MCQ Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 6

The slope of the tangent to the curve

$x=3t^2+1, y=t^3-1$ at

$x=1$ is

0%
$\dfrac{1}{2}$
0%
$0$
0%
$-2$
0%
$\infty$

Explanation

Given,

$x=3t^2+1, y=t^3-1$

Slope of the tangent to the given curve is

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx}$

$= 3t^2 \times \dfrac{1}{6t}$

$=\dfrac{t}{2}$

Since the slope has to be calculated at

$x = 1,$ i.e. at

$3t^2 + 1 = 1$ , we get

$t = 0$

Thus, the required slope is

$0.$

The function

$f(x)=\frac{x}{3}+\frac{3}{x}$ decreaes in the interval.

0%
(-3, 3)
0%
$(-\infty , 3)$
0%
$(3, \infty )$
0%
$(-9, 9)$
0%
$(-3,3) - \{0\}$

Explanation

We have

$f(x)=\dfrac{x}{3}+\dfrac{3}{x}$ , Domain is

$R-\{0\}$

Differentiate it,

$f'(x)=\dfrac{1}{3}-\dfrac{3}{x^2}$

Now for f to be decreasing

$f'(x)<0$

$\Rightarrow \dfrac{1}{3}-\dfrac{3}{x^2}<0$

$\Rightarrow \dfrac{x^2-9}{3x^2}<0$

$\Rightarrow x^2-9<0$ , since

$3x^2>0\forall x\in$ Domain

$\Rightarrow (x+3)(x-3)<0$

$\Rightarrow x\in (-3,3)-\{0\}$ , since

$0$ is not in the domain

Find the equation of the quadratic function

$f$ whose graph increases over the interval

$(-\infty, -2)$ and decreases over the interval

$(-2,+\infty)$ ,

$f(0)=23$ and

$f(1)=8$

0%
$f(x)=3({x+2)}^{2}+35$
0%
$f(x)=-3({x+2)}^{2}-35$
0%
$f(x)=-3({x-2)}^{2}+35$
0%
$f(x)=-3({x+2)}^{2}+35$

Explanation

Let the quadratic equation be

$f(x)=ax^{2}+bx+c$
Given

$f(0)=23$
Put

$x=0$ in

$f(x)$ we get

$c=23$
Given

$f(1)=8$
Put

$x=1$ in

$f(x)$ , we get

$a+b+23=8$

$a+b=-15$ ......(i)
Now we can see from the interval that maximum value of quadratic equation is at

$x=-2$

$f'(x)=0$ at

$x=-2$

$f'(x)=2ax+b$

Put

$x=-2$ , we get

$2a(-2)+b=0$

$b=4a$
Put it in equation (i), we get

$a+4a=-15$

$a=-3$
So

$b=4a=4(-3)=-12$

So quadratic equation is

$=-3x^{2}-12x+23$

$=-3(x^2+4x+4-4)+23$

$=-3((x+2)^2-4)+23$

$=-3(x+2)^2+12+23$

$=-3(x+2)^2+35$

For which region is

$f(x)=3x^2-2x+1$ strictly increasing?

0%
$(2,5)$
0%
$(\dfrac 13,\infty)$
0%
$(-1,\dfrac 13]$
0%
$(-\infty,\dfrac 13)$

Explanation

To find where the function is increasing

1) Find its derivatives

Given : $f\left( x \right) =3{ x }^{ 2 }-2x+1$

$f^{ 1 }\left( x \right) =6x-2$

Equate it to zero

$6x-2=0\quad$

$6x=2\quad \\$

$x=\dfrac { 1 }{ 3 }$ (So the values which make derivatives equal to $0$ is $\dfrac { 1 }{ 3 }$ )

$\rightarrow$ Split and separate the value between intervals $\left( -\infty ,\infty \right)$

$\rightarrow$ Which gives

$\left( -\infty ,\dfrac { 1 }{ 3 } \right) \left( \dfrac { 1 }{ 3 } ,\infty , \right)$

Since $(2,5)$ and $\left( \dfrac { 1 }{ 3 } ,\infty \right)$ falls under this intervals.
Hense both options A and B is the correct answers

The focal length of a mirror is given by

$\displaystyle \frac{2}{f}\, =\, \displaystyle \frac{1}{v}\, -\, \displaystyle \frac{1}{u}$ . In finding the values of u and v, the errors are equal and equal to 'p'. The the relative error in f is

0%
$\displaystyle \frac{p}{2}\, \left ( \displaystyle \frac{1}{u}\, +\, \displaystyle \frac{1}{v} \right )$
0%
$p \left ( \displaystyle \frac{1}{u}\, +\, \displaystyle \frac{1}{v} \right )$
0%
$\displaystyle \frac{p}{2}\, \left ( \displaystyle \frac{1}{u}\, -\, \displaystyle \frac{1}{v} \right )$
0%
$p \left ( \displaystyle \frac{1}{u}\, -\, \displaystyle \frac{1}{v} \right )$

Explanation

$\dfrac{2}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

$-2\dfrac{\Delta f}{f^2}=\dfrac{-\Delta V}{v^2}+\dfrac{\Delta u}{u^2}$

$\Rightarrow \dfrac{2\Delta f}{f^2}=\dfrac{\Delta v}{v^2}-\dfrac{\Delta v}{u^2}$

$\Rightarrow \dfrac{\Delta f}{f}=\dfrac{f P}{2}[\dfrac{1}{v}-\dfrac{1}{u}][\dfrac{1}{v}+\dfrac{1}{u}]$

$\dfrac{\Delta f}{f}=\dfrac{f p}{2}\times \dfrac{2}{f}[\dfrac{1}{v}+\dfrac{1}{u}]$

$\dfrac{\Delta f}{f} = p[\dfrac{1}{v}+\dfrac{1}{u}]$

Suppose that

$f$ is a polynomial of degree

$3$ and that

$f''(x)\neq 0$ at any of the stationary point. Then

0%
$f$ has exactly one stationary point
0%
$f$ must have no stationary point
0%
$f$ must have exactly $2$ stationary point
0%
$f$ has exactly $0$ or $2$ stationary point.

Explanation

The given information is a polynomial of degree

$3$ .

Let the function be

$f(x)=ax^3 +bx^2 + cx +d$ .

To find the stationary points we differentiate and equate it to zero.

