MCQ Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 1

The tangent at the point

$(2, -2)$ to the curve,

$x^2y^2-2x=4(1-y)$ does not pass through the point.

0%
$(8, 5)$
0%
$\left(4, \displaystyle\frac{1}{3}\right)$
0%
$(-2, -7)$
0%
$(-4, -9)$

Explanation

$x^2y^2-2x=4-4y$

$2xy^2+2y\cdot x^2\cdot\displaystyle\frac{dy}{dx}-2=-4\cdot\frac{dy}{dx}$

$\displaystyle\frac{dy}{dx}(2y\cdot x^2+4)=2-2x\cdot y^2$

$\left.\displaystyle\frac{dy}{dx}\right|_{2, -2}=\displaystyle\frac{2-2\times 2\times 4}{2(-2)\times 4+4}=\frac{+14}{+12}=\frac{7}{6}$
eq of tangent :

$(y+2)=\displaystyle\frac{7}{6}(x-2)\Rightarrow 7x-6y=26$ .

$(-2,-7)$ does not satisfy above eq.

If the tangent to the conic,

$y - 6 = x^2$ at (2, 10) touches the circle,

$x^2 + y^2 + 8x - 2y = k$ (for some fixed k) at a point

$(\alpha, \beta)$ ; then

$(\alpha, \beta)$ is;

0%
$\displaystyle \left( -\frac{4}{17}, \frac{1}{17} \right)$
0%
$\displaystyle \left( -\frac{7}{17}, \frac{6}{17} \right)$
0%
$\displaystyle \left( -\frac{6}{17}, \frac{10}{17} \right)$
0%
$\displaystyle \left( -\frac{8}{17}, \frac{2}{17} \right)$

Explanation

Given equation of conic is

$y - 6 = x^2$

$\dfrac{dy}{dx}=2x$
Slope of tangent at

$(2,10)$ is

$4.$

Equation of tangent to conic is

$y-10=4(x-2)$

$\Rightarrow y-10=4x-8$

$\Rightarrow y=4x+2$ .....(1)

Given equation of circle is

$x^{ 2 }+y^{ 2 }+8x-2y=k$

$\Rightarrow (x+4)^2+(y-1)^2=k+17$

Given

$(\alpha, \beta)$ is a point of tangency for fixed

$k$
Radius = Length of tangent from center

$(-4,1)$

$\Rightarrow \sqrt{k+17}=\dfrac{15}{\sqrt{17}}$

$\Rightarrow k+17=\dfrac{225}{17}$

So, equation of circle at

$(\alpha, \beta)$ is

$(\alpha+4)^2+(\beta-1)^2=\dfrac{225}{17}$

Now, eqn (1) is tangent to the circle for fixed

$k$ at

$(\alpha,\beta)$

$\Rightarrow (\alpha+4)^2+(4\alpha+1)^2=\dfrac{225}{17}$

$\Rightarrow 17\alpha^2+16\alpha+17=\dfrac{225}{17}$

$\Rightarrow 289\alpha^2+272\alpha+64=0$

$\Rightarrow \alpha =\dfrac{-8}{17}$ (Here

$D=0$ )

So, by (1),

$\beta =\dfrac{2}{17}$

Let b be a nonzero real number. Suppose

$f : R \rightarrow R$ is a differentiable function such that

$f(0) = 1$ .
If the derivative f' of f satisfies the equation

$f'(x) = \dfrac{f(x)}{b^2 + x^2}$ for all

$x \in R$ , then which of the following statements is/are TRUE?

0%
If $b > 0$ , then f is an increasing function
0%
If $b < 0$ , then f is a decreasing function
0%
$f\left( x \right) f\left( -x \right) =1$ for all $x\in R$
0%
$f(x) f(x) = 0$ for all $x \in R$

Explanation

$f'(x) = \dfrac{f(x)}{b^2 + x^2} \Rightarrow \dfrac{f'(x)}{f(x)} = \dfrac{1}{b^2 + x^2}$

$\Rightarrow ln f(x) = \dfrac{1}{b} \tan^{-1} \dfrac{x}{b} + c$

$\therefore f(0) = 1 \, \therefore 0 = c \, \Rightarrow c = 0$

$\therefore ln f(x) = \dfrac{1}{b} \tan^{-1} \dfrac{x}{b}$

$\Rightarrow f(x) = e^{\dfrac{1}{b} \tan^{-1} \dfrac{x}{b}}$

$\therefore f'(x) = e^{\dfrac{1}{b} \tan^{-1} \dfrac{x}{b}}. \dfrac{1}{b^2 + x^2 } > 0$ for

$x \in R$

$\Rightarrow f(x)$ is increasing option (A)

$f(x). f(-x) = (e^{\dfrac{1}{b} \tan^{-1} \dfrac{x}{b}} ) (e^{-\dfrac{1}{b} \tan^{-1} \dfrac{x}{b} } ) = 1$

What is the

$x$ -coordinate of the point on the curve

$f(x) = \sqrt {x}(7x - 6)$ , where the tangent is parallel to

$x$ -axis?

