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MATHS- 11th (PQRS & J) Partial Marking.pdf

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DPP. NO.-37 Select the correct alternative : (Only one is correct) Q.145/st.line The lines y = mx + b and y = bx + m intersect at the point (m – b, 9). The sum of the x-intercepts of the lines is 41 (A) 9 (B*) – 20 (D) can not be computed as data is insufficient [Sol. mx + b = bx + m 41 (C) 20 (m – b)x = m – b x = 1 (m b since for m = b, x coordinate will become zero) m – b = 1 (as x coordinate of the point of intersection of the lines is m – b) x = 1 y = b + m = 9 b + 1 + b = 9 (m = 1 + b) b = 4 and m = 5 sum of x-intercepts of lines is b + m 4 + 5 41 – b = – 4 = – 20 Ans. ] Q.213/st.line TS is the perpendicular bisector of AB with coordinate of A (0, 4) and B(p, 6) and the point S lies on the x-axis. If x-coordinate of S is an integer then the number of integral values of 'p' is (A) 0 (B) 1 (C) 2 (D*) 4 2 [Hint: mAB = p p ; mST = – 2 p ; mid point of AB = 2 , 5 p x p equation of TS y – 5 = – 2 10 p put y = 0 x = p + 2 Hence p = ± 2, ± 10 for x to be an integer ] Q.375/st.line The line x = c cuts the triangle with corners (0, 0); (1, 1) and (9, 1) into two regions. For the area of the two regions to be the same c must be equal to (A) 5/2 (B*) 3 (C) 7/2 (D) 3 or 15 1 9 [Sol. 2 · · c c 1 1 1 9 1 1 1 1 = 1 1 1 c 9 1 2 0 0 1 c = 3 Ans. ] SUBJECTIVE: Q.4 Find the roots of the equation (x – 3)3 + (x – 7)3 = (2x – 10)3. [Sol. Let A = x – 3; B = x – 7; C = 10 – 2x the equation can be written as (x – 3)3 + (x – 7)3 + (10 – 2x)3 = 0 A + B + C = 0 A3 + B3 + C3 = 3 ABC 3(x – 3)(x – 7)(10 – 2x) = 0 x = 3 or 5 or 7 Ans. ] [Ans. x = 3 or 5 or 7] Q.5174/5 Suppose that r1 r2 and r1r2 = 2 (r1 , r2 need not be real). If r1 and r2 are the roots of the biquadratic 1 i 7 x4 – x3 + ax2 – 8x – 8 = 0 find r , r and a. [Ans. r , r = 1 2 1 2 and a = – 4] [Sol. Since r1r2 = 2, r1 x2 + px + 2 = 0 r2 and r1r2r3r4 = – 8 r3r4 = – 4 x4 – x3 + ax2 – 8x – 8 = (x2 + px + 2)(x2 + qx – 4) compare coefficient of x3 and x p + q = – 1 (1) and 2q – 4p = – 8 q – 2p = – 4 (2) p = 1 and q = – 2 on comparing coefficient of x2; a = – 4 p = 1 x2 + x + 2 = 0 r1, 2 = 1 i 7 2 Ans. ] 11 Q.6 If tan = 2 where (0, /2). Find the value of cos ( 3) and sin ( 3). 11 0, [Sol. Given tan = ; 2 2 , 0 < 3 < 6 [Ans. sin 3 = ; cos = ] hence 3 tan( 3) tan3( 3) 11 1 3 tan2 ( 3) = 2 let tan ( 3) = t, we get the cubic as 2t3 – 33t2 – 6t + 11 = 0 t3(2t – 1) – 16t(2t – 1)) – 11(2t – 1) = 0 (2t – 1)(t2 – 16t – 11) = 0 t = 1/2 or t = 8 ± 10 tan( 3) = 1/2 (rejected as the value of would lie in the 2nd quadrant) sin = 3 and cos = Ans. ] Q.1102/2 The interior angle bisector of angle A for the triangle ABC whose coordinates of the vertices are A (–8, 5) ; B(–15, –19) and C(1, – 7) has the equation ax + 2y + c = 0. Find 'a' an

