36 + Interview Questions in DIFFERENTIALS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Find df for f(x) = x2 + 3x and evaluate it for (i) x = 2 and dx = 0.1 (ii) x = 3 and dx = 0.02
Answer» y = f(x) = x + 3x dy = (2x + 3) dx (i) dy {when x = 2 and ate = 0.1} = [2(2) + 3] (0.1) = 7(0.1) = 0.7 (ii) dy {when x = 3 and dx = 0.02} = [2(3) + 3] (0.0.2) = 9(0.02) = 0.18

1.

Find df for f(x) = x2 + 3x and evaluate it for (i) x = 2 and dx = 0.1 (ii) x = 3 and dx = 0.02

Answer»

y = f(x) = x + 3x

dy = (2x + 3) dx

(i) dy {when x = 2 and ate = 0.1}

= [2(2) + 3] (0.1)

= 7(0.1)

= 0.7

(ii) dy {when x = 3 and dx = 0.02}

= [2(3) + 3] (0.0.2)

= 9(0.02)

= 0.18

Discussion

2.	If there is an error of a% in measuring the edge of a cube, thenpercentage error in its surface is2a% (b)`a/2%`(c) `3a %`(d) none of theseA. `2a%`B. `a/2%`C. `3%`D. none of these
Answer» Correct Answer - A

Discussion

3.	If there is an error of a% in measuring the edge of a cube, thenpercentage error in its surface is2a% (b)`a/2%`(c) `3a %`(d) none of these
Answer» let x be the edge of cube `(dx)/x xx 100 = a` surface area of cube , `s= 6x^2` diff wrt x `(ds)/(dx) = 12x` `(ds)/s xx 100 = ( 12x xx dx)/(6x^2) xx 100` `= (12x xx (ax)/100)/(6x^2) xx 100` `= (12ax^2)/(6x^2 xx 100) xx 100` `= 2a` `(ds)/s xx 100 = 2a` `2a%`changes occur while measuring the edge option a is correct

Discussion

4.	If there is an error of 2% in measuring the length of simple pendulum,then percentage error in its period is:1% (b) 2%(c) 3% (d)4%A. `1%`B. `2%`C. `3%`D. `4%`
Answer» Correct Answer - A

Discussion

5.	If errors of 1% each are made in te base radius and height of a cylinder, then the percentage error in its volume, isA. `1%`B. `2%`C. `3%`D. none f these
Answer» Correct Answer - C Let V be the volume of the cone of base radius r and height h. Then, `V=pir^(2) h` `rArr " "dV=d(pir^(2) h)` `rArr dV=pi(h" "dr^(2)+ r^(2) dh)` `rArr" "dV=pi(2rhdr+r^(2)dh)` `rArr" "(dV)/V= (pi2rh" "dr+r^(2)dh)/(pir^(2)h)` `rArr" "(dV)/V=2/rdr+(dh)/h` `(dV)/V xx100=2(dr)/rxx100+(dh)/h xx100` `rArr" "(DeltaV)/V xx100=2 ((Delta r)/r xx100)+((Deltah)/h xx 100)2 xx1+1=3`

Discussion

6.	The height of acylinder is equal to the radius. If an error of `alpha%`is made in the height,then percentage error in its volume is`alpha%`(b) `2alpha%`(c) `3alpha%`(d) none of theseA. `a%`B. `2a%`C. `3a%`D. none of these
Answer» Correct Answer - C

Discussion

7.	The height of a cylinder is equal to the radius. If an error of `alpha%`is made in the height, then percentage error in its volume is`alpha%`(b) 2`alpha%`(c) 3`alpha%`(d) none of these
Answer» height of cyliner is h & radius is r so,`h=r` `(del h)/h xx 100 = a` `(dh)/h xx 100 = a` `dh = (ah)/100` volume of cylinder= `v= pi r^2 h= pi h^2 h = pi h^3` `v = pi h^3` diff wrt h `(dv)/(dh) = 3 pi h^2 ` `dv = 3 pi h^2 dh` `(dv)/v xx 100 = (dv)/v xx 100 = (3 pi h^2 dh)/(pi h^3) xx 100` `(dv)/v xx 100= 3a ` correct option is c