0%
$-\dfrac {1}{3}$
0%
$\dfrac {2}{7}$
0%
$\dfrac {6}{7}$
0%
$\dfrac {1}{2}$

Explanation

Given,

$f(x) = \sqrt {x}(7x - 6) = 7x^{3/2} - 6x^{1/2}$
Thus

$f'(x) = 7\times \dfrac {3}{2}x^{1/2} - 6\times \dfrac {1}{2} x^{-1/2}$
When tangent is parallel to

$x$ axis

$f'(x) = 0$ .
Therefore,

$\dfrac {21}{2}x^{1/2} - 3x^{-1/2} = 0$

$\Rightarrow \dfrac {21}{2}\sqrt {x} = \dfrac {3}{\sqrt {x}}$

$\Rightarrow 7x = 2$

$\Rightarrow x = \dfrac {2}{7}$

Consider the following statements in respect of the function

$f(x) = x^{3} - 1, \quad x\epsilon [-1, 1]$
I.

$f(x)$ is increasing in

$[-1, 1]$
II.

$f'(x)$ has no root in

$(-1, 1)$ .
Which of the statements given above is/ are correct?

0%
Only I
0%
Only II
0%
Both I and II
0%
Neither I nor II

Explanation

Given the function

$f(x) = x^{3} - 1, \quad x\epsilon [-1, 1]$

$f'(x)=3x^2$ which is positive for all

$x$

and at

$x=0$ ,

$f'(x)=0$ ,

$f''(x)=0$
I.

$f(x)$ is increasing in

$[-1, 1]$

$f(-1)=-2$ and

$f(1)=0$

Thus

$f(x)$ is increasing in the given interval.

II.

$f'(x)$ has a root ,

$x=0$ in

$(-1, 1)$ .

$\therefore$ only I is correct.

$\dfrac{x^2}{f(4a)}=\dfrac{y^2}{f(a^2-5)}$ respresents and ellipse with major axis as y-axis and

$f$ is a decreasing function, then

0%
$a \in (-\infty, 1)$
0%
$a \in (5, \infty)$
0%
$a \in (1, 4)$
0%
$a \in (-1, 5)$

Explanation

Since y-axis is major axis.

$\Rightarrow f(4a) < f(a^2-5)$

$\Rightarrow 4a > a^2-5$ (Q

$f$ is decreasing)

$\Rightarrow a^2-4a - 5 < 0$

$\Rightarrow a \in (-1, 5)$

The values of

$\mathrm{x}$ at which

$\mathrm{f}(\mathrm{x})=\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}$ is stationary are given by

0%
$\mathrm{n}\pi,\ \forall n\in Z$
0%
$(2\displaystyle \mathrm{n}+1)\frac{\pi}{2},\ \forall n\in Z$
0%
$\displaystyle \frac{\mathrm{n}\pi}{4},\ \forall n\in Z$
0%
$\displaystyle \frac{\mathrm{n}\pi}{2},\ \forall n\in Z$

Explanation

Stationary means

$f'(x)=0$

$f'(x)=cos x$

$f'(x)=0$ when

$x=(2n+1)\dfrac {\pi}{2}$

The number of stationary points of

$\mathrm{f}(\mathrm{x})=\mathrm{s}\mathrm{i}\mathrm{n} \mathrm{x}$ in

$[0, 2{\pi}]$ are

0%
1
0%
2
0%
3
0%
4

Explanation

$f'(x)=cos x$

$f'(x)=0$ when

$x=\dfrac {\pi}{2}, \dfrac {3\pi}{2}$

Find the equation of a line passing through

$(-2,3)$ and parallel to tangent at origin for the circle

$\displaystyle x^{2}+y^{2}+x-y=0$

0%
$x -2 y + 5 = 0$
0%
$x -4 y + 3 = 0$
0%
$x - y + 5 = 0$
0%
$2x - y + 6 = 0$

Explanation

Given,

$x^{2}+y^{2}+x-y=0$
Differentiating w.r.t

$x$ , we get

$2x+2y\cfrac{dy}{dx}+1-\cfrac{dy}{dx}=0$

$\Rightarrow \cfrac{dy}{dx}=\cfrac{1+2x}{1-2y}$
Thus slope of the tangent at origin is,

$m=\left(\cfrac{dy}{dx}\right)_{(0,0)}=1$
Hence required line through

$(-2,3)$ is,

$(y-3)=1(x+2)\Rightarrow x-y+5=0$

Stationary point of

$\displaystyle \mathrm{y}=\frac{\log \mathrm{x}}{\mathrm{x}}(\mathrm{x}>0)$ is

0%
$(1, 0)$
0%
$(\displaystyle \mathrm{e},\frac{1}{\mathrm{e}})$
0%
$(\displaystyle \frac{1}{\mathrm{e}},-\mathrm{e})$
0%
$(\displaystyle \frac{1}{\mathrm{e}'}\frac{1}{\mathrm{e}})$

