Continuity and Differentiability — Class 12 Maths Solution

ncert exercise SA NCERT Ex.5.8 ,Q.4,Page 186
Question

Verify Mean Value Theorem, if $f(x) = {x^2} - 4x - 3$ in the interval [a,b], where a $=$ 1 and b $=$ 4.

Step-by-step Solution

We have, $f(x) = {x^2} - 4x - 3,$ $x \in$[1, 4] which is a polynomial function. So

(i) f(x) is continuous in [1, 4]

(ii) f(x) is derivable in (1, 4) and, hence conditions of mean value theorem are satisfied, so there exists, at least one $c \in (1,4)$ such that

$f'(c) = \cfrac{{f(4) - f(1)}}{{4 - 1}} \Rightarrow 2c - 4 = \cfrac{{( - 3) - ( - 6)}}{3} \Rightarrow 2c - 4 = 1$

$\Rightarrow$ $2c = 5 \Rightarrow c = \cfrac{5}{2} \in (1,4)$
Hence, mean value theorem is verified.

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