Explore topic-wise InterviewSolutions in .

Correct choice is (a) increasing

Easy explanation: f(X) = e^2x.

f’(x) = 2e^2x.

 As we KNOW 2e^2x > 0, so it always has a value GREATER than zero.

Which shows that FUNCTION f is increasing for all x ∈ R.

Correct option is (d) 7.00714

Easiest EXPLANATION: Let y=\(\sqrt{49.1}\). Let x=49 and Δx=0.1

Then, Δy=\(\sqrt{x+Δx}-\sqrt{x}\)

 Δy=\(\sqrt{49.1}-\sqrt{49}\)

\(\sqrt{49.1}\)=Δy+7

dy is approximately EQUAL to Δy is equal to

dy=\(\FRAC{dy}{dx}\)Δx

dy=\(\frac{1}{(2\sqrt{x})}\).Δx

dy=\(\frac{1}{(2\sqrt{49})}\) (0.1)

dy=0.1/14=0.00714

∴ The approximate VALUE of \(\sqrt{49.1}\) is 7+0.00714=7.00714

The CORRECT choice is (B) 75.36 cm^2/s

To elaborate: The RATE of CHANGE of radius of the circle is \(\frac{dr}{dt}\)=6 cm/s

The area of a circle is A=πr^2

Differentiating w.r.t t we get,

\(\frac{dA}{dt}=\frac{d}{dt}\) (πr^2)=2πr \(\frac{dr}{dt}\)=2πr(6)=12πr.

\(\frac{dA}{dt}\)|r=2=24π= 24×3.14=75.36 cm^2/s

Right OPTION is (a) 0.18

For explanation I WOULD SAY: LET, y = f(x) = 2x^2 – 3x + 2

So, f’(x) = 4x – 3

Clearly, as x CHANGES from 3 to 3.02, the increment of x is:

3.02 – 3 = 0.02

So, increment in f(x) = f(3.02) – f(3)

= 2(3.02)^2 – 3(3.02) + 2 – 2(3)^2 – 3(3) + 2

= 0.1808

Now, differential of y is dy

We get, f’(3) * 0.02 = (4.3 – 3)*0.02

= 0.18

Right answer is (a) f’(X) > 0 ∀ x1, x2 ∈ (a,B)

The explanation: The condition for a function ‘f’ to be STRICTLY INCREASING is f’(x) > 0 ∀ x1, x2 ∈ (a,b) where the the function ‘f’ should be continuous and differentiable on (a,b).

The correct answer is (d) F’(x) ≥ 0 ∀ X1, X2 ∈ (a,b)

For explanation: If a FUNCTION f be continuous and DIFFERENTIABLE on (a,b), then f’(x) ≥ 0 ∀ x1, x2 ∈ (a,b) is the condition for the function f(x) to be increasing on (a,b).

Right answer is (d) –\(\FRAC{49}{24}\)

Explanation: GIVEN that, y=6x^2-7x

\(\frac{dy}{dx}\)=12x-7

It is given that the slope of tangent is 0

∴12x-7=0

⇒X=\(\frac{7}{12}\)

∴ y]x=\(\frac{7}{12}\)=6(\(\frac{7}{12}\))^2-7(\(\frac{7}{12}\))

6(\(\frac{49}{144})-\frac{49}{12}=49(-\frac{1}{24})=-\frac{49}{24}\)

Correct option is (a) 55/63 sq cm

To EXPLAIN: Let, y be the area of the triangle ABC.

Then, y = ½(AB)(sinC) where a and B are constants.

Now, dy = dy/dC * dC = ½(ab)(sinC) dC

Now, by problem, a = 25, b = 16, C = 60° and δC = (1/2)° = π/180 * ½ radian

Therefore, dy = 1/2 * 25 * 16 * cos60° * π/180 * ½ sq cm

=> dy = 25 * 8 * ½ * 22/7 * 1/180 * ½ sq cm

= 55/63 sq cm

Correct answer is (d) -25/54

Easiest explanation: Let the height on the wall be x and laser touches the ground at distance y from the wall. The length of the ladder is 20ft.