$f'(x) = 3ax^2 + 2bx + x$

The differentiated function is a quadratic equation which will give two real or two complex roots. If we get two real roots then we will have exactly

$2$ stationary points. Now if the roots are complex as they always occur in pair then we have no stationary points. Thus f has exactly

$0$ or

$2$ stationary points. .....Answer

Let

$f:R\rightarrow R$ be a differentiable function for all values of

$x$ and has the peoperty that

$f(x)$ and

$f'(x)$ have opposite signs for all values of

$x$ . Then,

0%
$f(x)$ is an increasing function
0%
$f(x)$ is a decreasing function
0%
$f ^2 (x)$ is a decreasing function
0%
$|f(x)|$ is an increasing function

Explanation

Given

$f(x)$ and

$f^{'} (x)$ have opposite signs

$\Rightarrow f(x) f^{'} (x) <0$

We cannot say f(x) is increasing or decreasing because we don't know the sign of f(x)

Let us take option C;

$f^{2}(x)$ is increasing

$\Rightarrow \dfrac {d} {dx} (f^{2}(x))>0$

$\Rightarrow 2f(x)f^{'}(x)>0$

$\Rightarrow f(x) f^{'} (x) >0$ which is the given condition

$\therefore$ option C is correct

The function

$f(x) = x^2$ is decreasing in

0%
$(- \infty, \infty)$
0%
$(- \infty, 0)$
0%
$(0, \infty)$
0%
$(-2, \infty)$

Explanation

$f(x)=x^{2}$ ,

$f'(x)=2x$ .

$f'(x)>0$ implies

$2x>0$ or

$x>0$ . Hence

$f(x)$ is increasing in the interval of

$(0,\infty)$ and decreasing in the interval

$(-\infty,0)$ .

Identify the correct statements
(a) Every constant function is an increasing function.
(b) Every constant function is a decreasing function.
(c) Every identify function is an increasing function.
(d) Every identify function is a decreasing function.

0%
$(a), (b)$ and $(c)$
0%
$(a)$ and $(c)$
0%
$(c)$ and $(d)$
0%
$(a), (c)$ and $(d)$

Explanation

Increasing function means

$f'\left( x \right) \ge 0$ and decreasing function means

$f'\left( x \right) \le 0$

For constant function , derivative will be zero.

Therefore constant function is both increasing and decreasing.

Identity function means

$f(x)=x$ , its derivative is always

$1 \ge 0$

Therefore identity function is increasing function

$a,b,c$ are correct

Therefore the correct option is

$A$

Identify the false statement:

0%
All the stationary numbers are critical numbers
0%
At the stationary point the first derivative is zero
0%
At critical numbers the first derivative need not exist
0%
All the critical numbers are stationary numbers

Explanation

The numbers where

$f'\left( x \right) =0$ are called stationary numbers

The numbers where

$f'\left( x \right) =0$ or

$f'\left( x \right)$ doesnt exist are called critical numbers

Therefore all critical numbers are not stationary numbers

So option

$D$ is false statement

The percentage error in the

$11^{th}$ root of the number

$28$ is approximately ____________ times the percentage error in

$28$

0%
$\dfrac { 1 }{ 28 }$
0%
$\dfrac { 1 }{ 11 }$
0%
$11$
0%
$28$

Explanation

Suppose

$y = x^{11}$

$dy = 11x^{10}dx$

Dividing by y on both sides, we have

$\cfrac{\Delta y}{y} = \cfrac{11x^{10}\Delta x}{y} = \cfrac{11x^{10}\Delta x}{x^{11}}$

$\therefore \cfrac{\Delta y}{y} = \cfrac{11 \Delta x}{x}$

Here, when

$y = 28$ ,

$x$ will be the

$11^{th}$ root of

$28$ .

$\therefore \cfrac{\Delta (28)}{28} = \cfrac{11 \Delta (\sqrt[11]{28})}{\sqrt[11]{28}}$

Hence, the percentage error in the

$11$ th root of

$28$ would approximately be

$\cfrac{1}{11}$ times the error in

$28$

The slope of the normal to the curve

$y = 3x^2$ at the point whose

$x$ -coordinate

$2$ is

0%
$\dfrac{1}{13}$
0%
$\dfrac{1}{14}$
0%
$\dfrac{-1}{12}$
0%
$\dfrac{1}{12}$

Explanation

$y=3x^{2}$ . Differentiating the equation of the curve with respect to

$x$ .

$\dfrac{dy}{dx}=6x$
Now
Slope

$_{x=2}=\dfrac{dy}{dx}|_{x=2}$

$=6(2)$

$=12$ .
Hence the slope of the tangent at

$x=2$ is

$12$ . Since the normal is perpendicular to the tangent, therefore, the

Slope

$_\text{normal}\times$ Slope

$_\text{tangent}=-1$ or Slope

$_\text{normal}=\dfrac{-1}{\text{Slope}_\text{tangent}}=\dfrac{-1}{12}$

The slope of the tangent to the curve

$y=3{ x }^{ 2 }+3\sin { x }$ at

$x=0$ is

0%
$3$
0%
$2$
0%
$1$
0%
$-1$

Explanation

$y=3x^2+3\sin x$

$\Rightarrow \dfrac{dy}{dx}=6x+3\cos x$

Thus slope of tangent to the given curve at

$x=0$ is

$=\dfrac{dy}{dx}\bigg|_{x=0}=(6x+3\cos x)\bigg|_{x=0}=6(0)+3\cos 0=0+3\cdot 1=3$

Consider the following in respect of the function

$f(x) = \left\{\begin{matrix}2+ x, & x \geq 0\\ 2 - x, & x < 0\end{matrix}\right.$

$\displaystyle \lim_{x \rightarrow 1} f(x)$ does not exist.
f(x) is differentiable at x = 0.
f(x) is continuous at x = 0.
Which of the above statements is/are correct?

0%
$1$ only
0%
$3$ only
0%
$2$ and $3$ only
0%
$1$ and $3$ only

Explanation

Left hand limit

$= \lim_{x\to 1^-}{f(x)}$

$= \displaystyle \lim_{h\to 0}{f(1-h)}$

$=\displaystyle \lim_{h\to 0}2+(1-h)=2+1=3$

Right hand limit

$= \displaystyle \lim_{x\to 1^+}{f(x)}$

$= \displaystyle \lim_{h\to 0}{f(1+h)}$

$=\displaystyle \lim_{h\to 0}2+(1+h)=2+1=3$

So, the limit exists at x=1

Left hand limit

$=\displaystyle \lim_{x\to 0^-}{f(x)}$

$= \displaystyle \lim_{h\to 0}{f(0-h)}$

$=\displaystyle \lim_{h\to 0} 2-(0-h)=2-0=2$

Right hand limit

$= \displaystyle \lim_{x\to 0^+}{f(x)}$

$= \displaystyle \lim_{h\to 0}{f(0+h)}$

$=\displaystyle \lim_{h\to 0}2+(0+h)=2+0=2$

So, LHL=RHL=f(0)

So, f(x) is continuous at x=0

Right hand limit

$= \displaystyle \lim_{h\to 0}\dfrac{f(0+h)-f(0)}{h}$

$= \displaystyle \lim_{h\to 0}\dfrac{(2+(0+h))-2}{h}$

$=\displaystyle \lim_{h\to 0}\dfrac{h}{h}=1$

Left hand limit

$= \displaystyle \lim_{h\to 0}\dfrac{f(0-h)-f(0)}{-h}$

$= \displaystyle \lim_{h\to 0}\dfrac{(2-(0-h))-(2)}{-h}$

$=\displaystyle \lim_{h\to 0}\dfrac{h}{-h}=-1$

LHL

$\neq$ RHL

So, f(x) is not differentiable at x=0

The answer is option (B)

Consider the following statements:

$y = \dfrac {e^{x} + e^{-x}}{2}$ is an increasing function on

$[0, \infty)$ .

