Practice Start Test

Application of Derivatives

JEE Mathematics · 188 questions · Page 4 of 19 · Click an option or "Show Solution" to reveal answer

Q31
The function f(x)=x2+2xf\left( x \right) = {x \over 2} + {2 \over x} has a local minimum at
A x=2x=2
B x=−2x=-2
C x=0x=0
D x=1x=1
Correct Answer
Option A
Solution
f(x)=x2+2x⇒f′(x)=12−2x2=0f\left( x \right) = {x \over 2} + {2 \over x} \Rightarrow f'\left( x \right) = {1 \over 2} - {2 \over {{x^2}}} = 0
⇒x2=4\Rightarrow {x^2} = 4

or

x=2,−2;x=2,-2;
     f′′(x)=4x3\,\,\,\,\,f''\left( x \right) = {4 \over {{x^3}}}
f′′(x)]x=2=+ve⇒f(x)f''{\left. {\left( x \right)} \right]_{x = 2}} = + ve \Rightarrow f\left( x \right)

has local min at

x=2.x=2.
Q32
Angle between the tangents to the curve y=x2−5x+6y = {x^2} - 5x + 6 at the points (2,0)(2,0) and (3,0)(3,0) is
A π\pi
B π2{\pi \over 2}
C π6{\pi \over 6}
D π4{\pi \over 4}
Correct Answer
Option B
Solution
dydx=2x−5{{dy} \over {dx}} = 2x - 5

∴\therefore

m1=(2x−5)(2,0)=−1,{m_1} = {\left( {2x - 5} \right)_{\left( {2,0} \right)}} = - 1,
m2=(2x−5)(3,0)=1⇒m1m2=−1{m_2} = {\left( {2x - 5} \right)_{\left( {3,0} \right)}} = 1 \Rightarrow {m_1}{m_2} = - 1

i.e. the tangents are perpendicular to each other.

Q33
A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length xx. The maximum area enclosed by the park is
A 32x2{3 \over 2}{x^2}
B x38\sqrt {{{{x^3}} \over 8}}
C 12x2{1 \over 2}{x^2}
D πx2\pi {x^2}
Correct Answer
Option C
Solution

Area

=12x2 sin⁡ θ= {1 \over 2}{x^2}\,\sin \,\theta

Maximum value of

sin⁡θ\sin \theta

is

11

at

θ=π2\theta = {\pi \over 2}
Amax⁡=12x2{A_{\max }} = {1 \over 2}{x^2}
Q34
If pp and qq are positive real numbers such that p2+q2=1{p^2} + {q^2} = 1, then the maximum value of (p+q)(p+q) is
A 12{1 \over 2}
B 12{1 \over {\sqrt 2 }}
C 2{\sqrt 2 }
D 22
Correct Answer
Option C
Solution

Given that

p2+q2=1{p^2} + {q^2} = 1

∴\therefore

p=cos⁡θp = \cos \theta

and

q=sin⁡θq = \sin \theta

Then

p+qp+q
=cos⁡θ+sin⁡θ= \cos \theta + \sin \theta

We know that

−a2+b2≤acos⁡θ+bsin⁡θ≤a2+b2- \sqrt {{a^2} + {b^2}} \le a\cos \theta + b\sin \theta \le \sqrt {{a^2} + {b^2}}

∴\therefore

−2≤cos⁡θ+sin⁡θ≤2- \sqrt 2 \le \cos \theta + \sin \theta \le \sqrt 2

Hence max. value of

p+qp + q

is

2\sqrt 2
Q35
A value of cc for which conclusion of Mean Value Theorem holds for the function f(x)=log⁡exf\left( x \right) = {\log _e}x on the interval [1,3]\left[ {1,3} \right] is
A log⁡3e{\log _3}e
B log⁡e3{\log _e}3
C 2  log⁡3e2\,\,{\log _3}e
D 12log⁡3e{1 \over 2}{\log _3}e
Correct Answer
Option C
Solution

Using Lagrange's Mean Value Theorem Let

f(x)f(x)

be a function defined on

[a,b]\left[ {a,b} \right]

then,

f′(c)=f(b)−f(a)b−a            ....(i)f'\left( c \right) = {{f\left( b \right) - f\left( a \right)} \over {b - a}}\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)
c  ∈[a,b]c\,\, \in \left[ {a,b} \right]

