Show that f(x)= ^ -1 ( + ) is an increasing function in (0, 4 ).

Question

Show that $\text{f(x)}=\tan^{-1}(\sin\text{x}+\cos\text{x})$ is an increasing function in $\Big(0,\frac{\pi}{4}\Big).$

✓

Answer

We have, $\text{f(x)}=\tan^{-1}(\sin\text{x}+\cos\text{x})$
$\therefore\ \text{f}'(\text{x})=\frac{1}{1+(\sin\text{x}+\cos\text{x})^2}\cdot(\cos\text{x}-\sin\text{x})$
$=\frac{1}{1+\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}.\cos\text{x}}(\cos\text{x}-\sin\text{x})$
$=\frac{1}{(2+\sin2\text{x})}(\cos\text{x}-\sin\text{x})$
$\big[\because\sin2\text{x}=2\sin\text{x}\cos\text{x and }\sin^2\text{x}+\cos^2\text{x}=1\big]$
For $\text{f}'(\text{x})\geq0.$
$\frac{1}{(2+\sin2\text{x})}\cdot(\cos\text{x}-\sin\text{x})\geq0$
$\Rightarrow\ \cos\text{x}-\sin\text{x}\geq0$ $\Big[\because(2+\sin2\text{x})\geq0\text{ in }\Big(0,\frac{\pi}{4}\Big)\Big]$
$\Rightarrow\ \cos\text{x}\geq\sin\text{x}$
Which is true, if $\text{x}\in\Big(0,\frac{\pi}{4}\Big)$
Hence, f(x) is an increasing function in $\Big(0,\frac{\pi}{4}\Big).$

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 "Case study (4 Marks)" from APPLICATION OF DERIVATIVES→APPLICATION OF DERIVATIVES — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

In a play zone, Aastha is playing crane game. It has 12 blue balls, 8 red balls, 10 yellow balls and 5 green balls. If Aastha draws two balls one after the other without replacement, then answer the following questions.

What is the probability that the first ball is blue and the second ball is green?

$\frac{5}{119}$
$\frac{12}{119}$
$\frac{6}{119}$
$\frac{15}{119}$

What is the probability that the first ball is yellow and the second ball is red?

$\frac{16}{119}$
$\frac{8}{119}$
$\frac{24}{119}$
None of these.

What is the probability that both the balls are red?

$\frac{4}{85}$
$\frac{24}{595}$
$\frac{12}{119}$
$\frac{64}{119}$

What is the probability that the first ball is green and the second ball is not yellow?

$\frac{10}{119}$
$\frac{6}{85}$
$\frac{12}{119}$
None of these.

What is the probability that both the balls are not blue?

$\frac{6}{595}$
$\frac{12}{85}$
$\frac{15}{17}$
$\frac{253}{595}$

→

Rohan, a student of class XII, visited his uncle's flat with his father. He observe that the window of the house is in the form of a rectangle surmounted by a semicircular opening having perimeter 10m as shown in the figure.

Based on the above information, answer the following questions.

If x and y represents the length and breadth of the rectangular region, then relation between x and y can be represented as.

$\text{x}+\text{y}+\frac{\pi}{2}=10$
$\text{x}+\text{2y}+\frac{\pi\text{x}}{2}=10$
$\text{2x}+\text{2y}=10$
$\text{x}+\text{2y}+\frac{\pi}{2}=10$

The area (A) of the window can be given by.

$\text{A}=\text{x}-\frac{\text{x}^3}{8}-\frac{\text{x}^2}{2}$
$\text{A}=\text{5x}-\frac{\text{x}^2}{8}-\frac{\pi\text{x}^2}{8}$
$\text{A}=\text{x}+\frac{\pi\text{x}^3}{8}-\frac{\text{3x}^2}{8}$
$\text{A}=\text{5x}+\frac{\text{x}^3}{2}+\frac{\pi\text{x}^2}{8}$

Rohan is interested in maximizing the area of the whole window, for this to happen, the value of x should be.

$\frac{10}{2-\pi}$
$\frac{20}{4-\pi}$
$\frac{20}{4+\pi}$
$\frac{10}{2+\pi}$

Maximum area of the window is.

$\frac{30}{4+\pi}$
$\frac{30}{4-\pi}$
$\frac{50}{4-\pi}$
$\frac{50}{4+\pi}$

For maximum value of A, the breadth of rectangular part of the window is.

$\frac{10}{4+\pi}$
$\frac{10}{4-\pi}$
$\frac{20}{4+\pi}$
$\frac{20}{4-\pi}$

→

A thermometer reading 80°F is taken outside. Five minutes later the thermometer reads 60°F. After another 5 minutes the thermometer reads 50°F. At any time t the thermometer reading be T°F and the outside temperature be S°F.

Based on the above information, answer the following questions.

If $\lambda$ is posinve constant of propornonality, then $\frac{\text{dn}}{\text{dt}}$ is:

$\lambda(\text{T - S})$
$\lambda(\text{T + S})$
$\lambda\text{T S}$
$-\lambda(\text{T - S})$

The value of T(S) is:

30°F
40°F
5D°F
60°F

The value of T(10) is:

50°F
60°F
80°F
90°F

Find the general solution of differential equation fanned in given situation.