Explanation

$y'=\frac {1}{x^2}-\frac {log x}{x^2}$

$=(1-log x)\frac {1}{x^2}$
and

$y'=0$ for stationary point

$\therefore x=e$

I: lf

$\mathrm{f}'(\mathrm{a})<0$ then the function

$\mathrm{f}$ is decreasing at

$\mathrm{x}=\mathrm{a}$
II: lf

$\mathrm{f}$ is decreasing at

$\mathrm{x}=\mathrm{a}$ then

$\mathrm{f}'(\mathrm{a})<0$
Which of the above statements are true ?

0%
onlyI
0%
only II
0%
both I and II
0%
neither I nor II

Explanation

(I)

$f'(a)<0$ Which implies f is decreasing at

$x=0$
(II) If f is decresing at

$x=a$
which implies

$f'(a)<0$
has slope is negative for decreasing function

The slope of tangent to the curve

$y=\int_{0}^{x}\displaystyle \frac{dx}{1+x^{3}}$ at the point where

$x=1$ is

0%
$\displaystyle \frac{1}{2}$
0%
$1$
0%
$\displaystyle \frac{1}{4}$
0%
none of these

Explanation

$y=\int _{ 0 }^{ x }{ \cfrac { dx }{ 1+{ x }^{ 3 } } } \\ \cfrac { dy }{ dx } =\cfrac { 1 }{ 1+{ x }^{ 3 } } =\cfrac { 1 }{ 2 }$

The value of

$\mathrm{x}$ at which

$\mathrm{f}(\mathrm{x})=$ cosx is stationary are given by

0%
$\mathrm{n}\pi,\ \forall n\in Z$
0%
$(2\displaystyle \mathrm{n}+1)\frac{\pi}{2},\ \forall n\in Z$
0%
$\displaystyle \frac{\mathrm{n}\pi}{4},\ \forall n\in Z$
0%
$\displaystyle \frac{\mathrm{n}\pi}{2},\ \forall n\in Z$

Explanation

$f'(x)=\sin x$ (

$f'(x)=0$ means stationary)

$f'(x)=0$
when

$x=n\pi$

The number of stationary points of

$\mathrm{f}(\mathrm{x})=\cos \mathrm{x}$ in

$[0, 2{\pi}]$ are

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$f'(x)=\sin x$

$f'(x)=0$ when

$x=0, \pi, 2\pi$

The curve

$\displaystyle y-e^{xy}+x=0$ has a vertical tangent at

0%
$(1, 1)$
0%
$(0, 1)$
0%
$(1, 0)$
0%
$(0,0)$

Explanation

Given,

$\displaystyle y-e^{xy}+x=0$
Differentiating w.r.t

$x$

$\cfrac{dy}{dx}-e^{xy}(y+x\cfrac{dy}{dx})+1=0$

$\Rightarrow \cfrac{dy}{dx}=\cfrac{1-ye^{xy}}{xe^{xy}-1}$
Thus for vertical tangent,

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

$\Rightarrow xe^{xy}-1=0$
Hence required point is

$(1,0)$

The stationary point of

$\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-10\mathrm{x}+43$ is

0%
(5, 18)
0%
(18, 5)
0%
(5, 5)
0%
(5, 15)

Explanation

$f'(x)=2x-10$

$f'(x)=0$ at

$x=5$

$f(5)=25-50+43$

$=18$
So stationary point

$=(5, 18)$

The point on the curve

$\displaystyle y=x^{2}-3x+2$ at which the tangent is perpendicular to the line

$y = x$ is -

0%
$(0, 2)$
0%
$(1, 0)$
0%
$(-1, 6)$
0%
$(2, -2)$

Explanation

Let the point be

$P(a,b)$
Now the given curve is

$y=x^2-3x+2$
Differentiating w.r.t

$x$

$\cfrac{dy}{dx}=2x-3$
Thus slope of tangent at P is

$=\left(\cfrac{dy}{dx}\right)_{(a,b)}=2a-3$
But given the tangent is perpendicular to line

$y=x$

$\Rightarrow 2a-3=-1\Rightarrow a=1$
Also the point P lies on the given curve,

$b=a^2-3a+2=0$
Therefore, the point P is

$(1,0)$
Hence, option 'B' is correct.

If tangent to curve at a point is perpendicular to

$x$ - axis then at that point -

0%
$\displaystyle \frac{dy}{dx}=0$
0%
$\displaystyle \frac{dx}{dy}=0$
0%
$\displaystyle \frac{dy}{dx}=1$
0%
$\displaystyle \frac{dy}{dx}=-1$

Explanation

We know slope of tangent to the any curve

$'y'$ is

$m=\cfrac{dy}{dx}$
Now if line is perpendicular to x-axis this means its slope is

$\infty$

$\Rightarrow \cfrac{dy}{dx} \to \infty \Rightarrow \cfrac{dx}{dy}=0$
Hence, option 'C' is correct.

