Q: Consider the function f(x)= |x-1| x^2 , then f(x) is

Given, f(x)= \ aligned x-1 x^2 ; & x 1 \\ -x+1 x^2 ; & x 1 aligned . aligned & f(x)= cases 1 x - 1 x^2 ; & x 1 \\ - 1 x + 1 x^2 ; & x 1 cases \\ & f^ (x)= cases -1 x^2 + 2 x^3 ; & x 1 \\ 1 x^2 - 2 x^3 ; & x 1 cases \\ & f^ (x)= cases 2-x x^3 ; & x 1 \\ x-2 x^3 ; & x 1 cases aligned By observation, f^ (x) will be Decreasing in (0,1) (2, ) and increasing in (- , 0) (1,2)

Question 1

The maximum area of rectangle inscribed in a circle of diameter R is

Accepted Answer

According to question, Diameter =R Diagonals, d_1=d_2=R aligned Max Area of rectangle (any quadrilateral) & = 1 2 d_1 d_2 \ = & 1 2 R R= R^2 2 aligned

Question 2

Consider the function f(x)= |x-1| x^2 , then f(x) is

Accepted Answer

Given, f(x)= \ aligned x-1 x^2 ; & x 1 \ -x+1 x^2 ; & x 1 aligned . aligned & f(x)= cases 1 x - 1 x^2 ; & x 1 \ - 1 x + 1 x^2 ; & x 1 cases \ & f^ (x)= cases -1 x^2 + 2 x^3 ; & x 1 \ 1 x^2 - 2 x^3 ; & x 1 cases \ & f^ (x)= cases 2-x x^3 ; & x 1 \ x-2 x^3 ; & x 1 cases aligned By observation, f^ (x) will be Decreasing in (0,1) (2, ) and increasing in (- , 0) (1,2)

Question 3

The maximum volume (in cu. units) of the cylinder which can be inscribed in a sphere of radius 12 units is

Accepted Answer

12^2=r^2+ ( h 2 )^2 V= r^2 h V= (144- h^2 4 ) h V=144 h- 4 h^3 d V d h =144 - 3 4 h^2 d V d h =0 144 = 3 4 h^2 h^2=48 4 h=8 3 12^2=r^2+48 r^2=96 Volume = r^2 h= 96 8 3 =768 3 ~cm ^3.

Question 4

If the angle made by the tangent at the point (x_0, y_0 ) on the curve x=12(t+ t t), y=12 (1+ t)^2, 0 t 2 , with the positive x -axis is 3 , then y_0 is equal to

Accepted Answer

Given equation of curve x=12(t+ t t), y=12(1+ t)^2 Differentiate w.r.t ' t ', aligned & d x d t =12 (1+ ^2 t- ^2 t ) \ & d x d t =12(1+ 2 t) and d y d t \ & =24(1+ t) t \ & d y d x = 2(1+ t) t 1+ 2 t aligned

Question 5

The altitude of a cone is 20 cm and its semivertical angle is 30^ . If the semi-vertical angle is increasing at the rate of 2^ per second, then the radius of the base is increasing at the rate of

Accepted Answer

Let be the semi-vertical angle and r be the radius of the cone at time t. the radius of the cone at time t. Then, r=20 d r d t =20 ^2 d d t d r d t =20 ^2 30^ 2 [ =30^ . and . d d t =2 ] d r d t =20 4 3 2 ~cm / s = 160 3 ~cm / s

Question 6

The point of inflexion for the curve y=(x-a)^n, where n is odd integer and n 3 is

Accepted Answer

Here d^2 y d x^2 =n(n-1)(x-a)^ n-2 Now, d^2 y d x^2 =0 x=a Differentiating equation of the curve n times, we get, d^n y d x^n =n ! at x=a, d^n y d x^n 0 and d^ n-1 y d x^ n-1 =0, where n is odd. Therefore (a, 0) is the point of inflexion

Question 7

The population p(t) at time t of a certain mouse species satisfies the differential equation d p(t) d t =0.5 p ( t )-450. If p(0)=850, then the time at which the population becomes zero is :

