Practice Start Test

Application of Derivatives

JEE Mathematics · 188 questions · Page 6 of 19 · Click an option or "Show Solution" to reveal answer

Q51
If ff and gg are differentiable functions in [0,1]\left[ {0,1} \right] satisfying f(0)=2=g(1),g(0)=0f\left( 0 \right) = 2 = g\left( 1 \right),g\left( 0 \right) = 0 and f(1)=6,f\left( 1 \right) = 6, then for some c∈]0,1[c \in \left] {0,1} \right[
A f′(c)=g′(c)f'\left( c \right) = g'\left( c \right)
B f′(c)=2g′(c)f'\left( c \right) = 2g'\left( c \right)
C 2f′(c)=g′(c)2f'\left( c \right) = g'\left( c \right)
D 2f′(c)=3g′(c)2f'\left( c \right) = 3g'\left( c \right)
Correct Answer
Option B
Solution

Since,

ff

and

gg

both are continuous function on

[0,1]\left[ {0,1} \right]

and differentiable on

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

then

∃c∈(0,1)\exists c \in \left( {0,1} \right)

such that

f′(c)=f(1)−f(0)1=6−21=4f'\left( c \right) = {{f\left( 1 \right) - f\left( 0 \right)} \over 1} = {{6 - 2} \over 1} = 4

and

g′(c)=g(1)−g(0)1=2−01=2g'\left( c \right) = {{g\left( 1 \right) - g\left( 0 \right)} \over 1} = {{2 - 0} \over 1} = 2

Thus, we get

f′(c)=2g′(c)f'\left( c \right) = 2g'\left( c \right)
Q52
If x=−1x=-1 and x=2x=2 are extreme points of f(x)=α log⁡∣x∣+βx2+xf\left( x \right) = \alpha \,\log \left| x \right|+\beta {x^2} + x then
A α=2,β=−12\alpha = 2,\beta = - {1 \over 2}
B α=2,β=12\alpha = 2,\beta = {1 \over 2}
C α=−6,β=12\alpha = - 6,\beta = {1 \over 2}
D α=−6,β=−12\alpha = - 6,\beta = -{1 \over 2}
Correct Answer
Option A
Solution

Let

f(x)=αlog⁡∣x∣+βx2+xf\left( x \right) = \alpha \log \left| x \right| + \beta {x^2} + x

Differentiating both sides,

f′(x)=αx+2βx+1f'\left( x \right) = {\alpha \over x} + 2\beta x + 1

Since

x=−1x=-1

and

x=2x=2

are extreme points therefore

f′(x)=0f'\left( x \right) = 0

at these points. Put

x=−1x = - 1

and

x=2x = 2

in

f′(x),f'\left( x \right),

we get

−α−2β+1=0- \alpha - 2\beta + 1 = 0
⇒α+2β=1  ...(i)\Rightarrow \alpha + 2\beta = 1\,\,...\left( i \right)
α2+4β+1=0{\alpha \over 2} + 4\beta + 1 = 0
⇒α+8β=−2  ...(ii)\Rightarrow \alpha + 8\beta = - 2\,\,...\left( {ii} \right)

On solving

(i)(i)

and

(ii)(ii)

, we get

6β=−3⇒β=−126\beta = - 3 \Rightarrow \beta = - {1 \over 2}

∴\therefore

    α=2\,\,\,\,\alpha = 2
Q53
Let f(x)f(x) be a polynomial of degree four having extreme values at x=1x=1 and x=2x=2. If lim⁡x→0[1+f(x)x2]=3\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3, then f(2)(2) is equal to :
A 00
B 44
C −8-8
D −4-4
Correct Answer
Option A
Solution
lim⁡x→0[1+f(x)x2]=3⇒Limx→0f(x)x2=2\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3 \Rightarrow \mathop {Lim}\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2

So,

f(x)f(x)

contains terms in

x2,x3x{}^2,{x^3}

and

x4{x^4}

Let

f(x)=a1x2+a2x3+a3x4f\left( x \right) = {a_1}{x^2} + {a_2}{x^3} + {a_3}{x^4}

Since

lim⁡x→0f(x)x2=2⇒a1=2\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2 \Rightarrow {a_1} = 2

Hence,

f(x)=2x2+a2x3+a3x4f\left( x \right) = 2{x^2} + {a_2}{x^3} + {a_3}{x^4}
f′(x)=4x+3a2x2+4a3x3f'\left( x \right) = 4x + 3{a_2}{x^2} + 4{a_3}{x^3}

