Practice Start Test

Application of Derivatives

JEE Mathematics · 188 questions · Page 5 of 19 · Click an option or "Show Solution" to reveal answer

Q41
Let f:R→Rf:R \to R be a continuous function defined by f(x)=1ex+2e−xf\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}}Statement−1: Statement - 1 : f(c)=13,f\left( c \right) = {1 \over 3},forsome for some c∈Rc \in R.Statement−2:. Statement - 2 : 0<f(x)≤122,0 < f\left( x \right) \le {1 \over {2\sqrt 2 }},forall for all x∈Rx \in R$
A Statement - 1 is true, Statement -2 is true; Statement - 2 is not a correct explanation for Statement - 1.
B Statement - 1 is true, Statement - 2 is false.
C Statement - 1 is false, Statement - 2 is true.
D Statement - 1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement - 1.
Correct Answer
Option D
Solution
f(x)=1ex+2e−x=exe2x+2f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}} = {{{e^x}} \over {{e^{2x}} + 2}}
f′(x)=(e2x+2)ex−2e2x.ex(e2x+2)2f'\left( x \right) = {{\left( {{e^{2x}} + 2} \right)e{}^x - 2{e^{2x}}.{e^x}} \over {{{\left( {{e^{2x}} + 2} \right)}^2}}}
f′(x)=0⇒e2x+2=2e2xf'\left( x \right) = 0 \Rightarrow {e^{2x}} + 2 = 2{e^{2x}}
e2x=2⇒ex=2{e^{2x}} = 2 \Rightarrow {e^x} = \sqrt 2

maximum

f(x)=24=122f\left( x \right) = {{\sqrt 2 } \over 4} = {1 \over {2\sqrt 2 }}
0<f(x)≤122  ∀x∈R0 < f\left( x \right) \le {1 \over {2\sqrt 2 }}\,\,\forall x \in R

Since

0<13<122⇒0 < {1 \over 3} < {1 \over {2\sqrt 2 }} \Rightarrow

for some

c∈Rc \in R
f(c)=13f\left( c \right) = {1 \over 3}
Q42
For x∈(0,5π2),x \in \left( {0,{{5\pi } \over 2}} \right), define f(x)=∫0xtsin⁡t dt.f\left( x \right) = \int\limits_0^x {\sqrt t \sin t\,dt.} Then ff has
A local minimum at π\pi and 2π2\pi
B local minimum at π\pi and local maximum at 2π2\pi
C local maximum at π\pi and local minimum at 2π2\pi
D local maximum at π\pi and 2π2\pi
Correct Answer
Option C
Solution
f′(x)=xsin⁡xf'\left( x \right) = \sqrt x \sin x

At local maxima or minima,

f′(x)=0f'\left( x \right) = 0
⇒x=0\Rightarrow x = 0

or

sinsin
x=0x=0
⇒x=2π,  π∈(0,5π2)\Rightarrow x = 2\pi ,\,\,\pi \in \left( {0,{{5\pi } \over 2}} \right)
f′′(x)=xcos⁡ x+12xsin⁡ xf''\left( x \right) = \sqrt x \cos \,x + {1 \over {2\sqrt x }}\sin \,x
=12x(2x cos⁡ x+sin⁡ x)= {1 \over {2\sqrt x }}\left( {2x\,\cos \,x + \sin \,x} \right)

At

x=π,x = \pi ,
f′′(x)<0f''\left( x \right) < 0

Hence, local maxima at

x=πx = \pi

At

x=2π,   f′′(x)>0x = 2\pi ,\,\,\,f''\left( x \right) > 0

Hence local minima at

x=2πx = 2\pi
Q43
The shortest distance between line y−x=1y-x=1 and curve x=y2x = {y^2} is
A 328{{3\sqrt 2 } \over 8}
B 832{8 \over {3\sqrt 2 }}
C 43{4 \over {\sqrt 3 }}
D 34{{\sqrt 3 } \over 4}
Correct Answer
Option A
Solution

Shortest distance between two curve occurred along - the common normal Slope of normal to

y2=x{y^2} = x

at point

P(t2,t)P\left( {{t^2},t} \right)

is

−2t-2t

and slope of line

y−x=1y - x = 1

is

1.1.