Discussion

8.	The value of `(127)^(1//3)` to four decimal places, isA. `5.0267`B. `5.4267`C. `5.5267`D. `5.001`
Answer» Correct Answer - A Let `y=x^(1//3), x=125 and x+ Delta x =127`. Then, `dy/dx=1/(3x^(2//3)) Delta x = 2` When, x =125, we have `y=5 and dy/dx=1/75` `:." "Deltay=dy/dx Deltax rArr Deltay= 1/75 xx2=2/75` `:." "(127)^(1//3) = y+ Deltay=5+2/75=5+8/3 xx1/100` `rArr" "(127)^(1//3) = 5+(2.6667)/100=5.02667 cong 5.0267`

Discussion

9.	Using differentials,find the approximate value of `sqrt(401)`A. `20.100`B. `20.025`C. `20.030`D. `20.125`
Answer» Correct Answer - B

Discussion

10.	The approximate value of `(1.0002)^3000`, isA. `1.2`B. `1.4`C. `1.6`D. `1.8`
Answer» Correct Answer - C Let `y=x^(3000), x=1 and x+Delta x = 1.0002` Then, `Delta x=1.0002-1=0.0002` Also,y=1 when x=1 Now, `y=x^(3000)` `rArr" "(dy)/dx=3000" "x^(2999) rArr((dy)/dx)_(x=1)=3000` `:." "Deltay=(dy)/dxDeltax` `rArr" "Deltay= 3000 xx0.0002 = 6/10=0.6` Hence `(1.0002)^(3000)= y+Deltay=1+0.6=1.6`

Discussion

11.	If `1^(@)=0.017` radians, then the approximate value of sin`46^(@)` , is
Answer» Correct Answer - C

Discussion

12.	Using differentials, the approximate value if `(627)^(1//4)`, isA. `5.002`B. `5.003`C. `5.005`D. `5.004`
Answer» Correct Answer - D

Discussion

13.	Use differentials to find the approximate value of `(log)_e(4. 01),`having given that `(log)_e4=1. 3863`.A. 1.3968`B. `1.3898`C. `1.3893`D. none of these
Answer» Correct Answer - C

Discussion

14.	A sphere of radius `100mm` shrinks to radius 98mm, then the approximatedecrease in its volume is(a)`12000pim m^3`(b) `80000 pim m^3`(C) `8000pim m^3`(d) `120pim m^3`
Answer» `v= (4/3 pi r^3)` `r= 100 mm` `del v = (4/3) pi (3r^2) del r ` `= 4 pi r^2 del r` `del r = 98-100 = -2` `r= 100mm` `del v = 4 pi r (100)^2 xx (-2) ` `= -80000 pi mm^3` `= 80000 pi mm^3` Answer

Discussion

15.	The radius of a sphere shrinks from 10 to 9.8 cm. Find approximatelythe decrease in its volume.
Answer» Volume of a sphere, `V = 4/3pir^3` `:. (dV)/(dr) = 4/3pi(3r^2)` `=> (dV)/(dr) = 4pir^2` Here, `r = 10cm, dr = 9.8-10 = -0.2cm` `:. (dV)/(-0.2) = 4pi(10)^2` `=>dV = 400pi**(-0.2) = -80pi`. So, decrease in volume of the sphere is `80pi cm^3`.

Discussion

16.	find the approximate volume of metal in a hollow spherical shell whoseinternal and external radii are 3cm and 3.0005cm, respectively.
Answer» `r= 3cm` `R= 3.0005 cm` `R= r + /_` `V= 4/3 pi R^3 - 4/3 pi r^3` `y = f(x) = 4/3 pi x^3` `/_ y = f(x + /_ x) - f(x) ` now,`r= 3cm , /_ r = 0.0005 cm` `/_ y = (dy/dx) /_ x` `= 4/3 pi (3x^2) xx 0.0005` `x=r` `= 4/3 pi (3^3) xx 0.0005` `= 0.018 pi ` `v= 0.018 pi cm^3` Answer

