Practice Start Test

Application of Derivatives

JEE Mathematics · 188 questions · Page 12 of 19 · Click an option or "Show Solution" to reveal answer

Q111
The sum of all the local minimum values of the twice differentiable function f : R →\to R defined by f(x)=x3−3x2−3f′′(2)2x+f′′(1)f(x) = {x^3} - 3{x^2} - {{3f''(2)} \over 2}x + f''(1) is :
A −-22
B 5
C −-27
D 0
Correct Answer
Option C
Solution
f(x)=x3−3x2−3f′′(2)2x+f′′(1)f(x) = {x^3} - 3{x^2} - {{3f''(2)} \over 2}x + f''(1)

..... (i)

f(x)=3x2−6x−32f′′(2)f(x) = 3{x^2} - 6x - {3 \over 2}f''(2)

..... (ii)

f′′(x)=6x−6f''(x) = 6x - 6

..... (iii) Now, is 3rd equation

f′′(2)=12−6=6f''(2) = 12 - 6 = 6
f′′(11=0)f''(11 = 0)

Use (ii)

f′(x)=3x2−6x−32f′′(2)f'(x) = 3{x^2} - 6x - {3 \over 2}f''(2)
f′(x)=3x2−6x−32×6f'(x) = 3{x^2} - 6x - {3 \over 2} \times 6
f′(x)=3x2−6x−9f'(x) = 3{x^2} - 6x - 9
f′(x)=0f'(x) = 0
3x2−6x−9=03{x^2} - 6x - 9 = 0

⇒\Rightarrow x = −-1 & 3 Use (iii)

f′′(x)=6x−6f''(x) = 6x - 6
f′′(−1)=−12<0f''( - 1) = - 12 < 0

maxima

f′′(3)=12>0f''(3) = 12 > 0

minima. Use (i)

f(x)=x3−3x2−32f′′(2)x+f′′(1)f(x) = {x^3} - 3{x^2} - {3 \over 2}f''(2)x + f''(1)
f(x)=x3−3x2−32×6×x+0f(x) = {x^3} - 3{x^2} - {3 \over 2} \times 6 \times x + 0
f(x)=x3−3x2−9xf(x) = {x^3} - 3{x^2} - 9x
f(3)=27−27−9×3=−27f(3) = 27 - 27 - 9 \times 3 = - 27
Q112
Let f : R →\to R be defined as f(x)={−43x3+2x2+3x,x>03xex,x≤0f(x) = \left\{ \begin{array}{ll}{ - {4 \over 3}{x^3} + 2{x^2} + 3x,} & {x > 0} \\ {3x{e^x},} & {x \le 0} \end{array} \right.. Then f is increasing function in the interval
A (−12,2)\left( { - {1 \over 2},2} \right)
B (0,2)
C (−1,32)\left( { - 1,{3 \over 2}} \right)
D (−-3, −-1)
Correct Answer
Option C
Solution
f′(x){−4x2+4x+3x>03ex(1+x)x≤0f'(x)\left\{ \begin{array}{ll}{ - 4{x^2} + 4x + 3} & {x > 0} \\ {3{e^x}(1 + x)} & {x \le 0} \end{array} \right.

For x > 0,

f′(x)=−4x2+4x+3f'(x) = - 4{x^2} + 4x + 3

f(x) is increasing in

(−12,32)\left( { - {1 \over 2},{3 \over 2}} \right)

For x ≤\le 0, f'(x) = 3ex(1 + x) f'(x) > 0 ∀\forall x

∈\in

(−-1, 0) ⇒\Rightarrow f(x) is increasing in (−-1, 0) So, in complete domain, f(x) is increasing in