As they are perpendicular to each other ∴\therefore

(−2t)=−1⇒t=12\left( { - 2t} \right) = - 1 \Rightarrow t = {1 \over 2}

∴\therefore

P(14,12)P\left( {{1 \over 4},{1 \over 2}} \right)

and shortest distance

=∣12−14−12∣= \left| {{{{1 \over 2} - {1 \over 4} - 1} \over {\sqrt 2 }}} \right|

So shortest distance between them is

328{{3\sqrt 2 } \over 8}
Q44
A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25 cm/sec, then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1 m above the ground is :
A 253{{25} \over 3}
B 25
C 253\sqrt 3
D 253{{25} \over {\sqrt 3 }}
Correct Answer
Option D
Solution

x2 + y2 = 4 ⇒\Rightarrow

2xdxdt+2ydydt=02x{{dx} \over {dt}} + 2y{{dy} \over {dt}} = 0
⇒dxdt=−yx.dydt\Rightarrow {{dx} \over {dt}} = - {y \over x}.{{dy} \over {dt}}

When upper end is 1m above the ground,

dxdt=−13.25=−253{{dx} \over {dt}} = - {1 \over {\sqrt 3 }}.25 = - {{25} \over {\sqrt 3 }}

cm/sec.

Q45
The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passes through the point :
A (12,12)\left( {{1 \over 2},{1 \over 2}} \right)
B (12,−13)\left( {{1 \over 2}, - {1 \over 3}} \right)
C (12,13)\left( {{1 \over 2},{1 \over 3}} \right)
D (−12,−13)\left( { - {1 \over 2}, - {1 \over 3}} \right)
Correct Answer
Option A
Solution

Given

y=x+6(x−2)(x−2)y = {{x + 6} \over {\left( {x - 2} \right)\left( {x - 2} \right)}}

At y-axis, x = 0 ⇒\Rightarrow y = 1 On differentiating, we get

dydx=(x2−5x+6)(1)−(x+6)(2x−5)(x2−5x+6)2{{dy} \over {dx}} = {{\left( {{x^2} - 5x + 6} \right)\left( 1 \right) - \left( {x + 6} \right)\left( {2x - 5} \right)} \over {{{\left( {{x^2} - 5x + 6} \right)}^2}}}
dydx=1{{dy} \over {dx}} = 1

at point (0, 1) ∴\therefore Slope of normal = – 1 Now equation of normal is y – 1 = –1 (x – 0) ⇒\Rightarrow y – 1 = – x x + y = 1 ......(

1) By checking each option you can see point

(12,12)\left( {{1 \over 2},{1 \over 2}} \right)

satisfy equation (1).

Q46
Let a,b∈Ra,b \in R be such that the function ff given by f(x)=In∣x∣+bx2+ax, x≠0f\left( x \right) = In\left| x \right| + b{x^2} + ax,\,x \ne 0 has extreme values at x=−1x=-1 and x=2x=2 Statement-1 : ff has local maximum at x=−1x=-1 and at x=2x=2. Statement-2 : a=12a = {1 \over 2} and b=−14b = {-1 \over 4}
A Statement - 1 is false, Statement - 2 is true.
B Statement - 1 is true , Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1.
C Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1.
D Statement - 1 is true, Statement - 2 is false.
Correct Answer
Option B
Solution

Given,

f(x)=ln⁡∣x∣+bx2+axf\left( x \right) = \ln \left| x \right| + b{x^2} + ax

∴\therefore

f′(x)=1x+2bx+af'\left( x \right) = {1 \over x} + 2bx + a

At

x=−1,x=-1,
f′(x)=−1−2b+a=0f'\left( x \right) = - 1 - 2b + a = 0
⇒a−2b=1                     ...(i)\Rightarrow a - 2b = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

At

x=2,x=2,
  f′(x)=12+4b+a=0\,\,f'\left( x \right) = {1 \over 2} + 4b + a = 0
⇒a+4b=−12                     ...(ii)\Rightarrow a + 4b = - {1 \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