Discussion

17.	The pressure p and the volume V of a gas are connected by the relation, `pV^(1//4) = k`, where k is a contant. Find the percentage increase in the pressure, corresponding to a diminution of 0.5% in the volumeA. `1/2`B. `1/4%`C. `1/8%`D. none of these
Answer» Correct Answer - C

Discussion

18.	If `y=x^n ,`then the ratio of relative errors in `ya n dx`is`1:1`(b) 2:1 (c)1:n (d) n:1A. `1:1`B. `2:1`C. `1:n`D. `n:1`
Answer» Correct Answer - D

Discussion

19.	Use differentials to approximate the cube root of 127.
Answer» `y = f(x) = (x)^(1/2)` `(127)^(1/3) = f(x + /_ x) ` when `x=125 & /_x=2` `f(x+ /_ x) = /_ y + f(x) ` `f(x) = f(125) = (125)^(1/3) = 5` `/_ y = dy = (dy/dx) /_/x` `= d/dx(x^(1/3)) xx 2` `= [1/(3x^(2/3))] xx 2` `= 2/(3x^(2/3))` `= 2/(3 xx25) = 2/75` `/_ y = 0.027` `f(x+ /_x ) = f(127) = (127)^(1/3) = 5 + 0.027= 5.027 ` Answer

Discussion

20.	Use differentials to find the approximate value of `(log)_e(4. 01),`having given that `(log)_e4=1. 3863`.
Answer» let `y= f(x) = log_e x` `x + /_ x = 4.01` `x + /_ x = 4 + 0.01` `x= 4 & /_ x = 0.01` `y= f(4) = log_e 4 = 1.3863` `/_ x = 0.01` `del x = 0.01` `dx = 0.01` now, `y= log_e x` diff wrt x `dy/dx = 1/x` at x=4`dy/dx = 1/4` `dy = dy/dx xx dx = 1/4 xx 0.01 = 0.0025` `/_ y = 0.0025` `log_e (4.01) = y + /_y ` `= 1.3868 + 0.0025` `= 1.3888` `log_e (4.01) = 1.3838` Answer

Discussion

21.	The pressure P and volume V of a gas are connected by the relation `P V^(1/4=)`constant. The percentage increase in the pressure corresponding to a deminition of % in the volume is`1/2%`(b) `1/4%`(c) `1/8%`(d) none of these
Answer» `pv^(1/4) = c` `p= cv^(-1/4)` `ln p = ln(cv^(-1/4))` `ln p = ln c + (-1/4)ln v` now, `(dp)/p = 0- 1/4(dv)/v` `(dp)/p = -1/4 (dv)/v` `(dp)/p = -1/4(-1/2%)` `= 1/8%` option c is correct

Discussion

22.	If `y=x^n ,`then the ratio of relative errors in `y and x`is(a) `1:1`(b) 2:1 (c)1:n (d) n:1
Answer» `y= x^n` `(del y)/y= (del x)/x` now, ` del y = n x^(n-1) del x` `(x xx nx^(n-1) del x)/(x^n del x)` `= n:1` Answer

Discussion

23.	In a right angled `DeltaΑ BC, cos^2A +cos^2B cos^2C=`A. 2RB. `pi`C. 0D. none of these
Answer» Correct Answer - C

Discussion

24.	Find the approximatevalue of `(log)_(10)1005`, given that `(log)_(10)e=0. 4343`
Answer» Here, we will use the following rule, `f(x+Deltax) = f(x) + f'(x)Deltax` Here, `f(x+Deltax) = log_10 1005, f(x) = log_10 x, x = 1000, Deltax = 5` `:. log_10(x+Deltax) = log_10 x+d/dx(log_10x)Deltax` `:. log(x+Deltax) = log_10x+d/dx(log_e x/log_e 10)Deltax` `=>log_10(1005) = log_10 (1000)+1/(log_e 10)(1/x)(5)` `= 3+log_10 e(1/1000)5``=3+0.43431/1000**5` `=3+0.0021` `=3.0021` `:. log_10(1005) = 3.0021`