$y = \dfrac {e^{x} - e^{-x}}{2}$ is an increasing function on

$(-\infty, \infty)$ .
Which of the above statements is/are correct?

0%
$1$ only
0%
$2$ only
0%
Both $1$ and $2$
0%
Neither $1$ nor $2$

Explanation

$y=\dfrac { { e }^{ x }+{ e }^{ -x } }{ 2 } =\cosh x$

$y=\dfrac { { e }^{ x }-{ e }^{ -x } }{ 2 } =\sinh x$

Both are increasing functions

Hence, both the statements are correct.

What is the slope of the tangent to the curve

$x = t^2 + 3t - 8, y = 2t^2 - 2t - 5$ at t = 2 ?

0%
$\dfrac{7}{6}$
0%
$\dfrac{6}{7}$
0%
1
0%
$\dfrac{5}{6}$

Explanation

Differentiating

$x,y$ w.r.t

$t,$ we get

$\dfrac { dy }{ dt } =\dfrac { d }{ dt } (2t^2-2t-5)=4t-2\\ \dfrac { dx }{ dt } =\dfrac { d }{ dt } (t^2+3t-8)=2t+3\\$

$\therefore \dfrac { dy }{ dx } =\dfrac { \left( \dfrac { dy }{ dt } \right) }{ \left( \dfrac { dx }{ dt } \right) } =\dfrac { 4t-2 }{ 2t+3 }$

$t=2$ derivative will be

$\dfrac {dy}{dx}=\dfrac {4\times 2-2}{2\times 2+3}=\dfrac {6}{7}$

Thus, the slope of tangent is

$\dfrac{6}{7}$ .

If the tangent to the function

$y = f(x)$ at

$(3, 4)$ makes an angle of

$\dfrac {3\pi}{4}$ with the positive direction of x-axis in anticlockwise direction then

$f'(3)$ is

0%
$-1$
0%
$1$
0%
$\dfrac {1}{\sqrt {3}}$
0%
$\sqrt {3}$

Explanation

Given

$y=f(x)$

Differentiating w.r.t x, we get

$y' = f'(x)$ which is the slope of the tangent

$f'(3) = \tan \dfrac {3\pi}{4} = -\tan \dfrac {\pi}{4} = -1$

Hence,

$f'(3)=-1$ .

How many tangents are parallel to x-axis for the curve

$y = x^2 - 4x + 3$ ?

0%
1
0%
2
0%
3
0%
No tangent is parallel to x-axis.

Explanation

Slope of the tangent is

$\dfrac {dy} {dx}$

So,

$\dfrac{dy}{dx}=2x-4$

Tangent parallel to x-axis. So, slope of tangent should be

$0.$

$\Rightarrow 2x-4=0$

$\Rightarrow x=2$ is the only point where slope is parallel to x-axis.

So, only 1 tangent exists.

The slope of the tangent to the curve given by

$x = 1 - \cos { \theta }$ ,

$y = \theta -\sin { \theta }$ at

$\theta = \dfrac { \pi }{ 2 }$ is

0%
$0$
0%
$-1$
0%
$1$
0%
Not defined

Explanation

$y=\theta -\sin\theta$

$\Longrightarrow \dfrac { dy }{ d\theta } =1-\cos\theta$

$x=1-\cos\theta$

$\Longrightarrow \dfrac { dx }{ d\theta } =sin\theta$

Slope of the tangent is

$\dfrac{dy}{dx}$

$\Longrightarrow \dfrac{dy}{dx} =\dfrac { 1-\cos\theta}{\sin\theta}$

$\theta =\dfrac { \pi }{ 2 }$

$\Longrightarrow \dfrac { dy }{ dx } =\dfrac { 1-\cos\frac { \pi }{ 2 } }{ \sin\frac { \pi }{ 2 } } =1$

What is the slope of the tangent to the curve

$y=sin^{-1}(sin^2x)$ at

$x=0$ ?

0%
0
0%
1
0%
2
0%
None of the above

Explanation

Slope of tangent to a curve f(x), at a point

$({x}_{o},{y}_{o})$ is given by

$f'(x)$ ,

where

$f'(x)$ is derivative of f(x) at

$x={x}_{o}$

$f(x) =\sin^{-1}({\sin ^{ 2 }{ x } })$

${x}_{o}=0$

$f'\left( x \right) =\cfrac { df\left( x \right) }{ dx } =\cfrac { df(x) }{ d\sin ^{ 2 }{ x } } \times \cfrac { d\sin ^{ 2 }{ x } }{ d\sin x } \times \cfrac { d\sin x }{ dx }$

$f'\left( x \right) =\cfrac { 1 }{ \sqrt { 1-\sin ^{ 2 }{ x } } } \times 2\sin x\times \cos x$

$f'\left( 0 \right) =0$

Slope of tangent to the curve is 0

Find the slope of the normal to the curve

$4x^3+6x^2-5xy-8y^2+9x+14=0$ T the point

$-2, 3$ .

