Download App

Mean Value Theorems — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMean Value Theorems5 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.
$f(x) = x(x + 4)^2$ on $[0,4]$
✓

Answer

Here, $f(x) = x(x + 4)^2$
$\Rightarrow f(x) = x(x^2+ 16 + 8x)$
$\Rightarrow f(x) = x^3 + 8x^2+ 16x$
Since $f(x)$ is a polynomial function which is everywhere continuous and differentiable.
Therefore, $f(x)$ is continuous on $[0, 4]$ and derivable on $(0, 4)$
Thus, both the conditions of Lagrange's theorem is satisfied.
Consequently, there exists some $\text{c}\in(0,4)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}=\frac{\text{f}(4)-\text{f}(0)}{4}$
Now, $f(x) = x^3 + 8x^2+ 16x$
$\Rightarrow f(x) = 3x^2+ 16x + 16,$
$\Rightarrow f(4) = 64+ 128 + 64 = 256,$
$⇒ f(0) = 0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
$\Rightarrow3\text{x}^2+16\text{x}+16=\frac{256}{4}$
$\Rightarrow3\text{x}^2+16\text{x}-48=0$
$\Rightarrow\text{x}=-\frac{4}{3}\big(2+\sqrt{13}\big),\frac{4}{3}\big(\sqrt{13}-2\big)$
Thus, $\text{c}=\frac{-8+4\sqrt{13}}{3}\in(0,4)$ such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
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 Mean Value Theorems→Mean Value Theorems — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

If $\text{A}=\begin{bmatrix}0&\text{c}&-\text{b}\\-\text{c}&0&\text{a}\\\text{b}&-\text{a}&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\text{a}^2&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2&\text{bc}\\\text{ac}&\text{bc}&\text{c}^2\end{bmatrix},$ show that $AB = BA = O_{3 \times 3}$
→
Find the distance of the point (1, -5, 9) from the plane x - y + z = 5 measured along the line x = y = z.
→
The odds against a certain event are 5 to 2 and the odds in favour of another event, independent to the former are 6 to 5. Find the probability that,
  1. At least one of the events will occur,
  2. None of the events will occur.
→
Differentiate the following functions with respect to x:
$(\log\text{x})^{\log\text{x}}$
→
If $\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=\text{a}(\text{x}-\text{y}),$ prove that $\frac{\text{dy}}{\text{dx}}=\sqrt{\frac{1-\text{y}^2}{1-\text{x}^2}}.$
→
Find the vector equation (in scalar product form) of the plane containing the line of intersection of the planes x - 3y + 2z - 5 = 0 and 2x - y + 3z - 1 = 0 and passing through (1, -2, 3).
→
$\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are the position vectors of points A, B and C respectively, prove that:
$\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}$ is a vector perpendicular to the plane of triangle ABC.
→
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=\text{x}^2\cot\text{x}+2\text{x}$
→
For the following matrices verify the associativity of multiplication i.e., (AB) C = A(BC):
$\text{A}=\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix},\text{B}=\begin{bmatrix}1&-1&1\\0&1&2\\2&-1&1\end{bmatrix}$ and $\text{C}=\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}$
→
Find the vector equation 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 $\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=6.$ 
→