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CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 9 - MCQExams.com
CBSE
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
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0%
$$\dfrac {22}{7}$$
0%
$$\dfrac {6}{7}$$
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$$-6$$
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 $$t=2$$
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?
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$$(3, 0), (-1, 0)$$
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$$(3, 0), (1, 2)$$
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$$(-1, 0), (1, 2)$$
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$$(1, 2), (1, -2)$$
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
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$$(0, 0)$$
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$$(2, 16)$$
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$$(3, 9)$$
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$$(6, 36)$$
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
so $$\left(\dfrac{dy}{dx}\right)_{x_1, y_1}=0$$
$$12-2x_1=0$$
$$2x_1=12$$
$$x_1=6$$
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
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$$x+y=3$$
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$$x-y=3$$
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$$x+y=1$$
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$$x-y=1$$
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
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$$\dfrac {1}{2}$$
0%
$$0$$
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$$-2$$
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$$\infty$$
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$$, where the normal to the curve makes equal intercepts with the axes is?
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$$(4, \pm 8/3)$$
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$$(-4, 8/3)$$
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$$(-4, -8/3)$$
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$$(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$$, if the value of m is?
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$$1$$
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$$2$$
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$$3$$
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$$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)$$ is
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0%
$$\dfrac {22}{7}$$
0%
$$\dfrac {6}{7}$$
0%
$$\dfrac {7}{6}$$
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$$-\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$$ increases along the entire number scale, then
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$$a=b$$
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$$a=\cfrac{1}{2}b$$
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$$a\le -\cfrac{1}{2}$$
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$$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}$$ is increasing, if
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$$\lambda< 1$$
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$$\lambda> 1$$
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$$\lambda< 2$$
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$$\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$$ satisfy
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$${a}^{2}-3b-15> 0$$
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$${a}^{2}-3b+15> 0$$
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$${a}^{2}-3b+15< 0$$
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$$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$$, then
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$$a\in \left (\dfrac {1}{2},\infty \right)$$
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$$a\in \left (-\dfrac {1}{2},\dfrac {1}{2}\right)$$
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$$a=\dfrac {1}{2}$$
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$$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$$, if
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$$a> 0$$
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$$a< 0$$
0%
$$0< a< 1$$
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$$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$$
or
$$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$$, then
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$$-1< k< 1$$
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$$k< -1$$ or $$k> 1$$
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$$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]$$, then
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$$k\in (2,\infty)$$
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$$k\in (-\infty,2)$$
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$$k\in (4,\infty)$$
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$$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$$, if
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$$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$$ is increasing on
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$$R$$
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$$(-\infty,0)$$
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$$(0,\infty)$$
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$${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$$ ?
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$$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$$ equal to ?
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$$0$$
0%
$$1$$
0%
$$2$$
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$$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$$
At $$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}}$$
At $$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$$ equal to ?
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$$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)$$
At $$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)}$$
At $$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$$ make with y-axis ?
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$$\tan^{-1}m^2$$
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$$\cot^{-1}(a+m^2)$$
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$$\sin^{-1}(\dfrac{1}{\sqrt{1+m^4}})$$
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$$\sec^{-1}\sqrt{1+m^4}$$
The function $$f(x)=4-3x+3x^2-x^3$$ is
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decreasing on $$R$$
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increasing on $$R$$
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strictly decreasing on $$R$$
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strictly increasing on $$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)$$, is
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$$-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)$$?
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$$f(x)$$ decreases for $$x$$ $$\epsilon$$ $$[2 n \pi, (2n + 1)\pi]$$, $$n$$ $$\epsilon$$ $$Z$$
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$$f(x)$$ decreases for $$x$$ $$\epsilon$$ $$\left [ (2n - 1)\frac{\pi}{2}, (2n + 1)\frac{\pi}{2}\right ]$$ $$n$$ $$\epsilon$$ $$Z$$
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$$f(x)$$ is non-monotonic function
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$$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$$, which are equally inclined to the axes, is
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0%
2
0%
1
0%
0
0%
4
If m is the slope of a tangent to the curve $$e^{y}=1+x^{2},$$ then
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$$\left | m \right |> 1$$
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$$m> 1$$
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$$m> -1$$
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$$\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$$ decreases in the interval
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0%
$$(-\infty , -2)$$
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$$(-2, -1)$$
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$$(1, 2)$$
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$$(2, \infty)$$
The function $$f(x)=3x+\cos 3x$$ is
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increasing on $$R$$
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decreasing on $$R$$
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strictly increasing on $$R$$
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strictly decreasing on $$R$$
$$f(x)=\sin x-kx $$ is decreasing for all $$x \in R$$, when
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$$k < 1$$
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$$k \le 1$$
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$$k > 1$$
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$$k \ge 1$$
For $$x > 1, y=\log_e x$$ satisfies the inequality
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$$x-1 > y$$
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$$x^2 -1 >y$$
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$$y > x-1$$
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$$\dfrac {x-1}{x} < y$$
The slope of the tangent to the curve $$y = \sqrt{4-x^{2}}$$ at the point, where the ordinate and the abscissa are equal , is
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-1
0%
1
0%
0
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None of these
Explanation
Putting $$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}$$, then n equals
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0%
1
0%
3
0%
2
0%
4
The curve given by $$x + y = e^{xy}$$ has a tangent parallel to the y-axis at the point
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$$(0,1)$$
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$$( 1, 0 )$$
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$$(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}$$ with the x-axis
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ln $$\left ( \dfrac{\sqrt{3}+\sqrt{7}}{7} \right )$$ and ln $$\left ( \dfrac{\sqrt{3}+\sqrt{5}}{2} \right )$$
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ln $$\left ( \dfrac{\sqrt{3}+\sqrt{7}}{2} \right )$$
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ln $$\left ( \dfrac{\sqrt{7}+\sqrt{3}}{2} \right )$$
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$$\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 $$ is
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0%
9
0%
-5
0%
7
0%
-7
At what points of curve $$y = \dfrac{2}{3}x^{3}+\dfrac{1}{2}x^{2}$$, the tangent makes the equal with the axis?
