MCQ Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 9

The slope of the tangent to the curve

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

$(2, -1)$ is

0%
$\dfrac {22}{7}$
0%
$\dfrac {6}{7}$
0%
$-6$
0%
$\dfrac {7}{6}$

Explanation

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

$(1)$

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

$(2)$

Differentiate

$(1)$ , we get

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

Differentiate

$(2)$ , we get

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

$m=\dfrac{dy}{dx}=\dfrac{4t-2}{2t+3}$

Given point is

$(2, -1)$

Put the point in original x and y, we get

$\Rightarrow x=t^2+3t-8$

$2=t^2+3t-8$

$t^2+3t-10=0$

$(t-2)(t+5)=0$

$t=2, t=-5$

$\Rightarrow y=2t^2-2t-5$

$(-1)=2t^2-2t-5$

$2t^2-2t-4=0$

$(t+1)(t-2)=0$

$t=-1, t=2$

Since,

$t=2$ common in both parts, so we take

$\dfrac{dy}{dx}=\dfrac{4t-2}{2t-3}$ at

$t=2$

$\dfrac{dy}{dx}=\dfrac{4(2)-2}{2(2)-3}=\dfrac{8-2}{4+3}=\dfrac{6}{7}$

$m=\dfrac{dy}{dx}=\dfrac{6}{7}$ .

At what points the slope of the tangent to the curve

$x^2+y^2-2x-3=0$ is zero?

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

Explanation

$x^2+y^2-2x-3=0$ is zero.

Differentiate w.r.t. x

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

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

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

$\dfrac{dy}{dx}=\dfrac{1-x}{y}$ ........

$(1)$

If line is parallel to x-axis

Angle with x-axis

$=\theta =0$

Slope of x-axis

$=\tan\theta =\tan 0^o=0$

Slope of tangent

$=$ Slope of x-axis

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

$\dfrac{1-x}{y}=0$

$x=1$

Finding y when

$x=1$

$x^2+y^2-2x-3=0\\$

$(1)^2+y^2-2(1)-3=0\\$

$1+y^2-2-3=0\\$

$y=\pm 2$

Hence, the points are

$(1, 2)$ and

$(1, -2)$ .

The point on the curve

$y=12x-x^2$ , where the slope of the tangent is zero will be

0%
$(0, 0)$
0%
$(2, 16)$
0%
$(3, 9)$
0%
$(6, 36)$

Explanation

Let the point be

$P(x, y)$

$y=12x-x^2$

$\dfrac{dy}{dx}=12-2x$

$\left(\dfrac{dy}{dx}\right)_{(x_1, y_1)}=12-2x_1$

since slope of tangent is zero

$\left(\dfrac{dy}{dx}\right)_{x_1, y_1}=0$

$12-2x_1=0$

$2x_1=12$

$x_1=6$

Also curve passing through tangent

$y_1=12x_1-x^2_1$

$y_1=12\times 6-36$

$y_1=72-36$

$y_1=36$

The points are

$(6, 36)$ .

The normal to the curve

$x^2=4y$ passing through

$(1, 2)$ is

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

Explanation

Curve is

$x^2=4y$

Diff wrt to x

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

Slope of normal

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

Let (h, k) be the point where normal and curve intersects

$\therefore$ Slope of normal at (h, k)

$=-2/h$

Equation of normal passes through (h, k)

$y-y_1=m(x-x_1)$

$y-k=\dfrac{-2}{h}(x-h)$

Since normal passes through

$(1, 2)$

$2-k=\dfrac{-2}{h}(1-h)$

$k=2+\dfrac{2}{h}(1-h)$ ..........

$(1)$

since (h, k) lies on the curve

$x^2=4y$

$h^2=4k$

$k=\dfrac{h^2}{4}$ .......

$(2)$

using

$(1)$ and

$(2)$

$2+\dfrac{2}{h}(1-h)=\dfrac{h^2}{4}$

$\dfrac{2}{h}=\dfrac{h^2}{4}$

$h=2$

Putting

$h=2$ in

$(2)$

$k=\dfrac{h^2}{4}, k=1$

$h=2$ and

$k=1$ putting in equation of normal

$\Rightarrow y-k=\dfrac{-2(k-h)}{h}$

$\Rightarrow y-1=\dfrac{-2(x-2)}{2}=y-1=-1(x-2)$

$\Rightarrow y-1=-x+2\\$

$\Rightarrow x+y=2+1\\$

$\Rightarrow x+y=3$ .

