{"id":541403,"date":"2021-09-27T04:19:12","date_gmt":"2021-09-27T04:19:12","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=541403"},"modified":"2022-12-26T09:30:51","modified_gmt":"2022-12-26T09:30:51","slug":"rd-sharma-solutions-for-class-12-maths-chapter-10-differentiability","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-12-maths-chapter-10-differentiability\/","title":{"rendered":"RD Sharma Solutions for Class 12 Maths Chapter 10\u2013Differentiability"},"content":{"rendered":"\n

Class 12: Maths Chapter 10 solutions. Complete Class 12 Maths Chapter 10 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 10\u2013Differentiability<\/h2>\n\n\n\n

RD Sharma 12th Maths Chapter 10, Class 12 Maths Chapter 10 solutions<\/p>\n\n\n\n

Exercise 10.1 Page No: 10.10<\/p>\n\n\n\n

1. Show that f (x) = |x \u2013 3| is continuous but not differentiable at x = 3.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

2. Show that f (x) = x 1\/3<\/sup> is not differentiable at x = 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Since, LHD and RHD does not exist at x = 0<\/p>\n\n\n\n

Hence, f(x) is not differentiable at x = 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Now we have to check differentiability of given function at x = 3<\/p>\n\n\n\n

That is LHD (at x = 3) = RHD (at x = 3)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 12<\/p>\n\n\n\n

Since, (LHD at x = 3) = (RHD at x = 3)<\/p>\n\n\n\n

Hence, f(x) is differentiable at x = 3.<\/p>\n\n\n\n

4. Show that the function f is defined as follows<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Is continuous at x = 2, but not differentiable thereat.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Since, LHL = RHL = f (2)<\/p>\n\n\n\n

Hence, F(x) is continuous at x = 2<\/p>\n\n\n\n

Now we have to differentiability at x = 2<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 5<\/p>\n\n\n\n

Since, (RHD at x = 2) \u2260 (LHD at x = 2)<\/p>\n\n\n\n

Hence, f (2) is not differentiable at x = 2.<\/p>\n\n\n\n

5. Discuss the continuity and differentiability of the function f (x) = |x| + |x -1| in the interval of (-1, 2).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

We know that a polynomial and a constant function is continuous and differentiable everywhere. So, f(x) is continuous and differentiable for x \u2208<\/p>\n\n\n\n

(-1, 0) and x \u2208 (0, 1) and (1, 2).<\/p>\n\n\n\n

We need to check continuity and differentiability at x = 0 and x = 1.<\/p>\n\n\n\n

Continuity at x = 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Since, f(x) is continuous at x = 1<\/p>\n\n\n\n

Now we have to check differentiability at x = 0<\/p>\n\n\n\n

For differentiability, LHD (at x = 0) = RHD (at x = 0)<\/p>\n\n\n\n

Differentiability at x = 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Since, (LHD at x = 0) \u2260 (RHD at x = 0)<\/p>\n\n\n\n

So, f(x) is differentiable at x = 0.<\/p>\n\n\n\n

Now we have to check differentiability at x = 1<\/p>\n\n\n\n

For differentiability, LHD (at x = 1) = RHD (at x = 1)<\/p>\n\n\n\n

Differentiability at x = 1<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Since, f(x) is not differentiable at x = 1.<\/p>\n\n\n\n

So, f(x) is continuous on (- 1, 2) but not differentiable at x = 0, 1<\/p>\n\n\n\n

Exercise 10.2 Page No: 10.16<\/p>\n\n\n\n

1. If f is defined by f (x) = x2<\/sup>, find f\u2019 (2).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

2. If f is defined by f (x) = x2<\/sup> \u2013 4x + 7, show that f\u2019 (5) = 2 f\u2019 (7\/2)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Hence the proof.<\/p>\n\n\n\n

3. Show that the derivative of the function f is given by f (x) = 2x3<\/sup> \u2013 9x2 <\/sup>+ 12 x + 9, at x = 1 and x = 2 are equal.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We are given with a polynomial function f(x) = 2x3<\/sup> \u2013 9x2<\/sup> + 12x + 9, and we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

4. If for the function \u00d8 (x) = \u03bb x2<\/sup> + 7x \u2013 4, \u00d8\u2019 (5) = 97, find \u03bb.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have to find the value of \u03bb given in the real function and we are given with the differentiability of the function f(x) = \u03bbx2<\/sup> + 7x \u2013 4 at x = 5 which is f \u2018(5) = 97, so we will adopt the same process but with a little variation.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

5. If f (x) = x3<\/sup> + 7x2<\/sup> + 8x \u2013 9, find f\u2019 (4).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We are given with a polynomial function f(x) = x3<\/sup> + 7x2<\/sup> + 8x \u2013 9, and we have to find whether it is derivable at x = 4 or not,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

6. Find the derivative of the function f defined by f (x) = mx + c at x = 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We are given with a polynomial function f(x) = mx + c, and we have to find whether it is derivable at x = 0 or not,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

This is the derivative of a function at x = 0, and also this is the derivative of this function at every value of x.<\/p>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 10: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 10\u2013Differentiability<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 12 Maths Chapter 10\u2013Differentiability PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 12 Maths :<\/strong><\/h2>\n\n\n\n