$y = f(x)$ be the equation of a parabola which is touched by the line

$y = x$ at the point where

$x = 1$ Then

0%
$f'(1) = 1$
0%
$f'(0) = f'(1)$
0%
$2f(0) = 1 - f'(0)$
0%
$f(0) + f'(0) + f"(0) = 1$

Explanation

Let,

$y=ax^2+bx+c$

$\therefore y'=2ax+b$
Given line

$y=x$ touches this parabola at

$x=1$

$\Rightarrow y(1)=1\Rightarrow a+b+c=1 ..(1)$
and

$y"(1)=1\Rightarrow 2a+b=1 ..(2)$
Solving (1) and (2) we get

$2c=1-b \Rightarrow 2f(0)=1-f'(0)$

The slope of the curve

$\displaystyle y=\sin x+\cos ^{2}x$ is zero at the point where -

0%
$\displaystyle x=\frac{\pi }{4}$
0%
$\displaystyle x=\frac{\pi }{2}$
0%
$\displaystyle x=\pi$
0%
No where

The slope of the tangent to the curve

$\displaystyle y=\sin x$ at point

$(0, 0)$ is

0%
$1$
0%
$0$
0%
$\displaystyle \infty$
0%
None of these

Explanation

Given

$y=\sin x$

$\Rightarrow \cfrac{dy}{dx}=\cos x$
Thus, slope of tangent at

$(0,0)$ is,

$m=\left (\cfrac{dy}{dx}\right )_{(0.0)}=\cos (0)=1$
Hence, option 'A' is correct.

If tangent at a point of the curve

$y = f(x)$ is perpendicular to

$2x - 3y = 5$ then at that point

$\displaystyle \dfrac{dy}{dx}$ equals

0%
$\dfrac 2 3$
0%
$-\dfrac 2 3$
0%
$\dfrac 3 2$
0%
$-\dfrac 3 2$

Explanation

We know slope of tangent to the curve is

$\displaystyle \frac{dy}{dx}$
Now slope of line

$2x - 3y = 5$ is

$\displaystyle \frac{2}{3}$
and tangent to

$y=f(x)$ is perpendicular to this line.

$\displaystyle \therefore$

$\displaystyle \frac{dy}{dx}=-\frac{3}{2}$
Hence, option 'D' is correct.

The inclination of the tangent w.r.t.

$x$ - axis to the curve

$\displaystyle x^{2}+2y=8x-7$ at the point

$x = 5$ is

0%
$\displaystyle\dfrac{ \pi }4$
0%
$\displaystyle\dfrac{ \pi }3$
0%
$\displaystyle\dfrac{3 \pi }4$
0%
$\displaystyle\dfrac{ \pi }2$

Explanation

Given,

$\displaystyle x^{2}+2y=8x-7$

$\Rightarrow 2y=8x-x^2-7$

$\therefore 2\cfrac{dy}{dx}=8-2x\Rightarrow \cfrac{dy}{dx}=4-x$
Thus slope of tangent to the given curve at

$x=5$ is,

$m=-1$
If

$'\theta '$ is the inclination of the tangent w.r.t

$x-axis$ then,

$\tan\theta=-1\Rightarrow \theta=\cfrac{3\pi}{4}$
Hence, option 'C' is correct.

The slope of the tangent to the curve

$\displaystyle y=-x^{3}+3x^{2}+9x-27$ is maximum when x equals.

0%
$1$
0%
$3$
0%
$\dfrac 12$
0%
$-\dfrac 12$

Explanation

Given,

$\displaystyle y=-x^{3}+3x^{2}+9x-27$
Slope at any point on thegiven curve is

$m= \cfrac{dy}{dx}=-3x^2+6x+9$
Now for maximum value of slope we must have

$\cfrac{dm}{dx}=0=-6x+6\Rightarrow x = 1$

At what point the tangent to the curve

$\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{a}$ is perpendicular to the

$x$ - axis

0%
$(0, 0)$
0%
$(a, a)$
0%
$(a, 0)$
0%
$(0, a)$

Explanation

Let the point be

$P\displaystyle \left ( x_{1},y_{1} \right )$
Now

$\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{a}$
Differentiating w.r.t

$x$

$\displaystyle \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\times \frac{dy}{dx}=0$

$\displaystyle \frac{dy}{dx}=-\sqrt{\frac{y}{x}}$
Thus slope of tangent at P is

$=\displaystyle \left ( \frac{dy}{dx} \right )_{\left ( x_{1},y_{1} \right )}=-\sqrt{\frac{y_{1}}{x_{1}}}$
if tangent

$\displaystyle \perp$ to x axis then

$\displaystyle \frac{dy}{dx}=\frac{1}{0}$

$\displaystyle \therefore x_1 = 0\Rightarrow y_1 = a$
Therefore, required point is

$(0, a)$
Hence, option 'D' is correct.

