{"id":542044,"date":"2021-09-27T10:41:19","date_gmt":"2021-09-27T10:41:19","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=542044"},"modified":"2022-12-26T09:36:29","modified_gmt":"2022-12-26T09:36:29","slug":"rd-sharma-solutions-for-class-12-maths-chapter-15-mean-value-theorems","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-12-maths-chapter-15-mean-value-theorems\/","title":{"rendered":"RD Sharma Solutions for Class 12 Maths Chapter 15\u2013Mean Value Theorems"},"content":{"rendered":"\n

Class 12: Maths Chapter 15 solutions. Complete Class 12 Maths Chapter 15 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 15\u2013Mean Value Theorems<\/h2>\n\n\n\n

RD Sharma 12th Maths Chapter 15, Class 12 Maths Chapter 15 solutions<\/p>\n\n\n\n

Exercise 15.1 Page No: 15.9<\/p>\n\n\n\n

1. Discuss the applicability of Rolle\u2019s Theorem for the following functions on the indicated intervals:<\/strong><\/p>\n\n\n\n

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

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

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

(ii) f (x) = [x] for -1 \u2264 x \u2264 1, where [x] denotes the greatest integer not exceeding x<\/strong><\/p>\n\n\n\n

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

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

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

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

(iv) f (x) = 2x2<\/sup> \u2013 5x + 3 on [1, 3]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = 2x2<\/sup> \u2013 5x + 3 on [1, 3]<\/p>\n\n\n\n

Since given function f is a polynomial. So, it is continuous and differentiable everywhere.<\/p>\n\n\n\n

Now, we find the values of function at the extreme values.<\/p>\n\n\n\n

\u21d2 f (1) = 2(1)2<\/sup>\u20135(1) + 3<\/p>\n\n\n\n

\u21d2 f (1) = 2 \u2013 5 + 3<\/p>\n\n\n\n

\u21d2 f (1) = 0\u2026\u2026 (1)<\/p>\n\n\n\n

\u21d2 f (3) = 2(3)2<\/sup>\u20135(3) + 3<\/p>\n\n\n\n

\u21d2 f (3) = 2(9)\u201315 + 3<\/p>\n\n\n\n

\u21d2 f (3) = 18 \u2013 12<\/p>\n\n\n\n

\u21d2 f (3) = 6\u2026\u2026 (2)<\/p>\n\n\n\n

From (1) and (2), we can say that, f (1) \u2260 f (3)<\/p>\n\n\n\n

\u2234 Rolle\u2019s Theorem is not applicable for the function f in interval [1, 3].<\/p>\n\n\n\n

(v) f (x) = x2\/3<\/sup> on [-1, 1]<\/strong><\/p>\n\n\n\n

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

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

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

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

2. Verify the Rolle\u2019s Theorem for each of the following functions on the indicated intervals:<\/strong><\/p>\n\n\n\n

(i) f (x) = x2<\/sup> \u2013 8x + 12 on [2, 6]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = x2<\/sup> \u2013 8x + 12 on [2, 6]<\/p>\n\n\n\n

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.<\/p>\n\n\n\n

Let us find the values at extremes:<\/p>\n\n\n\n

\u21d2 f (2) = 22<\/sup> \u2013 8(2) + 12<\/p>\n\n\n\n

\u21d2 f (2) = 4 \u2013 16 + 12<\/p>\n\n\n\n

\u21d2 f (2) = 0<\/p>\n\n\n\n

\u21d2 f (6) = 62<\/sup> \u2013 8(6) + 12<\/p>\n\n\n\n

\u21d2 f (6) = 36 \u2013 48 + 12<\/p>\n\n\n\n

\u21d2 f (6) = 0<\/p>\n\n\n\n

\u2234 f (2) = f(6), Rolle\u2019s theorem applicable for function f on [2,6].<\/p>\n\n\n\n

Now we have to find the derivative of f(x)<\/p>\n\n\n\n

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

(ii) f(x) = x2<\/sup> \u2013 4x + 3 on [1, 3]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = x2<\/sup> \u2013 4x + 3 on [1, 3]<\/p>\n\n\n\n

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R. Let us find the values at extremes:<\/p>\n\n\n\n

\u21d2 f (1) = 12<\/sup> \u2013 4(1) + 3<\/p>\n\n\n\n

\u21d2 f (1) = 1 \u2013 4 + 3<\/p>\n\n\n\n

\u21d2 f (1) = 0<\/p>\n\n\n\n

\u21d2 f (3) = 32<\/sup> \u2013 4(3) + 3<\/p>\n\n\n\n

\u21d2 f (3) = 9 \u2013 12 + 3<\/p>\n\n\n\n

\u21d2 f (3) = 0<\/p>\n\n\n\n

\u2234 f (1) = f(3), Rolle\u2019s theorem applicable for function \u2018f\u2019 on [1,3].<\/p>\n\n\n\n

Let\u2019s find the derivative of f(x)<\/p>\n\n\n\n

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

\u21d2 f\u2019(x) = 2x \u2013 4<\/p>\n\n\n\n

We have f\u2019(c) = 0, c \u03f5 (1, 3), from the definition of Rolle\u2019s Theorem.<\/p>\n\n\n\n

\u21d2 f\u2019(c) = 0<\/p>\n\n\n\n

\u21d2 2c \u2013 4 = 0<\/p>\n\n\n\n

\u21d2 2c = 4<\/p>\n\n\n\n

\u21d2 c = 4\/2<\/p>\n\n\n\n

\u21d2 C = 2 \u03f5 (1, 3)<\/p>\n\n\n\n

\u2234 Rolle\u2019s Theorem is verified.<\/p>\n\n\n\n

(iii) f (x) = (x \u2013 1) (x \u2013 2)2<\/sup> on [1, 2]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = (x \u2013 1) (x \u2013 2)2<\/sup> on [1, 2]<\/p>\n\n\n\n

Since, given function f is a polynomial it is continuous and differentiable everywhere that is on R.<\/p>\n\n\n\n

Let us find the values at extremes:<\/p>\n\n\n\n

\u21d2 f (1) = (1 \u2013 1) (1 \u2013 2)2<\/sup><\/p>\n\n\n\n

\u21d2 f (1) = 0(1)2<\/sup><\/p>\n\n\n\n

\u21d2 f (1) = 0<\/p>\n\n\n\n

\u21d2 f (2) = (2 \u2013 1)(2 \u2013 2)2<\/sup><\/p>\n\n\n\n

\u21d2 f (2) = 02<\/sup><\/p>\n\n\n\n

\u21d2 f (2) = 0<\/p>\n\n\n\n

\u2234 f (1) = f (2), Rolle\u2019s Theorem applicable for function \u2018f\u2019 on [1, 2].<\/p>\n\n\n\n

