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The correct option is (b) x1 < x2 ⇒ F(x1) ≥ f(x2) ∀ x1, x2 ∈ (a,b)

Easy explanation: The DEFINITION of monotonically decreasing FUNCTION is if a function f : (a,b) → R is said to be monotonically decreasing on (a,b) if x1 < x2 ⇒ f(x1) ≥ f(x2) ∀ x1, x2 ∈ (a,b). Hence, the mathematical expression is x1< x2 ⇒ f(x1) ≥ f(x2) ∀ x1, x2 ∈ (a,b).

Correct OPTION is (a) 10.08π cm^3

The explanation: LET x be the radius of the sphere.

Then, x=6cm and Δx=0.07cm

The VOLUME of a sphere is given by V=\(\frac{4}{3}\) πx^3

∴\(\frac{dV}{dx}=\frac{4}{3}\) π(3x^2)=4πx^2

dV=(\(\frac{dV}{dx}\))Δx=4πx^2 Δx

dV=4×π×6^2×0.07

dV=10.08π cm^3

Correct CHOICE is (c) 546.2

The EXPLANATION: Let x=4 and Δx=0.04

Then, f(x+Δx)=7(x+Δx)^3+7(x+Δx)^2-4(x+Δx)+3

Δy=f(x+Δx)-f(x)

∴f(x+Δx)=Δy+f(x)

Δy=f'(x)Δx

⇒f(x+Δx)=f(x)+f’ (x)Δx

Here, f'(x)=21x^2+12x-4

f(4.04)=(7(4)^3+6(4)^2-4(4)+3)+(21(4)^2+12(4)-4)(0.04)

f(4.04)=(448+96-16+3)+(336+48-4)(0.04)

f(4.04)=531+380(0.04)=546.2

Right OPTION is (b) (0,-1) and \((-\frac{1}{2},-\frac{15}{16})\)

EXPLANATION: GIVEN that, y=3x^4+2x^3-1

Differentiating w.r.t x, we GET

\(\frac{dy}{dx}\)=12x^3+6x^2

The tangent is parallel to x-axis, which implies that the slope \(\frac{dy}{dx}\) is 0.

∴12x^3+6x^2=0

6x^2 (2x+1)=0

⇒x=0,-\(\frac{1}{2}\)

If x=0

y=3(0)+2(0)-1=-1

If x=-\(\frac{1}{2}\)

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

y=\(\frac{3}{16}-\frac{2}{8}-1\)

y=-\(\frac{15}{16}\)

Hence, the points at which the tangent is parallel to the x-axis is (0,-1) and \((-\frac{1}{2},-\frac{15}{16})\).

Right answer is (a) 3.34

The best I can explain: Let y=\(\sqrt{X}\). Let x=9 and Δx=2

Then, Δy=\(\sqrt{x+Δx}-\sqrt{x}\)

 Δy=\(\sqrt{11}-\sqrt{9}\)

\(\sqrt{11}\)=Δy+3

dy is APPROXIMATELY equal to Δy is equal to

dy=\(\frac{dy}{dx}\)Δx

dy=\(\frac{1}{(2\sqrt{x})}\).Δx

dy=\(\frac{1}{(2\sqrt{9})}(2)\)

dy=2/6=0.34

∴ The approximate VALUE of \(\sqrt{11}\) is 3+0.34=3.34.

Correct answer is (b) Rs. 144.00

Easiest EXPLANATION: The marginal cost is given by the rate of change of revenue.

Hence, \(\frac{dN(X)}{DT}\)=0.18x^2-0.02x+10.

