Download App

Applications of Derivative — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplications of Derivative4 Marks
Question
In interval $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$, Find the difference in maximum value and minimum value of function $f(x)=\sin 2 x-x$.
✓

Answer

Given function
$f(x)=\sin 2 x-x$
So, $\quad f^{\prime}(x)=2 \cos 2 x-1$
For finding critical values :$
\begin{array}{rlrl} 
& & f^{\prime}(x) & =0 \\
\Rightarrow & 2 \cos 2 x-1 & =0 \\
\Rightarrow & & \cos 2 x & =\frac{1}{2} \\
\Rightarrow & & 2 x & =\frac{-\pi}{3}, \frac{\pi}{3} \\
& \Rightarrow & x & =\frac{-\pi}{6}, \frac{\pi}{6}
\end{array}
$
now $\begin{aligned} f\left(\frac{-\pi}{2}\right) & =\sin (-\pi)-\left(\frac{-\pi}{2}\right) \\ & =-\sin \pi+\frac{\pi}{2}=\frac{\pi}{2} \\ f\left(\frac{-\pi}{6}\right) & =\sin \frac{-\pi}{3}+\frac{\pi}{6}=\frac{-\sqrt{3}}{2}+\frac{\pi}{6} \\ f\left(\frac{\pi}{6}\right) & =\sin \frac{\pi}{3}-\frac{\pi}{6}=\frac{\sqrt{3}}{2}-\frac{\pi}{6} \\ \text { and } \quad f\left(\frac{\pi}{2}\right) & =\sin \pi-\frac{\pi}{2}=\frac{-\pi}{2}\end{aligned}$
Maximum value $=\frac{\pi}{2}$ and minimum value $=\frac{-\pi}{2}$ Hence such difference $=\frac{\pi}{2}-\left(\frac{-\pi}{2}\right)=\pi \quad$

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 "4 Marks" from Applications of Derivative→Applications of Derivative — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides where as the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is $\frac{31}{42},$ determine the value of n.
→
What are the values of 'a' for which f(x) = ax is decreasing on R?
→
Evaluate the following integrals:

$\int\frac{\sin2\text{x}}{\sqrt{\sin^4\text{x}+4\sin^2\text{x}-2}}\text{ dx}$

→
Evaluate the following integrals:
$\int\frac{\text{e}^{\text{m}\sin^{-1}\text{x}}}{\sqrt{1-\text{x}^2}}\text{ dx}$
→
Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+\text{y}=0,\text{y}(0)=1,\text{y}(0)=2$
Function $\text{y}=\text{xe}^\text{x}+\text{e}^{\text{x}}$
→
Given $\text{A}=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix},\text{B}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}$ , find BA and use this to solve the system of equations 

y + 2z = 7, x - y = 3, 2x + 3y + 4z = 17 

→
show that the differential equation of which $\text{y}=2(\text{x}^2-1)+\text{ce}^{-\text{x}^2}$ is a solution of the differential equation $\frac{\text{dy}}{\text{dx}}+2\text{xy}=4\text{x}^3$
→
Find the present worth of an ordinary annuity of Rs. 1,200 per annum for 10 years at 12% per annum, compounded annually.$\text[ {Use} : ( 1.12)^{-10} = 0.03221]$
→
Evalute the following integrals:
$\int\frac{\sin(\text{x}-\text{a})}{\sin(\text{x}-\text{b})}\text{dx}$
→
Let A $=\begin{bmatrix}1&-2&1\\-2&3&1\\1&1&5\end{bmatrix}.$ Verify that
  1. [adj. A]-1 = adj.(A-1)
  2. (A-1)-1 = A
→