On solving

(i)(i)

and

(ii)(ii)

we get

a=12,b=−14a = {1 \over 2},b = - {1 \over 4}

Thus,

f′(x)=1x−x2+12=2−x2+x2xf'\left( x \right) = {1 \over x} - {x \over 2} + {1 \over 2} = {{2 - {x^2} + x} \over {2x}}
=−x2+x+22x=−(x2−x−2)2x=−(x+1)(x−2)2x= {{ - {x^2} + x + 2} \over {2x}} = {{ - \left( {{x^2} - x - 2} \right)} \over {2x}} = {{ - \left( {x + 1} \right)\left( {x - 2} \right)} \over {2x}}

So maximum at

x=−1,2x=-1,2

Hence both the statements are true and statement

22

is a correct explanation for

1.1.
Q47
A spherical balloon is filled with 4500π4500\pi cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72π72\pi cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 4949 minutes after the leakage began is :
A 97{{9 \over 7}}
B 79{{7 \over 9}}
C 29{{2 \over 9}}
D 92{{9 \over 2}}
Correct Answer
Option C
Solution

Volume of spherical balloon

=V=43πr3= V = {4 \over 3}\pi {r^3}
⇒4500π=4πr33\Rightarrow 4500\pi = {{4\pi {r^3}} \over 3}

( as Given, volume

=4500πm3= 4500\pi {m^3}

) Differentiating both the sides,

w.r.t′t′w.r.t't'

we get,

dVdt=4πr2(drdt){{dV} \over {dt}} = 4\pi {r^2}\left( {{{dr} \over {dt}}} \right)

Now, it is given that

dVdt=72π{{dV} \over {dt}} = 72\pi

∴\therefore After

4949

min, Volume -

                       =(4500−49×72)π\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 49 \times 72} \right)\pi
                       =(4500−3528)π\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 3528} \right)\pi
                       =972πm3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 972\pi {m^3}
⇒V=972  πm3\Rightarrow V = 972\,\,\pi {m^3}

∴\therefore

972π=43πr3972\pi = {4 \over 3}\pi r{}^3
⇒r3=3×243=3×35=36=(32)3⇒r=9\Rightarrow {r^3} = 3 \times 243 = 3 \times {3^5} = {3^6} = {\left( {{3^2}} \right)^3} \Rightarrow r = 9

Also, we have

dVdt=72π{{dV} \over {dt}} = 72\pi

∴\therefore

72π=4π×9×9(drdt)⇒drdt=(29)72\pi = 4\pi \times 9 \times 9\left( {{{dr} \over {dt}}} \right) \Rightarrow {{dr} \over {dt}} = \left( {{2 \over 9}} \right)
Q48
A line is drawn through the point (1,2)(1, 2) to meet the coordinate axes at PP and QQ such that it forms a triangle OPQ,OPQ, where OO is the origin. If the area of the triangle OPQOPQ is least, then the slope of the line PQPQ is :
A −14-{1 \over 4}
B −4-4
C −2-2
D −12-{1 \over 2}
Correct Answer
Option C
Solution

Equation of a line passing through

(x1,y1)\left( {{x_1},{y_1}} \right)

having slope

mm

is given by

y−y1=m(x−x1)y - {y_1} = m\left( {x - {x_1}} \right)

Since the line

PQPQ

is passing through

(1,2)(1,2)

therefore its equation is

(y−2)=m(x−1)\left( {y - 2} \right) = m\left( {x - 1} \right)

where

mm

is the slope of the line

PQPQ

. Now, point

P(x,0)P\left( {x,0} \right)

will also satisfy the equation of

PQPQ

∴\therefore

y−2=m(x−1)y - 2 = m\left( {x - 1} \right)
⇒0−2=m(x−1)\Rightarrow 0 - 2 = m\left( {x - 1} \right)
⇒−2=m(x−1)\Rightarrow - 2 = m\left( {x - 1} \right)
⇒x−1=−2m\Rightarrow x - 1 = {{ - 2} \over m}
⇒x=−2m+1\Rightarrow x = {{ - 2} \over m} + 1

Also,

OP=(x−0)2+(0−0)2=x=−2m+1OP = \sqrt {\left( {x - 0} \right){}^2 + {{\left( {0 - 0} \right)}^2}} = x = {{ - 2} \over m} + 1