∴\(\frac{dN(x)}{dt}\)|_x=5=0.18(5)^2-0.02(5)+10

=4.5-0.1+10

=Rs. 14.4

The correct choice is (c) \(\frac{5}{4}\)

EASY explanation: Given that, x=4 cos^3⁡3θ and y=5 sin^3⁡3θ

\(\frac{DX}{dθ}\)=4(3)(3 cos^2⁡3θ)(-sin⁡3θ)

\(\frac{dy}{dθ}\)=5(3)(3 sin^2⁡3θ)(cos⁡3θ)

\(\frac{dy}{dx}\)=\(\frac{dy}{dθ}.\frac{dθ}{dx}=\frac{5(3)(3 sin^2⁡3θ)(cos⁡3θ)}{4(3)(3 cos^23θ)(-sin⁡3θ)}\)

\(\frac{dy}{dx}\)=-\(\frac{5 tan⁡3θ}{4}\)

\(\frac{dy}{dx}\)]θ=π/4=-\(\frac{5}{4} tan\frac{⁡3 \PI}{4}\)=-\(\frac{5}{4} (-1)=\frac{5}{4}\).

Correct OPTION is (a) 5.0267

For EXPLANATION: Let y=(x)^1/3. Let x=125 and Δx=2

Then, Δy=(x+Δx)^1/3-x^1/3

Δy=(127)^1/3-(125)^1/3

(127)^1/3=Δy+5

dy is APPROXIMATELY equal to Δy is equal to

dy=\(\frac{dy}{dx}\)Δx

dy=\(\frac{1}{3x^{2/3}}\).Δx

dy=\(\frac{1}{3×125^{2/3}} (2)\)

dy=2/75=0.0267

∴ The approximate value of (127)^1/3 is 5+0.0267=5.0267

Right option is (B) X/√(x^2 + 2) dx

To elaborate: Let, y = f(x) = √(x^2 + 2)

So, f(x) = (x^2 + 2)^1/2

On DIFFERENTIATING it we get,

f’(x) = d/dx[(x^2 + 2)^1/2]

f’(x) = 1/2 * 1/√(x^2 + 2) * 2x

So f’(x) = x/√(x^2 + 2)

So the differential EQUATION is:

DY = f’(x)dx

Hence, dy = x/√(x^2 + 2) dx

Right choice is (a) x1 < x2 ⇒ f(x1) < f(x2) ∀ x1, x2 ∈ (a,B)

The BEST EXPLANATION: A function f : (a,b) → R is said to be strictly INCREASING on (a,b) if x1< x2 ⇒ f(x1) < f(x2) ∀ x1, x2 ∈ (a,b). (a,b) MAY be replaced by [a,b] or any interval in the definition.

Correct answer is (c) 2

The explanation is: We have y = 2x/(x^2 + 1)

DIFFERENTIATING y with RESPECT to x, we get

dy/dx = d/dx(2x/(x^2 + 1))

= 2 * [(x^2 + 1)*1 – x * 2x]/(x^2 + 1)^2

= 2 * [1 – x^2]/(x^2 + 1)^2

Thus, the SLOPE of tangent to the curve at (0, 0) is,

[dy/dx](0, 0) = 2 * [1 – 0]/(0 + 1)^2

Thus [dy/dx](0, 0) = 2.

Correct CHOICE is (c) Strictly Increasing

Easiest explanation: Let X1 and X2 be any two numbers in R.

Then x1 7×1 < 7×2.

=> 7×1 – 4 < 7×2 – 4.

As F(x1) < f(x2), thus the function f is strictly increasing on R.

Correct answer is (b) 40π cm/s

Explanation: Let R be the radius and h be the HEIGHT of the CYLINDER. Then,

\(\frac{dr}{DT}\)=2 cm/s

The AREA of the cylinder is given by A=2πrh

\(\frac{dA}{dt}\)=2πh(\(\frac{dr}{dt}\))=4πh=4π(10)=40π cm/s.

The correct answer is (d) 1.5

Easiest explanation: 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

Thus, percentage error = relative error * 100 = 0.015 * 100 = 1.5

The correct choice is (B) 15 CU cm

The explanation: 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 cu cm.

