Application of Derivatives — Page 18 | JEE Mathematics

Q171

The sum of all local minimum values of the function

\mathrm{f}(x)=\left\{\begin{array}{lr} 1-2 x, & x2 \end{array}\right.

is

A

\dfrac{167}{72}

B

\dfrac{157}{72}

C

\dfrac{171}{72}

D

\dfrac{131}{72}

Correct Answer

Option B

Solution

$f(x)=\left\{\begin{array}{cc}1-2 x, & x<-1 \\ \dfrac{1}{3}(7-2 x), & -1 \leq x \leq 2 \\ \dfrac{1}{3}(7+2 x) & 0 \leq x<2 \\ \dfrac{11}{18}(x-4)(x-5), & x>2\end{array}\right.$

\begin{aligned} &\therefore \text{ Local minimum values at } \mathrm{A} \& \mathrm{~B}\\ &\begin{aligned} & \frac{7}{3}-\frac{11}{72} \\ & \Rightarrow \frac{168-11}{72} \Rightarrow \frac{157}{72} \end{aligned} \end{aligned}

Q172

Let the function

f(x) = \dfrac{x}{3} + \dfrac{3}{x} + 3, x \neq 0

be strictly increasing in

(-\infty, \alpha_1) \cup (\alpha_2, \infty)

and strictly decreasing in

(\alpha_3, \alpha_4) \cup (\alpha_4, \alpha_5)

. Then

\sum\limits_{i=1}^{5} \alpha_i^2

is equal to

A 48

B 40

C 36

D 28

Correct Answer

Option C

Solution

$$\begin{aligned} \mathrm{f}(\mathrm{x}) & =\frac{\mathrm{x}}{3}+\frac{3}{\mathrm{x}}+3, \mathrm{x} \neq 0 \\ \mathrm{f}^{\prime}(\mathrm{x}) & =\frac{1}{3}-\frac{3}{\mathrm{x}^2}=0 \quad \Rightarrow \mathrm{x}= \pm 3 \\ \mathrm{f}^{\prime}(\mathrm{x}) & =\frac{\mathrm{x}^2-3}{3 \mathrm{x}^2} \\ \mathrm{f}^{\prime}(\mathrm{x}) & >0 \forall(-\infty,-3) \cup(3, \infty) \rightarrow \text { increasing } \\ \mathrm{f}^{\prime}(\mathrm{x}) &

Q173

If the function

f(x)=2 x^3-9 a x^2+12 \mathrm{a}^2 x+1

, where

\mathrm{a}>0

, attains its local maximum and local minimum values at p and q , respectively, such that

\mathrm{p}^2=\mathrm{q}

, then

f(3)

is equal to :

A 55

B 37

C 10

D 23

Correct Answer

Option B

Solution

To determine the value of $f(3)$ for the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$ , where $a > 0$ , we follow these steps: First, find the critical points by setting the derivative equal to zero: $f'(x) = 6x^2 - 18ax + 12a^2 = 0$ Factoring gives: $6(x - a)(x - 2a) = 0$ Thus, the critical points are $x = a$ and $x = 2a$ . $x = a$ corresponds to a local maximum. $x = 2a$ corresponds to a local minimum.

According to the problem, $p^2 = q$ .

Substituting $p = a$ and $q = 2a$ gives: $a^2 = 2a$ Solving for $a$ gives: $a(a - 2) = 0$ Since $a > 0$ , we have $a = 2$ .

Now, substitute $a = 2$ back into the function: $f(x) = 2x^3 - 18x^2 + 48x + 1$ To find $f(3)$ : $f(3) = 2(3)^3 - 18(3)^2 + 48(3) + 1$ Calculate each term: $2(3)^3 = 54$ $18(3)^2 = 162$ $48(3) = 144$ Thus, $f(3) = 54 - 162 + 144 + 1 = 37$ So, $f(3) = 37$ .