0%
$\infty$
0%
$11$
0%
$\displaystyle\frac{9}{19}$
0%
$\displaystyle-\frac{19}{9}$

Explanation

The given curve is

$4x^3+6x^2-5xy-8y^2+9x+14=0$

Differentiating above equation w.r.t

$x$ , we get

$12{ x }^{ 2 }+12x-5x\cfrac { dy }{ dx } -5y-16y\cfrac { dy }{ dx } +9=0$

$\Rightarrow \cfrac { dy }{ dx } =\cfrac { 12x^{ 2 }+12x-5y+9 }{ \left( 5x+16y \right) }$

$\left( x,y \right) \equiv \left( -2,3 \right)$

$m=\cfrac { dy }{ dx } =\dfrac{12(4)+12(-2)-5(3)+9}{5(-2)+16(3)}=\dfrac{18}{38}=\dfrac{9}{19}$ is the slope of tangent to the curve

Slope of normal is

$-\dfrac{1}{m}=-\dfrac{19}{9}$

A mirror in the first quadrant is in the shape of a hyperbola whose equation is xy =A light source in the second quadrant emits a beam of light that hits the mirror at the point (2,1/2). If the reflected ray is parallel to the y-axis the slope of the incident beam is

0%
$\dfrac{13}{8}$
0%
$\dfrac{7}{4}$
0%
$\dfrac{15}{8}$
0%
$2$

Explanation

Slope of tangent at

$\left ( 2, \frac{1}{2} \right )$

$m=-\dfrac{1}{4}$

$tan \theta =-\dfrac{1}{y}$

$\theta =\phi -90^o$

$tan\theta =4$
Slope of incident ray

$= m$

$\left | \dfrac{m-\left ( -\dfrac{1}{4} \right )}{1+m\left ( -\dfrac{1}{4} \right )} \right |=tan \theta =4$

$\left | \dfrac{4m+1}{4-m} \right |=4$

$4m + 1 = 16 - 4m$

$m=\dfrac{15}{8}$

The value of

$K$ in order that

$f(x) = \sin x - \cos x - Kx + 5$ decreases for all positive real values of

$x$ is given by

0%
$K < 1$
0%
$K\geq 1$
0%
$K > \sqrt {2}$
0%
$K < \sqrt {2}$

Explanation

Given :

$f(x)=\sin x - \cos x - Kx+5$

$f(x)$ decrease for all

$x$ if

$f'(x)<0$

$f'(x) = \cos x + \sin x - K<0$

$\therefore k > \cos x + \sin x$

We know that,

$\cos x+\sin x=\sqrt{2}\left({\dfrac{1}{\sqrt2}\cos x+\dfrac{1}{\sqrt2}\sin x}\right)=\sqrt2\left({\sin\dfrac{\pi}{4}\cos x}+\cos\dfrac{\pi}{4}\sin x\right)=\sqrt2(\sin(\dfrac{\pi}{4}+x))$
and,

$-1\le\sin(\dfrac{\pi}{4}+x)\le1\Rightarrow -\sqrt2\le\sqrt2\cdot\sin(\dfrac{\pi}{4}+x)\le\sqrt2$

$\therefore -\sqrt2\le\cos x+\sin x\le\sqrt2$

$max(\cos x + \sin x) = \sqrt {2}$

$\therefore K > \sqrt {2}$

If the tangent to

$y^{2} = 4ax$ at the point

$(at^{2}, 2at)$ where

$|t| > 1$ is a normal to

$x^{2} - y^{2} = a^{2}$ at the point

$(a \sec \theta, a\tan \theta)$ , then

0%
$t = -cosec \theta$
0%
$t = -\sec \theta$
0%
$t = 2\tan \theta$
0%
$t = 2\cot \theta$

Explanation

We know equation of tangent is

$y=\dfrac{x}{t}+at$ and normal is

$xcos\theta+ycot\theta=2a$

So,

$(at^{2},2at)$ must satisfy the equation of normal

$\Rightarrow t^{2}cos\theta+2t\times cot\theta-2 =0$

So, the roots are

$t=-cosec\theta$ and

$t=2tan\theta$

The function

$f(x)=\cfrac { \sin { x } }{ x }$ is decreasing in the interval

0%
$\left( -\cfrac { \pi }{ 2 } ,0 \right)$
0%
$\left(0, \cfrac { \pi }{ 2 } \right)$
0%
$\left( -\cfrac { \pi }{ 4 } ,0 \right)$
0%
None of these

Explanation

$f(x)=\cfrac { \sin { x } }{ x }$

Let

$f'(x)=\cfrac { x\cos { x } -\sin { x } }{ { x }^{ 2 } }$

Let

$g(x)=x\cos { x } -\sin { x }$

$\quad \quad \quad =\cos { x(x-\tan { x) } }$

where

$\cos x>0$

$\quad \quad x-\tan { x } <0$

$\Rightarrow g(x)<0$

$\Rightarrow f'(x)<0$

$\therefore f(n)$ is a decreasing function in

$\left( 0,\cfrac { \pi }{ 2 } \right)$

The point on the curve

$y = \sqrt {x - 1}$ where the tangent is perpendicular to the line

$2x + y - 5 = 0$ is

0%
$(2, -1)$
0%
$(10, 3)$
0%
$(2, 1)$
0%
$(5, -2)$

Explanation

$\dfrac {dy}{dx} = \dfrac {1}{2\sqrt {x - 1}} = m_{1}$ is the slope of tangent to

$y=\sqrt{x-1}$
Slope of the line

$2x + y - 5 = 0$ is

$m_{2} = -2$
For lines are perpendicular

$m_{1} m_{2} = -1$

$\implies \left (\dfrac {1}{2\sqrt {x - 1}}\right )(-2) = -1$

$\implies \dfrac {2}{2\sqrt {x - 1}} = 1$

$\implies \sqrt {x - 1} = 1$
Squaring both sides,

$\implies x - 1 = 1$

$\implies x = 2$

$\therefore y = \sqrt {x - 1}$

$= \sqrt {2 - 1}$

$= \sqrt {1}$

$\therefore y = 1$

$\therefore (2, 1)$ is the point on the curve

$y=\sqrt{x-1}$

Consider the curve

$y = e^{2x}$ .Where does the tangent to the curve at (0, 1) meet the x-axis ?

0%
$(1, 0)$
0%
$(2, 0)$
0%
$\left(-\dfrac{1}{2}, 0\right)$
0%
$\left(\dfrac{1}{2}, 0\right)$

Explanation

Slope of tangent at any point to the curve is given by

$\left|{ \cfrac { df\left( x \right) }{ dx } }\right|_{ ({ x }_{ 0 }, { y }_{ 0 }) }$

Here

$f(x)=y={e}^{2x}$

$\Rightarrow \cfrac{dy}{dx}=2{e}^{2x}$

At point

$(0,1)$

$\cfrac{dy}{dx}=2{e}^{2(0)}$

$\cfrac{dy}{dx}=2$

Equation of tangent is at point

$({x}_{0},{y}_{0})$

$y-{y}_{0}=m(x-{x}_{0})$

So, tangent at (0,1) is

$y-1=2(x-0)$

$\therefore y=2x+1$

When this tangent meets x-axis,

$y$ co-ordinate becomes zero

Putting

$y$ in above equation as zero, we get

$0=2x+1$

$\Rightarrow x=\cfrac{-1}{2}$

Thus, the tangent to the curve meets at

$\left(-\dfrac{1}{2},0\right)$

The approximate value of

$f(x)={ x }^{ 3 }+5{ x }^{ 2 }-7x+9=0$ at

$x=1.1$ is

0%
$8.6$
0%
$8.5$
0%
$8.4$
0%
$8.3$

Explanation

Given,

$f(x)={ x }^{ 3 }+5{ x }^{ 2 }-7x+9=0$

Let

$x=1$ and

$\Delta x=0.1$

Approximate of a function

$f(x+\Delta x)\approx f(x)+f'(x)\Delta x$

$=f(1)+f'(1)\Delta x$

$=8+0.6=8.6$

The equation to the normal to the hyperbola

$\dfrac {x^{2}}{16} - \dfrac {y^{2}}{9} = 1$ at

$(-4, 0)$ is.