Let\u2019s find the derivative of f(x)<\/p>\n\n\n\n

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

(iv) f (x) = x (x \u2013 1)2<\/sup> on [0, 1]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = x(x \u2013 1)2<\/sup> on [0, 1]<\/p>\n\n\n\n

Since, given function f is a polynomial it is continuous and differentiable everywhere that is, on R.<\/p>\n\n\n\n

Let us find the values at extremes<\/p>\n\n\n\n

\u21d2 f (0) = 0 (0 \u2013 1)2<\/sup><\/p>\n\n\n\n

\u21d2 f (0) = 0<\/p>\n\n\n\n

\u21d2 f (1) = 1 (1 \u2013 1)2<\/sup><\/p>\n\n\n\n

\u21d2 f (1) = 02<\/sup><\/p>\n\n\n\n

\u21d2 f (1) = 0<\/p>\n\n\n\n

\u2234 f (0) = f (1), Rolle\u2019s theorem applicable for function \u2018f\u2019 on [0,1].<\/p>\n\n\n\n

Let\u2019s find the derivative of f(x)<\/p>\n\n\n\n

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

(v) f (x) = (x2 <\/sup>\u2013 1) (x \u2013 2) on [-1, 2]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = (x2<\/sup> \u2013 1) (x \u2013 2) on [\u2013 1, 2]<\/p>\n\n\n\n

Since, given function f is a polynomial it is continuous and differentiable everywhere that is on R.<\/p>\n\n\n\n

Let us find the values at extremes:<\/p>\n\n\n\n

\u21d2 f ( \u2013 1) = (( \u2013 1)2<\/sup> \u2013 1)( \u2013 1 \u2013 2)<\/p>\n\n\n\n

\u21d2 f ( \u2013 1) = (1 \u2013 1)( \u2013 3)<\/p>\n\n\n\n

\u21d2 f ( \u2013 1) = (0)( \u2013 3)<\/p>\n\n\n\n

\u21d2 f ( \u2013 1) = 0<\/p>\n\n\n\n

\u21d2 f (2) = (22<\/sup> \u2013 1)(2 \u2013 2)<\/p>\n\n\n\n

\u21d2 f (2) = (4 \u2013 1)(0)<\/p>\n\n\n\n

\u21d2 f (2) = 0<\/p>\n\n\n\n

\u2234 f (\u2013 1) = f (2), Rolle\u2019s theorem applicable for function f on [ \u2013 1,2].<\/p>\n\n\n\n

Let\u2019s find the derivative of f(x)<\/p>\n\n\n\n

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

f\u2019(x) = 3x2<\/sup> \u2013 4x \u2013 1<\/p>\n\n\n\n

We have f\u2019(c) = 0 c \u2208 (-1, 2), from the definition of Rolle\u2019s Theorem<\/p>\n\n\n\n

f'(c) = 0<\/p>\n\n\n\n

3c2<\/sup> \u2013 4c \u2013 1 = 0<\/p>\n\n\n\n

c = 4 \u00b1 \u221a[(-4)2<\/sup> \u2013 (4 x 3 x -1)]\/ (2 x 3) [Using the Quadratic Formula]<\/p>\n\n\n\n

c = 4 \u00b1 \u221a[16 + 12]\/ 6<\/p>\n\n\n\n

c = (4 \u00b1 \u221a28)\/ 6<\/p>\n\n\n\n

c = (4 \u00b1 2\u221a7)\/ 6<\/p>\n\n\n\n

c = (2 \u00b1\u221a7)\/ 3 = 1.5 \u00b1 \u221a7\/3<\/p>\n\n\n\n

c = 1.5 + \u221a7\/3 or 1.5 \u2013 \u221a7\/3<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

c = 1.5 \u2013 \u221a7\/3 since c \u2208 (-1, 2)<\/p>\n\n\n\n

\u2234 Rolle\u2019s Theorem is verified.<\/p>\n\n\n\n

(vi) f (x) = x (x \u2013 4)2<\/sup> on [0, 4]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = x (x \u2013 4)2<\/sup> on [0, 4]<\/p>\n\n\n\n

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.<\/p>\n\n\n\n

Let us find the values at extremes:<\/p>\n\n\n\n

\u21d2 f (0) = 0(0 \u2013 4)2<\/sup><\/p>\n\n\n\n

\u21d2 f (0) = 0<\/p>\n\n\n\n

\u21d2 f (4) = 4(4 \u2013 4)2<\/sup><\/p>\n\n\n\n

\u21d2 f (4) = 4(0)2<\/sup><\/p>\n\n\n\n

\u21d2 f (4) = 0<\/p>\n\n\n\n

\u2234 f (0) = f (4), Rolle\u2019s theorem applicable for function \u2018f\u2019 on [0,4].<\/p>\n\n\n\n

Let\u2019s find the derivative of f(x):<\/p>\n\n\n\n

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

(vii) f (x) = x (x \u2013 2)2<\/sup> on [0, 2]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = x (x \u2013 2)2<\/sup> on [0, 2]<\/p>\n\n\n\n

Since, given function f is a polynomial it is continuous and differentiable everywhere that is on R.<\/p>\n\n\n\n

Let us find the values at extremes:<\/p>\n\n\n\n

\u21d2 f (0) = 0(0 \u2013 2)2<\/sup><\/p>\n\n\n\n

\u21d2 f (0) = 0<\/p>\n\n\n\n

\u21d2 f (2) = 2(2 \u2013 2)2<\/sup><\/p>\n\n\n\n

\u21d2 f (2) = 2(0)2<\/sup><\/p>\n\n\n\n

\u21d2 f (2) = 0<\/p>\n\n\n\n

f (0) = f(2), Rolle\u2019s theorem applicable for function f on [0,2].<\/p>\n\n\n\n

Let\u2019s find the derivative of f(x)<\/p>\n\n\n\n

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

c = 12\/6 or 4\/6<\/p>\n\n\n\n

c = 2 or 2\/3<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

c = 2\/3 since c \u2208 (0, 2)<\/p>\n\n\n\n

\u2234 Rolle\u2019s Theorem is verified.<\/p>\n\n\n\n

(viii) f (x) = x2<\/sup> + 5x + 6 on [-3, -2]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = x2<\/sup> + 5x + 6 on [\u2013 3, \u2013 2]<\/p>\n\n\n\n

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R. Let us find the values at extremes:<\/p>\n\n\n\n

