Practice Start Test

Application of Derivatives

JEE Mathematics · 188 questions · Page 3 of 19 · Click an option or "Show Solution" to reveal answer

Q21
The maximum distance from origin of a point on the curve x=asin⁡t−bsin⁡(atb)x = a\sin t - b\sin \left( {{{at} \over b}} \right) y=acos⁡t−bcos⁡(atb),y = a\cos t - b\cos \left( {{{at} \over b}} \right), both a,b>0a,b > 0 is
A a−ba-b
B a+ba+b
C a2+b2\sqrt {{a^2} + {b^2}}
D a2−b2\sqrt {{a^2} - {b^2}}
Correct Answer
Option B
Solution

Distance of origin from

(x,y)=x2+y2\left( {x,y} \right) = \sqrt {{x^2} + {y^2}}
=a2+b2−2abcos⁡(t−atb);= \sqrt {{a^2} + {b^2} - 2ab\cos \left( {t - {{at} \over b}} \right)} ;
≤a2+b2+2ab\le \sqrt {{a^2} + {b^2} + 2ab}
[{cos⁡(t−atb)}min⁡=−1]\left[ {{{\left\{ {\cos \left( {t - {{at} \over b}} \right)} \right\}}_{\min }} = - 1} \right]
=a+b=a+b

∴\therefore Maximum distance from origin

=a+b=a+b
Q22
The number of points on the curve y=54x5−135x4−70x3+180x2+210xy=54 x^{5}-135 x^{4}-70 x^{3}+180 x^{2}+210 x at which the normal lines are parallel to x+90y+2=0x+90 y+2=0 is :
A 2
B 3
C 4
D 0
Correct Answer
Option C
Solution
y′=270x4−540x3−210x2+360x+210y'=270x^4-540x^3-210x^2+360x+210

Slope of normal

=−190=-\frac{1}{90}

∴\therefore Slope of tangent = 90 ∴\therefore Number of normal will be number of solutions of

270x4−540x3−210x2+360x+210=90270x^4-540x^3-210x^2+360x+210=90
⇒9x4−18x3−7x2+12x+4=0\Rightarrow 9x^4-18x^3-7x^2+12x+4=0
∴x=1,2,−13,−23\therefore x=1,2,-\frac{1}{3},-\frac{2}{3}

are roots

Q23
If the function f(x)=2x3−9ax2+12a2x+1,f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1, where a>0,a>0, attains its maximum and minimum at pp and qq respectively such that p2=q{p^2} = q , then aa equals
A 12{1 \over 2}
B 33
C 11
D 22
Correct Answer
Option D
Solution
f(x)=2x3−9ax2+12a2x+1f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1
f′(x)=6x2−18ax+12a2;f'\left( x \right) = 6{x^2} - 18ax + 12{a^2};
          f′′(x)=12x−18a\,\,\,\,\,\,\,\,\,\,f''\left( x \right) = 12x - 18a

For max. or min.

6x2−18ax+12a2=06{x^2} - 18ax + 12{a^2} = 0
⇒x2−3ax+2a2=0\Rightarrow {x^2} - 3ax + 2{a^2} = 0
⇒x=a\Rightarrow x = a

or

x=2a.x=2a.

At

x=ax=a
f′′(a)=12a−18a=−6a<0f''(a) = 12a - 18a = -6a < 0

At

x=ax=a

maximum As f''(

aa

) < 0. At

x=2ax=2a
f′′(a)=24a−18a=6a>0f''(a) = 24a - 18a = 6a > 0

At

x=2ax=2a

minimum As f''(2

aa

) > 0. ∴\therefore

p=ap=a

and

q=2aq=2a

As per question

p2=q{p^2} = q

∴\therefore

a2=2a{a^2} = 2a

⇒\Rightarrow

a(a−2)=0a(a - 2) = 0

∴\therefore

aa

= 0, 2 but

a>0,a > 0,

therefore,

a=2.a=2.
Q24
A point on the parabola y2=18x{y^2} = 18x at which the ordinate increases at twice the rate of the abscissa is
A (98,92)\left( {{9 \over 8},{9 \over 2}} \right)
B (2,−4)(2, -4)
C (−98,92)\left( {{-9 \over 8},{9 \over 2}} \right)
D (2,4)(2, 4)
Correct Answer
Option A
Solution
y2=18x⇒2ydydx=18⇒dydx=9y{y^2} = 18x \Rightarrow 2y{{dy} \over {dx}} = 18 \Rightarrow {{dy} \over {dx}} = {9 \over y}

Given

dydx=2⇒92=2⇒y=92{{dy} \over {dx}} = 2 \Rightarrow {9 \over 2} = 2 \Rightarrow y = {9 \over 2}

Puting in

y2=18x⇒x=98{y^2} = 18x \Rightarrow x = {9 \over 8}

∴\therefore Required point is

(98,92)\left( {{9 \over 8},{9 \over 2}} \right)
Q25
If 2a+3b+6c=02a+3b+6c=0, then at least one root of the equation ax2+bx+c=0a{x^2} + bx + c = 0 lies in the interval
A (1,3)(1, 3)
B (1,2)(1, 2)
C (2,3)(2, 3)
D (0,1)(0, 1)
Correct Answer
Option D
Solution