As given:

f′(1)=0f'\left( 1 \right) = 0

and

f′(2)=0f'\left( 2 \right) = 0

Hence,

4+3a2+4a3=0          ...(1)4 + 3{a_2} + 4{a_3} = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

and

8+12a2+32a3=0     ...(1)8 + 12{a_2} + 32{a_3} = 0\,\,\,\,\,...\left( 1 \right)

By

4x(eq1)−eq(2),4x\left( {eq1} \right) - eq\left( 2 \right),

we get

16+12a2+16a3−(8+12a2+32a3)=016 + 12{a_2} + 16{a_3} - \left( {8 + 12{a_2} + 32{a_3}} \right) = 0
⇒8−16a3=0⇒a3=1/2\Rightarrow 8 - 16{a_3} = 0 \Rightarrow {a_3} = 1/2

and by eqn.

(1),4+3a2+4/2=0⇒a2=−2\left( 1 \right),4 + 3{a_2} + 4/2 = 0 \Rightarrow {a_2} = - 2
⇒f(x)=2x2−2x3+12x4\Rightarrow f\left( x \right) = 2{x^2} - 2{x^3} + {1 \over 2}{x^4}
f(2)=2×4−2×8+12×16=0f\left( 2 \right) = 2 \times 4 - 2 \times 8 + {1 \over 2} \times 16 = 0
Q54
The normal to the curve, x2+2xy−3y2=0{x^2} + 2xy - 3{y^2} = 0, at (1,1)(1,1)
A meets the curve again in the third quadrant.
B meets the curve again in the fourth quadrant.
C does not meet the curve again.
D meets the curve again in the second quadrant.
Correct Answer
Option B
Solution

Given curve is

x2+2xy−3y2=0{x^2} + 2xy - 3{y^2} = 0

Difference

w.r.t.x,w.r.t.x,
2x+2xdydx+2y−6ydydx=02x + 2x{{dy} \over {dx}} + 2y - 6y{{dy} \over {dx}} = 0
(dydx)(1,1)=1{\left( {{{dy} \over {dx}}} \right)_{\left( {1,1} \right)}} = 1

Equation of normal at

(1,1)(1,1)

is

y=2−xy=2-x

Solving eq.

(1)(1)

and

(2),(2),

we get

x=1,3x=1,3

Point of intersection

(1,1),(3,−1)\left( {1,1} \right),\left( {3, - 1} \right)

Normal cuts the curve again in

44

th quadrant.

Q55
Consider : f (x)=tan⁡−1(1+sin⁡x1−sin⁡x),x∈(0,π2).\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin x} \over {1 - \sin x}}} } \right),x \in \left( {0,{\pi \over 2}} \right). A normal to y=y = f(x)\left( x \right) at x=π6x = {\pi \over 6} also passes through the point:
A (π6,0)\left( {{\pi \over 6},0} \right)
B (π4,0)\left( {{\pi \over 4},0} \right)
C (0,0)(0,0)
D (0,2π3)\left( {0,{{2\pi } \over 3}} \right)
Correct Answer
Option D
Solution
f(x)=tan⁡−1(1+sin⁡ x1−sin⁡x)f\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin \,x} \over {1 - \sin x}}} } \right)
=tan⁡−1((sin⁡x2+cos⁡x2)2(sin⁡xx−cos⁡x2)2)= {\tan ^{ - 1}}\left( {\sqrt {{{{{\left( {\sin {x \over 2} + \cos {x \over 2}} \right)}^2}} \over {{{\left( {\sin {x \over x} - \cos {x \over 2}} \right)}^2}}}} } \right)
=tan⁡−1(1+tan⁡x21−tan⁡x2)= {\tan ^{ - 1}}\left( {{{1 + \tan {x \over 2}} \over {1 - \tan {x \over 2}}}} \right)
=tan⁡−1(tan⁡(π4+x2))= {\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} + {x \over 2}} \right)} \right)
⇒y=π4+x2\Rightarrow y = {\pi \over 4} + {x \over 2}
⇒dydx=12\Rightarrow {{dy} \over {dx}} = {1 \over 2}

Slope of normal

=−1(dydx)=−2= {{ - 1} \over {\left( {{{dy} \over {dx}}} \right)}} = - 2

At

(π6,π4+π12)\left( {{\pi \over 6},{\pi \over 4} + {\pi \over {12}}} \right)
y−(π4+π12)=−2(x−π6)y - \left( {{\pi \over 4} + {\pi \over {12}}} \right) = - 2\left( {x - {\pi \over 6}} \right)
y−4π12=−2x+2π6y - {{4\pi } \over {12}} = - 2x + {{2\pi } \over 6}
y−π3=−2x+π3y - {\pi \over 3} = - 2x + {\pi \over 3}
y=−2x+2π3y = - 2x + {{2\pi } \over 3}