0%
$2x - 3y = 1$
0%
$x = 0$
0%
$x = 1$
0%
$y = 0$

Explanation

$\dfrac { { x }^{ 2 } }{ 16 } -\dfrac { { y }^{ 2 } }{ 9 } =1$

Differentiating above equation, we get

$\dfrac { { 2x } }{ 16 } -\dfrac { 2{ y } }{ 9 } .\dfrac { dy }{ dx } =0$

$\implies \dfrac{dy}{dx}=\dfrac{9x}{16y}$

$\implies m=\left|\dfrac { dy }{ dx }\right|_{(-4,0)} =\dfrac{9(-4)}{ 16(0)} =\infty$ is the slope of tangent

Slope of the normal is

$-\dfrac{1}{m} =0$

Slope of the normal is zero so it is parallel to

$y$ -axis.

Equation of normal is

$y-0=0(x-(-4))$

$\therefore y=0$ is the equation of normal.

Let

$f(x)=2{ x }^{ 3 }-5{ x }^{ 2 }-4x+3,\cfrac { 1 }{ 2 } \le x\le 3$ . The point at which the tangent to the curve is parallel to the X-axis is

0%
$(1,-4)$
0%
$(2,-9)$
0%
$(2,-4)$
0%
$(2,-1)$
0%
$(2,-5)$

Explanation

Given,

$f(x)=2{ x }^{ 3 }-5{ x }^{ 2 }-4x+3$

$f'(x)=6{x}^{2}-10x-4$

If tangent is parallel to X-axis, then

$\cfrac { dy }{ dx } =0\quad$

$\Rightarrow 6{ x }^{ 2 }-10x-4=0\quad$

$\Rightarrow x=\cfrac { 5\pm \sqrt { 25+24 } }{ 6 } =\cfrac { 5\pm 7 }{ 6 }$

$\Rightarrow x=2,\cfrac { -1 }{ 3 }$

$\quad \therefore y=f(x)=2{ (2) }^{ 3 }-5{ (2) }^{ 2 }-4(2)+3=-9$

The equation of the tangent to the curve

$y={ x }^{ 3 }-6x+5$ at

$(2,1)$ is

0%
$6x-y-11=0$
0%
$6x-y-13=0$
0%
$6x+y+11=0$
0%
$6x-y+11=0$

Explanation

The equation of the curve

$y={ x }^{ 3 }-6x+5$

$\Rightarrow \cfrac { dy }{ dx } =3{ x }^{ 2 }-6$

$\Rightarrow { \left( \cfrac { dy }{ dx } \right) }_{ 2,1 }=6$
Now, equation of the tangent at

$(2,1)$ is

$(y-1)=6(x-2)$

$\Rightarrow 6x-y-11=0$

The slope of the normal to the curve

$x=1-a\sin { \theta }$ ,

$y=b\cos ^{ 2 }{ \theta }$ at

$\theta =\dfrac { \pi }{ 2 }$ is

0%
$\dfrac { a }{ 2b }$
0%
$\dfrac { 2a }{ b }$
0%
$\dfrac { a }{ b }$
0%
$\dfrac { -a }{ 2b }$

Explanation

Given,

$x=1-a\sin { \theta }$ and

$y=b\cos ^{ 2 }{ \theta }$
On differentiating with respect to

$\theta$ , we get

$\dfrac { dx }{ d\theta } =-a\cos { \theta }$
and

$\dfrac { dy }{ d\theta } =2b\cos { \theta } \left( -\sin { \theta } \right)$
Then,

$\dfrac { dy }{ dx } =\dfrac { { dy }/{ d\theta } }{ { dx }/{ d\theta } } =\dfrac { 2b }{ a } \sin { \theta }$

$\therefore$ Slope of normal at the point

$\theta =\dfrac { \pi }{ 2 }$ is

$-\dfrac { dx }{ dy } =-\dfrac { 1 }{ { dy }/{ dx } }$

$=-\dfrac { 1 }{ \dfrac { 2b }{ a } \sin { \left( \dfrac { \pi }{ 2 } \right) } } =-\dfrac { a }{ 2b }$

If an edge of a cube measure

$2$ m with a possible error of

$0.5$ cm. Find the corresponding error in the calculated volume of the cube.

0%
$0.6\ m^{3}$
0%
$0.06\ m^{3}$
0%
$0.006\ m^{3}$
0%
$0.0006\ m^{3}$

Explanation

Let

$x$ be the length of an edge of a cube and

$V$ be the volume of that cube.
Thus

$V = x^{3}$
On differentiating w.r.t.

$x$ , we get

$\dfrac {dV}{dx} = 3x^{2}$
Let

$\delta V$ be error in

$V$ and corresponding error

$\delta x$ in

$x$ .

$\therefore \delta V = \dfrac {dV}{dx} \delta x = 3x^{2} \delta x$
Given that,

$x = 2$ m and

$\delta x = 0.5$ cm

$= \dfrac {0.5}{100}$ m

$\therefore \delta V = 3(2)^{2} \left (\dfrac {0.5}{100}\right )$

$= \dfrac {12\times 0.5}{100} = \dfrac {6}{100} = 0.06\ cm^{3}$

The tangents to curve

$y={ x }^{ 3 }-2{ x }^{ 2 }+x-2$ which are parallel to straight line

$y=x$ , are

0%
$x+y=2$ and $x-y=\dfrac { 86 }{ 27 }$
0%
$x-y=2$ and $x-y=\dfrac { 86 }{ 27 }$
0%
$x-y=2$ and $x+y=\dfrac { 86 }{ 27 }$
0%
$x+y=2$ and $x+y=\dfrac { 86 }{ 27 }$

Explanation

Given,

$y={ x }^{ 3 }-2{ x }^{ 2 }+x-2$

On differentiating both sides with respect to

$x$ , we get

$\dfrac { dy }{ dx } =3{ x }^{ 2 }-4x+1$

and

$y=x$

$\Rightarrow \dfrac { dy }{ dx } =1$

$\therefore$ Slope of tangent will be

$3{ x }^{ 2 }-4x+1$ .

