Tag: application of derivatives

Quizzes (2) Questions (52)

Questions Related to application of derivatives

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

The function $\displaystyle f\left( x \right) ={ e }^{ ax }+{ e }^{ -ax },a>0$ is monotonically increasing for

  1. $x = -1$
  2. $\displaystyle x<-1$
  3. $\displaystyle x>-1$
  4. $\displaystyle x>0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since, $\displaystyle f\left( x \right) ={ e }^{ ax }+{ e }^{ -ax }$
$\displaystyle \Rightarrow \quad f'\left( x \right) =a\left( { e }^{ ax }-{ e }^{ -ax } \right) $
$\displaystyle f\left( x \right) $ is monotonically increasing, if
$\displaystyle f'\left( x \right) >0$
$\displaystyle \Rightarrow \quad { e }^{ ax }-{ e }^{ -ax }>0$
$\displaystyle \Rightarrow \quad { e }^{ 2ax }>1$
$\displaystyle \Rightarrow \quad x>0$

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

The largest term in the sequence ${ a } { n }=\cfrac { { n }^{  } }{ { n }^{ 2 }+100 } $ is ______

  1. ${a} _{5}$
  2. ${a} _{7}$ or ${a} _{8}$
  3. ${a} _{4}$
  4. ${a} _{10}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $f(x)=\cfrac { { x }^ { } } { { x }^{ 2 }+100}$


For the largest term, we may calculate the maximum value of $f(x)$


$f'(x)=\cfrac { ( { x }^{ 2 }+100).1-{ x }^{ }.( 2x ) } { { ( { x }^{ 2 }+100 ) }^{ 2 } }=0$


$\Rightarrow \cfrac {100-{ x }^{ 2 } } { { ({ x }^{ 2 }+100) }^{ 2 }}=0$


$\Rightarrow { x }^{ 2 }=100$


$\Rightarrow x=10$ Since $x \neq -10$


To check for maxima or minima, we could evaluate $f''(x)$ at $x=10$

Alternatively, we could just check value of $f(x)$ at any other $x$ and compare its value with at $x=10$


$f(10)=\cfrac { 10 } { { 10 }^{ 2 }+100 }=\cfrac { 1 } { 20 }$


Checking at $x=1$,   $f(1)=\cfrac{ 1 } { { 1 }^{ 2 }+100 }=\cfrac { 1 } { 101 }$


f(10)$>$f(1),     $\therefore x=10$ is point of maxima


$\Rightarrow a _n(max)=\cfrac { 1 } { 20 } $


$\therefore$ Correct option is D

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

Let $g(x) =||x + 2| - 3|$. If a denotes the number of relative minima, $b$ denotes the number of relative maxima and $c$ denotes the product of the zeros. Then the value of $(a + 2b - c)$ is

  1. $-1$
  2. $-2$
  3. $8$
  4. $9$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This function has $a = 2$ relative minima at the x-intercepts, $(-5, 0)$ and $(1, 0), b = 1$ relative minima at $(-2, 3)$ and the product of the zeros is $c = (-5)(1) = - 5$. Thus $a + 2b - c = 9$.

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

Let p, q $\epsilon$ R be such that the function $f(x) = ln |x| + qx^2 + px, x \,\neq \,0$ has extreme values at x = - 1 and x = 2.
Statement-1 : f has local maximum at x = -1 and x = 2.
Statement-2 : $\displaystyle p =\frac{1}{2}$ and $\displaystyle q =\frac{-1}{4}.$

  1. Statement-1 is true, statement-2 is false.

  2. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

  3. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

  4. Statement-1 is false. statement-2 is true.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
On differentiating f(x) w.r.t x
We get
$f'(x) = \dfrac{1}{x} +2q\times  x^{} + p$

On putting x as -1
$f'(x) = -{1} -2q^{} + p$ ------(i)

On putting x as 2
$f'(x) = \dfrac{1}{2} +4q + p=0$ ------(ii)

On solving equation i and ii
We get :
$ p=\dfrac{1}{2}$
$ q=-\dfrac{1}{4}$

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

For what value of $x,x^{2} \ln (1/x)$ is maximum-

  1. $e^{-1/2}$
  2. $e^{1/2}$
  3. $e$
  4. $e^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $y=x^2\ln \dfrac{1}{x}$

$=x^2\ln (x^-)$
$=-x^2\ln (x)$
$\Rightarrow \dfrac{dy}{dx}=-2x\ln x-x^2.\dfrac{1}{x}$
$=-2x\ln x-x$
$=-x[2\ln x+1]=0$
$\Rightarrow x=0$ or $x=e^{-1/2}$
None $\dfrac{d^2y}{dx^2}=-2\ln x-2x\dfrac{1}{x}-1$
$=-2\ln x-3$
at $x=e^{-1/2}$     $\dfrac{d^2y}{dx^2}=-2<0$
$\Rightarrow $ maximum value is at $e^{-1/2}$