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 curves are,

$x=3t^2+1$ ---- ( 1 )

$y=t^3-1$ ---- ( 2 )

Substituting

$x=1$ in ( 1 ) we get,

$\Rightarrow$

$3t^2+1=1$

$\Rightarrow$

$3t^2=0$

$\Rightarrow$

$t=0$

Differentiate ( 1 ) w.r.t.

$t,$ we get

$\Rightarrow$

$\dfrac{dx}{dt}=6t$

Differentiate ( 2 ) w.r.t. $$t,$ we get

$\Rightarrow$

$\dfrac{dy}{dt}=3t^2$

$\Rightarrow$

$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{3t^2}{6t}$

$\therefore$

$\dfrac{dy}{dx}=\dfrac{t}{2}$

Slope of the tangent

$=\left(\dfrac{dy}{dx}\right)_{t=0}=\dfrac{0}{2}=0$

Mark the correct alternative of the following.
The point on the curve

$9y^2=x^3$ , where the normal to the curve makes equal intercepts with the axes is?

0%
$(4, \pm 8/3)$
0%
$(-4, 8/3)$
0%
$(-4, -8/3)$
0%
$(8/3, 4)$

Explanation

Let the required point be

$(x_1, y_1)$

equation of the curve is

$9y^2=x^2$

since

$(x_1, y_1)$ lies on the curve, therefore

$9y^2_1=x^3_1$ .......

$(1)$

Now

$9y^2=x^3$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{x^2}{6y}$

$\Rightarrow \left(\dfrac{dy}{dx}\right)_{(x_1, y_1)}=\dfrac{x^2}{6y^1}$

since normal to the curve at

$(x_1, y_1)$ makes equal intersepts with the coordinate axis, therefore slope of the normal

$=\pm 1$

$\Rightarrow \dfrac{1}{-(dy/dx)_{(x_1, y_1)}}=\pm 1$

$\Rightarrow (dy/dx)_{(x_1, y_1)}=\pm 1$

$\rightarrow \dfrac{x^2_1}{6y^1}=\pm 1$

$\Rightarrow x^4_1=36y^2_1=36\left(\dfrac{x^3_1}{9}\right)$ (using

$1$ )

$\Rightarrow x^4_1=4x^3_1\Rightarrow x^3_1(x_1-4)=0$

$\Rightarrow x_1=0, 4$

Putting

$x_1=0$ in

$(1)$ we get

$9y^2_1=0\Rightarrow y_1=0$

Putting

$x_1=4$ in

$(1)$ we get

$9y^2_1=(4)^3\Rightarrow y_1=\pm \dfrac{8}{3}$

But the line making equal intersepts with the coordinate axes cannot pass through origin.

Hence, the required points are

$\left(4, \dfrac{8}{3}\right)$ and

$\left(4, \dfrac{-8}{3}\right)$ .

Mark the correct alternative of the following.
The line

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

$y^2=4x$ , if the value of m is?

0%
$1$
0%
$2$
0%
$3$
0%
$1/2$

Explanation

Given equation of the tangent to the given curve

$y=mx+1$

Now substituting the value of y in

$y^2=4x$ , we get

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

$\Rightarrow m^2x^2+1+2mx-4x=0$

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

$(1)$

Since, a tangent touches the curve at one point, the root of equation

$(1)$ must be equal.

Thus, we get

Discriminant,

$D=b^2-4ac=0$

$(2m-4)^2-4(m^2)(1)=0$

$\Rightarrow 4m^2-16m+16-4m^2=0$

$\Rightarrow -16m+16=0$

$\Rightarrow m=1$ .