Accepted Answer

Given differential equation is d p(t) d t =0.5 p(t)-450 d p(t) d t = 1 2 p(t)-450 d p(t) d t = p(t)-900 2 2 d p(t) d t =-[900-p(t)] 2 d p(t) 900-p(t) =-d t Integrate both the side, we get : -2 d p(t) 900-p(t) = d t Let 900-p(t)=u -d p(t)=d u We have; 2 d u u = d t 2 u=t+c 2 [900-p(t)]=t+c when t=0, p(0)=850 2 (50)=c 2 [ ( 900-p(t) 50 ) ]=t 900-p(t)=50 e^ t 2 p(t)=900-50 e^ t 2 let p (t_1 )=0 0=900-50 e^ t_1 2 t_1=2 18

Question 8

The function f(x)= (1+x)- 2 x 2+x is increasing on

Accepted Answer

Given f(x)= (1+x)- 2 x 2+x aligned & f^ (x)= 1 1+x - (2+x)(2)-2 x (2+x)^2 \ & = 1 1+x - 4 (2+x)^2 \ & = (2+x)^2-4-4 x (1+x)(2+x)^2 = x^2 (1+x)(2+x)^2 >0 for all \ & x (0, ) aligned Thus, given function f ( x ) is increasing on (0, ).

Question 9

If f(x)= x, g(x)= 2 x, h(x)= 3 x and I(x) = x, then which of the following option is correct?

Accepted Answer

(a) Let f(x)= x, then f^ (x)=- x. In interval (0, 2 ), f^ (x) Therefore, f(x) is strictly decreasing on (0, 2 ) (b) Let f(x)= 2 x f^ (x)=-2 x 2 x In interval (0, 2 ), f^ (x) Because 2 x will either lie in the first or second quadrant which will give a positive value. Therefore, f(x) is strictly decreasing on (0, 2 ) (c) Let f^ (x)= 3 x f ^ ( x )=-3 3 x . In Interval (0, 3 ), f ^ ( x ) Because 3 x will either lie in the first or second quadrant which will give a positive value. Therefore, f ( x )

Question 10

The interval in which the function f(x)= 4 x^2+1 x is decreasing is :

Accepted Answer

Given f(x)= 4 x^2+1 x Thus f^ (x)=4- 1 x^2 f ( x ) will be decreasing if f ^ ( x ) Thus 4- 1 x^2 4 -1 2 Thus interval in which f(x) is decreasing, is (- 1 2 , 1 2 ).

Question 11

The curve given by x+y=e^ x y has a tangent parallel to the Y-axis at the point

Accepted Answer

x+y=e^ x y Differentiating w.r.t. x, we get 1+ d y d x =e^ x y [y+x d y d x ] d y d x (1-x e^ x y )=y e^ x y -1 d y d x = y e^ x y -1 1-x e^ x y d y d x = , as tangent is parallel to Y -axis 1-x e^ x y =0 xe ^ xy =1 This holds, when x=1 and y=0

Question 12

The function f(x)= ^ -1 ( x+ x) is an increasing function in

Accepted Answer

Since, f(x)= ^ -1 ( x+ x) f ^ ( x )= 1 1+( x + x )^2 ( x - x ) = 2 (x+ 4 ) 1+( x+ x)^2 f(x) is increasing if f^ (x)>0 (x+ 4 )>0 - 2 < x+ 4 < 2 - 3 2 < x+ 4

Question 13

A running track of 440 ft is to be laid out enclosing a football field, the shape of which is a rectangle with a semi-circle at each end. If the area of the rectangular portion is to be maximum, then

Accepted Answer

Perimeter =440 ft 2 x+ r+ r=440 2 x+2 r=440 aligned A & = Area of the rectangular portion =x 2 r \ A & =x (440-2 x) = 1 (440 x-2 x^2 ) aligned Let d A d x = 1 (440-4 x)=0 aligned & x=110 for which d^2 A d x^2 < 0 \ & A is maximum when x=110 \ & 2 r= 440-2 x = 440-220 22 / 7 =70 \ & r=35 ft and x=110 ft aligned

Question 14

A spherical balloon is filled with 4500 ~ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72 ~ cubic meters per minute then the rate (in meters per minute)

Accepted Answer

V= 4 3 r^3 4500 = 4 r^3 3 r=15 ~m After 49 =(4500-49.72) =972 m ^3 aligned & & 972 & = 4 3 r^3 \ & & r^3 & =3 243=3 3^5 aligned aligned r & =9 \ d v d t & =4 r^2 ( d r d t ) \ 72 & =4 9 9 ( d r d t ) \ d r d t & = ( 2 9 ) aligned