By Pythagoras THEOREM:

x^2 + y^2 = 400

Differentiating with respect to t:

2x(dx/dt) + 2y(dy/dt) = 0

Dividing throughout by 2:

x (dx/dt) + y (dy/dt) = 0

Now, dx/dt = -2 FT /s. negative because downwards

x(-2) + y (dy/dt) = 0………..(1)

When LOWER end is 12 ft from wall, let us find x:

x^2 + 12^2 = 400

x^2 = 400 – 144= 256

x = 16

x(2) + y (dy/dt) = 0 from (1)

16(-2) + 12 (dy/dt) = 0

-32 + 12(dy/dt) = 0

dy/dt = (32/12) = (8/3)

Thus, lower end moves on a horizontal floor when it is 12 ft from the wall at the rate of 8/3 ft/s

Now, assume that the ladder makes an angle θ with the horizontal plane at time t.

If, m be the slope of the ladder at time t, then,

 m = tanθ = x/y

Thus, dm/dt = d/dt(x/y) = [y(dx/dt) – x(dx/dt)]/y^2

Therefore, the rate of change of slope of ladder is,

[dm/dt]y = 2 = [12*(-2) – 16*(8/3)]/(12)^2

Now, putting the value of x = 16, when y = 12 and dx/dt = -2, dy/dt = 8/3

We GET, [dm/dt]y = 12 = [12(-2) – 16(8/3)]/(12)^2 = -25/54

The correct choice is (b) 20 sq cm

To elaborate: Let, r be the radius and s be the area of the surface of the sphere at TIME t.

By question, dr/dt = 1/2π

Now, s = 4πr^2;

Thus, ds/dt = 4π * 2r(dr/dt) = 8πr(dr/dt)

When r = 5CM and dr/dt = 1/2π

Then ds/dt = 8π*5*(1/2π) = 20

Thus, correct answer is 20 sq cm.

Right ANSWER is (b) f’(x) < 0 ∀ x1, x2 ∈ (a,b)

The explanation: The MATHEMATICAL EXPRESSION for strictly DECREASING function is f’(x) < 0 ∀ x1, x2 ∈ (a,b). This is the condition for strictly decreasing function and only possible when function ‘f’ is continuous and differentiable on (a,b).

The correct choice is (b) 76

Best EXPLANATION: The SLOPE of the tangent at x=2 is given by

\(\FRAC{dy}{dx}\)]x=2 = 21x^2-4x]x=2=21(2)^2-4(2)=84-8=76

Right choice is (c) 0.015

To explain: Let, v cubic cm be the volume of the CUBE of the cube of side X cm.

Then, v = x^3

Thus, dv/dx = d/dx(x^3) = 3x^2

Now, dv = dv/dx * δx = 3x^2 * δx

By problem, x = 10 and δx = 0.05

Therefore, APPROXIMATE ERROR =dv = 3 * 10^2 * 0.05 = 15 cc cm.

Hence, RELATIVE error = dv/v = 15/10^3 = 0.015

The correct CHOICE is (C) 1 SECOND

Best explanation: We have, x = t^3 + 6t^2 – 15T + 18

The particle stops when dx/dt = 0

And, dx/dt = d/dt(t^3 + 6t^2 – 15t + 18)

3t^2 + 12t – 15 = 0

=>t^2 + 4t –5 = 0

=>(t – 1)(t + 5)= 0

Thus, t = 1 or t = -5

Hence, the particle stops at the end of 1 second.

Correct answer is (b) True

The BEST explanation: Maximum and MINIMUM values of a function are represented by monotonicity. A MONOTONIC function on [a,b] is either a MONOTONICALLY increasing or monotonically DECREASING on [a,b].