The slope of the 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%
$\dfrac {7}{6}$
0%
$-\dfrac {6}{7}$

Explanation

Given curve

$x = t^2 + 3t - 8$ and

$y = 2t^2 - 2t - 5$

slope of tangent to the curve

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

$\therefore \dfrac{dx}{dt} = 2t + 3 , \, \dfrac{dy}{dt} = 4t - 2$

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{4t - 2}{2t + 3}$

$\left(\dfrac{dy}{dx} \right)_{t = 2}$

$\dfrac{4(2) - 2}{2(2) + 3} = \dfrac{6}{7}$

If the function

$f(x)=\cos\left| x \right| -2ax+b$ increases along the entire number scale, then

0%
$a=b$
0%
$a=\cfrac{1}{2}b$
0%
$a\le -\cfrac{1}{2}$
0%
$a> -\cfrac{3}{2}$

Explanation

$\because f(x)=\cos |x|-2ax+b$ increase in R

$=\cos x-2ax+b$ (as

$\cos (-x)=\cos x$ )

$f'(x)=-\sin x-2a$

for increasing

$f(x)$ ,

$f'(x)\geq 0$

$-\sin x-2a\geq 0$

$\sin x+2a\leq 0$

$2a\leq -\sin x$ (

$\because$ maximum value of

$\sin x=1$ )

$2a\leq -1$

$a\leq -\dfrac{1}{2}$ .

The function

$f(x)=\cfrac{\lambda \sin x+2\cos x}{\sin x+\cos x}$ is increasing, if

0%
$\lambda< 1$
0%
$\lambda> 1$
0%
$\lambda< 2$
0%
$\lambda> 2$

Explanation

$f(x) = \dfrac{\lambda \sin x + 2 \cos x}{\sin x + \cos x}$

for increasing

$f'(x) > 0$

$\Rightarrow \dfrac{(\sin x + \cos x) (\lambda \cos x - 2 \sin x) - (\lambda \sin x + 2 \cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2} > 0$

$\Rightarrow \dfrac{\lambda - 2}{(\sin x + \cos x)^2} > 0$

but

$(\sin x + \cos x)^2 > 0$

$\therefore \lambda - 2 > 0$

$\lambda > 2$

Let

$f(x)={x}^{3}+a{x}^{2}+bx+5{\sin}^{2}x$ be an increasing function on the set

$R$ . Then,

$a$ and

$b$ satisfy

0%
${a}^{2}-3b-15> 0$
0%
${a}^{2}-3b+15> 0$
0%
${a}^{2}-3b+15< 0$
0%
$a> 0$ and $b> 0$

Explanation

$f(x)=x^3+ax^2+bx+5\sin^2x$

$f^{\prime}(x)=3x^2+2ax+b+5.2\sin x\cos x>0$

$f^{\prime}(x)=3x^2+2ax+b+5\sin 2x >0$

$-1\le \sin 2x\le 1$

$\sin 2x=-1$

$f^{\prime}(x)=3x^2+2ax+b+5(-1)=3x^2+2ax+b-5 \ge 0$

$f^{\prime \prime}=6x+2a$

$6x+2a=0$ or,

$x=\dfrac{-2a}{6}=\dfrac{-a}{3}$

so now,

$3\left(\dfrac{-a}{3}\right)^2+2a.\left(\dfrac{-a}{3}\right)+b-5\ge 0$

$\dfrac{a^2}{3}-\dfrac{2a^2}{3}+b-5\ge 0$

$\dfrac{-a^2}{3}+b-5\ge 0$

$a^2-3b+15<0$

If the function

$f(x)=2\tan x+(2a+1)\log _{ e }{ \left| \sec { x } \right| } +(a-2)x$ is increasing on

$R$ , then

0%
$a\in \left (\dfrac {1}{2},\infty \right)$
0%
$a\in \left (-\dfrac {1}{2},\dfrac {1}{2}\right)$
0%
$a=\dfrac {1}{2}$
0%
$a\in R$

Explanation

$f(x)=2tanx+(2a+1)log_e \mid secx \mid +(a-2)x$

$f'(x)=2sec^2x+(2a+1)\dfrac{secxtanx}{secx}+(a-2)1$

$=2sec^2x+(2a+1)tanx+(a-2)$

Let

$tanx=t$

$=2(t^2+1)+(2a+1)t+(a-2)>0$

$=2t^2+2at+t+a>0$

$(2a+1)^2-4.2.a<0$

$4a^2+1+4a-8a<0$

$4a^2-4a+1<0$

$4(a-\dfrac{1}{2})^2)<0$

$a=\dfrac{1}{2}$

Function

$f(x)={a}^{x}$ is increasing on

$R$ , if

0%
$a> 0$
0%
$a< 0$
0%
$0< a< 1$
0%
$a> 1$

Explanation

$f(x) = a^x$

$f'(x) = a^x \log a$

function is increasing on R.