∴\therefore Given

f(x)=log⁡exf\left( x \right) = {\log _e}x

∴\therefore

f′(x)=1xf'\left( x \right) = {1 \over x}

∴\therefore equation

(i)(i)

become

1c=f(3)−f(1)3−1{1 \over c} = {{f\left( 3 \right) - f\left( 1 \right)} \over {3 - 1}}
⇒1c=log⁡e3−log⁡e12=log⁡e32\Rightarrow {1 \over c} = {{{{\log }_e}3 - {{\log }_e}1} \over 2} = {{{{\log }_e}3} \over 2}
⇒c=2log⁡e3⇒c=2 log⁡3e\Rightarrow c = {2 \over {{{\log }_e}3}} \Rightarrow c = 2\,{\log _3}e
Q36
The function f(x)=tan⁡−1(sin⁡x+cos⁡x)f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right) is an incresing function in
A (0,π2)\left( {0,{\pi \over 2}} \right)
B (−π2,π2)\left( { - {\pi \over 2},{\pi \over 2}} \right)
C (π4,π2)\left( { {\pi \over 4},{\pi \over 2}} \right)
D (−π2,π4)\left( { - {\pi \over 2},{\pi \over 4}} \right)
Correct Answer
Option D
Solution

Given

f(x)=tan⁡−1(sin⁡x+cos⁡x)f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)
f′(x)=11+(sin⁡x+cos⁡x)2.(cos⁡x−sin⁡x)f'\left( x \right) = {1 \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}.\left( {\cos x - \sin x} \right)
=2.(12cos⁡x−12sin⁡x)1+(sin⁡x+cos⁡x)2= {{\sqrt 2 .\left( {{1 \over {\sqrt 2 }}\cos x - {1 \over {\sqrt 2 }}\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}
=(cos⁡π4.cos⁡x−sin⁡π4.sin⁡x)1+(sin⁡x+cos⁡x)2= {{\left( {\cos {\pi \over 4}.\cos x - \sin {\pi \over 4}.\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}

∴\therefore

f′(x)=2cos⁡(x+π4)1+(sin⁡x+cos⁡x)2f'\left( x \right) = {{\sqrt 2 \cos \left( {x + {\pi \over 4}} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}

if

f′(x)>Of'\left( x \right) > O

then

f(x)f\left( x \right)

is increasing function. Hence

f(x)f(x)

is increasing, if

−π2<x+π4<π2- {\pi \over 2} < x + {\pi \over 4} < {\pi \over 2}
⇒−3π4<x<π4\Rightarrow - {{3\pi } \over 4} < x < {\pi \over 4}

Hence,

f(x)f(x)

is increasing when

n∈(−π2,π4)n \in \left( { - {\pi \over 2},{\pi \over 4}} \right)
Q37
Suppose the cubic x3−px+q{x^3} - px + q has three distinct real roots where p>0p>0 and q>0q>0. Then which one of the following holds?
A The cubic has minima at p3\sqrt {{p \over 3}} and maxima at −p3-\sqrt {{p \over 3}}
B The cubic has minima at −p3-\sqrt {{p \over 3}} and maxima at p3\sqrt {{p \over 3}}
C The cubic has minima at both p3\sqrt {{p \over 3}} and −p3-\sqrt {{p \over 3}}
D The cubic has maxima at both p3\sqrt {{p \over 3}} and −p3-\sqrt {{p \over 3}}
Correct Answer
Option A
Solution

Let

y=x3−px+qy = {x^3} - px + q
⇒dydx=3x2−p\Rightarrow {{dy} \over {dx}} = 3{x^2} - p

For

dydx=0⇒3x2−p=0{{dy} \over {dx}} = 0 \Rightarrow 3{x^2} - p = 0
⇒x=±p3\Rightarrow x = \pm \sqrt {{p \over 3}}
d2ydx2=6x{{{d^2}y} \over {d{x^2}}} = 6x
d2ydx2∣x=p3=+ve    {\left. {{{{d^2}y} \over {d{x^2}}}} \right|_{x = \sqrt {{p \over 3}} }} = + ve\,\,\,\,

and

          \,\,\,\,\,\,\,\,\,\,
   d2ydx2∣x=−p3=−ve{\left. {\,\,\,{{{d^2}y} \over {d{x^2}}}} \right|_{x = - \sqrt {{p \over 3}} }} = - ve