$\displaystyle \frac{x}{a}+\frac{y}{b}=1$ is a tangent to the curve

$\displaystyle x=Kt,y=\frac{K}{t},K> 0$ than

0%
$a>0, b>0$
0%
$a>0, b<0$
0%
$a<0, b>0$
0%
$a<0, b<0$

Explanation

$\displaystyle x=Kt,y=\frac{K}{t},K> 0$

$\cfrac{dx}{dt}=K, \cfrac{dy}{dt}=-\cfrac{K}{t^2}$
Hence slope of tangent at point 't' is,

$m=\cfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=-\cfrac{1}{t^2}< 0 \forall t \epsilon R$
Hence for line

$\cfrac{x}{a}+\cfrac{y}{b}=1$ to be tangent to the given curve
it slope should be negative

$\cfrac{-b}{a}<0\Rightarrow \cfrac{b}{a}>0$

$a>0, b> 0$ or

$a<0, b<0$

The line

$y = x + 1$ is a tangent to the curve

$y^2 = 4x$ at the point.

0%
$(1, 2)$
0%
$(2, 1)$
0%
$(1, 4)$
0%
$( 2, 2)$

Explanation

We have

$y=x+1, y^2=4x$
Solving these,

$\Rightarrow (x+1)^2=4x\Rightarrow x^2-2x+1=0$

$\Rightarrow (x-1)^2=0\Rightarrow x=1\therefore y = (x+1)_{x=1}=2$
Thus required point of contact is

$(1,2)$

If a tangent to the curve

$\displaystyle y=6x-{ x }^{ 2 }$ is parallel to the line

$\displaystyle 4x-2y-1=0$ , then the point of tangency on the curve is:

0%
(2, 8)
0%
(8, 2)
0%
(6, 1)
0%
(4, 2)

Explanation

$y=6x-x^2$

$\Rightarrow y' =6-2x\ as\ tangent \ is\ parallel \ to \ line 4x-2y-1=0 \$ slope

$=2$

$\Rightarrow 6-2x=2$

$\Rightarrow x=2$

$y=6\cdot 2-2^2=8$

Hence the point of tangency is

$(2,8)$

The slope of the tangent to the curve

$y = \int_{0}^{x} \dfrac {dt}{1 + t^{3}}$ at the point where

$x = 1$ is

0%
$\dfrac {1}{4}$
0%
$\dfrac {1}{3}$
0%
$\dfrac {1}{2}$
0%
$1$

Explanation

$m=\dfrac {dy}{dx} = \dfrac {1}{1 + x^{3}}$

$m = \left (\dfrac {dy}{dx}\right )_ {x = 1} = \dfrac {1}{1 + 1} = \dfrac {1}{2}$

If tangent to the curve

$\displaystyle x={ at }^{ 2 },y=2at$ is perpendicular to

$x$ -axis, then its point of contact is:

0%
$(a, a)$
0%
$(0, a)$
0%
$(0, 0)$
0%
$(a, 0)$

Explanation

We have,

$x=at^2\Rightarrow \dfrac{dx}{dt}=2at$

And

$y=2at\Rightarrow \dfrac{dy}{dt}=2a$

$\therefore \dfrac{dy}{dx}=\dfrac{1}{t}=\infty$ , for tangent to the curve perpendicular to

$x-$ axis

$\Rightarrow t = 0$

so the point of contact is

$(at^2,2at)=(0,0)$

The slope of the normal to the curve

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

$x = 0$ is.

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

Explanation

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

$\cfrac{dy}{dx} = 4x+3\cos x$
Thus slope of tangent at

$x=0$ is

$4(0)+3\cos 0=3$
Hence slope of normal at the same point is

$-\cfrac{1}{m}=-\cfrac{1}{3}$

The slope of the tangent to the curve

$y=\displaystyle\int_{0}^{x}\dfrac{dt}{1+t^3}$ at the point where x=1 is

0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$
0%
1

Explanation

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

$m=\left ( \dfrac{dy}{dx} \right )x=1=\dfrac{1}{1+1}=\dfrac{1}{2}$

Consider the curve

$y = e^{2x}$ .What is the slope of the tangent to the curve at (0, 1) ?