$a^x \log a > 0$

$a^x > 0 \, \& \, \log a > 0$

$a^x < 0 \, \& \, \log a < 0$

But log function is always positive

$\log a > 0$

$\Rightarrow a > 1$

If the function

$f(x)={x}^{3}-9k{x}^{2}+27x+30$ is increasing on

$R$ , then

0%
$-1< k< 1$
0%
$k< -1$ or $k> 1$
0%
$0< k< 1$
0%
$-1< k< 0$

Explanation

$f(x)=x^3-9kx^2+27x+30$ is increasing on R.

$f(x)=x^3-9kx^2+27x+30$

$f'(x)=3x^2-18kx+27$

$=3(x^2-6kx+9)$

Given

$f(x)$ is increasing on R

$\Rightarrow f'(x) > 0$ for all

$x\in R$

$\Rightarrow 3(x^2-6kx+9) > 0$

$(x^2-6kx+9) > 0$ all

$x\in R$

$ax^2+bx+c > 0$

so,

$(-6k)^2-4(1)(9) < 0$

$\Rightarrow 36k^2-36 < 0$

$(k+1)(k-1) < 0$

It can be possible when

$(k+1) < 0$ and

$d(k-1) > 0$

$\Rightarrow k < -1$ and

$k > 1$ (not possible)

or,

$(k+1) > 0$ and

$(k-1) < 0$

$k > -1$ and

$k < 1$

$-1 < k < 1$ so option A is correct.

If the function

$f(x)={x}^{2}-kx+5$ is increasing on

$[2,4]$ , then

0%
$k\in (2,\infty)$
0%
$k\in (-\infty,2)$
0%
$k\in (4,\infty)$
0%
$k\in (-\infty,4)$

Explanation

$f(x)=x^2-kx+5$ is increasing in

$x\in [2, 4]$

$f'(x)=2x-k > 0$

$2x > k$

$2\leq x\leq 4$

$k < 2x$

$4\leq 2x \leq 8$

k should less than the minimum value of

$2x$

$k < 4$

$k\in (-\infty, 4)$

Final Answer.

Function

$f(x)=\log _{ a }{ { x }_{ } }$ is increasing on

$R$ , if

0%
$0< a< 1$
0%
$a> 1$
0%
$a< 1$
0%
$a> 0$

Explanation

We have

$f(x)=log_ax$

Differentiate with respect x, we get

$f'(x)=\dfrac{1}{x log a}$

$\because$ function is increasing on R

$\dfrac{1}{x log a} > 0$

$a > 1$ .

The function

$f(x)={x}^{9}+3{x}^{7}+64$ is increasing on

0%
$R$
0%
$(-\infty,0)$
0%
$(0,\infty)$
0%
${R}_{0}$

Explanation

$f(x)=x^9+3x^7+64$

$f'(x)=9x^8+21x^6$

$=3x^6(3x^2+7)$

$\because$ function is increasing

$3x^6(3x^2+7) > 0$

$\Rightarrow$ function is increasing on R.

A curve

$y=me^{mx}$ where

$m > 0$ intersects y-axis at a point

$P$ .
What is the slope of the curve at the point of intersection

$P$ ?

0%
$m$
0%
$m^2$
0%
$2m$
0%
$2m^2$

Explanation

$y = m {e}^{mx}, \; m > 0$

Therefore,

Slope

$= \cfrac{dy}{dx} = {m}^{2} {e}^{mx}$

Substituting

$x = 0$ , we have

$Slope={m}^{2} {e}^{m \cdot 0} = {m}^{2}$

Consider the equation

$x^y=e^{x-y}$
What is

$\dfrac{d^2y}{dx^2}$ at

$x=1$ equal to ?

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

Explanation

${x}^{y} = {e}^{x - y}$

Taking

$\log$ both sides, we have

$y \log{x} = \left( x - y \right) \log{e}$

$y \log{x} = x - y ..... \left( 1 \right)$

$\dfrac{y}{x}=1+\log x$

$x = 1 \Rightarrow y = 1$

Differentiating equation

$\left( 1 \right)$ w.r.t.