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

The sixth term of an A.P is equal to 2. The value of the common difference of the A.P which makes the product $a _{1} a _{4} a _{5}$ least is given by 

  1. $\displaystyle \frac {8}{5}$
  2. $\displaystyle \frac {5}{4}$
  3. $\displaystyle \frac {2}{3}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given that sixth term of an A.P is 2.
$\Rightarrow { a } _{ 1 }+5d=2$
Consider, $p={ a } _{ 1 }{ a } _{ 4 }{ a } _{ 5 }$
$\Rightarrow p={ a } _{ 1 }({ a } _{ 1 }+3d)({ a } _{ 1 }+4d)$
Now substitute ${ a } _{ 1 }=2-5d$ in the above equation, we get;
$ p=(2-5d)(2-2d)(2-d)$
Now, solving the parentheses, we get;
$p=2[4-16d+17{ d }^{ 2 }-5{ d }^{ 3 }]$
Let, $S=-5{ d }^{ 3 }+17{ d }^{ 2 }-16d+4$
Now, taking the derivative of S w.r.t $d$, we get;
$S\prime =-15{ d }^{ 2 }+34{ d }-16$
Notice that, for $S\prime=0$, we get;
$d=\dfrac { 2 }{ 3 } ,\dfrac { 8 }{ 5 } $
Now, taking the derivative of $S\prime$ w.r.t $d$, we get;
$S\prime \prime =-30d+34$
At $d=\dfrac { 2 }{ 3 }$, we get;
$S\prime \prime =-20+34$
So, $S\prime \prime=14$ which is positive.
Therefore, $d=\dfrac { 2 }{ 3 }$ gives minimum value.
Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Let '$a$' and '$b$' are positive number. If $(x, y)$ is a point on the curve $\displaystyle ax^2 + by^2 = ab$ then the largest possible value of $xy$ is

  1. $\displaystyle \frac {\sqrt {ab}}{2}$
  2. $\displaystyle \sqrt {ab}$
  3. $\displaystyle \frac {ab}{a + b}$
  4. $\displaystyle \frac {2ab}{a + b}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The point (x,y) on the curve can be written in polar coordinates as 
$x=\sqrt{b} $cos$\theta$ and $y=\sqrt{a} $sin$\theta$

Thus, 
$(xy) _{max}= (\sqrt{ab}$sin$\theta $cos$\theta) _{max}$

$ = (\sqrt{ab}\dfrac{sin2\theta}{2}) _{max}$
$ = \dfrac{\sqrt{ab}}{2}        \because ($sin$2\theta) _{max}= 1 $

$\therefore$ Ans. is option A.
Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Let $g(x)=a _{0}+a _{1}x+a _{2}x^{2}+a _{3}x^{3}$ and $ f(x)=\sqrt{g(x)}$.
$f(x)$ has its non-zero local minimum and maximum values at $-3$ and $3$ respectively. If $a _{3}\in $ the domain of the function $ \displaystyle h(x)=\sin ^{-1}\left(\dfrac{1+x^{2}}{2x}\right)$. The value of $a _{0}$ is

  1. equal to $50$
  2. greater than $54$
  3. less than $54$
  4. less than $50$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation


$\displaystyle D _{h}=\left { -1, 1 \right }$, as only possible values in the domain of $h(x)$ is $1$ and $-1$
$\therefore  a _{3}=-1$
Now, $ g(x)=a _{0}+a _{1}x+a _{2}x^{2}-x^{3}$
$ {g}'(x)=a _{1}+2a _{2}x-3x^{2}$
$=-3(x-3)(x+3)$
$=-3x^{2}+27$
$\therefore  a _{1}=27, a _{2}=0$
$\therefore a _{1}+a _{2}=27$
Also, $g(-3)> 0$ and $g(3)> 0$
$\Rightarrow  a _{0}> 54$ and $a _{0}< -54$
$\therefore   a _{0}> 54$

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

Let $f(x) = ax^2+bx+c, a, b, c \in R.$ It is given $|f(x)| \le 1, \, |x| \le 1$ then the possible value of $|a+b|$, if $\dfrac{8}{3}a^2+2b^2$ is maximum, is given by

  1. $1$
  2. $0$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given |ax^2+bx+c| <= 1 for |x| <= 1, this is a classic problem related to Chebyshev polynomials. The maximum value of the expression 8/3*a^2 + 2*b^2 under these constraints occurs at specific coefficients, leading to |a+b| = 1.