\u21d2 f ( \u2013 3) = ( \u2013 3)2<\/sup> + 5( \u2013 3) + 6<\/p>\n\n\n\n

\u21d2 f ( \u2013 3) = 9 \u2013 15 + 6<\/p>\n\n\n\n

\u21d2 f ( \u2013 3) = 0<\/p>\n\n\n\n

\u21d2 f ( \u2013 2) = ( \u2013 2)2<\/sup> + 5( \u2013 2) + 6<\/p>\n\n\n\n

\u21d2 f ( \u2013 2) = 4 \u2013 10 + 6<\/p>\n\n\n\n

\u21d2 f ( \u2013 2) = 0<\/p>\n\n\n\n

\u2234 f (\u2013 3) = f( \u2013 2), Rolle\u2019s theorem applicable for function f on [ \u2013 3, \u2013 2].<\/p>\n\n\n\n

Let\u2019s find the derivative of f(x):<\/p>\n\n\n\n

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

3. Verify the Rolle\u2019s Theorem for each of the following functions on the indicated intervals:<\/strong><\/p>\n\n\n\n

(i) f (x) = cos 2 (x \u2013 \u03c0\/4) on [0, \u03c0\/2]<\/strong><\/p>\n\n\n\n

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

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

(ii) f (x) = sin 2x on [0, \u03c0\/2]<\/strong><\/p>\n\n\n\n

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

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

(iii) f (x) = cos 2x on [-\u03c0\/4, \u03c0\/4]<\/strong><\/p>\n\n\n\n

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

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

sin 2c = 0<\/p>\n\n\n\n

\u21d2 2c = 0<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

c = 0 as c \u2208 (-\u03c0\/4, \u03c0\/4)<\/p>\n\n\n\n

\u2234 Rolle\u2019s Theorem is verified.<\/p>\n\n\n\n

(iv) f (x) = ex<\/sup> sin x on [0, \u03c0]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = ex <\/sup>sin x on [0, \u03c0]<\/p>\n\n\n\n

We know that exponential and sine functions are continuous and differentiable on R. Let\u2019s find the values of the function at an extreme,<\/p>\n\n\n\n

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

(v) f (x) = ex<\/sup> cos x on [-\u03c0\/2, \u03c0\/2]<\/strong><\/p>\n\n\n\n

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

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

(vi) f (x) = cos 2x on [0, \u03c0]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = cos 2x on [0, \u03c0]<\/p>\n\n\n\n

We know that cosine function is continuous and differentiable on R. Let\u2019s find the values of function at extreme,<\/p>\n\n\n\n

\u21d2 f (0) = cos2(0)<\/p>\n\n\n\n

\u21d2 f (0) = cos(0)<\/p>\n\n\n\n

\u21d2 f (0) = 1<\/p>\n\n\n\n

\u21d2 f (\u03c0) = cos2(
\"https:\/\/gradeup-question-images.grdp.co\/liveData\/PROJ24056\/1544073308542287.png\")<\/p>\n\n\n\n

\u21d2 f (\u03c0) = cos(2 \u03c0)<\/p>\n\n\n\n

\u21d2 f (\u03c0) = 1<\/p>\n\n\n\n

We have f (0) = f (\u03c0), so there exist a c belongs to (0, \u03c0) such that f\u2019(c) = 0.<\/p>\n\n\n\n

Let\u2019s find the derivative of f(x)<\/p>\n\n\n\n

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

sin 2c = 0<\/p>\n\n\n\n

So, 2c = 0 or \u03c0<\/p>\n\n\n\n

c = 0 or \u03c0\/2<\/p>\n\n\n\n

But,<\/p>\n\n\n\n

c = \u03c0\/2 as c \u2208 (0, \u03c0)<\/p>\n\n\n\n

Hence, Rolle\u2019s Theorem is verified.<\/p>\n\n\n\n

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

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

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

(viii) f (x) = sin 3x on [0, \u03c0]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = sin3x on [0, \u03c0]<\/p>\n\n\n\n

We know that sine function is continuous and differentiable on R. Let\u2019s find the values of function at extreme,<\/p>\n\n\n\n

\u21d2 f (0) = sin3(0)<\/p>\n\n\n\n

\u21d2 f (0) = sin0<\/p>\n\n\n\n

\u21d2 f (0) = 0<\/p>\n\n\n\n

\u21d2 f (\u03c0) = sin3(\u03c0)<\/p>\n\n\n\n

\u21d2 f (\u03c0) = sin(3 \u03c0)<\/p>\n\n\n\n

\u21d2 f (\u03c0) = 0<\/p>\n\n\n\n

We have f (0) = f (\u03c0), so there exist a c belongs to (0, \u03c0) such that f\u2019(c) = 0.<\/p>\n\n\n\n

Let\u2019s find the derivative of f(x)<\/p>\n\n\n\n

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

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

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

(x) f (x) = log (x2<\/sup> + 2) \u2013 log 3 on [-1, 1]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = log(x2<\/sup> + 2) \u2013 log3 on [\u2013 1, 1]<\/p>\n\n\n\n

We know that logarithmic function is continuous and differentiable in its own domain. We check the values of the function at the extreme,<\/p>\n\n\n\n

\u21d2 f (\u2013 1) = log((\u2013 1)2<\/sup> + 2) \u2013 log 3<\/p>\n\n\n\n

\u21d2 f (\u2013 1) = log (1 + 2) \u2013 log 3<\/p>\n\n\n\n

\u21d2 f (\u2013 1) = log 3 \u2013 log 3<\/p>\n\n\n\n

\u21d2 f ( \u2013 1) = 0<\/p>\n\n\n\n

\u21d2 f (1) = log (12<\/sup> + 2) \u2013 log 3<\/p>\n\n\n\n

\u21d2 f (1) = log (1 + 2) \u2013 log 3<\/p>\n\n\n\n

\u21d2 f (1) = log 3 \u2013 log 3<\/p>\n\n\n\n

\u21d2 f (1) = 0<\/p>\n\n\n\n

We have got f (\u2013 1) = f (1). So, there exists a c such that c \u03f5 (\u2013 1, 1) such that f\u2019(c) = 0.<\/p>\n\n\n\n

Let\u2019s find the derivative of the function f,<\/p>\n\n\n\n

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

(xi) f (x) = sin x + cos x on [0, \u03c0\/2]<\/strong><\/p>\n\n\n\n

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

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

(xii) f (x) = 2 sin x + sin 2x on [0, \u03c0]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = 2sinx + sin2x on [0, \u03c0]<\/p>\n\n\n\n

