NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability Miscellaneous Exercise
Page No 191:
Question 1:

Answer:

Using chain rule, we obtain

Question 2:

Answer:

Question 3:

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

Question 4:

Answer:

Using chain rule, we obtain

Question 5:

Answer:

Question 6:

Answer:

Therefore, equation (1) becomes

Question 7:

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

Question 8:

Answer:

By using chain rule, we obtain

Question 9:

Answer:

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

Question 10:



Answer:


Differentiating both sides with respect to x, we obtain


Differentiating both sides with respect to x, we obtain

s = aa
Since a is constant, aa is also a constant.
∴

From (1), (2), (3), (4), and (5), we obtain

Question 11:


Answer:

Differentiating both sides with respect to x, we obtain

Differentiating with respect to x, we obtain

Also,

Differentiating both sides with respect to x, we obtain

Substituting the expressions of
in equation (1), we obtain


Question 12:
Find
, if 


Answer:

Question 13:
Find
, if 


Answer:

Question 14:
If
, for, −1 < x <1 font="" prove="" that="">1>


Answer:
It is given that,


Differentiating both sides with respect to x, we obtain

Hence, proved.
Question 15:
If
, for some
prove that



Answer:
It is given that,

Differentiating both sides with respect to x, we obtain


Hence, proved.
Page No 192:
Question 16:
If
with
prove that



Answer:

Then, equation (1) reduces to .png)
.png)
⇒sina+y-ydydx=cos2a+y⇒dydx=cos2a+ysina
Hence, proved.
Question 17:
If
and
, find 



Answer:


Question 18:
If
, show that
exists for all real x, and find it.


Answer:
It is known that, 

Therefore, when x ≥ 0, 

In this case,
and hence, 


When x < 0, 

In this case,
and hence, 


Thus, for
,
exists for all real x and is given by,



Question 19:
Using mathematical induction prove that
for all positive integers n.

Answer:

For n = 1,

∴P(n) is true for n = 1
Let P(k) is true for some positive integer k.
That is, 

It has to be proved that P(k + 1) is also true.

Thus, P(k + 1) is true whenever P (k) is true.
Therefore, by the principle of mathematical induction, the statement P(n) is true for every positive integer n.
Hence, proved.
Question 20:
Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.
Answer:

Differentiating both sides with respect to x, we obtain

Question 21:
Does there exist a function which is continuos everywhere but not differentiable at exactly two points? Justify your answer ?
Answer:
y=x -∞ It can be seen from the above graph that, the given function is continuos everywhere but not differentiable at exactly two points which are 0 and 1.
Question 22:
If
, prove that 


Answer:

Thus,

Question 23:
If
, show that 


Answer:
It is given that,