Correct choice is (a) 67.99

To EXPLAIN I would say: LET x=5 and Δx=0.03

Then, f(x+Δx)=4(x+Δx)^2-7(x+Δx)+2

Δy=f(x+Δx)-f(x)

∴f(x+Δx)=Δy+f(x)

Δy=f’ (x)Δx

⇒f(x+Δx)=f(x)+f’ (x)Δx

f(5.03)=(4(5)^2-7(5)+2)+(8(5)-7)(0.03) (∵f’ (x)=8x-7)

f(5.03)=(100-35+2)+(40-7)(0.03)

f(5.03)=67+33(0.03)

f(5.03)=67+0.99=67.99

Right answer is (C) 16

Easiest explanation: The SLOPE of the TANGENT at x=1 is given by

\(\frac{dy}{dx}\)]x=1= 20x^3-6x+2]x=1=20(1)^3-6(1)+2=20-6+2=16

The CORRECT answer is (a) -8

Easiest explanation: Let, y = f(x) = 16 – x^2

dy/dx = -2X

Now, [dy/dx]x = 4 = [-2x]x = 4

So, [dy/dx]x = 4 = -8

The correct answer is (c) 3π/4

Explanation: Let, Φ be the angle which the TANGENT to the curve y = f(X) at P makes with the positive direction of the x axis.

Then,

tanΦ = [f’(x)]p = -1

= -TAN(π/4)

So, it is CLEAR that this can be written as,

= tan(π – π/4)

= tan(3π/4)

So, Φ = 3π/4

Therefore, the required angle which the tangent at P to the curve y = f(x) makes with positive direction of x axis is 3π/4.

The correct answer is (a) 2 cm/s

For EXPLANATION: LET the side of the SQUARE be x.

A=x^2, where A is the area of the square

Given that, \(\frac{dA}{dt}\)=2x \(\frac{dx}{dt}\)=40 cm^2/s.

\(\frac{dx}{dt}=\frac{20}{x} \)cm/s

\(\frac{dx}{dt}=\frac{20}{10}\)=2 cm/s.

Right answer is (b) 0.043cm/minute

Best explanation: Let ‘R’ be the radius and ‘h’ be the height of the water LEVEL at time t.

Then the volume of water level ‘V’ at time t is given by,

V = 1/3 (πr^2h) ……….(1)

Given the radius of the base of the cone is = (OA)’ = 27/2 cm and its height = (OB)’ = 45cm.

Again at time t, the radius of the water level = r = (CD)’ and its height = h = (CB)’

Clearly, the triangle OAB and CBD are SIMILAR.

Therefore, (CD)’/(CB)’ = (OA)’/(OB)’

Or, r/h = (22/7)/45 = 3/10

or, r = 3h/10

Thus, from (1) we get,

V = 1/3 π (3h/10)^2 h = (3π/100)h^3

Thus, dV/dt = (3π/100)*3h^2(dh/dt)

= (9πh^2/100)(dh/dt)

When, h = 30cm, then,

11 = (9π/100) (30)^2 (dh/dt) [as for all the values of t we have, dV/dt = 11]

Or, dh/dt = (11/(9π*9)) = 0.043

Thus, the rising rate of rising is 0.043 cm/minute.

Right option is (C) 1.484

The EXPLANATION: Let, y = LOG10X

Then F’(x) = d/dx[log10x]

= d/dx[logex * log10e]

= 1/x(log10e)

Now, f(x + δx) = f(x) + f’(x) δx

Putting x = 30 and δx = 0.5 in the above equation we get,

f(30 + 0.5) = f(30) + f’(30) δx

=> f(30.5) = log1030 + (1/30) log10e * 0.5

Putting the VALUES we get,

log1030.5 = 1.4843 = 1.484

The correct answer is (b) x1 < x2 ⇒ f(x1) > f(x2) ∀ x1, x2 ∈ (a,b)

To EXPLAIN: There are two TYPES of DECREASING functions in maxima and minima. They are strictly decreasing and monotonically decreasing. The MATHEMATICAL EXPRESSION for strictly decreasing function is x1 < x2 ⇒ f(x1) > f(x2) ∀ x1, x2 ∈ (a,b).

