Tag: second derivative test

Questions Related to second derivative test

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

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

  1. $x = -1$

  2. $\displaystyle x<-1$

  3. $\displaystyle x>-1$

  4. $\displaystyle x>0$

Show answer
Correct Option: D
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
$\displaystyle f'\left( x \right) >0$
$\displaystyle \Rightarrow \quad { e }^{ ax }-{ e }^{ -ax }>0$
$\displaystyle \Rightarrow \quad { e }^{ 2ax }>1$
$\displaystyle \Rightarrow \quad x>0$

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

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

  1. ${a} _{5}$

  2. ${a} _{7}$ or ${a} _{8}$

  3. ${a} _{4}$

  4. ${a} _{10}$

Show answer
Correct Option: D
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

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

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

  1. $-1$

  2. $-2$

  3. $8$

  4. $9$

Show answer
Correct Option: D
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 $a + 2b - c = 9$.

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 $\displaystyle q =\frac{-1}{4}.$

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

  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.

Show answer
Correct Option: C
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}$

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

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

  1. $e^{-1/2}$

  2. $e^{1/2}$

  3. $e$

  4. $e^{-1}$

Show answer
Correct Option: A
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 $e^{-1/2}$

If $P = {x^3} - \frac{1}{{{x^3}}}$ and $Q = x - \frac{1}{x},$ $x \in \left( {0,x} \right)$ then minimum value of $P/{Q^2}$ is 

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

  1. $2\sqrt 3 $

  2. $-2\sqrt 3 $

  3. does not exist

  4. none of these

Show answer
Correct Option: B

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 

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

  1. $\displaystyle \frac {8}{5}$

  2. $\displaystyle \frac {5}{4}$

  3. $\displaystyle \frac {2}{3}$

  4. None of these

Show answer
Correct Option: C
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.

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

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

  1. $\displaystyle \frac {\sqrt {ab}}{2}$

  2. $\displaystyle \sqrt {ab}$

  3. $\displaystyle \frac {ab}{a + b}$

  4. $\displaystyle \frac {2ab}{a + b}$

Show answer
Correct Option: A
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.

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

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

  1. equal to $50$

  2. greater than $54$

  3. less than $54$

  4. less than $50$

Show answer
Correct Option: B
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 $a _{0}< -54$
$\therefore   a _{0}> 54$

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

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

  1. $1$

  2. $0$

  3. $2$

  4. $3$

Show answer
Correct Option: A