Q174

Let f : ℝ

\to

ℝ be a polynomial function of degree four having extreme values at x = 4 and x = 5. If

\lim\limits_{x \to 0} \dfrac{f(x)}{x^2} = 5

, then f(2) is equal to :

A 8

B 10

C 12

D 14

Correct Answer

Option B

Solution

\begin{aligned} & \lim _{x \rightarrow 0} \frac{f(x)}{x^2}=5 \\ & \lim _{x \rightarrow 0} \frac{\left.a x^4+b x^3+c x^2+d x+e\right)}{x^2}=5 \\ & c=5 \text{ and } d=e=0 \\ & f(x)=a x^4+b x^3+5 x^2 \\ & f^{\prime}(x)=4 a x^3+3 b x^2+10 x \\ & =x\left(4 a x^2+3 b x+10\right) \end{aligned}

has extremes at 4 and so $f^{\prime}(4)=0 \& f^{\prime}(5)=0$ so $\mathrm{a}=\dfrac{1}{8} \& \mathrm{~b}=\dfrac{-3}{2}$ so $f(2)=\dfrac{1}{8} \times 2^4-\dfrac{3}{2} \times 2^3+5 \times 2^2$

=2-12+20=10

Q175

Let

x=-1

and

x=2

be the critical points of the function

f(x)=x^3+a x^2+b \log _{\mathrm{e}}|x|+1, x \neq 0

. Let

m

and M respectively be the absolute minimum and the absolute maximum values of

f

in the interval

\left[-2,-\dfrac{1}{2}\right]

. Then

|\mathrm{M}+m|

is equal to

\left(\right.

Take

\left.\log _{\mathrm{e}} 2=0.7\right):

A 21.1

B 19.8

C 22.1

D 20.9

Correct Answer

Option A

Solution

$\left.\begin{array}{c}3-2 a-b=0 \\ 12+4 a+\dfrac{b}{2}=0\end{array}\right\} \begin{aligned} & a=\dfrac{-9}{2} \\ & b=12\end{aligned}$

\begin{aligned} & \therefore f(x)=x^3-\frac{9}{2} x^2+12 \ln |x|+1 \\ & f(-1)=-1-\frac{9}{2}+1=-\frac{9}{2}=-4.5 \\ & f(-2)=-8-18+12 \ln 2+1 \\ & \quad=-25+12 \ln 2=-16.6 \\ & f\left(-\frac{1}{2}\right)=-\frac{1}{8}-\frac{9}{8}+12 \ln \left(\frac{1}{2}\right)+1=-8.5 \\ & |M+m|=|-16.6-4.5|=21.1 \end{aligned}

Q176

The shortest distance between the curves

y^2=8 x

and

x^2+y^2+12 y+35=0

is:

A

2 \sqrt{3}-1

B

2 \sqrt{2}-1

C

3 \sqrt{2}-1

D

\sqrt{2}

Correct Answer

Option B

Solution

Equation of the Normal to the Parabola: The equation for the normal to the parabola $y^2 = 8x$ can be expressed as: $y = mx - 2am - am^3 \quad \text{where } a = 2$ Simplifying, we have: $y = mx - 4m - 2m^3$ Center and Radius of the Circle: The given second equation $x^2 + y^2 + 12y + 35 = 0$ can be rewritten to find the center and radius of the circle: Rewrite it as $x^2 + (y + 6)^2 = 1$ .

Thus, the center of the circle is $C(0, -6)$ and the radius $r = 1$ .

Finding the Slope of the Normal: To find the point $P$ on the parabola where the normal meets, equate: $-6 = -4m - 2m^3$ Solving for $m$ gives: $m = 1$ Determine Point $P$ on the Parabola: Substituting $m = 1$ back to find $P \left( am^2, -2am \right)$ : $P(2, -4)$ Calculate the Shortest Distance: The shortest distance from point $P(2, -4)$ to the center $C(0, -6)$ minus the radius is: $CP - r = \sqrt{(2 - 0)^2 + (-4 + 6)^2} - 1 = \sqrt{4 + 4} - 1$ $= \sqrt{8} - 1 = 2\sqrt{2} - 1$ Thus, the shortest distance between the given curves is $2\sqrt{2} - 1$ .