(−1,32)\left( { - 1,{3 \over 2}} \right)
Q113
Let f(x)=3sin⁡4x+10sin⁡3x+6sin⁡2x−3f(x) = 3{\sin ^4}x + 10{\sin ^3}x + 6{\sin ^2}x - 3, x∈[−π6,π2]x \in \left[ { - {\pi \over 6},{\pi \over 2}} \right]. Then, f is :
A increasing in (−π6,π2)\left( { - {\pi \over 6},{\pi \over 2}} \right)
B decreasing in (0,π2)\left( {0,{\pi \over 2}} \right)
C increasing in (−π6,0)\left( { - {\pi \over 6},0} \right)
D decreasing in (−π6,0)\left( { - {\pi \over 6},0} \right)
Correct Answer
Option D
Solution
f(x)=3sin⁡4x+10sin⁡3x+6sin⁡2x−3,x∈[−π6,π2]f(x) = 3{\sin ^4}x + 10{\sin ^3}x + 6{\sin ^2}x - 3,x \in \left[ { - {\pi \over 6},{\pi \over 2}} \right]
f′(x)=12sin⁡3xcos⁡x+30sin⁡2xcos⁡x+12sin⁡xcos⁡xf'(x) = 12{\sin ^3}x\cos x + 30{\sin ^2}x\cos x + 12\sin x\cos x
=6sin⁡xcos⁡x(2sin⁡2x+5sin⁡x+2)= 6\sin x\cos x(2{\sin ^2}x + 5\sin x + 2)
=6sin⁡xcos⁡x(2sin⁡x+1)(sin⁡+2)= 6\sin x\cos x(2\sin x + 1)(\sin + 2)

Decreasing in

(−π6,0)\left( { - {\pi \over 6},0} \right)
Q114
The local maximum value of the function f(x)=(2x)x2f(x) = {\left( {{2 \over x}} \right)^{{x^2}}}, x > 0, is
A (2e)1e{\left( {2\sqrt e } \right)^{{1 \over e}}}
B (4e)e4{\left( {{4 \over {\sqrt e }}} \right)^{{e \over 4}}}
C (e)2e{(e)^{{2 \over e}}}
D 1
Correct Answer
Option C
Solution
f(x)=(2x)x2f(x) = {\left( {{2 \over x}} \right)^{{x^2}}}

; x > 0

ln⁡f(x)=x2(ln⁡2−ln⁡x)\ln f(x) = {x^2}(\ln 2 - \ln x)
f′(x)=f(x){−x+(ln⁡2−ln⁡x)2x}f'(x) = f(x)\{ - x + (\ln 2 - \ln x)2x\}
f′(x)=f(x)⏟+ . x⏟+(2ln⁡2−2ln⁡x−1)⏟g(x)f'(x) = \underbrace {f(x)}_ + \,.\,\underbrace x_ + \underbrace {(2\ln 2 - 2\ln x - 1)}_{g(x)}
g(x)=2ln⁡2−2ln⁡x−1g(x) = 2{\ln ^2} - 2\ln x - 1
=ln⁡4x2−1=0⇒x=2e= \ln {4 \over {{x^2}}} - 1 = 0 \Rightarrow x = {2 \over {\sqrt e }}
LM=2eLM = {2 \over {\sqrt e }}

Local maximum value =

(22/e)4e=e2e{\left( {{2 \over {2/\sqrt e }}} \right)^{{4 \over e}}} = {e^{{2 \over e}}}
Q115
A wire of length 20 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is :
A 52+3{5 \over {2 + \sqrt 3 }}
B 102+33{{10} \over {2 + 3\sqrt 3 }}
C 53+3{5 \over {3 + \sqrt 3 }}
D 103+23{{10} \over {3 + 2\sqrt 3 }}
Correct Answer
Option D
Solution

Let the wire is cut into two pieces of length x and 20 −- x. Area of square =

(x4)2{\left( {{x \over 4}} \right)^2}

Area of regular hexagon =

6×34(20−x6)26 \times {{\sqrt 3 } \over 4}{\left( {{{20 - x} \over 6}} \right)^2}

Total area =

A(x)=x216+332(20−x)236A(x) = {{{x^2}} \over {16}} + {{3\sqrt 3 } \over 2}{{{{(20 - x)}^2}} \over {36}}
A′(x)=2x16+33×22×36(20−x)(−1)A'(x) = {{2x} \over {16}} + {{3\sqrt 3 \times 2} \over {2 \times 36}}(20 - x)( - 1)