We know that sine function continuous and differentiable over R.<\/p>\n\n\n\n

Let\u2019s check the values of function f at the extremes<\/p>\n\n\n\n

\u21d2 f (0) = 2sin(0) + sin2(0)<\/p>\n\n\n\n

\u21d2 f (0) = 2(0) + 0<\/p>\n\n\n\n

\u21d2 f (0) = 0<\/p>\n\n\n\n

\u21d2 f (\u03c0) = 2sin(\u03c0) + sin2(\u03c0)<\/p>\n\n\n\n

\u21d2 f (\u03c0) = 2(0) + 0<\/p>\n\n\n\n

\u21d2 f (\u03c0) = 0<\/p>\n\n\n\n

We have f (0) = f (\u03c0), so there exist a c belongs to (0, \u03c0) such that f\u2019(c) = 0.<\/p>\n\n\n\n

Let\u2019s find the derivative of function f.<\/p>\n\n\n\n

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

\u21d2 f\u2019(x) = 2cosx + 2cos2x<\/p>\n\n\n\n

\u21d2 f\u2019(x) = 2cosx + 2(2cos2<\/sup>x \u2013 1)<\/p>\n\n\n\n

\u21d2 f\u2019(x) = 4 cos2<\/sup>x + 2 cos x \u2013 2<\/p>\n\n\n\n

We have f\u2019(c) = 0,<\/p>\n\n\n\n

\u21d2 4cos2<\/sup>c + 2 cos c \u2013 2 = 0<\/p>\n\n\n\n

\u21d2 2cos2<\/sup>c + cos c \u2013 1 = 0<\/p>\n\n\n\n

\u21d2 2cos2<\/sup>c + 2 cos c \u2013 cos c \u2013 1 = 0<\/p>\n\n\n\n

\u21d2 2 cos c (cos c + 1) \u2013 1 (cos c + 1) = 0<\/p>\n\n\n\n

\u21d2 (2cos c \u2013 1) (cos c + 1) = 0<\/p>\n\n\n\n

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

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

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

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

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

(xv) f (x) = 4sin x<\/sup> on [0, \u03c0]<\/strong><\/p>\n\n\n\n

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

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

(xvi) f (x) = x2<\/sup> \u2013 5x + 4 on [0, \u03c0\/6]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = x2<\/sup> \u2013 5x + 4 on [1, 4]<\/p>\n\n\n\n

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.<\/p>\n\n\n\n

Let us find the values at extremes<\/p>\n\n\n\n

\u21d2 f (1) = 12<\/sup> \u2013 5(1) + 4<\/p>\n\n\n\n

\u21d2 f (1) = 1 \u2013 5 + 4<\/p>\n\n\n\n

\u21d2 f (1) = 0<\/p>\n\n\n\n

\u21d2 f (4) = 42<\/sup> \u2013 5(4) + 4<\/p>\n\n\n\n

\u21d2 f (4) = 16 \u2013 20 + 4<\/p>\n\n\n\n

\u21d2 f (4) = 0<\/p>\n\n\n\n

We have f (1) = f (4). So, there exists a c \u03f5 (1, 4) such that f\u2019(c) = 0.<\/p>\n\n\n\n

Let\u2019s find the derivative of f(x):<\/p>\n\n\n\n

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

(xvii) f (x) = sin4<\/sup> x + cos4<\/sup> x on [0, \u03c0\/2]<\/strong><\/p>\n\n\n\n

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

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

(xviii) f (x) = sin x \u2013 sin 2x on [0, \u03c0]<\/strong><\/p>\n\n\n\n

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

Given function is f (x) = sin x \u2013 sin2x on [0, \u03c0]<\/p>\n\n\n\n

We know that sine function is continuous and differentiable over R.<\/p>\n\n\n\n

Now we have to check the values of the function \u2018f\u2019 at the extremes.<\/p>\n\n\n\n

\u21d2 f (0) = sin (0)\u2013sin 2(0)<\/p>\n\n\n\n

\u21d2 f (0) = 0 \u2013 sin (0)<\/p>\n\n\n\n

\u21d2 f (0) = 0<\/p>\n\n\n\n

\u21d2 f (\u03c0) = sin(\u03c0) \u2013 sin2(\u03c0)<\/p>\n\n\n\n

\u21d2 f (\u03c0) = 0 \u2013 sin(2\u03c0)<\/p>\n\n\n\n

\u21d2 f (\u03c0) = 0<\/p>\n\n\n\n

We have f (0) = f (\u03c0). So, there exists a c \u03f5 (0, \u03c0) such that f\u2019(c) = 0.<\/p>\n\n\n\n

Now we have to find the derivative of the function \u2018f\u2019<\/p>\n\n\n\n

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

4. Using Rolle\u2019s Theorem, find points on the curve y = 16 \u2013 x2<\/sup>, x \u2208 [-1, 1], where tangent is parallel to x \u2013 axis.<\/strong><\/p>\n\n\n\n

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

Given function is y = 16 \u2013 x2<\/sup>, x \u03f5 [\u2013 1, 1]<\/p>\n\n\n\n

We know that polynomial function is continuous and differentiable over R.<\/p>\n\n\n\n

Let us check the values of \u2018y\u2019 at extremes<\/p>\n\n\n\n

\u21d2 y (\u2013 1) = 16 \u2013 (\u2013 1)2<\/sup><\/p>\n\n\n\n

\u21d2 y (\u2013 1) = 16 \u2013 1<\/p>\n\n\n\n

\u21d2 y (\u2013 1) = 15<\/p>\n\n\n\n

\u21d2 y (1) = 16 \u2013 (1)2<\/sup><\/p>\n\n\n\n

\u21d2 y (1) = 16 \u2013 1<\/p>\n\n\n\n

\u21d2 y (1) = 15<\/p>\n\n\n\n

We have y (\u2013 1) = y (1). So, there exists a c \u03f5 (\u2013 1, 1) such that f\u2019(c) = 0.<\/p>\n\n\n\n

We know that for a curve g, the value of the slope of the tangent at a point r is given by g\u2019(r).<\/p>\n\n\n\n

Now we have to find the derivative of curve y<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

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

\u21d2 y\u2019 = \u2013 2x<\/p>\n\n\n\n

We have y\u2019(c) = 0<\/p>\n\n\n\n

\u21d2 \u2013 2c = 0<\/p>\n\n\n\n

\u21d2 c = 0 \u03f5 (\u2013 1, 1)<\/p>\n\n\n\n

Value of y at x = 1 is<\/p>\n\n\n\n

\u21d2 y = 16 \u2013 02<\/sup><\/p>\n\n\n\n

\u21d2 y = 16<\/p>\n\n\n\n

\u2234 The point at which the curve y has a tangent parallel to x \u2013 axis (since the slope of x \u2013 axis is 0) is (0, 16).<\/p>\n\n\n\n