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- CLASS : XI (J-Batch) TIME : 40 Min. DPP. NO.-37 Select the correct alternative : (Only one is correct) Q.145/st.line The lines y= mx + b and y= bx + m intersect at the point (m – b, 9). The sum of the x-intercepts of thelinesis (A) 9 (B*) – 20 41 (C) 20 41 (D)cannot be computed as data is insufficient [Sol. mx + b = bx + m (m – b)x = m – b x = 1 (m b since for m = b, x coordinate will become zero) m – b = 1 (as x coordinate of the point of intersection of the lines is m – b) x = 1 y = b + m = 9 b + 1 + b = 9 (m = 1 + b) b = 4 and m = 5 sumofx-interceptsof linesis – b m m b = – 4 5 5 4 = – 20 41 Ans. ] Q.213/st.line TS is the perpendicular bisector ofABwith coordinate ofA(0, 4) and B(p, 6) and the point S lies on the x-axis. Ifx-coordinate of S is aninteger then the number ofintegral values of 'p'is (A) 0 (B) 1 (C) 2 (D*) 4 [Hint: mAB = p 2 ; mST = – 2 p ; mid point ofAB = 2 p , 5 equationofTS y – 5 = – 2 p 2 p x put y = 0 x = p 10 + 2 p Hence p = ± 2, ± 10 for x to be an integer ] Q.375/st.line The line x = c cuts the triangle with corners (0, 0); (1, 1) and (9, 1) into two regions. For the area of the two regions to be the same c must be equal to (A) 5/2 (B*) 3 (C) 7/2 (D) 3 or 15 [Sol. 2 · 2 1 · 1 9 c c 1 1 c 1 1 9 = 2 1 1 0 0 1 1 1 1 1 9 c = 3 Ans. ] SUBJECTIVE: Q.4 Find the roots of the equation (x – 3)3 + (x – 7)3 = (2x – 10)3. [Sol. Let A = x – 3; B = x – 7; C = 10 – 2x [Ans. x = 3 or 5 or 7] the equation canbe written as (x – 3)3 + (x – 7)3 + (10 – 2x)3 = 0 A + B + C = 0 A3 + B3 + C3 = 3 ABC 3(x – 3)(x – 7)(10 – 2x) = 0 x = 3 or 5 or 7 Ans. ]
- Q.5174/5 Suppose that r1 r2 and r1r2 = 2 (r1 , r2 need not be real). If r1 and r2 are the roots of the biquadratic x4 – x3 + ax2 – 8x – 8 = 0 find r1, r2 and a. [Ans. r1, r2 = 2 7 i 1 and a = – 4] [Sol. Since r1r2 = 2, x2 + px + 2 = 0 2 1 r r and r1r2r3r4 = – 8 r3r4 = – 4 x4 – x3 + ax2 – 8x – 8 = (x2 + px + 2)(x2 + qx – 4) comparecoefficient of x3 and x p + q = – 1 .....(1) and 2q – 4p = – 8 q – 2p = – 4 ....(2) p = 1 and q = – 2 on comparing coefficient of x2; a = – 4 p = 1 x2 + x + 2 = 0 r1, 2 = 2 7 i 1 Ans. ] Q.6 If tan = 2 11 where (0, /2). Find the value of cos 3 and sin 3 . [Sol. Given tan = 2 11 ; 2 , 0 , 0 < 3 < 6 [Ans. sin 3 = 5 1 ; cos = 5 2 ] hence ) 3 ( tan 3 1 ) 3 ( tan ) 3 tan( 3 2 3 = 2 11 let tan ) 3 ( = t, we get the cubic as 2t3 – 33t2 – 6t + 11 = 0 t3(2t – 1) – 16t(2t – 1)) – 11(2t – 1) = 0 (2t – 1)(t2 – 16t – 11) = 0 t = 1/2 or t = 8 ± 10 3 (rejected as the valueof would liein the 2nd quadrant) tan ) 3 ( = 1/2 sin 3 = 5 1 and cos = 5 2 Ans. ]