A'(x) = 0 at

x=4033+23x = {{40\sqrt 3 } \over {3 + 2\sqrt 3 }}

Length of side of regular Hexagon

=16(20−x)= {1 \over 6}(20 - x)
=16(20−4.33+23)= {1 \over 6}\left( {20 - {{4.\sqrt 3 } \over {3 + 2\sqrt 3 }}} \right)
=102+23= {{10} \over {2 + 2\sqrt 3 }}
Q116
A box open from top is made from a rectangular sheet of dimension a ×\times b by cutting squares each of side x from each of the four corners and folding up the flaps. If the volume of the box is maximum, then x is equal to :
A a+b−a2+b2−ab12{{a + b - \sqrt {{a^2} + {b^2} - ab} } \over {12}}
B a+b−a2+b2+ab6{{a + b - \sqrt {{a^2} + {b^2} + ab} } \over 6}
C a+b−a2+b2−ab6{{a + b - \sqrt {{a^2} + {b^2} - ab} } \over 6}
D a+b+a2+b2+ab6{{a + b + \sqrt {{a^2} + {b^2} + ab} } \over 6}
Correct Answer
Option C
Solution

V = l . b . h = (a −- 2x)(b −- 2x) x ⇒\Rightarrow V(x) = (2x −- a)(2x −- b) x ⇒\Rightarrow V(x) = 4x3 −- 2(a + b)x2 + abx

⇒ddxv(x)=12x2−4(a+b)x+ab\Rightarrow {d \over {dx}}v(x) = 12{x^2} - 4(a + b)x + ab
ddx(v(x))=0⇒12x2−4(a+b)x+ab=0<βα{d \over {dx}}(v(x)) = 0 \Rightarrow 12{x^2} - 4(a + b)x + ab = 0 < _\beta ^\alpha
⇒x=4(a+b)±16(a+b)2−48ab2(12)\Rightarrow x = {{4(a + b) \pm \sqrt {16{{(a + b)}^2} - 48ab} } \over {2(12)}}
=(a+b)±a2+b2−ab6= {{(a + b) \pm \sqrt {{a^2} + {b^2} - ab} } \over 6}

Let

x=α=(a+b)+a2+b2−ab6x = \alpha = {{(a + b) + \sqrt {{a^2} + {b^2} - ab} } \over 6}
β=(a+b)−a2+b2−ab6\beta = {{(a + b) - \sqrt {{a^2} + {b^2} - ab} } \over 6}

Now,

12(x−α)(x−β)=012(x - \alpha )(x - \beta ) = 0

∴\therefore x = β\beta

=a+b−a2+b2−abb= {{a + b - \sqrt {{a^2} + {b^2} - ab} } \over b}
Q117
The number of real roots of the equation e4x+2e3x−ex−6=0{e^{4x}} + 2{e^{3x}} - {e^x} - 6 = 0 is :
A 2
B 4
C 1
D 0
Correct Answer
Option C
Solution

Let

ex=t>0{e^x} = t > 0
f(t)=t4+2t3−t−6=0f(t) = {t^4} + 2{t^3} - t - 6 = 0
f′(t)=4t3+6t2−1f'(t) = 4{t^3} + 6{t^2} - 1
f′′(t)=12t2+12t>0f''(t) = 12{t^2} + 12t > 0
f(0)=−6,f(1)=−4,f(2)=24f(0) = - 6,f(1) = - 4,f(2) = 24

⇒\Rightarrow Number of real roots = 1

Q118
The function f(x)=x3−6x2+ax+bf(x) = {x^3} - 6{x^2} + ax + b is such that f(2)=f(4)=0f(2) = f(4) = 0. Consider two statements : Statement 1 : there exists x1, x2 ∈\in(2, 4), x1 < x2, such that f'(x1) = −-1 and f'(x2) = 0. Statement 2 : there exists x3, x4 ∈\in (2, 4), x3 < x4, such that f is decreasing in (2, x4), increasing in (x4, 4) and 2f′(x3)=3f(x4)2f'({x_3}) = \sqrt 3 f({x_4}). Then
A both Statement 1 and Statement 2 are true
B Statement 1 is false and Statement 2 is true
C both Statement 1 and Statement 2 are false
D Statement 1 is true and Statement 2 is false
Correct Answer
Option A
Solution
f(x)=x3−6x2+ax+bf(x) = {x^3} - 6{x^2} + ax + b
f(2)=8−24+2a+b=0f(2) = 8 - 24 + 2a + b = 0
2a+b=162a + b = 16