Correct option is (c) 8.01875

The explanation: Let y=\(\SQRT{x}\). Let x=64 and Δx=0.3

Then, Δy=\(\sqrt{x+Δx}-\sqrt{x}\)

Δy=\(\sqrt{64.3}-\sqrt{64}\)

\(\sqrt{64.3}\)=Δy+8

DY is APPROXIMATELY EQUAL to Δy is equal to:

dy=\(\frac{dy}{dx}\)Δx

dy=\(\frac{1}{2\sqrt{x}}\).Δx

dy=\(\frac{1}{2\sqrt{64}}\) (0.3)

dy=0.3/16=0.01875

∴ The approximate value of \(\sqrt{64.3}\) is 8+0.01875=8.01875

Right OPTION is (c) X>-1

The best explanation: f(x) = x^2+2x -5.

f’(x) = 4x – 3.

As we KNOW f’(x) = 0, x=-3/4, which shows that function f is INCREASING in interval (-3/4, ∞) for all x ∈ R.

The correct choice is (C) (\(\frac{1}{\sqrt{12}}\),2) and (-\(\frac{1}{\sqrt{12}}\),-2)

Easy explanation: Given that, y=\(\sqrt{4x^2+1}\)-2

\(\frac{DY}{dx}\)=\(\frac{1}{2\sqrt{4x^2+1}} (16x)=\frac{16x}{2\sqrt{4x^2+1}}\)

Given that the slope is 2

∴2=\(\frac{16x}{2\sqrt{4x^2+1}}\)

\(\sqrt{4x^2+1}\)=4x

4x^2+1=16x^2

12x^2=1

x=±\(\frac{1}{\sqrt{12}}\)

When x=+\(\frac{1}{\sqrt{12}}\)

y=\(\frac{16}{2\sqrt{12} (\sqrt{(1/3+1)})}=\frac{16}{2\sqrt{12} (2/\sqrt{3})}=2\)

When x=-\(\frac{1}{\sqrt{12}}\)

y=\(\frac{-16}{2\sqrt{12} (\sqrt{(1/3+1)})}=\frac{-16}{2√12 (2/√3))}=-2\)

Hence, the POINT on the curve is (\(\frac{1}{\sqrt{12}}\),2) and (-\(\frac{1}{\sqrt{12}}\),-2)

The correct option is (C) 2mph

Explanation: LET, AB be the lamp-post whose FOOT is A, and B is the source of light, and given (AB)’ = 12(½) ft.

Let MN denote the position of the man at time t where (MN)’ = 5ft.

Join BN and produce it to meet AM(produced) at P.

Then the length of man’s shadow= (MP)’

Assume, (AM)’ = x and (MP)’ = y. Then,

(PA)’ = (AM)’ + (MP)’ = x + y

And dx/dt = velocity of the man = 3

Clearly, TRIANGLES APB and MPN are similar.

Thus, (PM)’/(MN)’ = (PA)’/(AB)’

Or, y/5 = (x + y)/12(½)

Or, (25/2)y = 5X + 5y

Or, 3y = 2x

Or, y = (2/3)x

Thus, dy/dt = (2/3)(dx/dt)

As, dx/dt = 2,

= 2/3*3 = 2mph

Right OPTION is (a) –\(\frac{1}{26}\)

EASY explanation: The slope of the tangent at x=5 is given by:

\(\frac{dy}{DX}\)=8x-14

\(\frac{dy}{dx}\)]x=5=8(5)-14=40-14=26

∴ The slope of the NORMAL to the curve is

\(-\frac{1}{slope\, of \,the \,tangent \,at \,x=5}=-\frac{1}{26}\).

Correct OPTION is (b) False

To explain I would SAY: Monotonically increasing FUNCTIONS are usually referred to as increasing functions WHEREAS monotonically DECREASING functions are usually referred to as decreasing functions.

Correct choice is (c) f’(X) = 0 ∀ x1, x2 ∈ (a,b)

For explanation I would say: One of the properties of a function is to be constant. A function is said to be constant when it satisfies the condition f’(x) = 0 ∀ x1, x2 ∈ (a,b) where the function ‘f’ should be CONTINUOUS and DIFFERENTIABLE on (a,b).

The CORRECT choice is (b) 0.18x^3

For explanation I WOULD say: We KNOW that the volume V of a cube is given by

V=x^3

Differentiating w.r.t x, we get

\(\frac{dV}{dx}\)=3x^2

dV=(\(\frac{dV}{dx}\))Δx=3x^2 Δx

dV=3x^2 (6x/100)=0.18x^3

Therefore, the approximate change in volume is 0.18x^3.