Q177

Let

f: \mathrm{R} \rightarrow \mathrm{R}

be a function defined by

f(x)=||x+2|-2| x \|

. If

m

is the number of points of local minima and

n

is the number of points of local maxima of

f

, then

m+n

is

A 3

B 4

C 2

D 5

Correct Answer

Option A

Solution

\begin{aligned} & f(x)=\left\{\begin{array}{cc} |-x-2+2 x| & x \leq-2 \\ |x+2+2 x| & -2 \leq x \leq 0 \\ |x+2-2 x| & x \geq 0 \end{array}\right. \\ & f(x)=\left\{\begin{array}{lc} |-x-2+2 x| & x \leq-2 \\ |x+2+2 x| & -2 \leq x \leq 0 \\ |x+2-2 x| & x \geq 0 \end{array}\right. \\ & f(x)=\left\{\begin{array}{cc} |x-2| & x \leq-2 \\ |3 x+2| & -2< x \leq 0 \\ |2-x| & x>0 \end{array}\right. \end{aligned}

\begin{gathered} f(x)=\left\{\begin{array}{cc} 2-x & x \leq-2 \\ -3 x-2 & -2< x \leq-\frac{2}{3} \\ 3 x+2 & -\frac{2}{3}< x \leq 0 \end{array}\right. \\ f(x)=\left\{\begin{array}{cc} 2-x & x \leq-2 \\ -3 x-2 & -2< x \leq-\frac{2}{3} \\ 3 x+2 & -\frac{2}{3}< x \leq 0 \\ 2-x & 0< x<2 \\ x-2 & x \geq 2 \end{array}\right. \end{gathered}

\begin{aligned} & \text{ No. of maxima }=1 \\ & \text{ No. of minima }=2 \\ & m=2 \\ & n=1 \\ & m+n=3 \end{aligned}

Q178

The minimum distance of a point on the curve y = x2−4 from the origin is :

A

{{\sqrt {19} } \over 2}

B

\sqrt {{{15} \over 2}}

C

{{\sqrt {15} } \over 2}

D

\sqrt {{{19} \over 2}}

Correct Answer

Option C

Solution

Let point on the curve y = x2 $-$ 4 is ( $\alpha$ 2, $\alpha$ 2 $-$ 4) $\therefore$ Distance of the point ( $\alpha$ 2, $\alpha$ 2 $-$ 4) from origin, D =

\sqrt {{\alpha ^2} + {{\left( {{\alpha ^2} - 4} \right)}^2}}

$\Rightarrow$ D2 = $\alpha$ 2 + $\alpha$ 4 + 16 $-$ 8 $\alpha$ 2 $=$ $\alpha$ 4 $-$ 7 $\alpha$ 2 + 16 $\therefore$

{{d{D^2}} \over {d\alpha }}

= 4 $\alpha$ 3 $-$ 14 $\alpha$ Now,

{{d{D^2}} \over {d\alpha }}

= 0 $\Rightarrow$ 4 $\alpha$ 3 $-$ 14 $\alpha$ = 0 $\Rightarrow$ 2 $\alpha$ (2 $\alpha$ 2 $-$ 7) = 0 $\alpha$ = 0 or $\alpha$ 2 =

{7 \over 2}

{{{d^2}{D^2}} \over {d{\alpha ^2}}} = 12{\alpha ^2} - 14

$\therefore$

{\left( {{{{d^2}{D^2}} \over {d{\alpha ^2}}}} \right)_{at\,\,\alpha = 0}} = - 14 < 0

{\left( {{{{d^2}{D^2}} \over {d{\alpha ^2}}}} \right)_{at\,\,{\alpha ^2} = {7 \over 2}}} = 28 > 0