.... (1)

f(4)=64−96+4a+b=0f(4) = 64 - 96 + 4a + b = 0
4a+b=324a + b = 32

.... (2) Solving (1) and (2) a = 8, b = 0

f(x)=x3−6x2+8xf(x) = {x^3} - 6{x^2} + 8x
f′(x)=3x2−12x+8f'(x) = 3{x^2} - 12x + 8
f′′(x)=6x−12f''(x) = 6x - 12

⇒\Rightarrow f'(x) is ↑\uparrow for x > 2, and f'(x) is ↓\downarrow for x < 2

f′(2)=12−24+8=−4f'(2) = 12 - 24 + 8 = - 4
f′(4)=48−48+8=8f'(4) = 48 - 48 + 8 = 8
f′(x)=3x2−12x+8f'(x) = 3{x^2} - 12x + 8

vertex (2, −-4) f'(2) = −-4, f'(4) = 8, f'(3) = 27 −- 36 + 8 f'(x1) = −-1, then x1 = 3 f'(x2) = 0 Again f'(x) < 0 for x

∈\in

(2, x4) f'(x) > 0 for x

∈\in

(x4, 4) x4

∈\in

(3, 4) f(x) = x3 −- 6x2 + 8x f(3) = 27 −- 54 + 24 = −-3 f(4) = 64 −- 96 + 32 = 0 For x4(3, 4) f(x4) < −-3

3\sqrt 3

and f'(x3) > −-4 2f'(x3) > −-8 So, 2f'(x3) =

3\sqrt 3

f(x4) Correct Ans. (a).

Q119
A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is :
A 229+43{{22} \over {9 + 4\sqrt 3 }}
B 669+43{{66} \over {9 + 4\sqrt 3 }}
C 224+93{{22} \over {4 + 9\sqrt 3 }}
D 664+93{{66} \over {4 + 9\sqrt 3 }}
Correct Answer
Option B
Solution
4a+3b=224a + 3b = 22

Total area

=A=a2+34b2= A = {a^2} + {{\sqrt 3 } \over 4}{b^2}
A=(22−3b4)2+34b2A = {\left( {{{22 - 3b} \over 4}} \right)^2} + {{\sqrt 3 } \over 4}{b^2}
dAdB=2(22−3b4)(−34)+34 . 2b=0{{dA} \over {dB}} = 2\left( {{{22 - 3b} \over 4}} \right)\left( {{{ - 3} \over 4}} \right) + {{\sqrt 3 } \over 4}\,.\,2b = 0
⇒3b2=38(22−3b)\Rightarrow {{\sqrt {3b} } \over 2} = {3 \over 8}(22 - 3b)
⇒43b=66−9b\Rightarrow 4\sqrt 3 b = 66 - 9b
∴b=669+43\therefore b = {{66} \over {9 + 4\sqrt 3 }}
Q120
Let f : R →\to R be a function defined by f(x) = (x −- 3)n1 (x −- 5)n2, n1, n2 ∈\in N. Then, which of the following is NOT true?
A For n1 = 3, n2 = 4, there exists α\alpha ∈\in (3, 5) where f attains local maxima.
B For n1 = 4, n2 = 3, there exists α\alpha ∈\in (3, 5) where f attains local minima.
C For n1 = 3, n2 = 5, there exists α\alpha ∈\in (3, 5) where f attains local maxima.
D For n1 = 4, n2 = 6, there exists α\alpha ∈\in (3, 5) where f attains local maxima.
Correct Answer
Option C
Solution