Since, the tangent is parallel to line

$y=x$ .

$\therefore 3{ x }^{ 2 }-4x+1=1$

$\Rightarrow 3{ x }^{ 2 }-4x=0$

$\Rightarrow x\left( 3x-4 \right) =0$

$\Rightarrow x=0,\dfrac { 4 }{ 3 }$

When

$x=0$ , then

$y=-2$

When

$x=\dfrac { 4 }{ 3 }$ , then

$y=\dfrac { -50 }{ 27 }$

Now, equation of tangents at point

$\left( 0,-2 \right)$ is

$y-{ y }_{ 1 }=\dfrac { dy }{ dx } \left( x-{ x }_{ 1 } \right)$

$\Rightarrow y+2=1\left( x-0 \right)$

$\Rightarrow y+2=x$

$\Rightarrow x-y=2$ ...(i)

and equation of tangents at point

$\left( \dfrac { 4 }{ 3 } ,-\dfrac { 50 }{ 27 } \right)$ is

$y-{ y }_{ 1 }=\dfrac { dy }{ dx } \left( x-{ x }_{ 1 } \right)$

$y+\dfrac { 50 }{ 27 } =x-\dfrac { 4 }{ 3 }$

$\Rightarrow x-y=\dfrac { 50 }{ 27 } +\dfrac { 4 }{ 3 }$

$\Rightarrow x-y=\dfrac { 50+36 }{ 27 }$

$\Rightarrow x-y=\dfrac { 86 }{ 27 }$ ....(ii)

Hence, equations (i) and (ii) are required equations of the tangents.

The slope of tangent to the curve

$x=t^2 + 3t - 8, y = 2t^2 - 2t - 5$ at the point

$(2, -1)$ is :

0%
$\dfrac {22}{7}$
0%
$\dfrac {6}{7}$
0%
$-6$
0%
None of these

Explanation

Given curves are

$x=t^{2} + 3t - 8$ ...

$(i)$ and

$y = 2t^{2} - 2t - 5$ ...

$(ii)$

$(2,-1),$ From

$(i)$

$t^{2}+3t-10=0\implies t=2$ or

$t=-5$

From

$(ii)$

$2t^{2}-2t-4=0\implies t^{2}-t-2=0\implies t=2$ or

$t=-1$

From both the solutions, we get

$t=2$

Differentiating both the equations w.r.t.

$t$ , we get

$\dfrac{dx}{dt}=2t+3$ ........

$(iii)$

$\dfrac{dy}{dt}=4t-2$ .......

$(iv)$

Now,

$\dfrac{dy}{dx} = \dfrac {\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$

$= \dfrac {4t-2}{2t+3}$ .... From

$(iii)$ and

$(iv)$

$\therefore \dfrac{dy}{dx} = \dfrac {4t-2}{2t+3}$ is the slope of tangent to the given curve

$\therefore \left|\dfrac{dy}{dx}\right|_{(2,-1)} = \left|\dfrac {4t-2}{2t+3}\right|_{t=2}=\dfrac{8-2}{4+3}=\dfrac{6}{7}$ is the slope of tangent to the given curve at

$(2,-1)$

The points at which the tangent to the curve

$y = x^3 - 3x^2 - 9x + 7$ is parallel to the x-axis are

0%
$(3, - 20)$ and $(- 1, 12)$
0%
$(3, 20)$ and $(1, 12)$
0%
$(1, -10)$ and $(2, 6)$
0%
None of these

Explanation

Tangent to the curve is parallel to the axis is when slope of the tangent is 0.

$\therefore$ Equation of the curve is

$y=x^3-3x^2-9x+7=0$ ......

$(i)$

$\therefore \dfrac{dy}{dx}=3x^2-6x-9$

Now, the tangent is parallel to x-axis, then slope of the tangent is zero or we can say that

$\dfrac{dy}{dx}=0$ .

$\Rightarrow 3x^2-6x-9=0$

$\Rightarrow 3(x^2-2x-3)=0$

$\Rightarrow (x-3)(x+1)=0$

$\Rightarrow x=3, -1$

When

$x=3$ , then from Eq.

$(i),$ we get

$y=(3)^3-(3)\cdot (3)^2-9\cdot 3+7$

$=27-27-27+7=-20$

When

$x =-1,$ then from Eq.

$(i),$ we get

$y=(-1)^3-3(-1)^2-9(-1)+7$

$=-1-3+9+7=12$
Hence, the points at which the tangent is parallel to x-axis are

$(3, -20)$ and

$(-1, 12)$ .

The slope of the tangent to the curve

$y=3{ x }^{ 2 }-5x+6$ at

$\left( 1,4 \right)$ is

0%
$-2$
0%
$1$
0%
$0$
0%
$-1$
0%
$2$

Explanation

Given curve is

$y=3{ x }^{ 2 }-5x+6\Rightarrow \dfrac { dy }{ dx } =6x-5$

$\therefore$ Slope of tangent at

$\left( 1,4 \right) =6\times 1-5=1$

$y=8{ x }^{ 3 }-60{ x }^{ 2 }+144x+27$ is a strictly decreasing function in the interval

0%
$(-5,6)$
0%
$\left( -\infty ,2 \right)$
0%
$(5,6)$
0%
$\left( 3,\infty \right)$
0%
$(2,3)$

Explanation

Given,

$y=8{ x }^{ 3 }-60{ x }^{ 2 }+144x+27$

$\cfrac { dy }{ dx } =24{ x }^{ 2 }-120x+144\quad$

For function to be strictly decreasing

$\cfrac { dy }{ dx } <0$

$\Rightarrow 24{ x }^{ 2 }-120x+144<0$

$\Rightarrow \left( { x }^{ 2 }-5x+6 \right) <0$

$\Rightarrow \left( { x }^{ 2 }-2x-3x+6 \right) <0$

$\Rightarrow (x-3)(x-2)<0$

$\therefore x\in (2,3)$

If the tangent at

$(1,1)$ on

${ y }^{ 2 }=x{ (2-x) }^{ 2 }$ meets the curve again at

$P$ , then

$P$ is

0%
$(4,4)$
0%
$(-1,2)$
0%
$\left( \cfrac { 9 }{ 4 } ,\cfrac { 3 }{ 8 } \right)$
0%
$(1,2)$

Explanation

We have

$\Rightarrow { y }^{ 2 }={ x }^{ 3 }-4{ x }^{ 2 }+4x$ .....(i)

$\Rightarrow 2y\cfrac { dy }{ dx } =3{ x }^{ 2 }-8x+4$

$\Rightarrow \cfrac { dy }{ dx } =\cfrac { 3{ x }^{ 2 }-8x+4 }{ 2y }$

$\Rightarrow { \left[ \cfrac { dy }{ dx } \right] }_{ (1,1) }=\cfrac { 3-8+4 }{ 2 } =-\cfrac { 1 }{ 2 }$

The equation of the tangent at

$(1,1)$ is

$-1=-\cfrac { 1 }{ 2 } (x-1)$

$\Rightarrow x+2y-3=0$ .....(ii)

On solving Eqs. (i) and (ii) we get

$x=9/4$ and

$y=3/8$

Hence,the coordinates of

$P$ are

$\left( \cfrac { 9 }{ 4 } ,\cfrac { 3 }{ 8 } \right)$ .

