Download App

CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
f(x) = (x - 1)(x - 2)(x - 3) on [0, 4]
✓

Answer

f(x) = (x - 1)(x - 2)(x - 3) on [0, 4]
We know that, polynomial function is continuous and differentiable everywhere. So, f(x) is continuous in [0, 4] and differentiable in (0, 4). So Lagrange's mean value theorem is applicable. Thus, there exist a point $\text{c}\in(0,4)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
$\Rightarrow(\text{c}-1)(\text{c}-2)(\text{c}-3)+(\text{c}-1)(\text{c}-3)\\=\frac{(3)(2)(1)-(-1)(-2)(-3)}{4-1}$
$\Rightarrow\text{c}^2-3\text{c}+2+\text{c}^2+5\text{c}+6+\text{c}^2-4\text{c}+3=\frac{6+6}{4}$
$\Rightarrow3\text{c}^2-12\text{c}+11=3$
$\Rightarrow3\text{c}^2-12\text{c}+8=0$
$\Rightarrow\text{c}=\frac{-(-12)\pm\sqrt{144-4\times3\times8}}{6}$
$\Rightarrow\text{c}=\frac{12\pm\sqrt{48}}{6}$
$\Rightarrow\text{c}=2\pm\frac{2\sqrt3}{3}\in(0,4)$
$\Rightarrow\text{c}=2\pm\frac{2}{\sqrt3}\in(0,4)$
Hence, Lagrange's mean value theorem is verified.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from CONTINUITY AND DIFFERENTIABILITY→CONTINUITY AND DIFFERENTIABILITY — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Form the differential equation of the family of curve represented by $y^2 = (x - c)^3$
→
If $\text{y}\sqrt{1-\text{x}^2}+\text{x}\sqrt{1-\text{y}^2}=1,$ prove that $\frac{\text{dy}}{\text{dx}}=-\sqrt{\frac{1-\text{y}^2}{1-\text{x}^2}}$
→
Solve the equation 
$\cos^{-1}\frac{\text{a}}{\text{x}}-\cos^{-1}\frac{\text{b}}{\text{x}}=\cos^{-1}\frac{1}{\text{b}}-\cos^{-1}\frac{1}{\text{b}}-\cos^{-1}\frac{1}{\text{a}}$
→
Let $f: W \rightarrow W$ be defined as $f(n) = n – 1,$ if n is odd and $f (n) = n + 1,$ if $n$ is even. Show that $f$ is invertible. Find the inverse of $f.$ Here, $W$ is the set of all whole numbers.
→
Find the vector and cartesian equations of the line passing through $(1, 2, 3)$ and parallel to the planes $\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})=6$
→
$\text{if } \vec{\text{a}} = 2\hat{\text{i}} + \hat{\text{j}} - \hat{\text{k}}, \vec{\text{b}} = 4\hat{\text{i}} - 7\hat{\text{j}} + \hat{\text{k}}, \text{find a vector } \vec{\text{c}} \text{ such that } \vec{\text{a}} \times \vec{\text{c}}=\vec{\text{b }} \text{ and } \vec{\text{a }} . \vec{\text{c}} = 6.$
→
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\log\text{x}}+(\log\text{x}^\text{x})$
→
If  $\vec{\text{a}}=2\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}},\vec{\text{b}}=-3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}},$ compute $\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\vec{\text{c}}$ and $\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{c}}\big)$ and verify that these are not equal.
→
If $\text{y}=\cot^{-1}\Big\{\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\theta}}\Big\},$ show that $\frac{\text{dy}}{\text{dx}}$ is independent of x.
→
Show that the binary operation $\ast \text{ on A = R - {-1}}$ defined as a $\text{a} \ast \text{b} = \text{a + b + ab}$ for all $\text{a, b}\in \text{A}$ is communicative and associative on A. Also find the identity element of $\ast$ in A and prove that every element of a is invertible.
→