Given,

f(x)=(−3)n1(x−5)n2f(x) = {( - 3)^{{n_1}}}{(x - 5)^{{n_2}}}

Differentiating both side with respect to x, we get

f′(x)=(x−5)n2(n1(x−3)n1−1)+(x−3)n1(n2(x−5)n2−1)f'(x) = {(x - 5)^{{n_2}}}\left( {{n_1}{{(x - 3)}^{{n_1} - 1}}} \right) + {(x - 3)^{{n_1}}}\left( {{n_2}{{(x - 5)}^{{n_2} - 1}}} \right)
=(x−3)n1(x−5)n2[n1x−3+n2x−5]= {(x - 3)^{{n_1}}}{(x - 5)^{{n_2}}}\left[ {{{{n_1}} \over {x - 3}} + {{{n_2}} \over {x - 5}}} \right]
=(x−3)n1(x−5)n2[n1(x−5)+n2(x−3)(x−3)(x−5)]= {(x - 3)^{{n_1}}}{(x - 5)^{{n_2}}}\left[ {{{{n_1}(x - 5) + {n_2}(x - 3)} \over {(x - 3)(x - 5)}}} \right]
=(x−3)n1−1(x−5)n2−1[n1(x−5)+n2(x−3)]= {(x - 3)^{{n_1} - 1}}{(x - 5)^{{n_2} - 1}}[{n_1}(x - 5) + {n_2}(x - 3)]

Option A : When n1 = 3 and n2 = 4 then

f′(x)=(x−3)3−1(x−5)4−1[3(x−5)+4(x−3)]f'(x) = {(x - 3)^{3 - 1}}{(x - 5)^{4 - 1}}\left[ {3(x - 5) + 4(x - 3)} \right]
=(x−3)2(x−5)3(7x−27)= {(x - 3)^2}{(x - 5)^3}(7x - 27)

Critical point is at x = 3, 5 and

277{{27} \over 7}

When value of f'(x) is less than

275{{27} \over 5}

, f'(x) is positive means slope of f(x) graph is positive. And when value of f'(x) is greater than

277{{27} \over 7}

but less than 5 then f'(x) is negative means slope of f(x) is negative. So at

277{{27} \over 7}

between 3 and 5, f(x) attain local maxima. ∴\therefore Option A is correct. Option B : When n1 = 4 and n2 = 3 then

f′(x)=(x−3)4−1(x−5)3−1[4(x−5)+3(x−3)]f'(x) = {(x - 3)^{4 - 1}}{(x - 5)^{3 - 1}}\left[ {4(x - 5) + 3(x - 3)} \right]
=(x−3)3(x−5)2(7x−29)= {(x - 3)^3}{(x - 5)^2}(7x - 29)

Critical point is at x = 3, 5 and

297{{29} \over 7}

When f'(x) is less than

297{{29} \over 7}

but greater than 3 then f'(x) is negative means slope of graph of f(x) is negative. And when f'(x) is greater than

297{{29} \over 7}

but less than 5 then f'(x) is positive means slope of graph of f(x) is positive. So at

297{{29} \over 7}

between 3 and 5, f(x) attain local minima. ∴\therefore Option (B) is correct. Option C : When n1 = 3 and n2 = 5 then

f′(x)=(x−3)3−1(x−5)5−1[3(x−5)+5(x−3)]f'(x) = {(x - 3)^{3 - 1}}{(x - 5)^{5 - 1}}\left[ {3(x - 5) + 5(x - 3)} \right]
=(x−3)2(x−5)4(8x−30)= {(x - 3)^2}{(x - 5)^4}(8x - 30)

Critical point is at x = 3, 5 and

308{{30} \over 8}

When f'(x) is less than

308{{30} \over 8}

but greater than 3 then f'(x) is negative means slope of graph of f(x) is negative. And when f'(x) is greater than

308{{30} \over 8}

but less than 5 then f'(x) is positive means slope of graph of f(x) is positive. So, at

308{{30} \over 8}

between 3 and 5, f(x) attain local minima. ∴\therefore Option C is not true. Similarly you can check option D also.

Ready for a full JEE mock test? Timed · full syllabus · instant results
Take a Mock Test →