∴\therefore

yy

has ninima at

x=p3x = \sqrt {{p \over 3}}

and maxima at

x=−p3x = - \sqrt {{p \over 3}}
Q38
Given P(x)=x4+ax3+bx2+cx+dP\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d such that x=0x=0 is the only real root of P′ (x)=0.P'\,\left( x \right) = 0. If P(−1)<P(1),P\left( { - 1} \right) < P\left( 1 \right), then in the interval [−1,1]:\left[ { - 1,1} \right]:
A P(−1)P(-1) is not minimum but P(1)P(1) is the maximum of PP
B P(−1)P(-1) is the minimum but P(1)P(1) is not the maximum of PP
C Neither P(−1)P(-1) is the minimum nor P(1)P(1) is the maximum of PP
D P(−1)P(-1) is the minimum and P(1)P(1) is the maximum of PP
Correct Answer
Option A
Solution

We have

P(x)=x4+ax3+bx2+cx+dP\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d
⇒P′(x)=4 x3+3ax2+2bx+c\Rightarrow P'\left( x \right) = 4\,{x^3} + 3a{x^2} + 2bx + c

But

P′(0)=0⇒c=0P'\left( 0 \right) = 0 \Rightarrow c = 0

∴\therefore

P(x)=x4+ax3+bx2+dP\left( x \right) = {x^4} + a{x^3} + b{x^2} + d

As given that

P(−1)<P(a)P\left( { - 1} \right) < P\left( a \right)
⇒1−a+b+d  <  1+a+b+d⇒a>0\Rightarrow 1 - a + b + d\,\, < \,\,1 + a + b + d \Rightarrow a > 0

Now

P′(x)=4x3+3ax2+2bx=x(4x2+3ax+2b)P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx = x\left( {4{x^2} + 3ax + 2b} \right)

As

P′(x)=0,P'\left( x \right) = 0,

there is only one solution

x=0,x = 0,

therefore

4x2+3ax+2b=04{x^2} + 3ax + 2b = 0

should not have any real roots i.e.

D<0D < 0
⇒9a2−32b<0\Rightarrow 9{a^2} - 32b < 0
⇒b>9a232>0\Rightarrow b > {{9{a^2}} \over {32}} > 0

Hence

a,b>0⇒P′(x)=4x3+3ax2+2bx>0a,b > 0 \Rightarrow P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx > 0
∀x>0\forall x > 0

∴\therefore

P(x)P(x)

is an increasing function on

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

∴\therefore

P(0)<P(a)P\left( 0 \right) < P\left( a \right)

Similarly we can prove

P(x)P\left( x \right)

is decreasing on

(−1,0)\left( { - 1,0} \right)

∴\therefore

P(−1)>P(0)P\left( { - 1} \right) > P\left( 0 \right)

So we can conclude that Max

P(x)=P(1)P\left( x \right) = P\left( 1 \right)

and Min

P(x)=P(0)P\left( x \right) = P\left( 0 \right)
⇒P(−1)\Rightarrow P\left( { - 1} \right)

is not minimum but

P(1)P\left( 1 \right)

is the maximum of

P.P.
Q39
The equation of the tangent to the curve y=x+4x2y = x + {4 \over {{x^2}}}, that is parallel to the xx-axis, is
A y=1y=1
B y=2y=2
C y=3y=3
D y=0y=0
Correct Answer
Option C
Solution

Since tangent is parallel to

xx

-axis, ∴\therefore

dydx=0⇒1−8x3=0⇒x=2⇒y=3{{dy} \over {dx}} = 0 \Rightarrow 1 - {8 \over {{x^3}}} = 0 \Rightarrow x = 2 \Rightarrow y = 3

Equation of tangent is

y−3=0(x−2)⇒y=3y - 3 = 0\left( {x - 2} \right) \Rightarrow y = 3
Q40
Let f:R→Rf:R \to R be defined by f(x)={k−2x,  ifx≤−12x+3,  ifx>−1f\left( x \right) = \left\{ \begin{array}{ll}{k - 2x,\,\,if} & {x \le - 1} \\ {2x + 3,\,\,if} & {x > - 1} \end{array} \right.If If ffhasalocalminimumathas a local minimum at x=−1x=-1,thenapossiblevalueof, then a possible value of kk$ is
A 00
B −12 - {1 \over 2}
C −1-1
D 11
Correct Answer
Option C
Solution
f(x)={k−2x,    if    x≤−12x+3,    if    x>−1f\left( x \right) = \left\{ \begin{array}{ll}{k - 2x,\,\,\,\,if\,\,\,\,x \le - 1} \\ {2x + 3,\,\,\,\,if\,\,\,\,x > - 1} \end{array} \right.

This is true where

k=−1k=-1
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