Discussion

25.	A circular metal plate expands under heating so that its radiusincreases by 2%. Find the approximateincrease in the area of the plate if the radius of the plate before heatingis 10cm.
Answer» `x=`radius & y= area `y= pi x^2 ` `(/_x)/x xx 100 = 2` `(/_ x)/10 xx 100 = 2` `/_ x = 2/10= 0.2cm` `dx= 0.2cm` now,`y= pi x^2` `dy/dx = 2x xx pi` `dy/dx= 2 xx 10 xx pi = 20 pi` `dy= 20 pi xx dx` `dy= 20 xx pi xx 0.2= 4 pi` `:. dy~= /_ y, so, /_ y= 4pi` Answer

Discussion

26.	If f(x, y, z) = xy + yz + zx, then fx – fz is equal to …(a) z – x (b) y – z (c) x – z (d) y – x
Answer» (a) z – x f_x = y + z f_z = y + x f_x – f_z = y + z – y – x = z – x

26.

If f(x, y, z) = xy + yz + zx, then fx – fz is equal to …(a) z – x (b) y – z (c) x – z (d) y – x

Answer»

(a) z – x

f_x = y + z

f_z = y + x

f_x – f_z = y + z – y – x = z – x

Discussion

27.	Let V(x, y, z) = xy + yz + zx, x, y, z ∈ R. Find the differential dV.
Answer» V(x, y, z) = xy + yz + zx V_x = y + z V_y = x + z V_z = y + x The differential is dV = (y + z) dx + (x + z) dy + (y + x) dz

27.

Let V(x, y, z) = xy + yz + zx, x, y, z ∈ R. Find the differential dV.

Answer»

V(x, y, z) = xy + yz + zx

V_x = y + z

V_y = x + z

V_z = y + x

The differential is dV = (y + z) dx + (x + z) dy + (y + x) dz

Discussion

28.	The approximate change in the volume: V of a cube of side x metres caused by increasing the side by 1% is …(a) 0.3 xdx m3 (b) 0.03 x m3 (c) 0.03 x2 m3 (d) 0.03 x3 m3
Answer» (d) 0.03 x³m³ Let the side of the cube be x units v = x³ when dx = 0.01x dv = 3x²dx = 3x²(0.01 x) = 0.03 x³ m³

28.

The approximate change in the volume: V of a cube of side x metres caused by increasing the side by 1% is …(a) 0.3 xdx m3 (b) 0.03 x m3 (c) 0.03 x2 m3 (d) 0.03 x3 m3

Answer»

(d) 0.03 x³m³

Let the side of the cube be x units

v = x³

when dx = 0.01x

dv = 3x²dx

= 3x²(0.01 x)

= 0.03 x³ m³

Discussion

29.	If g(x, y) = 3x2 – 5y + 2y2, x(t) = et and y(t) = cos t, then dg/dt is equal to …(a) 6e2t + 5 sin t – 4 cos t sin t (b) 6e2t – 5 sin t + 4 cos t sin t (c) 3e2t + 5 sin t + 4 cos t sin t (d) 3e2t – 5 sin t + 4 cos t sin t
Answer» (a) 6e^2t + 5 sin t – 4 cos t sin t

Discussion

30.	If u(x,y) = ex^2 + y^2, then ∂u/∂x is equal to ...(a) ex^2 + y^2(b) 2xu(c) x2u (d) y2u
Answer» (b) 2xu ∂u/∂x = 2xe^{x^2 + y^2} = 2xu

30.

If u(x,y) = ex^2 + y^2, then ∂u/∂x is equal to ...(a) ex^2 + y^2(b) 2xu(c) x2u (d) y2u

Answer»

(b) 2xu

∂u/∂x = 2xe^{x^2 + y^2} = 2xu

Discussion

31.	If w (x, y) = xy, x > 0, then ∂w/∂x is equal to …(a) xy log x(b) y log x(c) yxy - 1(d) x log y
Answer» (c) yx^{y - 1} w(x,y) = x^y ∂w/∂x = yx^{y - 1}

31.