The correct option is (a) X+2y=11

The best explanation: Differentiating 2X^2+3y^2=3 w.r.t x, we get

4x+6y \(\frac{dy}{dx}\)=0

\(\frac{dy}{dx} = -\frac{2x}{3y}\)

\(\frac{dy}{dx}\)](3,4)=-\(\frac{2(3)}{3(4)}=-\frac{1}{2}\)

Therefore, the EQUATION of the tangent at (3,4) is

y-y0=m(x-x0)

y-4=-\(\frac{1}{2}\) (x-3)

2(y-4)=-x+3

2y-8=-x+3

x+2y=11

The CORRECT option is (b) X1 < x2 ⇒ F(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b)

BEST explanation: A FUNCTION f : (a,b) → R is said to be monotonically increasing on (a,b) if x1 < x2 ⇒ f(x1) ≤ f(x2) ∀ x1, x2 ∈ (a,b). A monotonically increasing function can also be called as non-decreasing function.

The CORRECT CHOICE is (b) (-∞, 2)

The best explanation: f(x) = x^2 – 4x + 5.

f’(x) = 2x – 4. Therefore f’(x) = 0 gives x = 2.

Now this POINT x=2 divides the line into two disjoint intervals and the interval namely (-∞, 2) is DECREASING on f(x).

The correct OPTION is (a) INCREASING

To explain: F(x) = x^3 – 3x^2 + 4x

f’(x) = 3x^2 – 6x + 4.

f’(x) = 3(x^2 – 2x + 1) + 1.

=> 3(x-1)^2 + 1>0, in every INTERVAL of R. Therefore the FUNCTION f is increasing on R.

Right answer is (d) \(\frac{1}{9}\)

Explanation: Let the VOLUME of cube be V.

V=x^3

\(\frac{DV}{dt}\)=3x^2 \(\frac{DX}{dt}\)

12=3x^2 \(\frac{dx}{dt}\)

\(\frac{dx}{dt}=\frac{4}{x^2}\)

\(\frac{dx}{dt}\)|_x=6=\(\frac{4}{6^2}=\frac{4}{36}=\frac{1}{9}\).

Right choice is (d) 30 cm^2/s

Explanation: The circumference of the CIRCLE is given by C=2πr, where r is the radius of the circle.

∴\(\FRAC{dC}{dt}\)=2π.\(\frac{DR}{dt}\)=5 cm/s

\(\frac{dr}{dt}\)=5/2π cm/s

\(\frac{dA}{dt}\)|r=6=2πr.\(\frac{dr}{dt}\)=2πr.\(\frac{5}{2 \pi}\)=5r=5(6)=30 cm^2/s.

Right choice is (c) 24cm/sec^2

Explanation: We have, X = t^3 + 6t^2 – 15t + 18

Let, ‘V’ be the velocity of the particle and ‘a’ be the acceleration of the particle at the end of t seconds. Then,

v = dx/dt = d/dt(t^3 + 6t^2 – 15t + 18)

So, v = 3t^2 + 12t – 15

Therefore, a = dv/dt = d/dt(3t^2 + 12t – 15)

So, a = 6t + 12

Thus, acceleration of the particle at the end of 2 seconds is,

[dv/dt]t = 2 = 6(2) + 12 = 24cm/sec^2.

Right option is (a) 2x/(x^2 + 4) dx

For EXPLANATION I would say: Let, y = f(x) = LOG(x^2 + 4)

So f(x) = log(x^2 + 4)

On DIFFERENTIATING it we get,

f’(x) = d/dx[log(x^2 + 4)]

So f’(x) = 2x/(x^2 + 4)

So the differential equation is:

DY = f’(x)dx

Hence, dy = 2x/(x^2 + 4) dx

Right CHOICE is (d) f’(x) ≤ 0 ∀ x1, X2 ∈ (a,b)

To EXPLAIN: If a function f be continuous and differentiable on (a,b), then f’(x) ≤ 0 ∀ x1, x2 ∈ (a,b) is the CONDITION for the function f(x) to be decreasing on (a,b).