The tangent to the curve

$y=a{ x }^{ 2 }+bx$ at

$\left( 2,-8 \right)$ is parallel to

$X$ -axis. Then,

0%
$a=2, b=-2$
0%
$a=2, b=-4$
0%
$a=2, b=-8$
0%
$a=4, b=-4$

Explanation

Given,

$y=a{ x }^{ 2 }+bx$
On differentiating with respect to

$x$ , we get

$\dfrac { dy }{ dx } =2ax+b$
At

$\left( 2,-8 \right) ,$

${ \left( \dfrac { dy }{ dx } \right) }_{ \left( 2,-8 \right) }=4a+b$
Since tangent is parallel to

$X$ -axis.
Thus

$\dfrac { dy }{ dx } =0\Rightarrow b=-4a$ ....(i)
Now, point

$\left( 2,-8 \right)$ is on the curve

$y=a{ x }^{ 2 }+bx$
Therefore,

$-8=4a+2b$ ....(ii)
On solving equations (i) and (ii), we get

$a=2,b=-8$

Find the critical points of the function

$f (x)= (x - 2)^{2/3} (2x + 1)$

0%
$-1$ and $2$
0%
$1$
0%
$1$ and $- 2$
0%
$1$ and $2$

Explanation

We have,

$f(x)=(x-2)^{2/3}(2x+1)$

$\Rightarrow f'(x)=\frac {2}{3}(x-2)^{-1/3}(2x+1)+(x-2)^{2/3}\cdot 2$

$=\dfrac {2(2x+1)}{3(x-2)^{1/3}}+2 (x-2)^{2/3}$

$= \dfrac {2(2x+1)+6(x-2)}{3(x-2)^{1/3}}$

$=\dfrac {4x+2+6x-12}{3(x-2)^{1/3}}$

$=\dfrac {10(x-1)}{3(x-2)^{1/3}}$

For critical points,

$f'(x)=0$

$\Rightarrow x=1$

and

$f'(x)$ is not defined at x= 2.

If the straight line

$y -2x +1=0$ is the tangent to the curve

$xy+ax+by=0$ at

$x=1,$ then the values of

$a$ and

$b$ are respectively :

0%
1 and 2
0%
1 and -1
0%
-1 and 2
0%
-1 and -2
0%
1 and -2

Explanation

Equation of tangent is $y-2x+1=0$

It is tangent at $x=1$ , so for $x=1$

$y-2(1)+1=0\\ \Rightarrow y=1$

So, its is tangent to the curve at $(1,1)$

Slope of tangent $= -\left (\dfrac{-2}{1}\right)=2$

$xy+ax+by=0\\ y+x\dfrac { dy }{ dx } +a+b\dfrac { dy }{ dx } =0$

Now $\dfrac { dy }{ dx } =2$ at $(1,1)$

$1+1(2)+a+b(2)=0\\ \Rightarrow a+2b=-3$ .....(i)

$(1,1)$ also lies on the curve $xy+ax+by=0$

$\Rightarrow 1(1)+a(1)+b(1)=0\\ \Rightarrow a+b=-1$ .......(ii)

Solving (i) and (ii), we get

$\Rightarrow a=1,b=-2$

So, option E is correct.

If the angle between the curves

$y = 2^x$ and

$y=3^x$ is

$\alpha,$ then the value of

$\tan \alpha$ is equal to :

0%
$\dfrac { \log \left( \dfrac {3}{2} \right) } { 1 + ( \log 2)( \log 3 ) }$
0%
$\dfrac {6}{7}$
0%
$\dfrac {1}{7}$
0%
$\dfrac { \log \left( 6 \right) } { 1 + ( \log 2)( \log 3 ) }$
0%
$0^o$

Explanation

Given curves are

$y = 2^x$ and

$y =3^x$
The point of intersection is

$3^x = 2^x \Rightarrow x = 0$
On differentiating w.r.t.

$x,$ we get

$\dfrac {dy}{dx} = 2^x \log 2 = m_1$
and

$\dfrac {dy}{dx} = 3^x \log 3 = m_2$
Therefore,

$\tan \alpha = \dfrac {m_2 - m_1}{1 + m_1m_2}$

$= \dfrac {3^x \log 3 - 2^x \log 2 }{1+3^x \times 2^x \log 3 \times \log 2}$
At

$x=0$ ,

$\tan \alpha = \dfrac {3^0 \log 3 - 2^0 \log 2 }{1 + 3^0 \times 2^0 \log 2 \log 3 }$

$= \dfrac { \log \dfrac {3}{2} } { 1 + \log 2 \log 3 }$

If the tangent at each point of the curve

$y=\cfrac { 2 }{ 3 } { x }^{ 3 }-2a{ x }^{ 2 }+2x+5$ makes an acute angle with positive direction of X-axis then

0%
$a\ge 1$
0%
$-1\le a\le 1$
0%
$a\le -1$
0%
None of these

Explanation

On differentiating, We get

$\dfrac{dy}{dx}=2x^2-4ax+2$

If Tangent makes acute angle with +ve x axis, then

$\dfrac{dy}{dx}\ge0$ For all x

$x^2-2ax+1\ge0\\ \Rightarrow 4a^2-4\le0\Rightarrow a^2\le1\\ -1\le a\le1$

The equation of the tangent to the curve

$\sqrt {\dfrac {x}{a}} + \sqrt {\dfrac {y}{b}} = 1$ at the point

$(x_{1}, y_{1})$ is

$\dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = k$ . Then, the value of

$k$ is

0%
$2$
0%
$1$
0%
$3$
0%
$7$
0%
$\sqrt {2}$

Explanation

Given curve is

$\sqrt {\dfrac {x}{a}} + \sqrt {\dfrac {y}{b}} = 1$ ... (i)
On differentiating w.r.t. to

$x$ , we get

$\dfrac {1}{\sqrt {a}} \cdot \dfrac {1}{2\sqrt {x}} + \dfrac {1}{\sqrt {b}} \cdot \dfrac {1}{2\sqrt {y}} \dfrac {dy}{dx} = 0$

