Show that for a 1, f(x)=3 - -2ax+b is decreasing in R.

Question

Show that for $\text{a}\geq1,\text{ f(x)}=\sqrt{3}\sin\text{x}-\cos\text{x}-2\text{ax}+\text{b}$ is decreasing in R.

✓

Answer

We have,$\text{ f(x)}=\sqrt{3}\sin\text{x}-\cos\text{x}-2\text{ax}+\text{b}$
$\Rightarrow\ \text{f(x)}=\sqrt{3}\cos\text{x}-(-\sin\text{x})-2\text{a}$
$=\sqrt{3}\cos\text{x}+\sin\text{x}-2\text{a}$
$=2\Big[\frac{\sqrt{3}}{2}\cdot\cos\text{x}+\frac{1}{2}\cdot\sin\text{x}\Big]-2\text{a}$
$=2\Big[\cos\frac{\pi}{6}\cdot\cos\text{x}+\sin\frac{\pi}{6}\cdot\sin\text{x}\Big]-2\text{a}$
$=2\cos\Big(\frac{\pi}{6}-\text{x}\Big)-2\text{a}$ $\big[\therefore\cos(\text{A}-\text{B})=\cos\text{A}\cdot\text{B}+\sin\text{A}\cdot\sin\text{B}\big]$
$=2\Big[\cos\Big(\frac{\pi}{6}-\text{x}\Big)-\text{a}\Big]$
Since, $\cos\text{x}\in[-1,1]\text{ and a}\geq1$
$\therefore\ 2\Big[\cos\Big(\frac{\pi}{6}-\text{a}\Big)-\text{a}\Big]\leq0$
We know tha if $\text{f(x)}\leq0,$ then f(x) is a decreasing function.
Thus, f(x) is a decreasing function in R.

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→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

In pre-board examination of class XII, commerce stream with Economics and Mathematics of a particular school, 50% of the students failed in Economics, 35% failed in Mathematics and 25% failed in both Economics and Mathematics. A student is selected at random from the class.

Based on the above information, answer the following questions.

The probability that the selected student has failed in Economics, if it is known that he has failed in Mathematics, is:

$\frac{3}{10}$
$\frac{12}{25}$
$\frac{1}{4}$
$\frac{5}{7}$

The probability that the selected student has failed in Mathematics, if it is known that he has failed in Economics, is:

$\frac{22}{25}$
$\frac{12}{25}$
$\frac{1}{2}$
$\frac{3}{25}$

The probability that the selected student has passed in at least one of the two subjects, is:

$\frac{1}{4}$
$\frac{1}{2}$
$\frac{3}{4}$
None of these.

The probability that the selected student has failed in at least one of the two subjects, is:

$\frac{3}{5}$
$\frac{22}{25}$
$\frac{2}{5}$
$\frac{43}{100}$

The probability that the selected student has passed in Mathematics, if it is known that he has failed in Economics, is:

$\frac{2}{5}$
$\frac{3}{4}$
$\frac{1}{3}$
$\frac{1}{2}$

→

If the equation is of the form $\frac{\text{dy}}{\text{dx}}=\frac{\text{f(x, y)}}{\text{g(x, y)}}$or $\frac{\text{dy}}{\text{dx}}=\text{F}\Big(\frac{\text{y}}{\text{x}}\Big),$ where f(x, y), g(x, y) are homogeneous functions of the same degree in x and y, then put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v+x}\Big(\frac{\text{dv}}{\text{dx}}\Big),$ so that the dependent variable y is changed to another variable v and then apply variable separable method.
Based on the above information, answer the following questions.

The general solution of $\text{x}^2\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{xy}+\text{y}^2$ is:

$\tan^{-1}\frac{\text{x}}{\text{y}}=\log|\text{x}|+\text{c}$
$\tan^{-1}\frac{\text{y}}{\text{x}}=\log|\text{x}|+\text{c}$
$\text{y}=\text{x}\log|\text{x}|+\text{c}$
$\text{x}=\text{y}\log|\text{y}|+\text{c}$

Solution of the differential equation $2\text{xy}\frac{\text{dy}}{\text{dx}}=\text{x}^2+3\text{y}^2$ is:

x³ + y² = cx²
$\frac{\text{x}^2}{2}+\frac{\text{y}^3}{3}=\text{y}^2+\text{c}$
x² + y³ = cx²
x² + y² = cx³

General solution of the differential equation (x² + 3xy + y²) dx - x² dy = 0 is:

$\frac{\text{x+y}}{\text{y}}-\log\text{x = c}$
$\frac{\text{x+y}}{\text{y}}+\log\text{x = c}$
$\frac{\text{x}}{\text{x+y}}-\log\text{x = c}$
$\frac{\text{x}}{\text{x+y}}+\log\text{x = c}$

General solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\bigg\{\log\Big(\frac{\text{y}}{\text{x}}\Big)+1\bigg\}$ is:

$\log(\text{xy})=\text{c}$
$\log\text{y}=\text{cx}$
$\log\frac{\text{y}}{\text{x}}=\text{cx}$
$\log\text{x}=\text{cy}$

Solution of the differential equation $\Big(\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\Big)\text{e}^\frac{\text{y}}{\text{x}}=\text{x}^2\ \cos\text{x}$ is:

$\text{e}^\frac{\text{y}}{\text{x}}-\sin\text{x = c}$
$\text{e}^\frac{\text{y}}{\text{x}}+\sin\text{x = c}$
$\text{e}^\frac{\text{-y}}{\text{x}}-\sin\text{x = c}$
$\text{e}^\frac{\text{-y}}{\text{x}}+\sin\text{x = c}$

→

A barge is pulled into harbour by two tug boats as shown in the figure.

Based on the above information, answer the following questions.

Position vector of A is:

$4\hat{\text{i}}+2\hat{\text{j}}$
$4\hat{\text{i}}+10\hat{\text{j}}$
$4\hat{\text{i}}-10\hat{\text{j}}$
$4\hat{\text{i}}-2\hat{\text{j}}$

Position vector of B is:

$4\hat{\text{i}}+4\hat{\text{j}}$
$6\hat{\text{i}}+6\hat{\text{j}}$
$9\hat{\text{i}}+7\hat{\text{j}}$
$3\hat{\text{i}}+3\hat{\text{j}}$

Find the vector $\overline{\text{AC}}$ in terms of $\hat{\text{i}},\hat{\text{j}}.$

$8\hat{\text{j}}$
$-8\hat{\text{j}}$
$8\hat{\text{i}}$
None of these

If $\vec{\text{A}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},$ then its unit vector is:

$\frac{\hat{\text{i}}}{\sqrt{14}}+\frac{2\hat{\text{j}}}{\sqrt{14}}+\frac{3\hat{\text{k}}}{\sqrt{14}}$
$\frac{3\hat{\text{i}}}{\sqrt{14}}+\frac{2\hat{\text{j}}}{\sqrt{14}}+\frac{\hat{\text{k}}}{\sqrt{14}}$
$\frac{2\hat{\text{i}}}{\sqrt{14}}+\frac{3\hat{\text{j}}}{\sqrt{14}}+\frac{\hat{\text{k}}}{\sqrt{14}}$
None of these

If $\vec{\text{A}}=4\hat{\text{i}}+3\hat{\text{j}}$ and $\vec{\text{B}}=3\hat{\text{i}}+4\hat{\text{j}},$ then $|\vec{\text{A}}|+|\vec{\text{B}}|=$

→

Two motorcycles A and Bare running at the speed more than allowed speed on the road along the lines $\vec{\text{r}}=\lambda(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$ and $\vec{\text{r}}=3\hat{\text{i}}+3\hat{\text{j}}+\mu(2\hat{\text{i}+\hat{\text{j}}+\hat{\text{k}}}),$ respectively.

Based on the above information, answer the following questions.

The cartesian equation of the line along which motorcycle A is running is:

$\frac{\text{x}+1}{1}=\frac{\text{y}+1}{2}=\frac{\text{z}-1}{-1}$
$\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{-1}$
$\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{1}$
None of these

The direction cosines of line along which motorcycle A is running, are:

< 1, -2, 1 >
< I, 2, -1 >
$<\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}}>$
$<\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}}>$

The direction ratios of line along which motorcycle Bis running, are:

< 1, 0, 2 >
< 2, 1, 0 >
< 1, 1, 2 >
< 2, 1, 1 >

The shortest distance between the gives lines is:

4 units
$2\sqrt{3}\text{ units}$
$3\sqrt{2}\text{ units}$
0 units

The motorcycles will meet with an accident at the point:

(-1, 1, 2)
(2, 1, -1)
(1, 2, -1)
Does not exist

→

Three schools A, B and C organized a mela for collecting funds for helping the rehabilitation of flood victims. They sold handmade fans, mats, and plates from recycled material at a cost of ₹ 25 , ₹ 100 and ₹ 50 each. The number of articles sold by school A, B, C are given below.

(i) Represent the sale of handmade fans, mats and plates by three schools A, B and C and the sale prices (in ₹) of given products per unit, in matrix form.

(ii) Find the funds collected by school A, B and C by selling the given articles.

(iii) If they increase the cost price of each unit by $20 \%$, then write the matrix representing new price.

Find the total funds collected for the required purpose after $20 \%$ hike in price.

→

Consider the following diagram, where the forces in the cable are given.