0%
0
0%
1
0%
2
0%
4

Explanation

We know that slope of tangent to a curve

$y=f(x)$ at a point

$(x_1,y_1)$ equals

$f^{'}(x_1)$

$\Rightarrow$ Slope of tangent to the curve

$y=e^{2x}$ equals

$\cfrac{d(e^{2x})}{dx} =2e^{2x}$

$\Rightarrow$ Slope of tangent at

$(0,1)=f^{'}(0)=2$

The gradient of the tangent line at the point

$(a cos \alpha, a sin \alpha)$ to the circle

$x^2 + y^2 = a^2$ , is

0%
$tan (\pi - \alpha)$
0%
$tan \alpha$
0%
$cot \alpha$
0%
- $cot \alpha$

Explanation

Equation of the circle,

$x^2+y^2=a^2$

At a point

$(a \cos \alpha, a \ sin \alpha)$

Gradient of radius

$= \dfrac{y_2-y_1}{x_2-x_1}$

$= \dfrac{a \ sin \alpha-0}{a \ cos \alpha-0}$

$=\ tan \alpha$

gradient of radius

$\times$ gradient of tangent = -1

$\ tan \alpha \times$ gradient of tangent = -1

$\Rightarrow$ gradient of tangent

$= -\ tan \alpha$

$\Rightarrow$ gradient of tangent

$= \ tan (\pi - \alpha)$

The function

$x^{x}$ is increasing, when

0%
$x > \dfrac {1}{e}$
0%
$x < \dfrac {1}{e}$
0%
$x < 0$
0%
For all $x$

Explanation

Let

$y = x^{x}$

$\Rightarrow \dfrac {dy}{dx} = x^{x}(1 + \log x)$
For increasing function,

$\dfrac {dy}{dx} > 0$

$\Rightarrow x^{x} (1 + \log x) > 0$

$\Rightarrow 1 + \log x > 0$

$\Rightarrow \log_{e} x > \log_{e} \dfrac {1}{e}$

$\Rightarrow x > \dfrac {1}{e}$
Therefore, the function is increasing, when

$x > \dfrac {1}{e}$ .

Find the approximate error in the volume of a cube with edge

$x$ cm, when the edge is increased by

$2\%$

0%
$4\%$
0%
$2\%$
0%
$6\%$
0%
$8\%$

Explanation

in a mutiplication while multiplying we have to add the percentage errors

so if the edge of the cube is

$x$ cm then the volume will be

${ x }^{ 3 }$

but for errors we have to add the percentage

therefore the total error percentage now becomes

$6\%$

therefore the error is increased by

$4\%$

Which one of the following be the gradient of the hyperbola

$xy=1$ at the point

$\left(t,\dfrac{1}{t}\right)$

0%
$-\dfrac{1}{t}$
0%
$-\dfrac{1}{t^2}$
0%
$\dfrac{1}{t}$
0%
$-\dfrac{2}{t^2}$

Explanation

Given the hyperbola

$xy=1$ .

Now differentiate both sides with respect to

$x$ we get,

$y+x\dfrac{dy}{dx}=0$

or,

$\dfrac{dy}{dx}=-\dfrac{y}{x}$

Now the gradient of the given hyperbola at

$\left(t,\dfrac{1}{t}\right)$ is

$\left.\dfrac{dy}{dx}\right|_{\left(t,\dfrac{1}{t}\right)}=-\dfrac{1}{t^2}$

If the product of the slope of tangent to curve at

$(x,y)$ and its y-co-ordinate is equal to the x-co-ordinate of the point, then it represent.

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circle
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parabola
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ellipse
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rectangular hyperbola

Explanation

Let slope of a tangent at point

$(x,y)$ to the curve is

$=\cfrac { dy }{ dx }$

$\therefore y\cfrac { dy }{ dx } =x$ (given)

$\therefore ydy=xdx\quad$
Integrating both side

$\int { y } dy=\int { x } dx$

$\therefore \cfrac { { y }^{ 2 } }{ 2 } =\cfrac { { x }^{ 2 } }{ 2 } +c'$

$\therefore { y }^{ 2 }-{ x }^{ 2 }+2c'$

$\therefore { y }^{ 2 }-{ x }^{ 2 }=c$
where

$2c'=c$
Which is a rectangular hyperbola

The slope of the tangent to the curve

$xy+ax-by=0$ at the point

$(1,1)$ is

$2$ , then value of

$a$ and

$b$ are respectively:

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$1,2$
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$2,1$
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$3,5$
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None of these

Explanation

$xy+ax-by=0$

$(1,1)$ lies on curve

$1+a-b=0$

$a-b=-1$ .......

$(i)$

$y+x\dfrac { dy }{ dx } +a-b\dfrac { dy }{ dx } =0$

$1+2+a-2b=0$

$a-2b=-3$ ........

$(ii)$

Solving

$(i)$ and

$(ii)$ , we get

$a=1,b=2$

The graph of the function

$f(x) = 2x^3 - 7$ goes :

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up to the right and down to the left
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down to the right and up to the left
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up to the right and up to the left
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down to the right and down to the left
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none of these ways.