$x$ , we have

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

$\Rightarrow \left( 1 + \log{x} \right) \cfrac{dy}{dx} = 1 - \cfrac{y}{x}$

$\Rightarrow \cfrac{dy}{dx} = \cfrac{x - y}{x \left( 1 + \log{x} \right)}$

$\Rightarrow \cfrac{dy}{dx} = \cfrac{y \log{x}}{x \left( 1 + \log{x} \right)}$

$\Rightarrow \cfrac{dy}{dx} = \cfrac{\log{x}}{{\left( 1 + \log{x} \right)}^{2}}$

Again differentiating above equation w.r.t.

$x$ , we have

$\cfrac{{d}^{2}y}{d{x}^{2}} = \cfrac{{\left( 1 + \log{x} \right)}^{2} \cdot \frac{1}{x} - \log{x} \cdot 2 \left( 1 + \log{x} \right) \frac{1}{x}}{{\left( 1 + \log{x} \right)}^{4}}$

$\Rightarrow \cfrac{{d}^{2}y}{d{x}^{2}} = \cfrac{\frac{1}{x} \left( 1 - \log{x} \right)}{{\left( 1 + \log{x} \right)}^{3}}$

$x = 1$ ,

$\cfrac{{d}^{2}y}{d{x}^{2}} = 1$

Consider the equation

$x^y=e^{x-y}$
What is

$\dfrac{dy}{dx}$ at

$x=1$ equal to ?

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

Explanation

${x}^{y} = {e}^{x - y}$

Taking

$\log$ both sides, we have

$y \log{x} = \left( x - y \right) \log{e}$

$y \log{x} = x - y ..... \left( 1 \right)$

$x = 1 \Rightarrow y = 1$

Differentiating equation

$\left( 1 \right)$ w.r.t.

$x$ , we have

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

$\Rightarrow \left( 1 + \log{x} \right) \cfrac{dy}{dx} = 1 - \cfrac{y}{x}$

$\Rightarrow \cfrac{dy}{dx} = \cfrac{x - y}{x \left( 1 + \log{x} \right)}$

$x = 1, \; y = 1$

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

A curve

$y=me^{mx}$ where

$m > 0$ intersects y-axis at a point

$P$ .
How much angle does the tangent at

$P$ make with y-axis ?

0%
$\tan^{-1}m^2$
0%
$\cot^{-1}(a+m^2)$
0%
$\sin^{-1}(\dfrac{1}{\sqrt{1+m^4}})$
0%
$\sec^{-1}\sqrt{1+m^4}$

The function

$f(x)=4-3x+3x^2-x^3$ is

0%
decreasing on $R$
0%
increasing on $R$
0%
strictly decreasing on $R$
0%
strictly increasing on $R$

Explanation

${f}' \left(x\right) = -3+6x-3x^{2} = -3\left(x^{2}-2x+1\right) = -3\left(x-1\right)^{2} \le 0.$

$\Rightarrow {f}' \left(x\right) \le 0$ for all

$x \in R \Rightarrow f\left(x\right)$ is decreasing on

$R$ .

The real value of

$k$ for which

$f(x)=x^2+kx+1$ is increasing on

$(1, 2)$ , is

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

Explanation

$f'(x)=(2x+k)$ .

$1 < x < 2\Rightarrow 2 < 2x , 4 \Rightarrow 2+k < 2x+k < 4+k \Rightarrow 2+k < f'(x) < 4+k$

$f(x)$ is increasing

$\Leftrightarrow (2x+k) \ge 0\Leftrightarrow 2+k\ge 0\Leftrightarrow k \ge -2$

$\therefore \$ least value of

$k$ is

$-2$

Consider the equation

$az^2 + z + 1 = 0$ having purely imaginary root where

$a$ = cos

$\theta + i$ sin

$\theta$ ,

$i = \sqrt{-1}$ and function

$f(x) = x^3 - 3x^2 + 3(1$ + cos

$\theta)x + 5$ , then answer the following questions.
Which of the following is true about

$f(x)$ ?

0%
$f(x)$ decreases for $x$ $\epsilon$ $[2 n \pi, (2n + 1)\pi]$ , $n$ $\epsilon$ $Z$
0%
$f(x)$ decreases for $x$ $\epsilon$ $\left [ (2n - 1)\frac{\pi}{2}, (2n + 1)\frac{\pi}{2}\right ]$ $n$ $\epsilon$ $Z$
0%
$f(x)$ is non-monotonic function
0%
$f(x)$ increases for $x$ $\epsilon$ $R$ .

