Objective Mathematics
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1 Chapter No - ( Area Bounded by Curves ). Normal at (, ) is given by : y y. f ( ) or f ( ). Area d ()() 7 Square units. Area (8)() 6 dy. ( ) d y c or f ( ) c f () c f ( ) As shown in figure, point P is given by Point P is ( ), f ( ) y tan and y cos meet at, 6 / 6 Area cos tan d sin ln(sec ) / 6 ln ln() ln() Area d ( ) ( ) ( ) ( ) ( ) 6 or 5 [9]
2 7. f ( ) min ( ), ( ) As shown in figure, if slope of line is +ve the also the bounded area can be 5. Let y and y meet at P. 6. Area ( y ) y dy y y y Square units. Area ( ) d ()() a ( ) d ( ) sin sin () Area Square units. 8. a a f ( ) d sin a cos a Apply Leibnitz rule of differentiation a f ( a) a (cos a) sin a sin a f 9. Bounded area 'A' is shown in figure Area () Square units e e A ln( y) dy or ln( e y) dy (By property) [5]
3 e y ln( y) y A Square units A area ( OABC) e d y e e d or e e dy. () () 6 Square units.. Solving the given two curves simultaneously cos cos cos or Area ( ) cos d cos d sin 8. In ABC PA PB PC, min,,. As shown in figure,. 56 ln ln A d f ( ) sin and f ( ) sin Area f ( ) f ( ) d ( sin ) d 8 ln y y e ; Let e A e ln y dy Let point 'P' be (, y) min y, ( ) y, ( ) ( y ) Hence point 'P' lies on the arc of unit radius circles as shown in figure. Area bounded by locus of point 'P' Area of ABC (area of unit radius circle) 6 from graph of ln y A ( e) e A e e [5]
4 y and y 5. Given curves are 7. t dt t dt. t t t t and are the locations for point of intersection. Let the bounded area be 'A' 6. /( ) A d A ( ) A. A. 6 A 6 6 Ama at Area / if R / sin sin cos cos tan tan d tan tan d from graph of 8 Area ln y f ( ), Area ln 8. Let the locus point be P(, y) Now, ( AP) ( BP) ( CP) ( DP) ( EP) ( FP) ( ) y y y Required area Sq. units., 9. Given curve are y and y 5 / tan d tan dt Let tan t d t [5]
5 ( ) A 56 Similarly, Area (A) > Area ( PTUQ) for ; ( ) ( ) d Area ()() ( ).. tan ( ) y sin ; y 8 Sq. units. from graph, Area () Required area Sq. units. y 8y ( y )( y 6) from figure, Area (A) < Area (PQRS) A ()(). A A graph of y log y 6 A log y dy Putting y dy d 8 8 / A log ( ).d ( log ) d 8 / A 6 log ( ) d 8 / A 6 log d / An (tan ) n d n / A A (tan ) ( tan ) d n / n (tan ) (tan ).(sec ) d n An An n If, n n, then tan n, n N, n tan (tan ) A n A Hence, A A A n n n n A n n A n n Similarly, An A A A A n n n n A n n n...(i) [5]
6 A n n from (i) and (ii), A n n n...(iii). ( y ) sin y sin and y sin f ( ) sin and g( ) sin Now, ( ) ( ) A f g d A sin d Putting t d dt from graph, An A n () OAB An An tan 8. f '() ( ) y Equation of tangent is ( ) A, and B, ( ) ( ) Area of AOB Now, ( ) A cos t dt cos t dt A cost dt (By property of even and odd functions) / A cos d. for,, sin sin sin d A d A ( ) Case : ( ) ( 8) ( ) Case : ( ) 8 ( ) 8 is only applicable from case () and (), or from graph, A sin / A (sin ) d. d [5]
7 5. f ( ) R 7. ; (n ) f ( ), where n I ; (n ) Area bounded by f ( ) with -ais 8 d Sq. units. ( ) 6. Area bounded by g() with y ( ) d Sq. units. Area bounded by g() with y ()() Sq. units. Area bounded by g( ) with y( ) d 7 Sq. units. Area ( 6 8) d Statement () and () are true but the eplanation is not appropriate. 8. Statements () and () are true and the reasoning is also correct. 9. y ( y ) ; y Area f ( ) f ( ) d / y y dy Area cos sin Area tan or cot 8 8 Both statements () and () are true and the eplanation is appropriate. ()() f ( ) d d sin ( ) [55]
8 Square units. Statement () is false.. Area d 5 A 7 Sq. units. ln ln As shown in graph, area bounded by y and y can be A or A and neither A nor A lie in between the lines and, hence statement () is false. Area which is calculated as 5 A i. e. 7 is the ln area bounded by y, y and the line. ( ) ( ),, Statement () is true. [56]