$\Rightarrow \dfrac {dy}{dx} = -\dfrac {\sqrt {b}\sqrt {y}}{\sqrt {a}\sqrt {x}} \Rightarrow \left [\dfrac {dy}{dx}\right ]_{(x_{1}, y_{1})} = \dfrac {-\sqrt {b}\sqrt {y_{1}}}{\sqrt {a}\sqrt {x_{1}}}$
Equation of tangent passing through the point

$(x_{1}, y_{1})$ is

$(y - y_{1}) = \dfrac {-\sqrt {b}\sqrt {y_{1}}}{\sqrt {a}\sqrt {x_{1}}} (x - x_{1})$

$\dfrac {y}{\sqrt {by_{1}}} - \dfrac {y_{1}}{\sqrt {by_{1}}} = -\dfrac {x}{\sqrt {ax_{1}}} + \dfrac {x_{1}}{\sqrt {ax_{1}}}$

$\Rightarrow \dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = \dfrac {x_{1}}{\sqrt {ax_{1}}} + \dfrac {y_{1}}{\sqrt {by_{1}}} = \sqrt {\dfrac {x_{1}}{a}} + \sqrt {\dfrac {y_{1}}{b}}$

$\Rightarrow \dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = 1$ [from Eq. (i)]

$\left [\because at (x_{1}, y_{1}), \sqrt {\dfrac {x_{1}}{a}} + \sqrt {\dfrac {y_{1}}{b}} = 1\right ]$
But

$\dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = k$ (given)
Therefore,

$k = 1$

The slope of the normal to the curve

$y = x^2 - \dfrac{1}{x^2}$ at

$(-1, 0)$ is

0%
$\dfrac{1}{4}$
0%
$- \dfrac{1}{4}$
0%
$4$
0%
$-4$
0%
$0$

Explanation

Given curve is

$y = x^2 - \dfrac{1}{x^2}$
On differentiating both sides w.r.t.

$x$ , we get

$\dfrac{dy}{dx} = 2x + \dfrac{2}{x^3}$
At point

$(-1, 0)$ .

$\dfrac{dy}{dx} = 2(-1) + \dfrac{2}{(-1)^3} = -4$
Therefore, slope of normal to the curve

$= - \dfrac{1}{dy/dx}$

$= - \dfrac{1}{-4} = \dfrac{1}{4}$

The point on the curve

$y = 5 + x - x^{2}$ at which the normal makes equal intercepts is

0%
$(1, 5)$
0%
$(0, -1)$
0%
$(-1, 3)$
0%
$(0, 3)$
0%
$(0, 5)$

Explanation

Given curve is

$y = 5 + x - x^{2}$ ..... (i)
On differentiating w.r.t.

$x$ , we get

$\dfrac {dy}{dx} = 1 - 2x$
Slope of normal

$= \dfrac {-1}{(dy/dx)} = \dfrac {-1}{1 - 2x}$

$= \dfrac {1}{2x - 1}$ .... (ii)
Since, normal makes equal intercepts.

$\therefore \theta = 135^{\circ}$
From Eq. (ii), we have

$\dfrac {1}{2x - 1} = \tan 135^{\circ} = -1$

$\Rightarrow -2x + 1 = 1\Rightarrow x = 0$
Then, from Eq. (i),

$y = 5$
So, the required point is

$(0, 5)$ .

The function

$f(x) = 2x^3 - 15 x^2 + 36 x + 6$ is strictly decreasing in the interval

0%
$(2, 3)$
0%
$( - \infty, 2)$
0%
$(3, 4)$
0%
$(- \infty, 3) \cup (4, \infty)$
0%
$(- \infty, 2) \cup (3, \infty)$

Explanation

$f(x)=2x^3-15x^2+36x+6$

$f'(x)=6x^2-30x+36$

$f'(x)=0$

$6x^2-30x+36=0$

$\Rightarrow x^2-5x+6=0$

$\Rightarrow (x-2)(x-3)=0$

$\Rightarrow x=2$ or

$x=3$

Since, from figure in interval

$(2,3)$ is negative it means

$f(x)$ is strictly decreasing in this interval.

Hence, A is the correct option.

A normal to parabola, whose inclination is

$30^o$ , cuts it again at an angle of.

0%
$\tan^{-1}\left(\displaystyle\frac{\sqrt{3}}{2}\right)$
0%
$\tan^{-1}\left(\displaystyle\frac{2}{\sqrt{3}}\right)$
0%
$\displaystyle\tan^{-1}\cdot(2\sqrt{3})$
0%
$\displaystyle\tan^{-1}\left(\displaystyle\frac{1}{2\sqrt{3}}\right)$

Explanation

Equation of normal in terms of parameter

To parabola

$y^2=4ax$ at

$(at^2, 2at)$ , the equation of normal is

$y=-tx+2at+at^3$

To parabola

$x^2=4ay$ at

$(2at, at^2)$ the equation of normal is

$x+ty=2at+at^3$

Let the normal

$P(at_1,2at_1)$ by

$y= -t_1x + 2at_1+at_1^3$

$\theta$

$=-t_1=$ slope of normal

It meets the curve at

$\theta$ say

$(at_2^2,2at_2)$

Now angle between the normal and parabola

$=$ Angle the normal and tangent at

$\theta$ (i.e)

$t_2= x + at_2^2$

$\tan \phi$ =

$\dfrac{m_1-m_2}{1+m_1m_2}$

$=\dfrac{-t_1-\dfrac{1}{t_2}}{1+\dfrac{(-t_1)}{(-t_2)}}$

$=\dfrac{-(t_1t_2+1)}{t_2-t_1}$

$=\dfrac{-(t_1^2+1)}{t_2-t_1}$

$=\dfrac{-(-t_1^2-1)}{\dfrac{-2(1+t_1^2)}{(t_1)}}$

$=\dfrac{-t_1}{2}$

${\tan}\phi=\dfrac{\tan\theta}{2}$

hence,

$\phi$ =

$\tan^{-1}\dfrac{\tan30^.}{2} = \tan^{-1}(\dfrac{1}{2\times3^{\tfrac{1}{2}}})$

If the slope of the tangent to the curve

$y=a{ x }^{ 3 }+bx+4$ at

$(2,14) = 21$ , then the values of

$a$ and

$b$ are respectively

0%
$2,-3$
0%
$3,-2$
0%
$-3,-2$
0%
$2,3$

Explanation

$(2,14)$ lies on the curve

$y=ax^3+bx+4$

$8a+2b+4=14$

$4a+b=5$ ----(1)

$\dfrac{dy}{dx}=3ax^2+b|_{x=2}=21$

$12a+b=21$ --(2)

Solving equations (1) and (2), we get

$a=2, b=-3$

CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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