Exercise 15.2 Page No: 15.17<\/p>\n\n\n\n

1. Verify Lagrange\u2019s mean value theorem for the following functions on the indicated intervals. In each case find a point \u2018c\u2019 in the indicated interval as stated by the Lagrange\u2019s mean value theorem:<\/strong><\/p>\n\n\n\n

(i) f (x) = x2<\/sup> \u2013 1 on [2, 3]<\/strong><\/p>\n\n\n\n

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

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

(ii) f (x) = x3<\/sup> \u2013 2x2<\/sup> \u2013 x + 3 on [0, 1]<\/strong><\/p>\n\n\n\n

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

Given f (x) = x3<\/sup> \u2013 2x2<\/sup> \u2013 x + 3 on [0, 1]<\/p>\n\n\n\n

Every polynomial function is continuous everywhere on (\u2212\u221e, \u221e) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [0, 1] and differentiable in (0, 1). So both the necessary conditions of Lagrange\u2019s mean value theorem is satisfied.<\/p>\n\n\n\n

Therefore, there exist a point c \u2208 (0, 1) such that:<\/p>\n\n\n\n

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

f (x) = x3<\/sup> \u2013 2x2<\/sup> \u2013 x + 3<\/p>\n\n\n\n

Differentiating with respect to x<\/p>\n\n\n\n

f\u2019(x) = 3x2<\/sup> \u2013 2(2x) \u2013 1<\/p>\n\n\n\n

= 3x2<\/sup> \u2013 4x \u2013 1<\/p>\n\n\n\n

For f\u2019(c), put the value of x=c in f\u2019(x)<\/p>\n\n\n\n

f\u2019(c)= 3c2<\/sup> \u2013 4c \u2013 1<\/p>\n\n\n\n

For f (1), put the value of x = 1 in f(x)<\/p>\n\n\n\n

f (1)= (1)3<\/sup> \u2013 2(1)2<\/sup> \u2013 (1) + 3<\/p>\n\n\n\n

= 1 \u2013 2 \u2013 1 + 3<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

For f (0), put the value of x=0 in f(x)<\/p>\n\n\n\n

f (0)= (0)3<\/sup> \u2013 2(0)2<\/sup> \u2013 (0) + 3<\/p>\n\n\n\n

= 0 \u2013 0 \u2013 0 + 3<\/p>\n\n\n\n

= 3<\/p>\n\n\n\n

\u2234 f\u2019(c) = f(1) \u2013 f(0)<\/p>\n\n\n\n

\u21d2 3c2<\/sup> \u2013 4c \u2013 1 = 1 \u2013 3<\/p>\n\n\n\n

\u21d2 3c2<\/sup> \u2013 4c = 1 + 1 \u2013 3<\/p>\n\n\n\n

\u21d2 3c2<\/sup> \u2013 4c = \u2013 1<\/p>\n\n\n\n

\u21d2 3c2<\/sup> \u2013 4c + 1 = 0<\/p>\n\n\n\n

\u21d2 3c2<\/sup> \u2013 3c \u2013 c + 1 = 0<\/p>\n\n\n\n

\u21d2 3c(c \u2013 1) \u2013 1(c \u2013 1) = 0<\/p>\n\n\n\n

\u21d2 (3c \u2013 1) (c \u2013 1) = 0<\/p>\n\n\n\n

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

(iii) f (x) = x (x \u2013 1) on [1, 2]<\/strong><\/p>\n\n\n\n

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

Given f (x) = x (x \u2013 1) on [1, 2]<\/p>\n\n\n\n

= x2<\/sup> \u2013 x<\/p>\n\n\n\n

Every polynomial function is continuous everywhere on (\u2212\u221e, \u221e) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [1, 2] and differentiable in (1, 2). So both the necessary conditions of Lagrange\u2019s mean value theorem is satisfied.<\/p>\n\n\n\n

Therefore, there exist a point c \u2208 (1, 2) such that:<\/p>\n\n\n\n

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

f (x) = x2<\/sup> \u2013 x<\/p>\n\n\n\n

Differentiating with respect to x<\/p>\n\n\n\n

f\u2019(x) = 2x \u2013 1<\/p>\n\n\n\n

For f\u2019(c), put the value of x=c in f\u2019(x):<\/p>\n\n\n\n

f\u2019(c)= 2c \u2013 1<\/p>\n\n\n\n

For f (2), put the value of x = 2 in f(x)<\/p>\n\n\n\n

f (2) = (2)2<\/sup> \u2013 2<\/p>\n\n\n\n

= 4 \u2013 2<\/p>\n\n\n\n

= 2<\/p>\n\n\n\n

For f (1), put the value of x = 1 in f(x):<\/p>\n\n\n\n

f (1)= (1)2<\/sup> \u2013 1<\/p>\n\n\n\n

= 1 \u2013 1<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

\u2234 f\u2019(c) = f(2) \u2013 f(1)<\/p>\n\n\n\n

\u21d2 2c \u2013 1 = 2 \u2013 0<\/p>\n\n\n\n

\u21d2 2c = 2 + 1<\/p>\n\n\n\n

\u21d2 2c = 3<\/p>\n\n\n\n

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

(iv) f (x) = x2<\/sup> \u2013 3x + 2 on [-1, 2]<\/strong><\/p>\n\n\n\n

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

Given f (x) = x2<\/sup> \u2013 3x + 2 on [\u2013 1, 2]<\/p>\n\n\n\n

Every polynomial function is continuous everywhere on (\u2212\u221e, \u221e) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [\u2013 1, 2] and differentiable in (\u2013 1, 2). So both the necessary conditions of Lagrange\u2019s mean value theorem is satisfied.<\/p>\n\n\n\n

Therefore, there exist a point c \u2208 (-1, 2) such that:<\/p>\n\n\n\n

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

f (x) = x2<\/sup> \u2013 3x + 2<\/p>\n\n\n\n

Differentiating with respect to x<\/p>\n\n\n\n

f\u2019(x) = 2x \u2013 3<\/p>\n\n\n\n

For f\u2019(c), put the value of x = c in f\u2019(x):<\/p>\n\n\n\n

f\u2019(c)= 2c \u2013 3<\/p>\n\n\n\n

For f (2), put the value of x = 2 in f(x)<\/p>\n\n\n\n

f (2) = (2)2<\/sup> \u2013 3 (2) + 2<\/p>\n\n\n\n

= 4 \u2013 6 + 2<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

For f (\u2013 1), put the value of x = \u2013 1 in f(x):<\/p>\n\n\n\n

f (\u2013 1) = (\u2013 1)2<\/sup> \u2013 3 (\u2013 1) + 2<\/p>\n\n\n\n

= 1 + 3 + 2<\/p>\n\n\n\n

= 6<\/p>\n\n\n\n

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

\u21d2 2c = 1<\/p>\n\n\n\n

\u21d2 c = \u00bd \u2208 (-1, 2)<\/p>\n\n\n\n

Hence, Lagrange\u2019s mean value theorem is verified.<\/p>\n\n\n\n

(v) f (x) = 2x2<\/sup> \u2013 3x + 1 on [1, 3]<\/strong><\/p>\n\n\n\n

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

Given f (x) = 2x2<\/sup> \u2013 3x + 1 on [1, 3]<\/p>\n\n\n\n

Every polynomial function is continuous everywhere on (\u2212\u221e, \u221e) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [1, 3] and differentiable in (1, 3). So both the necessary conditions of Lagrange\u2019s mean value theorem is satisfied.<\/p>\n\n\n\n