This equation is satisfied only by the point

(0,2π3)\left( {0,{{2\pi } \over 3}} \right)
Q56
A wire of length 22 units is cut into two parts which are bent respectively to form a square of side =x=x units and a circle of radius =r=r units. If the sum of the areas of the square and the circle so formed is minimum, then:
A x=2rx=2r
B 2x=r2x=r
C 2x=(π+4)r2x = \left( {\pi + 4} \right)r
D (4−π)x=π  r\left( {4 - \pi } \right)x = \pi \,\, r
Correct Answer
Option A
Solution
4x+2πr=24x + 2\pi r = 2
   \,\,\,
⇒2x+πr=1\Rightarrow 2x + \pi r = 1
S=x2+πr2S = {x^2} + \pi {r^2}
S=(1−πr2)2+πr2S = {\left( {{{1 - \pi r} \over 2}} \right)^2} + \pi {r^2}
dSdr=2(1−πr2)(−π2)+2πr{{dS} \over {dr}} = 2\left( {{{1 - \pi r} \over 2}} \right)\left( {{{ - \pi } \over 2}} \right) + 2\pi r
⇒−π2+π2r2+2πr=0\Rightarrow {{ - \pi } \over 2} + {{{\pi ^2}r} \over 2} + 2\pi r = 0
⇒r=1π+4\Rightarrow r = {1 \over {\pi + 4}}
⇒x=2π+4 \Rightarrow x = {2 \over {\pi + 4}}\,
⇒x=2r\Rightarrow x = 2r
Q57
A tangent to the curve, y = f(x) at P(x, y) meets x-axis at A and y-axis at B. If AP : BP = 1 : 3 and f(1) = 1, then the curve also passes through the point :
A (13,24)\left( {{1 \over 3},24} \right)
B (12,4)\left( {{1 \over 2},4} \right)
C (2,18)\left( {2,{1 \over 8}} \right)
D (3,128)\left( {3,{1 \over 28}} \right)
Correct Answer
Option C
Solution

We have

(y−y2)(x−x1)=f′(x1){{(y - {y_2})} \over {(x - {x_1})}} = f'({x_1})
⇒y−y1=f′(x1)(x−x1)\Rightarrow y - {y_1} = f'({x_1})(x - {x_1})

∙\bullet When y = 0:

−y1f′(x1)=x−x1{{ - {y_1}} \over {f'({x_1})}} = x - {x_1}
⇒x=x1−y1f′(x1)\Rightarrow x = {x_1} - {{{y_1}} \over {f'({x_1})}}

Therefore, point A is

A(x1−y1f′(x1),0)A\left( {{x_1} - {{{y_1}} \over {f'({x_1})}},0} \right)

∙\bullet When x = 0:

y−y1=f′(x) . (−x1)y - {y_1} = f'(x)\,.\,( - {x_1})
⇒y=y1−x1f′(x1)\Rightarrow y = {y_1} - {x_1}f'({x_1})

Therefore, point B is

B(0,y1−x1f′(x1))B(0,{y_1} - {x_1}f'({x_1}))

Point P divides AB in the ratio 1 : 3.

x1=[3(x1−y1f′(x1))]4{x_1} = {{\left[ {3\left( {{x_1} - {{{y_1}} \over {f'({x_1})}}} \right)} \right]} \over 4}
y1=y1−x1f′(x1)4{y_1} = {{{y_1} - {x_1}f'({x_1})} \over 4}

Therefore,

4y1=y1−x1f′(x1)4{y_1} = {y_1} - {x_1}f'({x_1})
⇒f′(x1)=−3y1x1⇒f′(x)=−3yx\Rightarrow f'({x_1}) = {{ - 3{y_1}} \over {{x_1}}} \Rightarrow f'(x) = {{ - 3y} \over x}

Now,

dydx=−3yx⇒dyy=−3dxx{{dy} \over {dx}} = {{ - 3y} \over x} \Rightarrow {{dy} \over y} = {{ - 3dx} \over x}

On integrating, we get

ln⁡y=−3ln⁡x+C⇒y=kx−3\ln y = - 3\ln x + C \Rightarrow y = k{x^{ - 3}}
y(1)=1⇒k=1y(1) = 1 \Rightarrow k = 1
y=1x3y = {1 \over {{x^3}}}

When we substitute the values from the given options, only option (c) satisfies the above equation.

