For the function f(x) = x + x ,x , the value of c for mean value theorem is

1 - c 2 = 3 ] - 3 - 1 - 1 c 2 = 3 - 2 2 - 1 c 2 = 3 2 = 3 3 ( c 2 - 1 ) = 2 c 2 3 c 2 - 2 c 2 = 3 = 3 c = 3

Continuity and Differentiability — Class 12 Maths Solution

exemplar objective MCQ NCERT Exemp. Ex.5.3 ,Q.96,Page 116

Question

For the function $f(x) = x + \frac{1}{x},x \in [1,3],$ the value of $c$ for mean value theorem is

Step-by-step Solution

$\Rightarrow$ $1 - \frac{1}{{{c^2}}} = \frac{{\left[ {3 + \frac{1}{3}} \right] - \left[ {1 + \frac{1}{1}} \right]}}{{3 - 1}}$

$\Rightarrow$ $\frac{{{c^2} - 1}}{{{c^2}}} = \frac{{\frac{{10}}{3} - 2}}{2}$

$\Rightarrow$ $\frac{{{c^2} - 1}}{{{c^2}}} = \frac{4}{{3 \times 2}} = \frac{2}{3}$

$\Rightarrow$ $3\left( {{c^2} - 1} \right) = 2{c^2}$

$\Rightarrow$ $3{c^2} - 2{c^2} = 3$

$\Rightarrow$ ${c^2} = 3 \Rightarrow c = \pm \sqrt 3$

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Continuity and Differentiability. Curated by Sachin Sharma. Free for all students.