Question 15

The points at which the tangent passes through the origin for the curve y=4 x^ 3 -2 x^ 5 are

Accepted Answer

The equation of the given curve is array l y=4 x^ 3 -2 x^ 5 \ dy dx =12 x^ 2 -10 x^ 4 array Therefore, the slope of the tangent at point (x, y) is 12 x^ 2 -10 x^ 4 . The equation of the tangent at (x, y) is given by Y -y= (12 x^ 2 -10 x^ 4 )( X -x) (i) When, the tangent passes through the origin (0,0), then X = Y =0 Therefore, eq. (i) reduce to array l -y= (12 x^ 2 -10 x^ 4 )(-x) \ y=12 x^ 3 -10 x^ 5 array Also, we have y=4 x^ 3 -2 x^ 5 array l 12 x^ 3 -10 x^ 5 \ =4 x^ 3 -2 x^ 5 \ 8 x^ 5 -8

Question 16

If the radius of a sphere is measured as 9 ~cm with an error of 0.03 ~cm , then find the approximating error in calculating its volume.

Accepted Answer

Let r be the radius of the sphere and r be the error in measuring the radius. Then, r=9 ~cm and r=0.03 ~cm Let V be the volume of the sphere. Then, array l V = 4 3 r^ 3 \ dV dr =4 r^ 2 \ ( dV dr )_ r=9 =4 9^ 2 =324 array Let V be the error in V due to error r in r. Then, array l V = dV dr r \ V =324 0.03=9.72 cm ^ 3 array

Question 17

f(x)= ( e^ 2 x -1 e^ 2 x +1 ) is

Accepted Answer

array l f (x)= ( e^ 2 x -1 e^ 2 x +1 ) \ f (-x)= e^ -2 x -1 e^ -2 x +1 = 1-e^ 2 x 1+e^ 2 x \ f (-x)= - (e^ 2 x -1 ) e^ 2 x +1 =- f (x) array f (x) is an odd function. Again, f ^ (x)= 4 e^ 2 x (1+e^ 2 x )^ 2 >0, x R f (x) is an increasing function.

Question 18

If the tangent at P(1,1) on y^ 2 =x(2-x)^ 2 meets the curve again at Q, then Q is

Accepted Answer

array l y^ 2 =x(2-x)^ 2 y^ 2 =x^ 3 -4 x^ 2 +4 x (i) \ 2 y dy dx =3 x^ 2 -8 x+4 \ dy dx = 3 x^ 2 -8 x+4 2 y \ [ dy dx ]_ P = 3-8+4 2 =- 1 2 array y^ 2 =x(2-x)^ 2 y^ 2 =x^ 3 -4 x^ 2 +4 x (i) 2 y dy dx =3 x^ 2 -8 x+4 dy dx = 3 x^ 2 -8 x+4 2 y [ dy dx ]_ P = 3-8+4 2 =- 1 2 Equation of tangent at P is: y-1=- 1 2 (x-1) x+2 y-3=0 x+2 y-3=0 Using y= 3-x 2 in (i), we get: ( 3-x 2 )^ 2 array l =x^ 3 -4 x^ 2 +4 x \ 4 x^ 3 -17 x^ 2 +22 x-9=0 ...(ii) array which has two roots 1,1 (Because of (ii) being t

Question 19

Consider the function f(x)=|x-1| /x^ 2 , then f(x) is

Accepted Answer

f(x)= |x-1| x^ 2 = \ array l 1-x x^ 2 , x 1 array . Clearly, f(x) is continuous for all x R except at x=0 f^ (x)= \ array l x-2 x^ 3 , x 1 array . f^ (x)>0 x 2 Hence, f(x) is increasing in (- , 0) (1,2) and decreasing in (0,1) (2, ).

Question 20

Let x+y=3- 4 and x-y=4 2 then the greatest of x y is

Accepted Answer

x= 3- 4 +4 2 2 = 3- (1- ^ 2 2 )+4 2 2 =(1+ 2 )^ 2 y= 3- 4 -4 2 2 = 3- (1- ^ 2 2 )+4 2 2 =(1- 2 )^ 2 x y= (1- ^ 2 2 )^ 2 = ^ 4 2 1

Application of Derivatives — Previous Year Questions & Solutions