Tag: application of derivatives - iii

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Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

The value of $a$ for which the function $f(x)=a\ \sin x+\dfrac{1}{3}\sin 3x$ has an extremum at $x=\dfrac{\pi}{3}$ is

  1. $1$
  2. $-1$
  3. $0$
  4. $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$f\left( x \right) = a\sin x + \dfrac{1}{3}\sin 3x$
$ \Rightarrow f'\left( x \right) = a\cos x + \dfrac{1}{3}\cos 3x \times 3$
$ \Rightarrow f'\left( x \right) = a\cos x + \cos 3x$
For extremum at ${\dfrac{\pi }{3}}$
$f'\left( {\dfrac{\pi }{3}} \right) = 0$
$ \Rightarrow a\cos \left( {\dfrac{\pi }{3}} \right) + \cos 3\left( {\dfrac{\pi }{3}} \right) = 0$
$ \Rightarrow \dfrac{a}{2} - 1 = 0$
$\Rightarrow a = 2$
Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

The function $f\left( x \right)\, = \,\dfrac{x}{2}\, + \,\dfrac{2}{x}\,$ has a local minimum at

  1. $ x = -2$
  2. $x= 0$
  3. $x = 1$
  4. $x = 2$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

first find the first derivative of function

$=\dfrac{d}{dx}(\dfrac{x}{2}+\dfrac{2}{x})$
$=\dfrac{1}{2}-\dfrac{2}{x^2}$
now equate the $f^1(x)$ to zero we get the local minimum values
$\dfrac{1}{2}-\dfrac{2}{x^2}=0$
$\dfrac{x^2-4}{2x^2}=0$
$x^2=4$
$x=2,-2$

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

If $p$ and $q$ are positive real numbers such that ${p}^{2}+{q}^{2}=1$, then the maximum value of $(p+q)$ is

  1. $2$
  2. $\cfrac{1}{2}$
  3. $\cfrac{1}{\sqrt{2}}$
  4. $\sqrt{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$AM\geq GM\implies \dfrac{p+q}{2}\geq \sqrt{pq}$

squaring on both sides 
$(p+q)^{2}\geq {4}p{q}$
$p^{2}+q^{2}+2{p}{q}\geq 4{p}{q}$
$1\geq 2{p}{q}\implies  {p}{q}\leq \dfrac{1}{2}$
$(p+q)^{2}=1+2{p}{q}\leq 1+1$
$(p+q)^{2}\leq 2\implies p+q\in[-\sqrt{2},\sqrt{2}]$
The maximum value of $p+q$ is $\sqrt{2}$

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Let $A = (3,-4), B = (1,2)$ .Let $P = (2k-1,2k+1)$ be  a variable point  such that PA+PB is the minimum. then $k$ is

  1. $\dfrac 79$
  2. $0$
  3. $\dfrac 78$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To minimize PA + PB, P must lie on the line segment AB. The slope of AB is (2 - (-4)) / (1 - 3) = 6 / -2 = -3. The equation of line AB is y - 2 = -3(x - 1) => y = -3x + 5. Substituting P(2k-1, 2k+1): 2k+1 = -3(2k-1) + 5 => 2k+1 = -6k + 3 + 5 => 8k = 7 => k = 7/8.

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Let f(x) = tan $(\pi /4-x)/cot 2x(x\neq \pi /4)$. The value which should be assigned to f at $x=\pi /4$. So that it is continuous every where, is

  1. 1/2

  2. 1

  3. 2

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

f(x) = tan(pi/4 - x) / cot(2x). Using tan(pi/4 - x) = (1 - tan x) / (1 + tan x) and cot(2x) = (1 - tan^2 x) / (2 tan x), the limit as x approaches pi/4 is 1.

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Let x and y be two varibles such that $\displaystyle x> 0$ and $xy=1$. Find the minimum value of $x+y$.

  1. $ 2 $
  2. $ \dfrac {1}{2}$
  3. $ \dfrac {2}{3}$
  4. $ 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let,  $\displaystyle z= x+y= x+\dfrac{1}{x}$
for minimum value of $z$
$\cfrac{dz}{dx}=0\Rightarrow 1-\cfrac{1}{x^2}=0\Rightarrow x=\pm 1$
but given $x>0, \Rightarrow x=1$
Hence minimum value of $z$ is 2.

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

In a GP, first term is $1$. If $4T _2 + 5T _3$ is minimum,then its common ratio is.

  1. $\dfrac {2}{5}$
  2. $-\dfrac {2}{5}$
  3. $\dfrac {3}{5}$
  4. $-\dfrac {3}{5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given first term of the G.P is $1.$


Let the G.P be, $1,r,r^2,r^3,.....$

Thus $4T _2+5T _3= 4r+5r^2 = y$ (say)

For minimum value of $y$

$\cfrac{dy}{dr}=0\Rightarrow 4+10r=0\Rightarrow r=-\cfrac{2}{5}$
Hence, option 'B' is correct.

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Let $<\,a _n\,>$ be an $A.P.$ whose first term is $1\;and\;<\,b _n\,>$ is any $G.P.$ whose first term is $2$. If common difference of $A.P.$ is twice of common ratio of $G.P.$ then minimum value of $(a _1b _1+a _2b _2+1)$ is

  1. $3$
  2. $\displaystyle\frac{11}{3}$
  3. $0$
  4. $\displaystyle\frac{11}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let common ratio of $G.P.$ be $r$ & common difference of $A.P.$ be $2r$
$\Rightarrow a _1b _1+a _2b _2+1=1\times2+(1+2r)2r+1$
$=4r^2+2r+3=z$ (say)
Now for minimum value of $z$
$\cfrac{dz}{dr}=0\Rightarrow 8r+2=0\Rightarrow r=-\cfrac{1}{4}$
$\therefore z _{min}=4(\dfrac 1{16})-2(\dfrac 14)+3=\cfrac{11}{4}$

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

If 'x' is real, then maximum value of $\dfrac{3x^2+9x+17}{3x^2+9x+7}$ is - 

  1. $41$
  2. $1$
  3. $\dfrac{17}{7}$
  4. $-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Maximise: $\cfrac{3{ x }^{ 2 }+9x+17}{3{ x }^{ 2 }+9x+7}$

Now we can see that coefficient of ${x}^{2}$ and $x$ are same in ${N}^{x}$ and ${D}^{x}$ so
$\Rightarrow$ $\cfrac{3{ x }^{ 2 }+9x+10+7}{3{ x }^{ 2 }+9x+7}$
$y=1+\cfrac{10}{3{ x }^{ 2 }+9x+7}$
We want to maximise $y$ so we need to minimize $3{ x }^{ 2 }+3x+7$
$y=1+\cfrac{10}{min(3{ x }^{ 2 }+9x+7)}$
$y=1+40=41$ ($\because$ we know min value of quadratic is $\cfrac{-D}{ya}$)