If w (x, y) = xy, x > 0, then ∂w/∂x is equal to …(a) xy log x(b) y log x(c) yxy - 1(d) x log y

Answer»

(c) yx^{y - 1}

w(x,y) = x^y

∂w/∂x = yx^{y - 1}

Discussion

32.	Find ∆f and df for the function f for the indicated values of x, ∆x and compare (i) f (x) = x3 – 2x2; x = 2, ∆x = dx = 0.5 (ii) f(x) = x2 + 2x + 3; x = -0.5, ∆x = dx = 0.1
Answer» (i) y = f(x) = x³ – 2x² dy = (3x² – 4x) dx dy (when x = 2 and dx = 0.5) = [3(2²) – 4(2)] (0.5) = (12 – 8)(0.5) = 4(0.5) = 2 (i.e.,) df = 2 Now ∆f = f(x + ∆x) – f(x) Here x = 2 and ∆x = 0.5 f(x) = x³ – 2x² So f(x + ∆x) = f(2 + 0.5) = f(2.5) = (2.5)³ – 2(2.5)² = (2.5)² [2.5 – 2] = 6.25 (0.5) = 3.125 f(x) = f(2) = 2³ – 2(2²) = 8 – 8 = 0 So ∆f = 3.125 – 0 = 3.125 (ii) y = f(x) = x² + 2x + 3 dy = (2x + 2) dx dy (when x = – 0.5 and dx = 0.1) = [2(-0.5) + 2] (0.1) = (-1 + 2) (0.1) = (1) (0.1) = 0.1 (i.e.,) df = 0.1 Now ∆f = f(x + ∆x) – f(x) Here x = -0.5 and ∆x = 0.1 x² + 2x + 3 f(x + ∆x) = f(-0.5 + 0.1) = f(-0.4)² = (-0.4) + 2(-0.4) + 3 = 0.16 – 0.8 + 3 = 3.16 – 0.8 = 2.36 f(x) = f(-0.5) = (-0.5)² + 2(-0.5) + 3 = 0.25 – 1 + 3 = 3.25 – 1 = 2.25 So ∆ f = f(x + ∆x) – f(x) = 2.36 – 2.25 = 0.11

32.

Find ∆f and df for the function f for the indicated values of x, ∆x and compare (i) f (x) = x3 – 2x2; x = 2, ∆x = dx = 0.5 (ii) f(x) = x2 + 2x + 3; x = -0.5, ∆x = dx = 0.1

Answer»

(i) y = f(x) = x³ – 2x²

dy = (3x² – 4x) dx

dy (when x = 2 and dx = 0.5)

= [3(2²) – 4(2)] (0.5)

= (12 – 8)(0.5) = 4(0.5) = 2

(i.e.,) df = 2

Now ∆f = f(x + ∆x) – f(x)

Here x = 2 and ∆x = 0.5

f(x) = x³ – 2x²

So f(x + ∆x) = f(2 + 0.5)

= f(2.5) = (2.5)³ – 2(2.5)²

= (2.5)² [2.5 – 2] = 6.25 (0.5) = 3.125

f(x) = f(2) = 2³ – 2(2²) = 8 – 8 = 0

So ∆f = 3.125 – 0 = 3.125

(ii) y = f(x) = x² + 2x + 3

dy = (2x + 2) dx

dy (when x = – 0.5 and dx = 0.1)

= [2(-0.5) + 2] (0.1)

= (-1 + 2) (0.1)

= (1) (0.1)

= 0.1

(i.e.,) df = 0.1

Now ∆f = f(x + ∆x) – f(x)

Here x = -0.5 and ∆x = 0.1

x² + 2x + 3

f(x + ∆x) = f(-0.5 + 0.1)

= f(-0.4)² = (-0.4) + 2(-0.4) + 3

= 0.16 – 0.8 + 3 = 3.16 – 0.8 = 2.36

f(x) = f(-0.5) = (-0.5)² + 2(-0.5) + 3

= 0.25 – 1 + 3 = 3.25 – 1 = 2.25

So ∆ f = f(x + ∆x) – f(x) = 2.36 – 2.25 = 0.11

Discussion

33.	A sphere of radius 100 mm shrinks to radius 98 mm, then the approximate decrease in its volume is`12000 pi m m^3`(b) `800 pi m m^3`(c) `80000 pi m m^3`(d) `120 pi m m^3`A. `12000 pi mm^(3)`B. `800 pi mm^(3)`C. `80000 pi mm^(3)`D. `120 pi mm^(3)`
Answer» Correct Answer - C

