Question No: 24 ( Marks: 10 ) parallel, perpendicular or none of these?

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1 ALL IN ONE Mega File MTH0 Midterm PAPERS, MCQz & subjective Created BY Farhan & Ali BS (cs) rd sem Hackers Group Mandi Bahauddin Remember us in your prayers Question No: 4 ( Marks: 0 ) Evaluate the following limit Q # Whether the given lines are parallel, perpendicular or none of these? ( ) 8 y and y L ( y ) L 8 y First we have to calculate the slope.

2 L ( y ) y ( ) y 6 y 5 Comparing it with equation of line y = m + c Slope of L = m = L 8 y y y y ( ) = - Comparing it with equation of line y = m + c Slope of L = m = - / Hence the given two lines are perpendicular because m m = () Q # Let f( ) and g() Find whether ( f g)( ) and ( g f )( ) are equal or not?

3 Here f( ) and g() ( f g)( ) f ( g( ) ) ( f g)( ) f ( ) ( f g)( ) Also ( g f )( ) g( f ( ) ) ( g f )( ) g( ) ( g f )( ) ( ) ( g f )( ) 69 Hence ( f g)( ) ( g f )( ) Q # Determine whether the equation represents a circle, if the equation represents a circle, find the center and radius? + + y - 4y = 5 First, group the -terms, group the y-terms, and take the

4 constant to the right side: ( + ) + ( y - 4y ) = y -4y +4=5++4 Then write the left side as squares and add up the right side and you get (+) +(y-) =0, and now you can find the center and radius and graph it. Here the center would be (-,) and the radius would be. So to graph it, all you need to do is find the point (-,) and then plot the points up, down, to the right, and to the left of it, and draw the circle through them. Q # 4 Solve the inequality multiplyby, weget 84 subtractingby4 divideby or Hencethe requiredsolutionis,

5 Q # Whether the given lines are parallel, perpendicular or none of these? ( ) 8 y and y L ( y ) L 8 y First we have to calculate the slope. L ( y ) y ( ) y 6 y 5 Comparing it with equation of line y = m + c Slope of L = m = L 8 y y y y Comparing it with equation of line y = m + c

6 Slope of L = m = - / ( ) = - Hence the given two lines are perpendicular because m m = () Q # Let f( ) and g() Find whether ( f g)( ) and ( g f )( ) are equal or not? Here f( ) and g() ( f g)( ) f ( g( ) ) ( f g)( ) f ( ) ( f g)( ) Also

7 ( g f )( ) g( f ( ) ) ( g f )( ) g( ) ( g f )( ) ( ) ( g f )( ) 69 Hence ( f g)( ) ( g f )( ) Q # Determine whether the equation represents a circle, if the equation represents a circle, find the center and radius? + + y - 4y = 5 First, group the -terms, group the y-terms, and take the constant to the right side: ( + ) + ( y - 4y ) = y -4y +4=5++4 Then write the left side as squares and add up the right side and you get (+) +(y-) =0, and now you can find the center and radius and graph it. Here the center would be (-,) and the radius would be. So to graph it, all you need to do is find the point (-,) and then plot the points up,

8 down, to the right, and to the left of it, and draw the circle through them. Q # 4 Solve the inequality multiplyby, weget 84 subtractingby4 divideby or Hencethe requiredsolutionis, Q # Whether the given lines are parallel, perpendicular or none of these? ( ) 8 y and y

9 L ( y ) L 8 y First we have to calculate the slope. L ( y ) y ( ) y 6 y 5 Comparing it with equation of line y = m + c Slope of L = m = L 8 y y y y ( ) = - Comparing it with equation of line y = m + c Slope of L = m = - / Hence the given two lines are perpendicular because m m = () Q # Let

10 f( ) and g() Find whether ( f g)( ) and ( g f )( ) are equal or not? Here f( ) and g() ( f g)( ) f ( g( ) ) ( f g)( ) f ( ) ( f g)( ) Also ( g f )( ) g( f ( ) ) ( g f )( ) g( ) ( g f )( ) ( ) ( g f )( ) 69 Hence ( f g)( ) ( g f )( ) Q # Determine whether the equation represents a circle, if the equation represents a circle, find the center and radius?