Correct option is (b) [0, π/4) and (5π/4, 2π]

To explain I would say: f(x) = sinx+cosx.

f’(x) = cosx – sinx. Now f’(x) = 0 GIVES sinx = cosx which gives that x= π/4, 5π/4 as 0 ≤ x ≤ 2π.

The points x = π/4 and x = 5π/4 divide the interval [0, 2π] into THREE DISJOINT INTERVALS which are

[0, π/4), (π/4, 5π/4) and (5π/4, 2π].

Therefore on checking the values we get f is increasing in [0, π/4) and (5π/4, 2π].

The correct CHOICE is (d) Rs.6

Best explanation: The Marginal COST is the rate of change of REVENUE w.r.t the no. of units produced, we GET

\(\frac{dP(X)}{dt}\)=0.8x+2

cost(MC)=\(\frac{dP(x)}{dt}|\)x=5=0.8x+2=0.8(5)+2=4+2=6.

Correct OPTION is (c) 1

The best I can EXPLAIN: Given that, \(\FRAC{dy}{DT}=4.\frac{dx}{dt}\)

y=3x^2-2x+7

\(\frac{dy}{dt}\)=(6x-2) \(\frac{dx}{dt}\)

4.\(\frac{dx}{dt}\)=(6x-2) \(\frac{dx}{dt}\)

4=6x-2

6x=6

⇒x=1

The correct OPTION is (d) 0.24x^2

Easy EXPLANATION: Let the edge of the CUBE be x. Given that dx or Δx is equal to 0.02x(2%).

The surface area of the cube is A=6x^2

Differentiating w.r.t x, we GET

\(\frac{dA}{dx}\)=12x

dA=(\(\frac{dA}{dx}\))Δx=12x(0.02x)=0.24x^2

Hence, the approximate change in volume is 0.24x^2.

The correct choice is (c) 6

To explain: GIVEN that, x=12 cosecθ and y=2 sec⁡θ

\(\frac{dx}{dθ}\)=-12 COSEC θ cot⁡θ

\(\frac{dy}{dθ}\)=2 tan⁡θ sec⁡θ

\(\frac{dy}{dx}\)=\(\frac{dy}{dθ}.\frac{dθ}{dx}=\frac{2 \,tan⁡θ \,sec⁡θ}{-12 \,cosec θ \,cot⁡θ}\)=-\(\frac{1}{6} \frac{sin⁡θ}{cos^2⁡θ} × \frac{cos⁡θ}{sin^2⁡⁡θ}\) = –\(\frac{cot⁡θ}{6}\)

\(\frac{dy}{dx}]_{x=\frac{\pi}{4}}\)=\(-\frac{\frac{cot\pi}{4}}{6}=-\frac{1}{6}\)

Hence, the slope of NORMAL at θ=π/4 is

–\(\frac{1}{slope \,of \,tangent \,at \,θ=\pi/4}=-\frac{-1}{\frac{1}{6}}\)=6

Correct option is (a) (π/4, 5π/4)

To EXPLAIN: f(x) = sinx+COSX.

f’(x) = cosx – sinx. Now f’(x) = 0 gives sinx = cosx which gives that x=π/4, 5π/4 as 0 ≤ x ≤ 2π.

Therefore on CHECKING the values we get f is DECREASING in (π/4, 5π/4).

The correct choice is (d) 1 cm/s

The explanation is: Let the LENGTH be l, width be B and the area be A.

The Area is given by A=lb

\(\FRAC{dA}{dt}\)=l.\(\frac{DB}{dt}\)+b.\(\frac{DL}{dt}\) -(1)

Given that, \(\frac{dl}{dt}\)=4cm/s and \(\frac{dA}{dt}\)=8 cm/s

Substituting in the above equation, we get

8=l.\(\frac{db}{dt}\)+4b

Given that, l=4 cm and b=1 cm

∴8=4(\(\frac{db}{dt}\))+4(1)

8=4(\(\frac{db}{dt}\))+4

\(\frac{db}{dt}\)=1 cm/s.