Differentiability MCQ Quiz in मल्याळम - Objective Question with Answer for Differentiability - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

A function f(x) is said to be increasing in the given interval if it’s first order differential  holds for every point in the given interval.

Calculation:

Function I:

Therefore, the function is non increasing.        

Function II:

Therefore, this function is also non increasing.

Furthermore, since cosine function is periodic function, it cannot be strictly increasing in the given range

Function III:

0\)

Therefore, the function is increasing in the given interval.

Therefore  function is the only increasing everywhere in [0, 1]

Hence Option(1) is the correct answer.

Both the functions are continuous in  and differentiable in .

By Cauchy's Mean Value theorem we should have 

Therefore  can be any value between . 

Important Points

Since LHS = RHS = -1 This is true for any value of c.

But in Question specific range is mentioned 9i.e. [1,2] ). So "c" can be any value in 1 & 2. 

Concept:

Formula used:

Calculation:

f(x) = x2  +  4x + 3

Differentiate f(x) with respect to 'x'.

⇒ f'(x) = 2x + 4

Put x = 1,

So, f'(1) = 2 × 1 + 4 = 6

∴The value of f'(1) is 6.

Concept:

For a function to be monotonically increasing over an interval, its slope must always be positive in the interval i.e.

f'(x) > 0

For a function to be monotonically decreasing over an interval, its slope must always be negative over that interval. i.e.

f'(x)

Application:

Given:

The slope of the function will be:

In the given interval of -3 to 3, we observe that the slope is always positive.

∴ The given function f(x) is monotonically increasing in the interval -3 to 3

Explanation

From mean value theorem,

we know that 

 Here, a = 0, b = 2

 and 

 Substituting the values 

 = f'(x), x ϵ[0, 2]

 f(2) = 2 × f'(x) + f(0) 

f(2) = 2 ×  + 4

lower bound of f(2) =

 = 1 + 4 = 5

Concept:

Jacobian of (y1, y2, y3) w.r.t. (x1, x2, x3) is equal to

Calculation:

Given:

x = uv and 

We know that

Rolle’s mean value theorem:

Suppose f(x) is a function that satisfies all of the following

1. f(x) is continuous on the closed interval [a, b]

2. f(x) is differentiable on the closed interval (a, b)

3. f(a) = f(b)

Then there is a number c such that a

Let h(x) = f(x) – 2g(x)

Given that f(x) and g(x) are continuous as well as differentiable

Hence h(x) is also continuous as well as differentiable.

h(0) = f(0) – 2g(0) = 2 – 2(0) = 2

h(1) = f(1) – 2g(1) = 6 – 2(2) = 2

h(0) = h(1)

hence h(x) satisfies all the conditions of Rolle’s theorem

Now, h'(c) = 0

⇒ f'(c) – 2g'(c) = 0

⇒ f'(c) = 2g'(c)

Concept:

Continuity of a function:

We say f(x) is continuous at x = c if

LHL = RHL = value of f(c)

i.e., 

Differentiability of a Function:

A function f(x) is differentiable at a point x = a, in its domain if its derivative is continuous at a.

This means that f'(a) must exist, or equivalently:

 .

Calculation:-

To check the continuity,

∴ the function is continuous at x = 0

hence, the function can be differentiable or not differentiable.

To check the differentiability

LHD = 

= 0

RHD = 

= 4

∴ LHD ≠ RHD

Concept:

If x and y are independent variables, then  and 

Calculation:

Given:

r = x2 + y - z     ----(1)

z3 - xy + yz + y3 = 1      ----(2)

Since y is an independent variables, derivative of ‘y’ w.r.t. ‘x’ is .

From equation (1):

Differentiate w.r.t to 'x'

      ----(3)

From equation (2):

Differentiate w.r.t to 'x'

z³ – xy + yz + y3 = 1

Differentiate w.r.t x

     ----(4)

Substitute (4) in (3)

At, (1, -1, 1)

∴ 2.5 is the required answer.

Concept:

Function f(x) is laid to be differentiable at x = a, if

 exist.

And value of this limit is called derivative of f(x) at x = a & it is denoted by f’(a)

i.e.  

Now,

LHD 

RHD  

Shortcut method to find LHD and RHD,

If LHL   &  RHL  

Then,

LHD = LHL of  

RHD = RHL of  

Calculation:

\frac{2}{3}}\\ {0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x = \frac{2}{3}}\\ {\left( {2 - 3x} \right)\;\;\;\;\;\;x  

Exact value of f(x) = 0

L.H.L.  = 0

R.H.L  = 0

∵ L.H.L = R.H.L = Exact value

∴ f(x) is continuous at 

Check for differentiability;

\frac{2}{3}}\\ {0\;\;\;\;\;\;\;;\;\;\;\;\;\;\;x = 0}\\ {2 - 3x\;\;\;;\;\;x  

\frac{2}{3}\;}\\ {0\;\;\;\;\;\;\;\;\;;\;\;\;\;\;\;\;x = \frac{2}{3}}\\ { - 3\;\;\;\;\;\;\;\;;\;\;\;\;\;\;\;x  

∵ LHD ≠ RHD

∴ at , f’(x) is Non-Differentiable.

Method II:

Graph of f(x) = |2 – 3x| is

From the graph, at , function is continuous but since there is a kink at  , the function is not Differentiable at