Practice Start Test

Application of Derivatives

JEE Mathematics · 188 questions · Page 2 of 19 · Click an option or "Show Solution" to reveal answer

Q11
Area of the greatest rectangle that can be inscribed in the ellipse x2a2+y2b2=1{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1
A 2ab2ab
B abab
C ab\sqrt {ab}
D ab{a \over b}
Correct Answer
Option A
Solution

Area of rectangle

ABCDABCD
=2a cos⁡ θ= 2a\,\cos \,\theta
(2b sin⁡ θ)=2ab sin⁡ 2θ\left( {2b\,\sin \,\theta } \right) = 2ab\,\sin \,2\theta

⇒\Rightarrow Area of greatest rectangle is equal to

2ab2ab

When

sin⁡ 2θ=1.\sin \,2\theta = 1.
Q12
How many real solutions does the equation x7+14x5+16x3+30x−560=0{x^7} + 14{x^5} + 16{x^3} + 30x - 560 = 0 have?
A 77
B 11
C 33
D 55
Correct Answer
Option B
Solution

Let

f(x)=x7+14x5+16x3+30x−560f\left( x \right) = {x^7} + 14{x^5} + 16{x^3} + 30x - 560
⇒f′(x)=7x6+70x4+48x2+30>0, ∀x∈R\Rightarrow f'\left( x \right) = 7{x^6} + 70{x^4} + 48{x^2} + 30 > 0,\,\forall x \in R
⇒f\Rightarrow f

is an increasing function on

RR

Also

lim⁡x→∞  f(x)=∞\mathop {\lim }\limits_{x \to \infty } \,\,f\left( x \right) = \infty

and

lim⁡x→−∞  f(x)=−∞\mathop {\lim }\limits_{x \to - \infty } \,\,f\left( x \right) = - \infty

⇒\Rightarrow The curve

y=f(x)y = f\left( x \right)

crosses

xx

-axis only once. ∴\therefore

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

has exactly one real root.

Q13
The function f(x)=xex(1−x),x∈Rf(x)=x \mathrm{e}^{x(1-x)}, x \in \mathbb{R}, is :
A increasing in (−12,1)\left(-\dfrac{1}{2}, 1\right)
B decreasing in (12,2)\left(\dfrac{1}{2}, 2\right)
C increasing in (−1,−12)\left(-1,-\dfrac{1}{2}\right)
D decreasing in (−12,12)\left(-\dfrac{1}{2}, \dfrac{1}{2}\right)
Correct Answer
Option A
Solution
f(x)=xex(1−x), x∈Rf(x) = x{e^{x(1 - x)}},\,x \in R
f′(x)=xex(1−x) . (1−2x)+ex(1−x)f'(x) = x{e^{x(1 - x)}}\,.\,(1 - 2x) + {e^{x(1 - x)}}
=ex(1−x)[x−2x2+1]= {e^{x(1 - x)}}[x - 2{x^2} + 1]
=−ex(1−x)[2x2−x−1]= - {e^{x(1 - x)}}[2{x^2} - x - 1]
=−ex(1−x)(2x+1)(x−1)= - {e^{x(1 - x)}}(2x + 1)(x - 1)

∴\therefore

f(x)f(x)

is increasing in

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

and decreasing in

(−∞, −12)∪(1,∞)\left( { - \infty ,\, - {1 \over 2}} \right) \cup \left( {1,\infty } \right)
Q14
Let f:→R→(0,∞)f: \rightarrow \mathbb{R} \rightarrow(0, \infty) be strictly increasing function such that lim⁡x→∞f(7x)f(x)=1\lim \limits_{x \rightarrow \infty} \dfrac{f(7 x)}{f(x)}=1. Then, the value of lim⁡x→∞[f(5x)f(x)−1]\lim \limits_{x \rightarrow \infty}\left[\dfrac{f(5 x)}{f(x)}-1\right] is equal to
A 0
B 4
C 1
D 7/5
Correct Answer
Option A
Solution
f:R→(0,∞)lim⁡x→∞f(7x)f(x)=1\begin{aligned} & f: R \rightarrow(0, \infty) \\ & \lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1 \end{aligned}
∵f\because \mathrm{f}

is increasing $$\begin{aligned} & \therefore \mathrm{f}(\mathrm{x})

Q15
Let f(x) = xa2+x2−d−xb2+(d−x)2,  {x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }},\,\, x  ∈\, \in R, where a, b and d are non-zero real constants. Then :
A f is an increasing function of x
B f is neither increasing nor decreasing function of x
C f ' is not a continuous function of x
D f is a decreasing function of x
Correct Answer
Option A
Solution
f(x)=xa2+x2−d−xb2+(d−x)2f\left( x \right) = {x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }}
f′(x)=a2(a2+x2)3/2+b2(b2+(d−x)2)3/2>0∀x∈Rf'\left( x \right) = {{{a^2}} \over {{{\left( {{a^2} + {x^2}} \right)}^{3/2}}}} + {{{b^2}} \over {{{\left( {{b^2} + {{\left( {d - x} \right)}^2}} \right)}^{3/2}}}} > 0\forall x \in R

f(x) is an increasing function.

Q16
If S1 and S2 are respectively the sets of local minimum and local maximum points of the function, ƒ(x) = 9x4 + 12x3 – 36x2 + 25, x ∈ \in R, then :
A S1 = {–1}; S2 = {0, 2}
B S1 = {–2}; S2 = {0, 1}
C S1 = {–2, 0}; S2 = {1}
D S1 = {–2, 1}; S2 = {0}
Correct Answer
Option D
Solution

ƒ(x) = 9x4 + 12x3 – 36x2 + 25 ƒ'(x) = 36x3 + 36x2 – 72x ƒ'(x) = 36x(x2 + x – 2) ƒ'(x) = 36x(x + 2)(x - 1) While moving left to right on x-axis whenever derivative changes sign from negative to positive, we get local minima, and whenever derivative changes sign from positive to negative, we get local maxima.