Therefore, there exist a point c \u2208 (1, 3) such that:<\/p>\n\n\n\n

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

f (x) = 2x2<\/sup> \u2013 3x + 1<\/p>\n\n\n\n

Differentiating with respect to x<\/p>\n\n\n\n

f\u2019(x) = 2(2x) \u2013 3<\/p>\n\n\n\n

= 4x \u2013 3<\/p>\n\n\n\n

For f\u2019(c), put the value of x = c in f\u2019(x):<\/p>\n\n\n\n

f\u2019(c)= 4c \u2013 3<\/p>\n\n\n\n

For f (3), put the value of x = 3 in f(x):<\/p>\n\n\n\n

f (3) = 2 (3)2<\/sup> \u2013 3 (3) + 1<\/p>\n\n\n\n

= 2 (9) \u2013 9 + 1<\/p>\n\n\n\n

= 18 \u2013 8 = 10<\/p>\n\n\n\n

For f (1), put the value of x = 1 in f(x):<\/p>\n\n\n\n

f (1) = 2 (1)2<\/sup> \u2013 3 (1) + 1<\/p>\n\n\n\n

= 2 (1) \u2013 3 + 1<\/p>\n\n\n\n

= 2 \u2013 2 = 0<\/p>\n\n\n\n

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

(vi) f (x) = x2<\/sup> \u2013 2x + 4 on [1, 5]<\/strong><\/p>\n\n\n\n

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

Given f (x) = x2<\/sup> \u2013 2x + 4 on [1, 5]<\/p>\n\n\n\n

Every polynomial function is continuous everywhere on (\u2212\u221e, \u221e) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [1, 5] and differentiable in (1, 5). So both the necessary conditions of Lagrange\u2019s mean value theorem is satisfied.<\/p>\n\n\n\n

Therefore, there exist a point c \u2208 (1, 5) such that:<\/p>\n\n\n\n

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

f (x) = x2<\/sup> \u2013 2x + 4<\/p>\n\n\n\n

Differentiating with respect to x:<\/p>\n\n\n\n

f\u2019(x) = 2x \u2013 2<\/p>\n\n\n\n

For f\u2019(c), put the value of x=c in f\u2019(x):<\/p>\n\n\n\n

f\u2019(c)= 2c \u2013 2<\/p>\n\n\n\n

For f (5), put the value of x=5 in f(x):<\/p>\n\n\n\n

f (5)= (5)2<\/sup> \u2013 2(5) + 4<\/p>\n\n\n\n

= 25 \u2013 10 + 4<\/p>\n\n\n\n

= 19<\/p>\n\n\n\n

For f (1), put the value of x = 1 in f(x)<\/p>\n\n\n\n

f (1) = (1)2<\/sup> \u2013 2 (1) + 4<\/p>\n\n\n\n

= 1 \u2013 2 + 4<\/p>\n\n\n\n

= 3<\/p>\n\n\n\n

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

(vii) f (x) = 2x \u2013 x2<\/sup> on [0, 1]<\/strong><\/p>\n\n\n\n

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

Given f (x) = 2x \u2013 x2<\/sup> on [0, 1]<\/p>\n\n\n\n

Every polynomial function is continuous everywhere on (\u2212\u221e, \u221e) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [0, 1] and differentiable in (0, 1). So both the necessary conditions of Lagrange\u2019s mean value theorem is satisfied.<\/p>\n\n\n\n

Therefore, there exist a point c \u2208 (0, 1) such that:<\/p>\n\n\n\n

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

\u21d2 f\u2019(c) = f(1) \u2013 f(0)<\/p>\n\n\n\n

f (x) = 2x \u2013 x2<\/sup><\/p>\n\n\n\n

Differentiating with respect to x:<\/p>\n\n\n\n

f\u2019(x) = 2 \u2013 2x<\/p>\n\n\n\n

For f\u2019(c), put the value of x = c in f\u2019(x):<\/p>\n\n\n\n

f\u2019(c)= 2 \u2013 2c<\/p>\n\n\n\n

For f (1), put the value of x = 1 in f(x):<\/p>\n\n\n\n

f (1)= 2(1) \u2013 (1)2<\/sup><\/p>\n\n\n\n

= 2 \u2013 1<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

For f (0), put the value of x = 0 in f(x):<\/p>\n\n\n\n

f (0) = 2(0) \u2013 (0)2<\/sup><\/p>\n\n\n\n

= 0 \u2013 0<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

f\u2019(c) = f(1) \u2013 f(0)<\/p>\n\n\n\n

\u21d2 2 \u2013 2c = 1 \u2013 0<\/p>\n\n\n\n

\u21d2 \u2013 2c = 1 \u2013 2<\/p>\n\n\n\n

\u21d2 \u2013 2c = \u2013 1<\/p>\n\n\n\n

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

(viii) f (x) = (x \u2013 1) (x \u2013 2) (x \u2013 3)<\/strong><\/p>\n\n\n\n

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

Given f (x) = (x \u2013 1) (x \u2013 2) (x \u2013 3) on [0, 4]<\/p>\n\n\n\n

= (x2<\/sup> \u2013 x \u2013 2x + 2) (x \u2013 3)<\/p>\n\n\n\n

= (x2<\/sup> \u2013 3x + 2) (x \u2013 3)<\/p>\n\n\n\n

= x3<\/sup> \u2013 3x2<\/sup> + 2x \u2013 3x2<\/sup> + 9x \u2013 6<\/p>\n\n\n\n

= x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6 on [0, 4]<\/p>\n\n\n\n

Every polynomial function is continuous everywhere on (\u2212\u221e, \u221e) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [0, 4] and differentiable in (0, 4). So both the necessary conditions of Lagrange\u2019s mean value theorem is satisfied.<\/p>\n\n\n\n