- CLASS : XI (J-Batch) TIME : 45 Min. DPP. NO.-38 Q.1102/2 The interior angle bisector of angle A for the triangle ABC whose coordinates of the vertices are A (–8, 5) ; B(–15, –19) and C(1, – 7) has the equation ax + 2y + c = 0. Find 'a' and 'c'. [Ans. a = 11 , c = 78] [Sol. AB = 576 49 = 25 AC = 144 81 = 15 coordinate of the point D 8 ) 5 45 ( , 8 ) 35 57 ( 2 23 , 5 equationofAD: y – 5 = 5 8 2 23 5 (x + 8) –3(y – 5) = 2 33 (x + 8) – 6y + 30 = 33x + 264 33x + 6y + 234 = 0 ....(1) given ax + 2y + c = 0 ....(2) comparing(1)and (2) 33 a = 6 2 = 234 c a = 11 ; c = 3 234 = 78 a = 11 and c = 78 ] Q.2110/2 Find the combined equation of the pair of lines through (2, 3) perpendicular to the lines 3x2 – 8xy + 5y2 = 0. [Ans. 5x2 + 8xy + 3y2 – 44x – 34y + 95 = 0] [Sol. 3x2 – 8xy + 5y2 = 0 ....(1) Let m x y 5m2 – 8m + 3 = 0 5m2 – 5m – 3m + 3 = 0 m = 1, 5 3 nowequationof straight lineperpendicularto (1) and passingthrough (2, 3) y – 3 = – 1 (x – 2) x + y – 5 = 0 ....(2) y – 3 = – 3 5 (x – 2) 3y – 9 = – 5x + 10 5x + 3y – 19 = 0 ....(3) thenequationofrequiredline (x + y – 5) (5x + 3y – 19) = 0 5x2 + 8xy + 3y2 – 44x – 34y + 95 = 0 ]
- Q.3111/2 Each sideof a squareisof length 4. Thecentre of the squareis (3, 7) andone of its diagonals is parallel to y= x. Find the coordinates of its vertices. [Ans. (1, 5) ; (1, 9) ; (5, 9) ; (5, 5) ] [Sol. Let lineAC is parallel to y= x AC2 = 16 + 16 AC = 2 4 sin 7 y cos 3 x = 2 2 x = 3 + 2 2 cos ( tan = 1, then sin = 2 1 , cos = 2 1 ) y= 7 + 2 2 sin x = 3 + 2 = 5 ; y = 9 C (5, 9) To findcoordinates ofA x = 3 – 2 2 cos x = 3 – 2 = 1 y= 7 – 2 2 sin y = 7 – 2 = 5 A (1, 5) Then B (5, 5) and D (1, 9) ] Q.4116/2Avariable straight linewhoselengthis Cmoves in such awaythat one ofits end lies on the x-axis and the otheron the y-axis. Show that the locus of the feet of theperpendicular from origin on the variable linehastheequation (x2 + y2)3 = C2x2y2 [Sol. equationofAB y – k = ) h x ( k h hx + ky = h2 + k2 ........(1) OA = h k h 2 2 OB = k k h 2 2 Hence 2 2 2 2 2 2 2 C k k h h k h (h2 + K2)2 2 2 2 2 k h k h = C2 (h2 + K2)3 = C2h2K2 or (x2 + y2)3 = C2x2y2 ] Q.5120/2 Arayof light is sent along the line x 2y 3 = 0. Upon reaching the line 3x 2y 5 = 0, the rayis reflectedfromit.Find theequationofthelinecontainingthereflected ray. [Ans: 29x 2y= 31] [ REE ’90, 6+6 ]
- [Sol: Equation of the reflected rayBP can be takens as x – 2y – 3 + (3x – 2y – 5) = 0 ....(1) Now perpendicular from Q (x1, y1) are (1) and x – 2y – 3 = 0 should be equal 2 2 1 1 1 1 1 1 ) 1 ( 4 ) 1 3 ( ) 5 y 2 x 3 ( 3 y 2 x 5 3 y 2 x or (3 + 1)2 + 4 ( + 1)2 = 5] 92 + 1 + 6 + 42 + 8 + 4 = 5 132 + 14 = 0 = – 13 14 (as 0) requiredequationis x – 2y – 3 – 13 14 (3x – 2y – 5) = 0 13 x – 26 y – 39 = 42 x – 28 y – 70 29 x – 2 y = 31 Ans.] Q.6136/2 Find the equations of the sides of a square whose each side is of length 4 units and centre is (1, 1) . Given that one pair of sides is parallel to 3 x 4y = 0 . [ Ans. : 3 x 4y + 11 = 0 ; 3 x 4 y 9 = 0 ; 4 x + 3 y + 3 = 0 ; 4 x + 3 y 17 = 0 ]

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