Let us define a function

f(x)=ax33+bx22+cxf\left( x \right) = {{ax{}^3} \over 3} + {{b{x^2}} \over 2} + cx

Being polynomial, it is continuous and differentiable, also,

f(0)=0 f\left( 0 \right) = 0\,

and

  f(1)=a3+b2+c\,\,f\left( 1 \right) = {a \over 3} + {b \over 2} + c
⇒f(1)=2a+3b+6c6=0\Rightarrow f\left( 1 \right) = {{2a + 3b + 6c} \over 6} = 0

(given) ∴\therefore

f(0)=f(1)f\left( 0 \right) = f\left( 1 \right)

∴\therefore

f(x)f(x)

satisfies all conditions of Rolle's theorem therefore

f′(x)=0f'\left( x \right) = 0

has a root in

(0,1)\left( {0,1} \right)

i.e.

ax2+bx+c=0a{x^2} + bx + c = 0

has at lease one root in

(0,1)(0, 1)
Q26
The normal to the curve x = a(1 + cos θ\theta ), y=asin⁡θy = a\sin \theta at ′θ′'\theta ' always passes through the fixed point
A (a,a)(a, a)
B (0,a)(0, a)
C (0,0)(0, 0)
D (a,0)(a, 0)
Correct Answer
Option D
Solution
dxdθ=−asin⁡θ{{dx} \over {d\theta }} = - a\sin \theta

and

dydθ=acos⁡θ{{dy} \over {d\theta }} = a\cos \theta

∴\therefore

dydx=−cot⁡θ.{{dy} \over {dx}} = - \cot \theta .

∴\therefore The slope of the normal at θ\theta =

−1−cot⁡θ- {1 \over { - \cot \theta }}
=tan⁡θ= \tan \theta

∴\therefore The equation of the normal at θ\theta is

y−asin⁡θ=tan⁡θ(x−a−acos⁡θ)y - a\sin \theta = \tan \theta \left( {x - a - a\cos \theta } \right)

⇒\Rightarrow

y−asin⁡θ=y - a\sin \theta =
sin⁡θcos⁡θ{{\sin \theta } \over {\cos \theta }}
(x−a−acos⁡θ)\left( {x - a - a\cos \theta } \right)
⇒ycos⁡θ−asin⁡θcos⁡θ\Rightarrow y\cos \theta - a\sin \theta \cos \theta
           \,\,\,\,\,\,\,\,\,\,\,
=xsin⁡θ−asin⁡θ−asin⁡θcos⁡θ= x\sin \theta - a\sin \theta - a\sin \theta \cos \theta
⇒xsin⁡θ−ycos⁡θ=asin⁡θ\Rightarrow x\sin \theta - y\cos \theta = a\sin \theta
⇒y=(x−a)tan⁡θ\Rightarrow y = \left( {x - a} \right)\tan \theta

which always passes through

(a,0)(a, 0)
Q27
A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?
A Interval Function (- ∞\infty , ∞\infty ) x3 - 3x2 + 3x + 3
B Interval Function [2, ∞\infty ) 2x3 - 3x2 - 12x + 6
C Interval Function (−∞,13]\left( { - \infty ,{1 \over 3}} \right] 3x2 - 2x + 1
D Interval Function (−∞ - \infty , - 4 ) x3 + 6x2 + 6
Correct Answer
Option C
Solution

Clearly function

f(x)=3x2−2x+1f\left( x \right) = 3{x^2} - 2x + 1

is increasing when

f′(x)=6x−2≥0⇒     x∈[1/3,∞)f'\left( x \right) = 6x - 2 \ge 0 \Rightarrow \,\,\,\,\,x \in \left[ {1/3,\left. \infty \right)} \right.

∴\therefore

f(x)f(x)

is incorrectly matched with

(−∞,13)\left( { - \infty ,{1 \over 3}} \right)
Q28
If the equation anxn+an−1xn−1+...........+a1x=0{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........... + {a_1}x = 0 a1≠0,n≥2,{a_1} \ne 0,n \ge 2, has a positive root x=αx = \alpha , then the equation nanxn−1+(n−1)an−1xn−2+...........+a1=0n{a_n}{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + ........... + {a_1} = 0 has a positive root, which is
A greater than α\alpha
B smaller than α\alpha
C greater than or equal to smaller than α\alpha
D equal to smaller than α\alpha
Correct Answer
Option B
Solution

Let

f(x)=anxn+an−1xn−1+...........+a1x=0f\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........... + {a_1}x = 0

The other given equation,

nanxn−1+(n−1)an−1xn−2+....+a1=0=f′(x)na{}_n{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + .... + {a_1} = 0 = f'\left( x \right)

Given

a1≠0⇒f(0)=0{a_1} \ne 0 \Rightarrow f\left( 0 \right) = 0

Again

f(x)f(x)

has root

α,   ⇒f(α)=0\alpha ,\,\,\, \Rightarrow f\left( \alpha \right) = 0

∴\therefore

     f(0)=f(α)\,\,\,\,\,f\left( 0 \right) = f\left( \alpha \right)