Therefore, there exist a point c \u2208 (0, 4) such that:<\/p>\n\n\n\n

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

f (x) = x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6<\/p>\n\n\n\n

Differentiating with respect to x:<\/p>\n\n\n\n

f\u2019(x) = 3x2<\/sup> \u2013 6(2x) + 11<\/p>\n\n\n\n

= 3x2<\/sup> \u2013 12x + 11<\/p>\n\n\n\n

For f\u2019(c), put the value of x = c in f\u2019(x):<\/p>\n\n\n\n

f\u2019(c) = 3c2<\/sup> \u2013 12c + 11<\/p>\n\n\n\n

For f (4), put the value of x = 4 in f(x):<\/p>\n\n\n\n

f (4) = (4)3<\/sup> \u2013 6(4)2<\/sup> + 11 (4) \u2013 6<\/p>\n\n\n\n

= 64 \u2013 96 + 44 \u2013 6<\/p>\n\n\n\n

= 6<\/p>\n\n\n\n

For f (0), put the value of x = 0 in f(x):<\/p>\n\n\n\n

f (0) = (0)3<\/sup> \u2013 6 (0)2<\/sup> + 11 (0) \u2013 6<\/p>\n\n\n\n

= 0 \u2013 0 + 0 \u2013 6<\/p>\n\n\n\n

= \u2013 6<\/p>\n\n\n\n

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

3c2<\/sup> \u2013 12c + 11 = [6 \u2013 (-6)]\/ 4<\/p>\n\n\n\n

3c2<\/sup> \u2013 12c + 11 = 12\/4<\/p>\n\n\n\n

3c2<\/sup> \u2013 12c + 11 = 3<\/p>\n\n\n\n

3c2<\/sup> \u2013 12c + 8 = 0<\/p>\n\n\n\n

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

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

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

(x) f (x) = tan-1<\/sup>x on [0, 1]<\/strong><\/p>\n\n\n\n

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

Given f (x) = tan \u2013 1<\/sup> x on [0, 1]<\/p>\n\n\n\n

Tan \u2013 1<\/sup> x has unique value for all x between 0 and 1.<\/p>\n\n\n\n

\u2234 f (x) is continuous in [0, 1]<\/p>\n\n\n\n

f (x) = tan \u2013 1<\/sup> x<\/p>\n\n\n\n

Differentiating with respect to x:<\/p>\n\n\n\n

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

x2<\/sup> always has value greater than 0.<\/p>\n\n\n\n

\u21d2 1 + x2<\/sup> > 0<\/p>\n\n\n\n

\u2234 f (x) is differentiable in (0, 1)<\/p>\n\n\n\n

So both the necessary conditions of Lagrange\u2019s mean value theorem is satisfied. Therefore, there exist a point c \u2208 (0, 1) such that:<\/p>\n\n\n\n

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

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

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

(xii) f (x) = x (x + 4)2<\/sup> on [0, 4]<\/strong><\/p>\n\n\n\n

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

Given f (x) = x (x + 4)2<\/sup> on [0, 4]<\/p>\n\n\n\n

= x [(x)2 <\/sup>+ 2 (4) (x) + (4)2<\/sup>]<\/p>\n\n\n\n

= x (x2<\/sup> + 8x + 16)<\/p>\n\n\n\n

= x3<\/sup> + 8x2<\/sup> + 16x on [0, 4]<\/p>\n\n\n\n

Every polynomial function is continuous everywhere on (\u2212\u221e, \u221e) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [0, 4] and differentiable in (0, 4). So both the necessary conditions of Lagrange\u2019s mean value theorem is satisfied. Therefore, there exist a point c \u2208 (0, 4) such that:<\/p>\n\n\n\n

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

f (x) = x3<\/sup> + 8x2<\/sup> + 16x<\/p>\n\n\n\n

Differentiating with respect to x:<\/p>\n\n\n\n

f\u2019(x) = 3x2<\/sup> + 8(2x) + 16<\/p>\n\n\n\n

= 3x2<\/sup> + 16x + 16<\/p>\n\n\n\n

For f\u2019(c), put the value of x = c in f\u2019(x):<\/p>\n\n\n\n

f\u2019(c)= 3c2<\/sup> + 16c + 16<\/p>\n\n\n\n

For f (4), put the value of x = 4 in f(x):<\/p>\n\n\n\n

f (4)= (4)3<\/sup> + 8(4)2<\/sup> + 16(4)<\/p>\n\n\n\n

= 64 + 128 + 64<\/p>\n\n\n\n

= 256<\/p>\n\n\n\n

For f (0), put the value of x = 0 in f(x):<\/p>\n\n\n\n

f (0)= (0)3<\/sup> + 8(0)2<\/sup> + 16(0)<\/p>\n\n\n\n

= 0 + 0 + 0<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

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

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

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

(xiv) f (x) = x2<\/sup> + x \u2013 1 on [0, 4]<\/strong><\/p>\n\n\n\n

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

Given f (x) = x2<\/sup> + x \u2013 1 on [0, 4]<\/p>\n\n\n\n

Every polynomial function is continuous everywhere on (\u2212\u221e, \u221e) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [0, 4] and differentiable in (0, 4). So both the necessary conditions of Lagrange\u2019s mean value theorem is satisfied. Therefore, there exist a point c \u2208 (0, 4) such that:<\/p>\n\n\n\n

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

f (x) = x2<\/sup> + x \u2013 1<\/p>\n\n\n\n

Differentiating with respect to x:<\/p>\n\n\n\n

f\u2019(x) = 2x + 1<\/p>\n\n\n\n

For f\u2019(c), put the value of x = c in f\u2019(x):<\/p>\n\n\n\n

f\u2019(c) = 2c + 1<\/p>\n\n\n\n

For f (4), put the value of x = 4 in f(x):<\/p>\n\n\n\n

f (4)= (4)2<\/sup> + 4 \u2013 1<\/p>\n\n\n\n

= 16 + 4 \u2013 1<\/p>\n\n\n\n

= 19<\/p>\n\n\n\n

For f (0), put the value of x = 0 in f(x):<\/p>\n\n\n\n

f (0) = (0)2<\/sup> + 0 \u2013 1<\/p>\n\n\n\n

= 0 + 0 \u2013 1<\/p>\n\n\n\n

= \u2013 1<\/p>\n\n\n\n

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

(xv) f (x) = sin x \u2013 sin 2x \u2013 x on [0, \u03c0]<\/strong><\/p>\n\n\n\n

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

Given f (x) = sin x \u2013 sin 2x \u2013 x on [0, \u03c0]<\/p>\n\n\n\n

Sin x and cos x functions are continuous everywhere on (\u2212\u221e, \u221e) and differentiable for all arguments. So both the necessary conditions of Lagrange\u2019s mean value theorem is satisfied. Therefore, there exist a point c \u2208 (0, \u03c0) such that:<\/p>\n\n\n\n