11 + + y - 4y = 5 First, group the -terms, group the y-terms, and take the constant to the right side: ( + ) + ( y - 4y ) = y -4y +4=5++4 Then write the left side as squares and add up the right side and you get (+) +(y-) =0, and now you can find the center and radius and graph it. Here the center would be (-,) and the radius would be. So to graph it, all you need to do is find the point (-,) and then plot the points up, down, to the right, and to the left of it, and draw the circle through them. Q # 4 Solve the inequality 4 6

12 4 6 multiplyby, weget 84 subtractingby4 divideby or Hencethe requiredsolutionis, y y if y 5 lim g( y) where, g( y) y if y ALL IN ONE Mega File MTH0 Midterm PAPERS, MCQz & subjective Created BY Farhan & Ali BS (cs) rd sem Hackers Group Mandi Bahauddin Remember us in your prayers Mindhacker4@gmail.com

13 Question No. Use implicit differentiation to find / d if y y d( y) d( y) ( y) ) ( y)( ) d( ) d d d ( y) ( y)( ) ( y)( ) d d ( y) y y y y y y d d d d ( y) y d ( y) ( y) y d ( y) y d ( y) y d ( y) y d ( y) y d Some other method can be used to solve this.

14 Question No.. Find the slope of the tangent line to the given curve at the specified point ( y ) 5( y ); (,) ( y ) 5( y ) differentiating with respect to ' ' d( y ) d( y ) ( y ) 5 d d d d d d d d y y (5 5 y ) d d y y y y d d y y y d d y y y y d d y y(4 4 y ) 5 y ( 4 4y 5) d d y(4 4y 5) ( 4 4y 5) d ( 4 4y 5) d y(4 4y 5) 4( y )( y ) 5( y ) 4( y ).( y ) (5 5 y ) 8( y )( y ) (5 5 y ) 4( )( ) 4 ( ) 4 ( ) ( ) 5 4 ( ) 5 4 ( ) 5 (4 4 ) 5 At point (,)

15 y d y(4 4y 5) d d ( 4 4 5) (,) ( 4 4 5) (,) (,) (4 4 5) ( 6 4 5) Question No. A 0 f t ladder is leaning against a wall. If the top of the ladder slips down the wall at the rate of - ft/sec, how fast will the foot be moving away from the wall when the top is 6 ft above the ground? Let = Distance in feet between wall and foot of the ladder y = Distance in feet between floor and top of the ladder t = number of seconds after the ladder starts to slip. ft / sec dt It is negaive because y is decreasing as time increases, since the ladder is slipping down the wall.

16 d? dt Using Pythagoras Theorem, y 0 differentiating w. r. t ' t ' d y 0 dt dt d y dt dt when y 6ft d 6...( i) dt dt When thetop is 6 ft abovethe ground put in () i we get d 6 dt 8 dt d 6 ( ) dt 8 8 ft / sec ALL IN ONE Mega File MTH0 Midterm PAPERS, MCQz & subjective Created BY Farhan & Ali BS (cs) rd sem Hackers Group

17 Mandi Bahauddin Remember us in your prayers Question No. : Find the relative etreme values of the function Since, f() = a sin + b cos f() = a sin + b cos

18 f ( ) a cos bsin...( i) put f ( ) 0 a cos bsin 0 a cos bsin a tan b a b By Pythagorus theorem tan tan ( ) a b sin, cos a b a b Now, taking second derivative a b f ( ) asin bcos......( ii) put sin a, cos b in ( ii) a b a b f ( ) a a b b a b a b f ( ) ( a b ) a b 0 a b a b So f ( ) has relative ma ima at sin a and cos b a b a b and sin ce bothsin and cos are positve in first quadrant so we say f has relative ma ima at a tan ( ) b put sin a b, cos a b a b in( ii) f ( ) a a b b a b a b f ( ) ( a b ) a b 0 a b a b So f ( ) has relative min ima at sin a b and cos a b a b and sin cebothsin and cos are positvein third quadrant so we say f has relative min ima at a b tan ( )

19 Question No.: Let, if f( ), if Does Mean Value Theorem hold for f on,? To check whether f satisfies M.V.T I. Clearly f is continuous on,. II. To check f is differentiable on, We check whether f () eists. f ( ) f () L. H. Limt f () lim 0 lim 0 lim 0 f ( ) f () R. H. Limtf () lim 0 lim 0 Thus L. H. Limt f () R. H. Limtf (). Therefore, f () does not eist and the M. V. T. does not hold on,. Question No. : Integrate the following