Q58
The function f defined by f(x) = x3 −- 3x2 + 5x + 7 , is :
A increasing in R.
B decreasing in R.
C decreasing in (0, ∞\infty ) and increasing in (−- ∞\infty , 0)
D increasing in (0, ∞\infty ) and decreasing in (−- ∞\infty , 0)
Correct Answer
Option A
Solution

The given function is

f(x)=x2−3x2+5x+7f(x) = {x^2} - 3{x^2} + 5x + 7
f′(x)=3x2−6x+5f'(x) = 3{x^2} - 6x + 5

The discriminant of the above quadratic equation is

Δ=36−4(3)(5)=36−60Therefore,\Delta = 36 - 4(3)(5) = 36 - 60 Therefore,

f'(x) > 0\,\forall x \in {R^ + }

Also,Also,

f'(x) > 0\,\forall x \in {R^ - }$$ Therefore, the given function f is increasing in R.

Q59
Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is :
A 10
B 25
C 30
D 12.5
Correct Answer
Option B
Solution

We have Total length = r + r + rθ\theta = 20 ⇒\Rightarrow 2r + rθ\theta = 20 ⇒\Rightarrow

θ=20−2rr\theta = {{20 - 2r} \over r}

.......(1) A = Area =

θ2π×πr2{\theta \over {2\pi }} \times \pi {r^2}

=

12r2θ{1 \over 2}{r^2}\theta

=

12r2(20−2rr){1 \over 2}{r^2}\left( {{{20 - 2r} \over r}} \right)

⇒\Rightarrow A = 10r – r2 For A to be maximum

dAdr=0{{dA} \over {dr}} = 0

⇒\Rightarrow 10 – 2r = 0 ⇒\Rightarrow r = 5

d2Adr2=−2<0{{{d^2}A} \over {d{r^2}}} = - 2 < 0

∴\therefore For r = 5, A is maximum From (1)

θ=20−2(5)r{\theta = {{20 - 2\left( 5 \right)} \over r}}

= 2 A =

22π×π(5)2{2 \over {2\pi }} \times \pi {\left( 5 \right)^2}

= 25 sq. m

Q60
Let M and m be respectively the absolute maximum and the absolute minimum values of the function, f(x) = 2x3 −- 9x2 + 12x + 5 in the interval [0, 3]. Then M −-m is equal to :
A 5
B 9
C 4
D 1
Correct Answer
Option B
Solution

To determine the absolute maximum (M) and absolute minimum (m) of the function f(x)=2x3−9x2+12x+5 f(x) = 2x^3 - 9x^2 + 12x + 5 over the interval [0,3][0, 3], we need to examine its critical points and endpoints.

First, we find the derivative of the function, f′(x) f'(x) , to locate the critical points:

f′(x)=ddx(2x3−9x2+12x+5)=6x2−18x+12f'(x) = \frac{d}{dx} (2x^3 - 9x^2 + 12x + 5) = 6x^2 - 18x + 12

Next, we set the derivative equal to zero to find the critical points:

6x2−18x+12=06x^2 - 18x + 12 = 0

Simplifying this equation by dividing by 6:

x2−3x+2=0x^2 - 3x + 2 = 0

We solve this quadratic equation using the factorization method:

(x−1)(x−2)=0(x - 1)(x - 2) = 0

So, the critical points are:

x=1andx=2x = 1 \quad \text{and} \quad x = 2

We now evaluate the function f(x) f(x) at the critical points and at the endpoints of the interval [0, 3]: 1.

At x=0 x = 0 :

f(0)=2(0)3−9(0)2+12(0)+5=5f(0) = 2(0)^3 - 9(0)^2 + 12(0) + 5 = 5

2. At x=1 x = 1 :

f(1)=2(1)3−9(1)2+12(1)+5=2−9+12+5=10f(1) = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 2 - 9 + 12 + 5 = 10

3. At x=2 x = 2 :

f(2)=2(2)3−9(2)2+12(2)+5=16−36+24+5=9f(2) = 2(2)^3 - 9(2)^2 + 12(2) + 5 = 16 - 36 + 24 + 5 = 9

4. At x=3 x = 3 :

f(3)=2(3)3−9(3)2+12(3)+5=54−81+36+5=14f(3) = 2(3)^3 - 9(3)^2 + 12(3) + 5 = 54 - 81 + 36 + 5 = 14

We now identify the absolute maximum (M) and absolute minimum (m) from the above values: Maximum value M=14 M = 14 at x=3 x = 3 Minimum value m=5 m = 5 at x=0 x = 0 Thus, the difference M−m M - m is:

M−m=14−5=9M - m = 14 - 5 = 9

Therefore, the correct answer is: Option B: 9

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