Explanation

Given

$f(x) =2x^{3}-7$

$\Rightarrow f^{'} (x) =6x^{2}>0\Rightarrow$ strictly increasing function

$\therefore$ Function is strictly increasing function

And

$f(0)=0$ when

$x=\sqrt[3]{\dfrac{7}{2} }$

$\Rightarrow$ Graph will go to infinity to the right as

$x\rightarrow \infty$

$\Rightarrow$ Graph will go to minus infinity to the left as

$x\rightarrow - \infty$

$\therefore$ Graph goes up to the right and down to the left

Option 'A' is correct

A curve with equation of the form

$y=a{x}^{4}+b{x}^{3}+cx+d$ has zero gradient at the point

$(0,1)$ and also touches the x-axis at the point

$(-1,0)$ then

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$a=3$
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$b=4$
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$c+d=1$
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for $x< -1$ the curve has a negative gradient

Explanation

given

$y=ax^4+bx^3+cx+d$

$y_{}{'}=4ax^3+3bx^2+c=0$ at (0,1)

$c=0$

curve passes through point (-1,0)

$0=a-b-c+d=>a+d=b$

here

$y_{}{'}<0$ for

$x<-1$

The local maximum value of

$x{(1-x)}^{2},0\le x\le 2$ is

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$2$
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$\dfrac {4}{27}$
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$5$
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$2,\dfrac {4}{27}$

Explanation

$\begin{array}{l} Let\, f\left( x \right) =x{ \left( { 1-x } \right) ^{ 2 } } \\ f'\left( x \right) -1{ \left( { 1-x } \right) ^{ 2 } }+x\times 2\left( { 1-x } \right) \times -1 \\ ={ \left( { 1-x } \right) ^{ 2 } }-2x\left( { 1-x } \right) \end{array}$

$\begin{array}{l} f''\left( x \right) =-2\left( { 1-x } \right) -2\left[ { 1-x+x\times -1 } \right] \\ =-2+2x-2\left[ { 1-x-x } \right] \\ =-2+2x-2+4x \\ =6x-4 \end{array}$

For critical points,

$F'\left( x \right) = 0$

$\begin{array}{l} \Rightarrow { \left( { 1-x } \right) ^{ 2 } }-2x\left( { 1-x } \right) =0 \\ \Rightarrow \left( { 1-x } \right) \left( { 1-x-2x } \right) =0 \\ \Rightarrow \left( { 1-x } \right) \left( { 1-3x } \right) =0 \\ \Rightarrow \left( { x-1 } \right) \left( { 3x-1 } \right) =0 \\ x=1 ,\quad \frac { 1 }{ 3 } \end{array}$

$f'\left( 1 \right) =6\left( 1 \right) -4=6-4=2>0\, \min imum$

$f''\left( { \dfrac { 1 }{ 3 } } \right) =6\times \dfrac { 1 }{ 3 } -4=2-4=-2<0\, { { miximum } }$

$f\left( { \dfrac { 1 }{ 3 } } \right) =\dfrac { 1 }{ 3 } { \left( { 1-\dfrac { 1 }{ 3 } } \right) ^{ 2 } }=\dfrac { 1 }{ 3 } \times \dfrac { 4 }{ 9 } =\dfrac { 4 }{ { 27 } }$

So,

$x = \dfrac{1}{3}$ is the point of local Maximum and the local maximum value is

$\dfrac{4}{{27}}$

Function

$f(x)=x-\ell nx$ is decreasing, when

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$x \in (0,1)$
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$x \in (-1,1)$
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$x \in (1,\infty)$
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$None\ of\ these$

Explanation

$f\left( x \right) = x - \ln x$

$f'\left( x \right) = 1 - \frac{1}{x}$

$= \frac{{x - 1}}{x}$

$f'\left( x \right)\,is\,not\,define\,at\,x = 0$

$For\,critical\,po{\mathop{\rm int}} s\,$

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

$= \frac{{x - 1}}{x} = 0$

$= x - 1 = 0$

$x = 1$

$So,\,critical\,po{\mathop{\rm int}} s\,are\,x = 1\,and\,x = 0$

$f'\left( x \right) = \frac{{x - 1}}{x} > 0\ for\ all x \in \left( {1,\infty } \right)$

$and\,f'\left( x \right) = \frac{{x - 1}}{x} < 0 \ for\ all \left( {0,1} \right)$

$so,\,f\left( x \right)\,is\,decreasing\,\forall x \in \left( {0,1} \right)$

If the curves

${y}^{2}=6x,9{x}^{2}+b{y}^{2}=16$ intersect each other at right angles, then the values of

$b$ is

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$6$
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$\cfrac{7}{2}$
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$4$
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$\cfrac{9}{2}$

Explanation

given the curves intersect at right angles so the tangents to curves at intersection points are perpendicular

consider curve 1

$y^2=6x$

$2y.y_{}{'}=6=>y_{}{'}=\dfrac{6}{2y}$

now slope of tangent to second curve is

$18x+2by.y_{}{'}=0=>y_{}{'}=\dfrac{-9x}{by}$

now product of slopes is

$-1$

$\dfrac{6}{2y}\times \dfrac{-9x}{by}=-1$

but

$6x=y^2$

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

$b=\dfrac{9}{2}$

The interval in which the function

$f(x) = {x^3}$ increases less rapidly than

$\,g(x) = 6{x^2} + 15x + 5$ is :