\therefore\,\,\,

Distance is minimum at $\alpha$ 2 =

{7 \over 2}

$\therefore$ Minimum distance D =

\sqrt {{{49} \over 4} - {{49} \over 4} + 16}

=

{{\sqrt {15} } \over 2}

Q179

If the tangent at a point P, with parameter t, on the curve x = 4t2 + 3, y = 8t3−1, t

\in

R, meets the curve again at a point Q, then the coordinates of Q are :

A (t2 + 3, − t3 −1)

B (4t2 + 3, − 8t3 −1)

C (t2 + 3, t3 −1)

D (16t2 + 3, − 64t3 −1)

Correct Answer

Option A

Solution

Given, x = 4t2 + 3 and y = 8t3 $-$ 1 $\therefore$ P $\equiv$ (4t2 + 3, 8t3 $-$ 1)

{{dx} \over {dt}} = 8t

and

{{dy} \over {dt}}

$=$ 24t2 Slope of tangent at P $=$

{{dy} \over {dx}}

$=$

{{dy/dt} \over {dx/dt}}

$=$ 3t Assume Q $=$ (4 $\lambda$ 2 + 3, 8 $\lambda$ 3 $-$ 1) $\therefore$ Slope of PQ $=$ 3t $\Rightarrow$

{{8{\lambda ^3} - 8{t^3}} \over {4{\lambda ^2} - 4{t^2}}} = 3t

$\Rightarrow$

{{8\left( {\lambda - t} \right)\left( {{\lambda ^2} + {t^2} + \lambda t} \right)} \over {4\left( {\lambda - t} \right)\left( {\lambda + t} \right)}} = 3t

$\Rightarrow$ 2 $\lambda$ 2 + 2t2 + 2 $\lambda$ t $=$ 3t (t + $\lambda$ ) $\Rightarrow$ t2 + $\lambda$ t $-$ 2 $\lambda$ 2 = 0 $\Rightarrow$ (t $-$ $\lambda$ ) (t + 2 $\lambda$ ) $=$ 0 $\therefore$ $\lambda$ $=$ t or $\lambda$ $=$ $-$

{t \over 2}

$\therefore$ Q $=$ (4t2 + 3, 8t3 $-$ 1) Or Q $=$

\left( {4 \times {{\left( { - {t \over 2}} \right)}^2} + 3, - 8 \times {{{t^3}} \over 8} - 1} \right)

$=$ (t2 + 3, $-$ t3 $-$ 1)

Q180

The tangent at the point (2,

-

2) to the curve, x2y2

-

2x = 4(1

-

y) does not pass through the point :

A

\left( {4,{1 \over 3}} \right)

B (8, 5)

C (

-

4,

-

9)

D (

-

2,

-

7)

Correct Answer

Option D

Solution

As,

{{dy} \over {dx}}

= $-$

\left[ {{{{{\delta f} \over {\delta x}}} \over {{{\delta f} \over {\delta y}}}}} \right]

{{{\delta f} \over {\delta x}}}

= y2 $\times$ 2x $-$ 2

{{{\delta f} \over {\delta y}}}

= x2 $\times$ 2y + 4

\therefore\,\,\,

{{dy} \over {dx}}

= $-$

\left( {{{2x{y^2} - 2} \over {2{x^2}y + 4}}} \right)

{\left[ {{{dy} \over {dx}}} \right]_{(2, - 2)}}

= $-$

\left( {{{2 \times 2 \times 4 - 2} \over {2 \times 4 \times ( - 2) + 4}}} \right)

= $-$

\left( {{{14} \over { - 12}}} \right)

=

{7 \over 6}

\therefore\,\,\,

Slope of tangent to the curve =

{7 \over 6}

Equation of tangent passes through (2, $-$ 2) is y + 2 =

{7 \over 6}

(x $-$ 2) $\Rightarrow$

\,\,\,

7x $-$ 6y = 26 . . . . .(

1) Now put each option in equation (1) and see which one does not satisfy the equation.

By verifying each points you can see ( $-$ 2, $-$ 7) does not satisfy the equation.