Explanation

We have,

$az^2 + z + 1 = 0$ (1)

_________________

$\Rightarrow$

$az^2 + z + 1 = 0$ (taking conjugate of both sides)

$\Rightarrow$

$\bar a z^2 - z + 1 = 0$ (2)

[since

$z$ is purely imaginary

$\bar z = -z$ ]

Eliminating

$z$ from (1) and (2) by cross-multiplication rule,

$(\bar a - a)^2 + 2(a + \bar a) = 0 \Rightarrow \left ( \frac{\bar a - a}{2}\right )^2$ +

$\frac{a + \bar a}{2} = 0$

$\Rightarrow$ -

$\left ( \frac{a - \bar a}{2i}\right)^2 + \left ( \frac{a + \bar a}{2i}\right) = 0$

$\Rightarrow$ -

$sin^2 \theta + cos \theta = 0$

$\Rightarrow$

$cos \theta = sin^2 \theta$ (3)

Now,

$f(x) = x^3 - 3x^2 + 3(1$ + cos

$\theta)x + 5$

${f}'(x) = 3x^3 - 6x + 3(1$ + cos

$\theta)$

Its discriminant is

$36 - 36(1 + cos \theta) = -36 cos \theta = - 36 sin^2 \theta$ < 0

$\Rightarrow$

$f(x)$ > 0

$\forall$

$x$

$\epsilon$

$R$

Hence,

$f(x)$ is increasing

$\forall$

$x$

$\epsilon$

$R$ . Also,

$f(0)$ = 5, then

$f(x)$ = 0 has one negative root. Now,

$cos 2\theta = cos \theta \Rightarrow 1 - 2 sin^2 \theta = cos \theta$

$\Rightarrow$

$1 - 2 cos \theta = cos \theta$

$\Rightarrow$

$cos \theta = 1/3$

which has four roots for

$\theta$

$\epsilon$

$[0,4\pi]$ .

The number of tangents to the cure

$x^{3/2}+y^{3/2}=2a^{3/2}, a> 0$ , which are equally inclined to the axes, is

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

If m is the slope of a tangent to the curve

$e^{y}=1+x^{2},$ then

0%
$\left | m \right |> 1$
0%
$m> 1$
0%
$m> -1$
0%
$\left | m \right |\leq 1$

Explanation

Differentiating w.r.t.x, we get

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

$\Rightarrow \dfrac{dy}{dx} = \dfrac{2x}{1+x^{2}} (\because e^{y} = 1 +x^{2})$

$\Rightarrow m = \dfrac{2x}{1 + x^{2}} or \left | m \right |= \dfrac{2\left | x \right |}{1 + \left | x \right |^{2}}$

But

$1 + \left | x \right |^{2} - 2\left | x \right |=(1-\left | x \right |)^{2}\geq 0$

$\Rightarrow 1 + \left | x \right |^{2} \geq 2\left | x \right |$

$\therefore \left | m \right |\leq 1$

Let

$f(x)=\displaystyle \int e^x (x-1)(x-2)dx$ . Then

$f$ decreases in the interval

0%
$(-\infty , -2)$
0%
$(-2, -1)$
0%
$(1, 2)$
0%
$(2, \infty)$

The function

$f(x)=3x+\cos 3x$ is

0%
increasing on $R$
0%
decreasing on $R$
0%
strictly increasing on $R$
0%
strictly decreasing on $R$

$f(x)=\sin x-kx$ is decreasing for all

$x \in R$ , when

0%
$k < 1$
0%
$k \le 1$
0%
$k > 1$
0%
$k \ge 1$

For

$x > 1, y=\log_e x$ satisfies the inequality

0%
$x-1 > y$
0%
$x^2 -1 >y$
0%
$y > x-1$
0%
$\dfrac {x-1}{x} < y$

The slope of the tangent to the curve

$y = \sqrt{4-x^{2}}$ at the point, where the ordinate and the abscissa are equal , is

0%
-1
0%
1
0%
0
0%
None of these

Explanation

Putting

$y=x$ in

$y = \sqrt{4-x^{2}}$ , we get

$x = \sqrt{2}, -\sqrt{2}$ .

So, the point is

$(\sqrt{2}, \sqrt{2})$ .