$\log\text{T}=\text{St + c}$
$\log(\text{T}-\text{S})=\lambda{\text{t + c}}$
$\log\text{T}=\text{tT + c}$
$\log(\text{T + S})=\lambda{\text{t + c}}$

Find the value of constant of integration c in the solution of differential equation formed in given situation.

$\log(60\ -\ \text{S})$
$\log(80\ +\ \text{S})$
$\log(80\ -\ \text{S})$
$\log(60\ +\ \text{S})$

→

If y = f(u) is a differentiable function of u and u = g(x) is a differentiable function of x, then y = f(g(x)] is a differentiable function of x and $\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{du}}\times\frac{\text{du}}{\text{dx}}.$ This rule is also known as CHAIN RULE.
Based on the above information, find the derivative of functions w.r.t. x in the following questions.

$\cos\sqrt{\text{x}}$

$\frac{-\sin\sqrt{\text{x}}}{2\sqrt{\text{x}}}$
$\frac{\sin\sqrt{\text{x}}}{2\sqrt{\text{x}}}$
$\sin\sqrt{\text{x}}$
$-\sin\sqrt{\text{x}}$

$7^{\text{x}+\frac{1}{\text{x}}}$

$\Big(\frac{\text{x}^2-1}{\text{x}^2}\Big)\cdot7^{\text{x}+\frac{1}{\text{x}}}\cdot\log7$
$\Big(\frac{\text{x}^2+1}{\text{x}^2}\Big)\cdot7^{\text{x}+\frac{1}{\text{x}}}\cdot\log7$
$\Big(\frac{\text{x}^2-1}{\text{x}^2}\Big)\cdot7^{\text{x}-\frac{1}{\text{x}}}\cdot\log7$
$\Big(\frac{\text{x}^2+1}{\text{x}^2}\Big)\cdot7^{\text{x}-\frac{1}{\text{x}}}\cdot\log7$

$\sqrt\frac{{1-\cos\text{x}}}{1+\cos\text{x}}$

$\frac{1}{2}\sec^2\frac{\text{x}}{2}$
$-\frac{1}{2}\sec^2\frac{\text{x}}{2}$
$\sec^2\frac{\text{x}}{2}$
$-\sec^2\frac{\text{x}}{2}$

$\frac{1}{\text{b}}\tan^{-1}\Big(\frac{\text{x}}{\text{b}}\Big)+\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)$

$\frac{-1}{\text{x}^2+\text{b}^2}+\frac{1}{\text{x}^2+\text{a}^2}$
$\frac{1}{\text{x}^2+\text{b}^2}+\frac{1}{\text{x}^2+\text{a}^2}$
$\frac{1}{\text{x}^2+\text{b}^2}-\frac{1}{\text{x}^2+\text{a}^2}$
None of these.

$\sec^{-1}\text{x}+\text{cosec}^{-1}\frac{\text{x}}{\sqrt{\text{x}^2-1}}$

$\frac{2}{\sqrt{\text{x}^2-1}}$
$\frac{-2}{\sqrt{\text{x}^2-1}}$
$\frac{1}{|\text{x}|\sqrt{\text{x}^2-1}}$
$\frac{2}{|\text{x}|\sqrt{\text{x}^2-1}}$

→

Read the following passage and answer the questions given below: elation between the height of the plant $\left(y^{\prime}\right.$ in cm $)$ with respect to its exposure to is governed by the following equation $y=4 x-\frac{1}{2} x^2$, where ' $x$ ' is the number of days exposed to the sunlight, for $x \leq 3$

(i) Find the rate of growth of the plant with respect to the number of days exposed to the sunlight.

(ii) Does the rate of growth of the plant increase or decrease in the first three days?
What will be the height of the plant after 2 days?

→

To teach the application of probability a maths teacher arranged a surprise game for 5 of his students namely Archit, Aadya, Mivaan, Deepak and Vrinda. He took a bowl containing tickets numbered 1 to 50 and told the students go one by one and draw two tickets simultaneously from the bowl and replace it after noting the numbers.

Based on the above information, answer the following questions.

Teacher ask Vrinda, what is the probability that both tickets drawn by Arch it shows even number?

$\frac{1}{50}$
$\frac{12}{49}$
$\frac{13}{49}$
$\frac{15}{49}$

Teacher ask Mivaan, what is the probability that both tickets drawn by Aadya shows odd number?

$\frac{1}{50}$
$\frac{2}{49}$
$\frac{12}{49}$
$\frac{5}{49}$

Teacher ask Deepak, what is the probability that tickets drawn by Mivaan, shows a multiple of 4 on one ticket and a multiple 5 on other ticket?

$\frac{14}{245}$
$\frac{16}{245}$
$\frac{24}{245}$
None of these.

Teacher ask Arch it, what is the probability that tickets are drawn by Deepak, shows a prime number on one ticket and a multiple of 4 on other ticket?