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$( - \infty , - 1)\,\,\,\,$
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$( - 5,1)\,\,\,\,$
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$( - 1,5)$
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$(5,\infty )$

Explanation

${\text{Increasing}}\;{\text{rate = }}\dfrac{d}{{dx}}\left( {{f_1}{x_1}} \right)$

${\text{f}}\left( x \right) = 6{x^2} + 15x + 5$

${\text{f}}\left( x \right) = 12x + 15$

${\text{ }}\dfrac{d}{{dx}}{\left( x \right)^3} < \;\dfrac{d}{{dx}}\left( {6{x^2} + 15x + 5} \right)$

${\text{ 3}}{{\text{x}}^2}{\text{ < 12x + 15}}$

${\text{3}}{{\text{x}}^2} - 12x - 15 < 0$

${x^2} - 4x - 5 < 0$

${x^2} - 5x + x - 5 < 0$

$\left( {x + 1} \right)\left( {x - 5} \right) < 0$

$x \in \left( { - 1,5} \right)$

The values of

$x$ for which the tangents to the curves

$y=x\cos{x},y=\cfrac{\sin{x}}{x}$ are parallel to the axis of

$x$ are roots of (respectively)

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$\sin{x}=x,\tan{x}=x$
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$\cot{x}=x,\sec{x}=x$
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$\cot{x}=x,\tan{x}=x$
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$\tan{x}=x,\cot{x}=x$

Explanation

$y=x\cos x$

$\dfrac{dy}{dx}=\cos x-x\sin x=0$

$\implies x=\cot x$

$y=\dfrac{\sin x}{x}$

$\dfrac{dy}{dx}=\dfrac{x\cos x-\sin x}{x^{2}}=0$

$\implies x=\tan x$

Among all the critical points of a function f(x)=(4-x)|2-x|. Let 'a' and 'b' be the maximum and minimum values of their abscissate respectively then match the correct option.

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a+2b=7
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2a+b=7
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2a+b=5
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2a-b=5

The slope of the tangent to the curve

$y=sinx$ where it crosses the

$x-axis$ is

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$1$
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$-1$
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$\pm 1$
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$\pm 2$

Explanation

Consider the given equation of the curve.

$y=\sin x$ ……(1)

On differentiating equation (1) with respect to $x$ , we get

$\dfrac{dy}{dx}=\cos x$

When given curve crosses the x-axis then.

$y=0$

$\sin x=0$

$x=0$

Therefore,

${{\left( \dfrac{dy}{dx} \right)}_{y=0}}=\cos 0$

${{\left( \dfrac{dy}{dx} \right)}_{y=0}}=1$

Hence, this is the answer.

The equation of normal to the curve

$y=\left| { x }^{ 2 }-\left| x \right| \right|$ at

$x=-2$ is

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$3y=2x+10$
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$3y=x+8$
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$2y=x+6$
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$2y=3x+10$

Explanation

$y=\mid x^{2}-\mid x \mid \mid=(\mid x\ \mid )(\mid \mid x\mid-1\mid )$

$x=-2,y=2$

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

$\bigg(\dfrac{dy}{dx}\bigg)_{x=-2}=-1-2=-3$

The equation of normal is

$y=\dfrac{x}{3}+\bigg(2+\dfrac{2}{3}\bigg)$

$3{y}=x+8$

The Point (s) on the cure

${ y }^{ 3 }+{ 3x }^{ 2 }=12y$ where the tangent is vertical (parallel to y-axis), is/are.

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$\left[ \pm \dfrac { 4 }{ \sqrt { 3 } } ,-2 \right]$
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$\left( \pm \dfrac { \sqrt { 11 } }{ 3 } ,1 \right)$
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$(0,0)$
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$\left( \pm \dfrac { 4 }{ \sqrt { 3 } } ,2 \right)$

Explanation

curve given

$y^3+3x^2=12y$

Thus differentiating w.r.t

$x$

$3y^2\dfrac {dy}{dx}+6x=12\dfrac {dy}{dx}$

$\dfrac {dy}{dx}(y^2-4)+2x=0$

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

For tangent to be

$\parallel$ to

$y-$ axis

$\dfrac {dy}{dx}\rightarrow \infty$

Thus

$y^2-4=0$

$y=\pm 2$

Putting

$y=+2$ in curve we get

$x=\pm \dfrac {4}{\sqrt 3}$

Putting

$y=-2$ in curve

$x$ does not exit

Thus

$y=-2$ not in range as

$x^2 < 0$

Thus points where tangent is

$\parallel$ to

$y-$ axis is

$\left [\pm \dfrac {4}{\sqrt {3}},\ 2 \right]$ Answer.

CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 1 - 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 1 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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