Discussion

34.	The approximatevalue of `(33)^(1//5)`is(a) 2.0125 (b) 2.1 (c) 2.01 (d) none of theseA. `2.0125`B. `2.1`C. `2.01`D. none of these
Answer» Correct Answer - A

Discussion

35.	A sphere of radius 100mm shrinks to radius 98 mm, then the approximate decrease in its volume is`12000 pi m m^3`(b) `800 pi m m^3`(c) `80000 pi m m^3`(d) `120 pi m m^3`A. `12000 pi mm^(3)`B. `800 pi mm^(3)`C. `80000 pi mm^(3)`D. `120 pi mm^(3)`
Answer» Correct Answer - C

Discussion

36.	The circumference of a circle is measured as 28cm with an error of0.01cm. The percentage error in the area is
Answer» let say `c=`circumference `A=` area `r=` radius `c= 28 = 2 pi r` `r = 14/pi` `del c = 2 pi del r= 0.01` `del r= 001/(2 pi)` `A= pi r^2` `del A = 2 pi r del r ` `del A/A = (2 pi r del r)/(pi r^2)= (2 del r )/r xx 100` `%` error = `(2 xx (0.01/(2 pi)) xx 100)/(14/pi)` `= 1/14` Answer

Discussion

Explore topic-wise InterviewSolutions in .

Find df for f(x) = x2 + 3x and evaluate it for (i) x = 2 and dx = 0.1 (ii) x = 3 and dx = 0.02

If there is an error of a% in measuring the edge of a cube, thenpercentage error in its surface is2a% (b)`a/2%`(c) `3a %`(d) none of theseA. `2a%`B. `a/2%`C. `3%`D. none of these

If there is an error of a% in measuring the edge of a cube, thenpercentage error in its surface is2a% (b)`a/2%`(c) `3a %`(d) none of these

If there is an error of 2% in measuring the length of simple pendulum,then percentage error in its period is:1% (b) 2%(c) 3% (d)4%A. `1%`B. `2%`C. `3%`D. `4%`

If errors of 1% each are made in te base radius and height of a cylinder, then the percentage error in its volume, isA. `1%`B. `2%`C. `3%`D. none f these

The height of acylinder is equal to the radius. If an error of `alpha%`is made in the height,then percentage error in its volume is`alpha%`(b) `2alpha%`(c) `3alpha%`(d) none of theseA. `a%`B. `2a%`C. `3a%`D. none of these

The height of a cylinder is equal to the radius. If an error of `alpha%`is made in the height, then percentage error in its volume is`alpha%`(b) 2`alpha%`(c) 3`alpha%`(d) none of these

The value of `(127)^(1//3)` to four decimal places, isA. `5.0267`B. `5.4267`C. `5.5267`D. `5.001`

Using differentials,find the approximate value of `sqrt(401)`A. `20.100`B. `20.025`C. `20.030`D. `20.125`

The approximate value of `(1.0002)^3000`, isA. `1.2`B. `1.4`C. `1.6`D. `1.8`

If `1^(@)=0.017` radians, then the approximate value of sin`46^(@)` , is

Using differentials, the approximate value if `(627)^(1//4)`, isA. `5.002`B. `5.003`C. `5.005`D. `5.004`

Use differentials to find the approximate value of `(log)_e(4. 01),`having given that `(log)_e4=1. 3863`.A. 1.3968`B. `1.3898`C. `1.3893`D. none of these

A sphere of radius `100mm` shrinks to radius 98mm, then the approximatedecrease in its volume is(a)`12000pim m^3`(b) `80000 pim m^3`(C) `8000pim m^3`(d) `120pim m^3`

The radius of a sphere shrinks from 10 to 9.8 cm. Find approximatelythe decrease in its volume.

find the approximate volume of metal in a hollow spherical shell whoseinternal and external radii are 3cm and 3.0005cm, respectively.