Similarly, point

Q(0,y)Q\left( {0,y} \right)

will satisfy equation of

PQPQ

∴\therefore

y−2=m(x−1)y - 2 = m\left( {x - 1} \right)
⇒y−2=m(−1)⇒y=2−m\Rightarrow y - 2 = m\left( { - 1} \right) \Rightarrow y = 2 - m

b and

OQ=y=2−mOQ = y = 2 - m

Area of

ΔPOQ=12(OP)(OQ)=12(1−2m)(2−m)\Delta POQ = {1 \over 2}\left( {OP} \right)\left( {OQ} \right) = {1 \over 2}\left( {1 - {2 \over m}} \right)\left( {2 - m} \right)

( As Area of

Δ=12×\Delta = {1 \over 2} \times

base

  ×  \,\, \times \,\,

height )

=12[2−m−4m+2]=12[4−(m+4m)]= {1 \over 2}\left[ {2 - m - {4 \over m} + 2} \right] = {1 \over 2}\left[ {4 - \left( {m + {4 \over m}} \right)} \right]
=2−m2−2m= 2 - {m \over 2} - {2 \over m}

Let Area

=f(m)=2−m2−2m= f\left( m \right) = 2 - {m \over 2} - {2 \over m}

Now,

f′(m)=−12+2m2f'\left( m \right) = {{ - 1} \over 2} + {2 \over {{m^2}}}

Put

f′(m)=0f'\left( m \right) = 0
⇒m2=4⇒m=±2\Rightarrow {m^2} = 4 \Rightarrow m = \pm 2

Now,

f′′(m)=−4m3f''\left( m \right) = {{ - 4} \over {{m^3}}}
f′′(m)∣m=2=−12<0{\left. {f''\left( m \right)} \right|_{m = 2}} = - {1 \over 2} < 0
f′′(m)∣m=−2=12>0{\left. {f''\left( m \right)} \right|_{m = - 2}} = {1 \over 2} > 0

Area will be least at

m=−2m=-2

Hence, slope of

PQPQ

is

−2.-2.
Q49
The real number kk for which the equation, 2x3+3x+k=02{x^3} + 3x + k = 0 has two distinct real roots in [0, 1]\left[ {0,\,1} \right]
A lies between 1 and 2
B lies between 2 and 3
C lies between −1 - 1 and 0
D does not exist.
Correct Answer
Option D
Solution
f(x)=2x3+3x+kf\left( x \right) = 2{x^3} + 3x + k
f′(x)=6x2+3>0f'\left( x \right) = 6{x^2} + 3 > 0
∀x∈R\forall x \in R
      \,\,\,\,\,\,

(as

      \,\,\,\,\,\,
x2>0{x^2} > 0

)

⇒f(x)\Rightarrow f\left( x \right)

is strictly increasing function

⇒f(x)=0\Rightarrow f\left( x \right) = 0

has only one real root, so two roots are not possible.

Q50
The intercepts on xx-axis made by tangents to the curve, y=∫0x∣t∣dt,x∈R,y = \int\limits_0^x {\left| t \right|dt,x \in R,} which are parallel to the line y=2xy=2x, are equal to :
A ±1 \pm 1
B ±2 \pm 2
C ±3 \pm 3
D ±4 \pm 4
Correct Answer
Option A
Solution

Since,

y=∫0x∣t∣dt,x∈Ry = \int\limits_0^x {\left| t \right|} dt,x \in R

therefore

dydx=∣x∣{{dy} \over {dx}} = \left| x \right|

But from

y=2x,dydx=2y = 2x,{{dy} \over {dx}} = 2
⇒∣x∣=2⇒x=±2\Rightarrow \left| x \right| = 2 \Rightarrow x = \pm 2

Points

y=∫0±2∣t∣dt=±2y = \int\limits_0^{ \pm 2} {\left| t \right|dt} = \pm 2

∴\therefore equation of tangent is

y−2=2(x−2)y - 2 = 2\left( {x - 2} \right)

or

y+2=2(x+2)y + 2 = 2\left( {x + 2} \right)

⇒\Rightarrow

xx

-intercept

=±1.= \pm 1.
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