9 . n An.tan ( ) d Refer Hints/Solutions of chapter-9 (Eecrcise ()/Q- 9) ( n ) An ( n ) An ( n ). ( n ) An ( n ) An n Putting n = and n = A A...(i) A 5A...(ii) Adding (i) and (ii) 7 A A A 5A 7 ( r ) Ar r. ( n ) An ( n ) An n Putting n and n A A...(i) A 5A...(ii) Subtracting (i) from (ii) 5A A 5A A A 5 (tan ) d tan ln( ) 5 5 ln ln A sin f ( ) d ( a )cos a sin a. a differentiate w.r.t. a sin a f ( a) ( a )( sin a) cos a cos a sin a f ( a) asin a sin a cos a cos a f ( a) a sin a f ( ) sin for point A, sin a a sin a a A ( sin sin ) d ( ) sin d A sin() A sin sin d A 7. A 7 y sin f '() ( y sin) cos sin sin for ( k, ), cos sin k cos k cos sin k tan [57]
10 7. 8. p q lim p q f ( ) Now, ( ) f f R f ( ) f ( ) q q q f ( ) 5 f ( ) g( ) f ( ) from graph it can seen that no of points of discontinuity are ( ) 8 t Let point 'B' be t, ; t t t Area of R t t 9. t Area t t t Maimum Area 5 ; R lim f ( ) f () and lim f ( ) f () lim k k f () k k k k ( k )( k ) k R (, ). d( P, AB) min d( P, BC), d( P, CD), d( P, AD) Let the moving point P be (, y) d( P, AB) y, d( P, BC) ( ) d( P, CD) ( y), d( P, AD) ( ) Case : If min d( P, BC), d( P, CD), d( P, AD) d( P, BC) y and y and Now d( P, AB) d( P, BC) y y according to case () :, y, y Similarly, consider two other cases when d( P, CD) and d( P, AD ) is minimum as shown in figure. Required aes = area of trapezium (EFBA) 6 () 8 Sq. units. [58]
11 . y( a a ) y( a)( a) a A d 6 a A a a a a a a a da a a New, da a a a is location of local minima a for least value of area 'A'.. Tangent to curve at (, ) y ( y ) (, ) y or (y ) Required area ( y ) (y ) dy y 6 y 9 dy ( y ) dy. A ( ) d ( ) d A ( ) d ( ) d. A ( ) y y ( ) ( ) A d d Put t A ( t ) t( t) dt. ( y ) 9 Sq. units. 5 t t (t t ) dt 5 A Sq. units. 5 Least integer greater than A = 5. (a) sin ( ) y y sin ( ) y sin ( ) and [59]
12 (b) area d y sin ( ) d Sq. units ( e e ) d for, ; (sin cos ) (sin cos ) d A d A, (c) f ( ), (c) (d) 8 e e ( ) d e ( ) e d e ( ) = Square units. y y, where Area () Sq. units. 6. (a) f ( ), (b) [, ] d A d A f ( ) ( ), (sin cos ) f ( ) d A d sin A sin A. 6 A, 6 (d) f ( ), 6 for 6, ; d A d ln A sin ( ) ln A A ln, 7. Let point 'P' be (, ), where (, ) area of region ( OPR) f ( ) d y area of region ( OPQ) y dy [6]
13 f d y dy y ( ) differentiate w.r.t. f ( ) ( ) f ( ) f ( ) f ( ) (a) f '( ) ( ) is location of local minima global minima f f 7 (b) area (c) d Sq. units. 6 ; g( ) ; 7 / area d 7 8 Sq. units (d) area d d 8 Sq. units. [6]
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. A tangent to the ellipse is intersected by a b the tangents at the etremities of the major ais at 'P' and 'Q' circle on PQ as diameter always passes through : (a) one fied point two fied points (c) four
C) 2 D) 4 E) 6. ? A) 0 B) 1 C) 1 D) The limit does not exist.
. The asymptotes of the graph of the parametric equations = t, y = t t + are A) =, y = B) = only C) =, y = D) = only E) =, y =. What are the coordinates of the inflection point on the graph of y = ( +
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. f() = 4 cosec 4 +, where is in radians. (a) Show that there is a root α of f () = 0 in the interval [.,.3]. Show that the equation f() = 0 can be written in the form = + sin 4 Use the iterative formula
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ANSWER KEY 1. [A] 2. [C] 3. [B] 4. [B] 5. [C] 6. [A] 7. [B] 8. [C] 9. [A] 10. [A] 11. [D] 12. [A] 13. [D] 14. [C] 15. [B] 16. [C] 17. [D] 18.
![ANSWER KEY 1. [A] 2. [C] 3. [B] 4. [B] 5. [C] 6. [A] 7. [B] 8. [C] 9. [A] 10. [A] 11. [D] 12. [A] 13. [D] 14. [C] 15. [B] 16. [C] 17. [D] 18.](/thumbs/89/97771760.jpg "ANSWER KEY 1. [A] 2. [C] 3. [B] 4. [B] 5. [C] 6. [A] 7. [B] 8. [C] 9. [A] 10. [A] 11. [D] 12. [A] 13. [D] 14. [C] 15. [B] 16. [C] 17. [D] 18.")
ANSWER KEY. [A]. [C]. [B] 4. [B] 5. [C] 6. [A] 7. [B] 8. [C] 9. [A]. [A]. [D]. [A]. [D] 4. [C] 5. [B] 6. [C] 7. [D] 8. [B] 9. [C]. [C]. [D]. [A]. [B] 4. [D] 5. [A] 6. [D] 7. [B] 8. [D] 9. [D]. [B]. [A].
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