

Therefore f(x) is not continuous at x=2. As such f(x) can not be differentiable at x=2. Hence f(x) is continuous and differentiable at all points except at x=2.

12) If

_{ } Then find a) Constant term b) Coefficient of x.

Solution:

Here _{ }

Putting x =0 we get

_{ }

Ãž A=0

Again Differentiating (1) w.r.t x we get,

_{ }

Therefore Coefficient of constant term = Coefficient of x = 0.

(MEDIUM TYPE)

1) Let f(x) be a real function not identically zero such that_{ } , nÃŽN and x, y are any real numbers and f^{1}(0) ³ 0. Find the values of f (5) and f^{11111}(10).

Solution:

Here _{ } ------ (1)

Putting x=0, y=0 we get f (0) = f (0) + {f (0)}^{2n+1}.

\ f (0) = 0.

_{ } _{ }

\ If x ³ 0, f(x) ³ 0 -------- (2)

Putting x = 0, y = 1 in (1)

_{ }

Putting y=1 in (1) for all real x,

f (x+1) = f(x) + {f (1)}^{2n+1}------ (3)

\f (1) = 0 Ãž f(x+1) = f(x)

Ãž f (1) = f (2) = f (3) = ------ = 0

i.e. f(x) is identically zero.

\ f (1) ¹ 0 Hence f (1) =1.

So from (3) f(x+1) = f (x) +1 ---- (4)

\ f (5) = f (4) +1 = {f (3) + 1}+ 1

= {f (2) +1} +2

= {f (1) +1} +3

=f (1) +4

=1+4

=5

Also (4)Ãž f(x) is a function whose value increases by 1.When variable x is increased by 1.

\f (x) = x \f^{1}(x) = 1

\ f^{1}(10) =1.

Dumb Question:

1) Why does the equation _{ } has the solution as f (1) = 0, 1? Some complex solution could also be possible.

Ans: It is mentioned in the question that function f is real valued function, so the value of f (1) has to be real and cannot be complex.

2) If f be a function such that f (xy) = f (x). f (y), " yÃŽR and f (1+x) = 1+x (1+ g(x)). Where_{ } , find the value of _{ }

Solution:

We know,

_{ }

_{ }

To find _{ }

_{ }