20 a ( i) ( ii) ln a () i d dt t t a dt a t [ ( ln (ln ) t (ln ) dt t a a a...)] t t t!! ( a ln a (ln a) (ln a)...)!! t [ ln (ln ) t (ln ) a a a...)] t!! put t d dt d t (ln ln (ln ) t (ln ) t t a a a...) 4 8 ( ) ( ) (ln ln a (ln a) (ln a)...) 4 8 ln ( ii) d put ln t d dt t tdt c By back substitution tdt (ln ) c

21 Question Marks 8 Find area of region enclosed by given curves y y 4, 0, 0,. y y 4, 0, 0,. y = 0 is the -ais, = 0 is y-ais and is a line parallel to y-ais crossing -ais at point. So we have to find out the area bounded by the curve y= -4 + and the -ais over the interval [0,] Graph of y 4 is below We have to find the area of shaded region. It is clear from the graph that total area, A, under the curve in interval [0,] is divided into two regions A and A. In [0,], A is above -ais and in [,], A is below -ais. So A A A d d

22 NOTE: Since 4 A d is below -ais, so we will get negative value of integral. That s why we subtract A from A. Even if we don t have the graph, we can come to this conclusion by first finding the zero of a function that lie between given interval [0,]. Zero of a function is a value of which makes a function f() equal to zero. 4 0 ( 4 ) 0 ( ) 0 { ( ) ( )} 0 ( )( ) 0 0,, So zero of a function 4 is 0, and Now lie between our given interval [0, ], which means that total area is divided into two regions. One in interval [0,] and other in [,] So

23 4 4 A A A d d Question Evaluate Marks () () 0 4(0) (0) 4 4 4() () 4() () e d 0 e d lim t t 0 e d

24 t lim ( ) e t 0 d lim e t [0 ] t 0 Question Marks 0 Evaluate d ( )( ) 0 a b Cons ider, A B C D... () ( a )( b ) a b ( A B)( b ) ( C D)( a ) A B Ab Bb C D Ca Da ( AC) ( B D) ( Ab Ca ) ( Bb Da ) Comparing coefficients AC 0, B D 0 Ab Ca 0, Bb Da

25 Here C A Ab Aa, 0 thus Bb Ba B ( ) C 0 Since B D 0 D B, So () becomes A b a A if b a D ( b a ) b a ) B ( b a ) (0) /( b a ) (0) /( b a ) ( a )( b ) a b int egrating both sides, ( ( b a )( a ) ( b a )( b ) b a a b d d d ( a )( b ) a b ( b ) ( a ) t t d d lim t a b 0 ( b ) 0 ( a ) a t lim tan ( ) tan ( ) b t b b a a 0 t t lim tan ( ) tan ( ) a b t b b a a lim a b t b a ab ( a b ) Question No. Marks 0 Find the volume of the solid that results when the region enclosed by the given curve is revolved about the -ais y y 5,

26 We know the volume formula, when we revolve given by b a ([ ( )] [ ( )] ) f g d y f about -ais, is Here given that y 5 and y. Now to find the initial and final limits of integrations we shall solve both of curves simultaneously to find their points of intersection. On equating both functions we have , 4 Hence given a 4, and b 4, so we have b a 4 ([ f ( )] [ g( )] ) d ([ 5 ] [] ) d 5 9 d 6 d / Question No. Marks 8 Use cylindrical shells to find the volume of the solid generated when the region enclosed by the given curves is revolved about the y-ais.

27 y, 4, 9, y 0 To find the volume when we revolve around y-ais, we integrate with respect to i.e. V b a the curve is y b 9 4 f d so V f d [ ] d a Here a 4, and b 9 Thus V [ ] d [ / 5 b a / d Question No. Find the area of the surface generated by revolving the given curve about the y ais 6 y, 0 y 5 Here given that

28 g( y) 6 y, 0 y 5 g( y) 6 y g( y) g( y) 5 0 4(6 y) 4(6 y) 65 4y 4(6 y) g( y) Since 65 4y 4(6 y) S g( y) So, by above values y S 6 y 0 4(6 y) ( ) 6 ALL IN ONE Mega File MTH0 Midterm PAPERS, MCQz & subjective Created BY Farhan & Ali BS (cs) rd sem Hackers Group Mandi Bahauddin Remember us in your prayers Mindhacker4@gmail.com

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