Differentiating

$y^{2}+x^{2} = 4$ w.r.t. x,

$2y \dfrac{dy}{dx}+ 2x = 0$ or

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

$\Rightarrow at(\sqrt{2}, \sqrt{2}), \dfrac{dy}{dx} = -1$

At the point

$P(a, a^{n})$ on the graph of

$y = x^{n}(n \epsilon n)$ in the first quadrant, a normal is drawn. the normal intersects the y-axis at the point (0, b) . if

$\underset{a\rightarrow b}{lim}b=\dfrac{1}{2}$ , then n equals

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

The curve given by

$x + y = e^{xy}$ has a tangent parallel to the y-axis at the point

0%
$(0,1)$
0%
$( 1, 0 )$
0%
$(1, 1)$
0%
None of these

Explanation

Differentiating w.r.t.x, we get

$1 + \dfrac{dy}{dx} = e^{xy}\left ( y + x\dfrac{dy}{dx} \right )$ or

$\dfrac{dy}{dx}=\dfrac{ye^{xy}-1}{1-xe^{xy}}$

As the tangent is parallel along

$y-axis$

$\dfrac{dy}{dx} = \infty=\dfrac{ye^{xy}-1}{1-xe^{xy}}$

$\Rightarrow 1-xe^{xy}=0$ This holds for x = 1, y = 0

The abscissa of points P and Q in the curve

$y = e^{x}+e^{-x}$ such that tangents at P and Q make

$60^{o}$ with the x-axis

0%
ln $\left ( \dfrac{\sqrt{3}+\sqrt{7}}{7} \right )$ and ln $\left ( \dfrac{\sqrt{3}+\sqrt{5}}{2} \right )$
0%
ln $\left ( \dfrac{\sqrt{3}+\sqrt{7}}{2} \right )$
0%
ln $\left ( \dfrac{\sqrt{7}+\sqrt{3}}{2} \right )$
0%
$\pm$ ln $\left ( \dfrac{\sqrt{3}+\sqrt{7}}{2} \right )$

If x=4 y = 14 is a normal to the curve

$y^{2}=ax^{3}-\beta$ at (2,3) then the value of

$\alpha +\beta$ is

0%
9
0%
-5
0%
7
0%
-7

At what points of curve

$y = \dfrac{2}{3}x^{3}+\dfrac{1}{2}x^{2}$ , the tangent makes the equal with the axis?

0%
$(\dfrac{1}{2},\dfrac{5}{24})$ and $\left ( -1,\dfrac{-1}{6} \right )$
0%
$(\dfrac{1}{2},\dfrac{4}{9})$ and $( -1,0)$
0%
$\left ( \dfrac{1}{3},\dfrac{1}{7} \right )$ and $\left ( -3, \dfrac{1}{2} \right )$
0%
$\left ( \dfrac{1}{3},\dfrac{4}{47} \right )$ and $\left ( -1, \dfrac{1}{2} \right )$

The curve represented parametrically by the equations x = 2 in

$\cot t+1$ and

$y=\tan t+\cot t$

0%
tanfent and normal intersect at the point (2, 1)
0%
normal at $t = \pi /4$ is parallel to the y-axis
0%
tangent at $t = \pi /4$ is parallel to the line y = x
0%
tangent at $t = \pi /4$ is parallel to the x-axis

If a variable tangent to the curve

$x^{2}y=c^{3}$ makes intercepts a, b on x-and y-axes, respectively, then the value of

$a^{2}b$ is

0%
$27c^{3}$
0%
$\dfrac{4}{27}c^{3}$
0%
$\dfrac{27}{4}c^{3}$
0%
$\dfrac{4}{9}c^{3}$

The x-intercept of the tangent at any arbitrary point of the curve

$\dfrac{a}{x^{2}}+\dfrac{b}{y^{2}}=1$ is proportion to

0%
square of the abscissa of the point of tangency
0%
square root of the abscissa of the point of tangency
0%
cube of the abscissa pf the point of tangency
0%
cube root of the abscissa of the point of tangency

The angle between the tangent to the curves

$y = x^{2}$ and

$x = y^{2}$ at (1, 1) is

0%
$\cos ^{-1}\dfrac{4}{5}$
0%
$\sin ^{-1}\dfrac{3}{5}$
0%
$\tan ^{-1}\dfrac{3}{4}$
0%
$\tan ^{-1}\dfrac{1}{3}$

