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NCERT solution class 12 chapter 5 Continuity and Differentiability exercise 5.9 mathematics part 1

EXERCISE 5.9


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:

for some constant a and b.

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:

, for some fixed and 

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:

, for 

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, prove that

Answer:

It is given that,

Differentiating both sides with respect to x, we obtain

Hence, proved.

Question 15:

If, for some  prove that

 is a constant independent of a and b.

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 

⇒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, forexists 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           -∞<x≤1    2-x         1≤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,


 

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