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

(xvi) f (x) = x3<\/sup> \u2013 5x2<\/sup> \u2013 3x on [1, 3]<\/strong><\/p>\n\n\n\n

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

Given f (x) = x3<\/sup> \u2013 5x2<\/sup> \u2013 3x on [1, 3]<\/p>\n\n\n\n

Every polynomial function is continuous everywhere on (\u2212\u221e, \u221e) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [1, 3] and differentiable in (1, 3). So both the necessary conditions of Lagrange\u2019s mean value theorem is satisfied.<\/p>\n\n\n\n

Therefore, there exist a point c \u2208 (1, 3) such that:<\/p>\n\n\n\n

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

f (x) = x3<\/sup> \u2013 5x2<\/sup> \u2013 3x<\/p>\n\n\n\n

Differentiating with respect to x:<\/p>\n\n\n\n

f\u2019(x) = 3x2<\/sup> \u2013 5(2x) \u2013 3<\/p>\n\n\n\n

= 3x2<\/sup> \u2013 10x \u2013 3<\/p>\n\n\n\n

For f\u2019(c), put the value of x=c in f\u2019(x):<\/p>\n\n\n\n

f\u2019(c)= 3c2<\/sup> \u2013 10c \u2013 3<\/p>\n\n\n\n

For f (3), put the value of x = 3 in f(x):<\/p>\n\n\n\n

f (3)= (3)3<\/sup> \u2013 5(3)2<\/sup> \u2013 3(3)<\/p>\n\n\n\n

= 27 \u2013 45 \u2013 9<\/p>\n\n\n\n

= \u2013 27<\/p>\n\n\n\n

For f (1), put the value of x = 1 in f(x):<\/p>\n\n\n\n

f (1)= (1)3<\/sup> \u2013 5 (1)2<\/sup> \u2013 3 (1)<\/p>\n\n\n\n

= 1 \u2013 5 \u2013 3<\/p>\n\n\n\n

= \u2013 7<\/p>\n\n\n\n

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

\u21d2 3c2<\/sup> \u2013 10c \u2013 3 = \u2013 10<\/p>\n\n\n\n

\u21d2 3c2<\/sup> \u2013 10c \u2013 3 + 10 = 0<\/p>\n\n\n\n

\u21d2 3c2<\/sup> \u2013 10c + 7 = 0<\/p>\n\n\n\n

\u21d2 3c2<\/sup> \u2013 7c \u2013 3c + 7 = 0<\/p>\n\n\n\n

\u21d2 c (3c \u2013 7) \u2013 1(3c \u2013 7) = 0<\/p>\n\n\n\n

\u21d2 (3c \u2013 7) (c \u2013 1) = 0<\/p>\n\n\n\n

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

2. Discuss the applicability of Lagrange\u2019s mean value theorem for the function f(x) = |x| on [\u2013 1, 1].<\/strong><\/p>\n\n\n\n

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

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

3. Show that the Lagrange\u2019s mean value theorem is not applicable to the function f(x) = 1\/x on [\u20131, 1].<\/strong><\/p>\n\n\n\n

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

Given
<\/p>\n\n\n\n

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

Here, x \u2260 0<\/p>\n\n\n\n

\u21d2 f (x) exists for all values of x except 0<\/p>\n\n\n\n

\u21d2 f (x) is discontinuous at x=0<\/p>\n\n\n\n

\u2234 f (x) is not continuous in [\u2013 1, 1]<\/p>\n\n\n\n

Hence the Lagrange\u2019s mean value theorem is not applicable to the function f (x) = 1\/x on [-1, 1]<\/p>\n\n\n\n

4. Verify the hypothesis and conclusion of Lagrange\u2019s mean value theorem for the function<\/strong><\/p>\n\n\n\n

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

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

Given<\/p>\n\n\n\n

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

5. Find a point on the parabola y = (x \u2013 4)2<\/sup>, where the tangent is parallel to the chord joining (4, 0) and (5, 1).<\/strong><\/p>\n\n\n\n

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

Given f(x) = (x \u2013 4)2<\/sup> on [4, 5]<\/p>\n\n\n\n

This interval [a, b] is obtained by x \u2013 coordinates of the points of the chord.<\/p>\n\n\n\n

Every polynomial function is continuous everywhere on (\u2212\u221e, \u221e) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [4, 5] and differentiable in (4, 5). So both the necessary conditions of Lagrange\u2019s mean value theorem is satisfied.<\/p>\n\n\n\n

Therefore, there exist a point c \u2208 (4, 5) such that:<\/p>\n\n\n\n

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

\u21d2 f\u2019(x) = 2 (x \u2013 4) (1)<\/p>\n\n\n\n

\u21d2 f\u2019(x) = 2 (x \u2013 4)<\/p>\n\n\n\n

For f\u2019(c), put the value of x=c in f\u2019(x):<\/p>\n\n\n\n

f\u2019(c) = 2 (c \u2013 4)<\/p>\n\n\n\n

For f (5), put the value of x=5 in f(x):<\/p>\n\n\n\n

f (5) = (5 \u2013 4)2<\/sup><\/p>\n\n\n\n

= (1)2<\/sup><\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

For f (4), put the value of x=4 in f(x):<\/p>\n\n\n\n

f (4) = (4 \u2013 4)2<\/sup><\/p>\n\n\n\n

= (0)2<\/sup><\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

f\u2019(c) = f(5) \u2013 f(4)<\/p>\n\n\n\n

\u21d2 2(c \u2013 4) = 1 \u2013 0<\/p>\n\n\n\n

\u21d2 2c \u2013 8 = 1<\/p>\n\n\n\n

\u21d2 2c = 1 + 8<\/p>\n\n\n\n

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

We know that, the value of c obtained in Lagrange\u2019s Mean value Theorem is nothing but the value of x \u2013 coordinate of the point of the contact of the tangent to the curve which is parallel to the chord joining the points (4, 0) and (5, 1).<\/p>\n\n\n\n

Now, put this value of x in f(x) to obtain y:<\/p>\n\n\n\n

y = (x \u2013 4)2<\/sup><\/p>\n\n\n\n

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

RD Sharma Solutions for Class 12 Maths Chapter 15: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 15\u2013Mean Value Theorems<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 12 Maths Chapter 15\u2013Mean Value Theorems PDF<\/a><\/p>\n\n\n\n

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