Point on the curve

$f(x)=\dfrac{x}{1-x^{2}}$ where the tangent is inclined at an angle of

$\dfrac{\pi }{4}$ ot the x-axis are

0%
(0, 0)
0%
$\left ( \sqrt{3},\dfrac{-\sqrt{3}}{2} \right )$
0%
$\left ( -2 ,\dfrac{2}{3}\right )$
0%
$\left (- \sqrt{3},\dfrac{\sqrt{3}}{2} \right )$

If the tangent at any point

$P(4m^{2}, 8m^{3})$ of

$x^{3}-y^{3}=0$ is also a normal to the curve

$x^{3}-y^{3}=0$ , then value of m is

0%
$m = \dfrac{\sqrt{2}}{3}$
0%
$m = -\dfrac{\sqrt{2}}{3}$
0%
$m = \dfrac{3}{\sqrt{2}}$
0%
$m = -\dfrac{3}{\sqrt{2}}$

A curve passes through

$(2,1)$ and is such that the square of the ordinate is twice the contained by the abscissa and the intercept of the normal. Then the equation of curve is

0%
$x^2 +y^2=9x$
0%
$4x^2 +y^2=9x$
0%
$4x^2 +2y^2=9x$
0%
None of these

The tangent to the curve

$y = e^{x}$ drawn at the point

$(c, e^{c})$ intersects the line joining the points

$(c-1, e^{c-1})$ and

$(c+1, e^{c+1})$

0%
on the left of x =c
0%
on the right of x = c
0%
at no point
0%
at all point

Let

$f:[1, \infty) \rightarrow R$ and

$f(x)=x \int_{1}^{x} \dfrac{e^{t}}{t} d t-e^{x},$ then

0%
$f(x)$ is an increasing function
0%
$\lim _{x \rightarrow \infty} f(x) \rightarrow \infty$
0%
$f^{\prime}(x)$ has a maxima at $x=e$
0%
$f(x)$ is a decreasing function

Explanation

$f(x)=x \int_{1}^{x} \dfrac{e^{t}}{t} d t-e^{x}\\$

$\Rightarrow f^{\prime}(x)=x \dfrac{e^{x}}{x}+\int_{1}^{x} \dfrac{e^{t}}{t} d t-e^{x}\\$

$\Rightarrow f^{\prime}(x)=\int_{1}^{x} \dfrac{e^{t}}{t} d t>0[\because x \in[1, \infty)]\\$

$\Rightarrow f(x)$ is an increasing function.

If the line ax +by + c = 0 is a normal to the curve xy = 1, then

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

Consider the following statement is

$S$ and

$R$

$S$ . Both

$\sin x$ and

$\cos x$ are decreasing function in the interval

$\left(\dfrac {\pi}{2}, \pi \right)$

$R:$ If a differentiable function decreases in an interval

$(a, b)$ then its derivative also decreases in

$(a, b)$ , which of the following is true?

0%
Both $S$ and $R$ are wrong
0%
Both $S$ and $R$ are correct but $R$ is not the correct explanation of $S$
0%
$S$ is the correct and $R$ is the correct explanation of $S$
0%
$S$ is the correct and $R$ is the wrong

Explanation

From the graph, it is clear that both

$\sin x$ and

$\cos x$ in the internal

$(\pi /2, \pi)$ are the decreasing functions.
Therefore,

$S$ is correct.
To disprove

$R$ let us consider the counter example,

$f(x)=\sin x$ in

$(0, \pi/2)$
so that

$f'(x)=\cos x$
again from the graph, it is clear that

$f(x)$ is increasing in

$(0, \pi /2)$ , but

$f'(x)$ is decreasing in

$(0, \pi /2)$
Therefore,

$R$ is wrong. Therefore, d is the correct option.

The slope of the tangent to the curve

$y = f(x)$ at

$\left [ x, f(x) \right ]$ is 2x +If the curve passes through the point (1, 2)then the area bounded by the curve, the x-axis and the line x = 1 is

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

The point(s) on the curve

$y^{3} + 3x^{2} = 12y,$ where the tangent is vertical, is (are)

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

The normal to the curve

$x = a (\cos 0 + 0\sin 0), y= a (\sin 0- 0\cos 0)$ at any point 0 is such that

0%
it makes a constant angle with x-axis
0%
it passes through the origin
0%
it is at a constant distance from the origin
0%
none of these

CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 9 - 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 9 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

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