∴\therefore By Roll's theorem

f′(x)=0f'(x)=0

has root be- tween

(0,α)\left( {0,\alpha } \right)

Hence

f′(x)f'(x)

has a positive root smaller than

α.\alpha .
Q29
The normal to the curve x=a(cos⁡θ+θsin⁡θ),y=a(sin⁡θ−θcos⁡θ)x = a\left( {\cos \theta + \theta \sin \theta } \right),y = a\left( {\sin \theta - \theta \cos \theta } \right) at any point θ ′\theta\, ' is such that
A it passes through the origin
B it makes an angle π2+θ{\pi \over 2} + \theta with the xx-axis
C it passes through (aπ2,−a)\left( {a{\pi \over 2}, - a} \right)
D it is at a constant distance from the origin
Correct Answer
Option D
Solution
x=a(cos⁡θ+θsin⁡θ)x = a\left( {\cos \theta + \theta \sin \theta } \right)
⇒dxdθ=a(−sin⁡θ+sin⁡θ+θcos⁡θ)\Rightarrow {{dx} \over {d\theta }} = a\left( { - \sin \theta + \sin \theta + \theta \cos \theta } \right)
⇒dxdθ=aθcos⁡θ                ...(1)\Rightarrow {{dx} \over {d\theta }} = a\theta \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)
y=a(sin⁡θ−θcos⁡θ)y = a\left( {\sin \theta - \theta \cos \theta } \right)
dydθ=a[cos⁡θ−cos⁡θ+θsin⁡θ]{{dy} \over {d\theta }} = a\left[ {\cos \theta - \cos \theta + \theta \sin \theta } \right]
⇒dydθ=aθsin⁡θ                ...(2)\Rightarrow {{dy} \over {d\theta }} = a\theta \sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)

From equations

(1)(1)

and

(2),(2),

we get

dydx=tan⁡θ⇒{{dy} \over {dx}} = \tan \theta \Rightarrow

Slope of normal

=−cot⁡ θ= - \cot \,\theta

Equation of normal at

′θ′'\theta '

is

y−a(sin⁡θ−θcos⁡θ)y - a\left( {\sin \theta - \theta \cos \theta } \right)
=−cot⁡θ(x−a(cos⁡θ+θsin⁡θ)= - \cot \theta \left( {x - a\left( {\cos \theta + \theta \sin \theta } \right)} \right.
⇒ysin⁡θ−asin⁡2θ+a  θcos⁡θsin⁡θ\Rightarrow y\sin \theta - a{\sin ^2}\theta + a\,\,\theta \cos \theta \sin \theta
=−xcos⁡θ+acos⁡2θ+a θsin⁡θcos⁡θ= - x\cos \theta + a{\cos ^2}\theta + a\,\theta \sin \theta \cos \theta
⇒xcos⁡θ+ysin⁡θ=a\Rightarrow x\cos \theta + y\sin \theta = a

Clearly this is an equation of straight line - which is at a constant distance

′a′'a'

from origin.

Q30
A lizard, at an initial distance of 21 cm behind an insect moves from rest with an acceleration of 2 cm/s22 \mathrm{~cm} / \mathrm{s}^2 and pursues the insect which is crawling uniformly along a straight line at a speed of 20 cm/s20 \mathrm{~cm} / \mathrm{s}. Then the lizard will catch the insect after :
A 20 s
B 1 s
C 21 s
D 24 s
Correct Answer
Option C
Solution

The motion of the lizard, which starts from rest and accelerates at a rate of a=2 cm/s2a = 2 \, \text{cm/s}^2, can be described by the equation of motion : Dl=12at2D_l = \dfrac{1}{2} a t^2 where DlD_l is the distance the lizard travels, aa is its acceleration, and tt is the time.

The insect, moving at a constant speed of v=20 cm/sv = 20 \, \text{cm/s}, has a motion that can be simply described as: Di=vtD_i = v t where DiD_i is the distance the insect travels, vv is its velocity, and tt is the time.

Since the lizard starts 21 cm behind the insect and needs to catch up, it must travel the distance the insect travels plus an additional 21 cm.

Equating these two distances gives us : 12at2=vt+21\dfrac{1}{2} a t^2 = v t + 21 Substituting the given values into this equation gives : t2=(vt+21)⋅2at^2 = \dfrac{(v t + 21) \cdot 2}{a} Substituting a=2 cm/s2a = 2 \, \text{cm/s}^2 and v=20 cm/sv = 20 \, \text{cm/s}, we simplify to : t2=20t+21t^2 = 20t + 21 This simplifies to a quadratic equation : t2−20t−21=0t^2 - 20t - 21 = 0 Solving this quadratic equation for tt gives the solutions t=21 st = 21 \, \text{s} and t=−1 st = -1 \, \text{s}.

Since time cannot be negative, we discard the -1 s solution.

Therefore, the lizard will catch the insect after 21 seconds.

So, the correct answer is Option C : 21 s.

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