∴\therefore S1 = {–2, 1} S2 = {0}

Q17
Let f be any function defined on R and let it satisfy the condition : ∣f(x)−f(y)∣ ≤ ∣(x−y)2∣,∀(x,y)∈R|f(x) - f(y)|\, \le \,|{(x - y)^2}|,\forall (x,y) \in R If f(0) = 1, then :
A f(x) can take any value in R
B f(x)<0,∀x∈Rf(x) < 0,\forall x \in R
C f(x)>0,∀x∈Rf(x) > 0,\forall x \in R
D f(x)=0,∀x∈Rf(x) = 0,\forall x \in R
Correct Answer
Option C
Solution
∣f(x)−f(y)∣ ≤ ∣(x−y)2∣|f(x) - f(y)|\, \le \,|{(x - y)^2}|
⇒∣f(x)−f(y)x−y∣ ≤ ∣x−y∣\Rightarrow \left| {{{f(x) - f(y)} \over {x - y}}} \right|\, \le \,|x - y|
⇒∣lim⁡x→yf(x)−f(y)x−y∣ ≤ ∣lim⁡x→y(x−y)∣\Rightarrow \left| {\mathop {\lim }\limits_{x \to y} {{f(x) - f(y)} \over {x - y}}} \right|\, \le \,|\mathop {\lim }\limits_{x \to y} (x - y)|
⇒∣f′(x)∣ ≤0\Rightarrow |f'(x)|\, \le 0
⇒f′(x)=0\Rightarrow f'(x) = 0
⇒f(x)\Rightarrow f(x)

is constant function. ∵\because

f(0)f(0)

= 1 then f(x) = 1

Q18
Let f(x) = sin4x + cos4 x. Then f is an increasing function in the interval :
A ]0,π4[] 0, \dfrac{\pi}{4}[
B ]π4,π2[] \dfrac{\pi}{4}, \dfrac{\pi}{2}[
C ]π2,5π8[] \dfrac{\pi}{2}, \dfrac{5 \pi}{8}[
D ]5π8,3π4[] \dfrac{5 \pi}{8}, \dfrac{3 \pi}{4}[
Correct Answer
Option B
Solution

f(x) = sin4x + cos4x ∴\therefore f'(x) = 4sin3x cosx + 4cos3x (−- sinx) = 4sinx cosx (sin2x −- cos2x) = −- 2sin2x cos2x = −- sin4x As, f(x) is increasing function when f'(x) > 0 ⇒\Rightarrow −- sin4x > 0 ⇒\Rightarrow sin4x < 0 ∴\therefore π\pi < 4x < 2π\pi

π4<x<π2{\pi \over 4} < x < {\pi \over 2}

∴\therefore x

∈\in
(π4,π2)\left( {{\pi \over 4},{\pi \over 2}} \right)
Q19
Let the function f(x)=2x3+(2p−7)x2+3(2p−9)x−6f(x) = 2{x^3} + (2p - 7){x^2} + 3(2p - 9)x - 6 have a maxima for some value of x0x 0. Then, the set of all values of p is
A (−92,92)\left( { - {9 \over 2},{9 \over 2}} \right)
B (92,∞)\left( {{9 \over 2},\infty } \right)
C (0,92)\left( {0,{9 \over 2}} \right)
D (−∞,92)\left( { - \infty ,{9 \over 2}} \right)
Correct Answer
Option D
Solution

f′(x)=6x2+2x(2p−7)+3(2p−9)f^{\prime}(x)=6 x^{2}+2 x(2 p-7)+3(2 p-9) x10x_{1}0 ⇒f′(0)<0\Rightarrow f^{\prime}(0)<0 ⇒p<92\Rightarrow p<\dfrac{9}{2}

p∈(−∞,92)p \in \left( { - \infty ,{9 \over 2}} \right)
Q20
A function y=f(x)y=f(x) has a second order derivative f′′(x)=6(x−1).f''\left( x \right) = 6\left( {x - 1} \right). If its graph passes through the point (2,1)(2, 1) and at that point the tangent to the graph is y=3x−5y = 3x - 5, then the function is :
A (x+1)2{\left( {x + 1} \right)^2}
B (x−1)3{\left( {x - 1} \right)^3}
C (x+1)3{\left( {x + 1} \right)^3}
D (x−1)2{\left( {x - 1} \right)^2}
Correct Answer
Option B
Solution
f′′(x)=6(x−1).f''\left( x \right) = 6\left( {x - 1} \right).

Inegrating, we get

f′(x)=3x2−6x+cf'\left( x \right) = 3{x^2} - 6x + c

Slope at

(2,1)=f′(2)=c=3\left( {2,1} \right) = f'\left( 2 \right) = c = 3
[  \left[ {\,\,} \right.

As slope of tangent at

(2,1)(2, 1)

is

33
  ]\left. {\,\,} \right]

∴\therefore

f′(x)=3x2−6x+3=3(x−1)2f'\left( x \right) = 3{x^2} - 6x + 3 = 3{\left( {x - 1} \right)^2}

Inegrating again, we get

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

The curve passes through

(2,1)(2,1)
⇒1=(2−1)3+D⇒D=0\Rightarrow 1 = {\left( {2 - 1} \right)^3} + D \Rightarrow D = 0

∴\therefore

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