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$$(\dfrac{1}{2},\dfrac{5}{24})$$ and $$\left ( -1,\dfrac{-1}{6} \right )$$
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$$(\dfrac{1}{2},\dfrac{4}{9})$$ and $$ ( -1,0)$$
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$$\left ( \dfrac{1}{3},\dfrac{1}{7} \right )$$ and$$ \left ( -3, \dfrac{1}{2} \right )$$
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$$\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$$
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tanfent and normal intersect at the point (2, 1)
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normal at $$t = \pi /4$$ is parallel to the y-axis
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tangent at $$t = \pi /4$$ is parallel to the line y = x
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tangent at $$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$$ is
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$$27c^{3}$$
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$$\dfrac{4}{27}c^{3}$$
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$$\dfrac{27}{4}c^{3}$$
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$$\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$$ is proportion to
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square of the abscissa of the point of tangency
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square root of the abscissa of the point of tangency
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cube of the abscissa pf the point of tangency
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cube root of the abscissa of the point of tangency
The angle between the tangent to the curves $$y = x^{2}$$ and $$x = y^{2}$$ at (1, 1) is
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$$\cos ^{-1}\dfrac{4}{5}$$
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$$\sin ^{-1}\dfrac{3}{5}$$
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$$\tan ^{-1}\dfrac{3}{4}$$
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$$\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}$$ ot the x-axis are
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(0, 0)
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$$\left ( \sqrt{3},\dfrac{-\sqrt{3}}{2} \right )$$
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$$\left ( -2 ,\dfrac{2}{3}\right )$$
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$$\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$$ , then value of m is
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$$m = \dfrac{\sqrt{2}}{3}$$
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$$m = -\dfrac{\sqrt{2}}{3}$$
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$$m = \dfrac{3}{\sqrt{2}}$$
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$$m = -\dfrac{3}{\sqrt{2}}$$
A curve passes through $$(2,1)$$ and is such that the square of the ordinate is twice the contained by the abscissa and the intercept of the normal. Then the equation of curve is
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$$x^2 +y^2=9x$$
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$$4x^2 +y^2=9x$$
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$$4x^2 +2y^2=9x$$
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None of these
The tangent to the curve $$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})$$
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on the left of x =c
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on the right of x = c
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at no point
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at all point
Let $$f:[1, \infty) \rightarrow R$$ and $$f(x)=x \int_{1}^{x} \dfrac{e^{t}}{t} d t-e^{x},$$ then
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$$f(x)$$ is an increasing function
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$$\lim _{x \rightarrow \infty} f(x) \rightarrow \infty$$
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$$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
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$$a > 0, b> 0$$
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$$a > 0, b < 0$$
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$$a < 0, b > 0$$
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$$a < 0, b < 0$$
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?
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Both $$S$$ and $$R$$ are wrong
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Both $$S$$ and $$R$$ are correct but $$R$$ is not the correct explanation of $$S$$
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$$S$$ is the correct and $$R$$ is the correct explanation of $$S$$
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$$S$$ 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
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0%
$$\dfrac{5}{6}$$
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$$\dfrac{6}{5}$$
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$$\dfrac{1}{6}$$
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6
The point(s) on the curve $$y^{3} + 3x^{2} = 12y,$$ where the tangent is vertical, is (are)
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$$\left ( \pm \dfrac{4}{\sqrt{3}}, -2 \right )$$
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$$\left ( \pm \sqrt{\dfrac{11}{3}}, 1 \right )$$
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(0, 0)
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$$\left ( \pm \dfrac{4}{\sqrt{3}}, 2 \right )$$
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
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it makes a constant angle with x-axis
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it passes through the origin
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it is at a constant distance from the origin
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none of these
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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