The pressure p and the volume V of a gas are connected by the relation, `pV^(1//4) = k`, where k is a contant. Find the percentage increase in the pressure, corresponding to a diminution of 0.5% in the volumeA. `1/2`B. `1/4%`C. `1/8%`D. none of these

If `y=x^n ,`then the ratio of relative errors in `ya n dx`is`1:1`(b) 2:1 (c)1:n (d) n:1A. `1:1`B. `2:1`C. `1:n`D. `n:1`

Use differentials to approximate the cube root of 127.

Use differentials to find the approximate value of `(log)_e(4. 01),`having given that `(log)_e4=1. 3863`.

The pressure P and volume V of a gas are connected by the relation `P V^(1/4=)`constant. The percentage increase in the pressure corresponding to a deminition of % in the volume is`1/2%`(b) `1/4%`(c) `1/8%`(d) none of these

If `y=x^n ,`then the ratio of relative errors in `y and x`is(a) `1:1`(b) 2:1 (c)1:n (d) n:1

In a right angled `DeltaΑ BC, cos^2A +cos^2B cos^2C=`A. 2RB. `pi`C. 0D. none of these

Find the approximatevalue of `(log)_(10)1005`, given that `(log)_(10)e=0. 4343`

A circular metal plate expands under heating so that its radiusincreases by 2%. Find the approximateincrease in the area of the plate if the radius of the plate before heatingis 10cm.

If f(x, y, z) = xy + yz + zx, then fx – fz is equal to …(a) z – x (b) y – z (c) x – z (d) y – x

Let V(x, y, z) = xy + yz + zx, x, y, z ∈ R. Find the differential dV.

The approximate change in the volume: V of a cube of side x metres caused by increasing the side by 1% is …(a) 0.3 xdx m3 (b) 0.03 x m3 (c) 0.03 x2 m3 (d) 0.03 x3 m3

If g(x, y) = 3x2 – 5y + 2y2, x(t) = et and y(t) = cos t, then dg/dt is equal to …(a) 6e2t + 5 sin t – 4 cos t sin t (b) 6e2t – 5 sin t + 4 cos t sin t (c) 3e2t + 5 sin t + 4 cos t sin t (d) 3e2t – 5 sin t + 4 cos t sin t

If u(x,y) = ex^2 + y^2, then ∂u/∂x is equal to ...(a) ex^2 + y^2(b) 2xu(c) x2u (d) y2u

If w (x, y) = xy, x &gt; 0, then ∂w/∂x is equal to …(a) xy log x(b) y log x(c) yxy - 1(d) x log y

Find ∆f and df for the function f for the indicated values of x, ∆x and compare (i) f (x) = x3 – 2x2; x = 2, ∆x = dx = 0.5 (ii) f(x) = x2 + 2x + 3; x = -0.5, ∆x = dx = 0.1

A sphere of radius 100 mm shrinks to radius 98 mm, then the approximate decrease in its volume is`12000 pi m m^3`(b) `800 pi m m^3`(c) `80000 pi m m^3`(d) `120 pi m m^3`A. `12000 pi mm^(3)`B. `800 pi mm^(3)`C. `80000 pi mm^(3)`D. `120 pi mm^(3)`

The approximatevalue of `(33)^(1//5)`is(a) 2.0125 (b) 2.1 (c) 2.01 (d) none of theseA. `2.0125`B. `2.1`C. `2.01`D. none of these

A sphere of radius 100mm shrinks to radius 98 mm, then the approximate decrease in its volume is`12000 pi m m^3`(b) `800 pi m m^3`(c) `80000 pi m m^3`(d) `120 pi m m^3`A. `12000 pi mm^(3)`B. `800 pi mm^(3)`C. `80000 pi mm^(3)`D. `120 pi mm^(3)`

The circumference of a circle is measured as 28cm with an error of0.01cm. The percentage error in the area is

If w (x, y) = xy, x > 0, then ∂w/∂x is equal to …(a) xy log x(b) y log x(c) yxy - 1(d) x log y