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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
23. Two numbers are selected at random (withoutreplacement) from the first five positive integers. LetX denote the larger of the two numbers obtained.Find the mean and variance of X |
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Answer» Let X denote the larger of two numbers The two positive integers can be selected from the first six positive integers without replacement in 5x4 = 20 ways X represents the larger of the two numbers obtained. Therefore, X can take the value of 2, 3, 4, 5 For X = 2, the possible observations are (1, 2) and (2, 1).P(X=2) = 2/20 For X = 3, the possible observations are (1, 3), (2, 3), (3, 1), and (3, 2).P(X=3) = 4/20 For X = 4, the possible observations are (1, 4), (2, 4), (3, 4), (4, 3), (4, 2), and (4, 1).P(X=4) = 6/20 For X = 5, the possible observations are (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5, 2), and (5, 1).P(X=5) = 8/20 Mean[E(X)] = 2*2/20 + 3*4/20 + 4*6/20 + 5*8/20= 4/20 + 12/20 + 24/20 + 40/20= 80/20 = 4 E(X^2) = 2^2 *2/20 + 3^2*4/20 + 4^2*6/20 + 5^2*8/20= 8/20 + 36/20 + 96/20 + 200/20= 340/20 = 17 Variance = E(X^2) - [E(X)]^2= 17 - 16= 1 your answer is wrong mean come 4 and var... come 1 |
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| 2. |
In an A.P.of 50 terms,the sum of first 10 terms is 210 and sum of its last 15 terms is 2565. Find the A.P. |
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Answer» Consider a and d as the first term and the common difference of an A.P. respectively. n th term of an A.P., an= a + ( n – 1)d Sum of n terms of an A.P., S n = n/ 2 [2a + (n – 1)d] Given that the sum of the first 10 terms is 210. ⇒ 10 / 2 [2a + 9d ] = 210 ⇒ 5[ 2a + 9 d ] = 210 ⇒2a + 9d = 42 ----------- (1) 15 th term from the last = ( 50 – 15 + 1 ) th = 36 th term from the beginning ⇒ a36= a + 35d Sum of the last 15 terms = 15/2 [2a36+ ( 15 – 1)d ] = 2565 ⇒ 15 / 2 [ 2(a + 35d) + 14d ] = 2565 ⇒ 15 [ a + 35d + 7d ] = 2565 ⇒a + 42d = 171 ----------(2) From (1) and (2), we have d = 4 and a = 3. Therefore, the terms in A.P. are 3, 7, 11, 15 . . . and 199. |
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| 3. |
If an AP of 50 terms ,the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565 find the AP |
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Answer» Consider a and d as the first term and the common difference of an A.P. respectively. n th term of an A.P., an= a + ( n – 1)d Sum of n terms of an A.P., S n = n/ 2 [2a + (n – 1)d] Given that the sum of the first 10 terms is 210. ⇒ 10 / 2 [2a + 9d ] = 210 ⇒ 5[ 2a + 9 d ] = 210 ⇒2a + 9d = 42 ----------- (1) 15 th term from the last = ( 50 – 15 + 1 ) th = 36 th term from the beginning ⇒ a36= a + 35d Sum of the last 15 terms = 15/2 [2a36+ ( 15 – 1)d ] = 2565 ⇒ 15 / 2 [ 2(a + 35d) + 14d ] = 2565 ⇒ 15 [ a + 35d + 7d ] = 2565 ⇒a + 42d = 171 ----------(2) From (1) and (2), we have d = 4 and a = 3. Therefore, the terms in A.P. are 3, 7, 11, 15 . . . and 199 Thanks Where r u from |
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| 4. |
9.If sin (x-20)" = cos (3x-10), then find the value of x. |
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Answer» x-20=90-(3x-10)=>x-20=90-3x+10=>4x=120°=>x=30° |
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| 5. |
If Pa II ST, ZPQR-110 and ZRSM-130, then ZQRS is equal to(A) 40°(B 60°(C) 70°(D) 80°18.130110' |
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| 6. |
25. In an AP of 50 terms, the sum of first 10 terns is 210 and the sum of its last 15 terms is 2565.Find the A.P |
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Answer» Like if it is useful |
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| 7. |
e, the line segment ST is parallel to side PR of APOR and ia divides the triangle into two partsof equal area. Find the ratioPO |
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| 8. |
e that the length of tangents drawn from an externalSt to a circle are equal. Making use of the above, provethe following:m an external point P two tangents PA and PB are drawnto a circle with centre O as shown in figure. Show that OP isperpendicular bisector of AB. |
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| 9. |
In figure, the line segment ST is parallel to side PR of APQR and it divides the triangle into two partsPSPQof equal area. Find the ratio |
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Answer» Step-by-step explanation:Sol: In ΔABC, XY || AC and area of ΔBXY = area of quadrilateral XYCA ⇒ ar (ΔABC) = 2.ar (ΔBXY) ----------------(1) XY || AC and BA is a transversal. ⇒ ∠BXY = ∠BAC --------------------------- (2) ΔBAC and ΔBXY, ∠XBY = ∠ABC (common angle) ∠BXY = ∠BAC [from equation (2)] ⇒ ΔBAC ~ ΔBXY ⇒ ar(ΔBAC) / ar(ΔBXY) = BA2 / BX2 ⇒ BA = √2 BX ⇒ BA = √2 (BA – AX) ⇒ (√2 – 1) BA = √2 AX ⇒ AX/XB = (√2 – 1) / √2 |
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| 10. |
Find the values:\lim _{x \rightarrow 1}\left[\frac{1}{\log x}-\frac{1}{x-1}\right] |
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| 11. |
The potentiometer wire of length 200 cm has a resistance of 2000.it is connected in series with aance ofion and a source cell of emf 6 V having negligible internal resistance .a cell of 2.4 V is balancedst lenth of the potentiometer wire.find the length I. R2104 1.6HEVORT |
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| 12. |
29. 1gret?30. A cord of length 126m has been cut into 46 pieces of equal length. What is the lengthof each piece?75 |
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Answer» 252.5 is the length of thd cord46 piece so252.5/46=5.48 |
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| 13. |
4. As shown in the figure, four napkins all of the samesize were made from a square piece of cloth of length1 m. What length of lace will be required to trim allfour sides of all the napkins ? &m |
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| 14. |
\lim _{x \rightarrow 0} \frac{x^{2}-log(1+x)}{x^{2}} |
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Answer» given limit is in 0/0 form use L'Hopitals {e^x + xe^x - 1/(1+x)}lim --------------------------------------- 2xAgain 0/0 form lim (e^x -1) --------------- = 0 2 Ans : 0 can you please explain me which part you don't understand you apply here LH rule can you please explain how to apply |
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| 15. |
\lim _ { x \rightarrow 0 } \frac { \log x ^ { 2 } } { 6 + x ^ { 2 } }\\ { \ text { by L Hospital's rule}} |
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| 16. |
\lim _{x \rightarrow 0} \frac{\log (1+x)}{3^{x}-1} |
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| 17. |
\operatorname { lim } _ { x \rightarrow 0 } \frac { x \operatorname { cos } x - \operatorname { log } ( 1 + x ) } { x ^ { 2 } } |
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Answer» nice solution |
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| 18. |
\lim _ { x \rightarrow 0 } \frac { \sin 3 x } { \log _ { e } ( 1 + 7 x ) } |
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Answer» if we substitute x=0 we will get 0/0 form sk using L'hospital rule lim x->0 3cos3x/(7/(1+7x)) = 3/7 any easy methods? |
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| 19. |
\lim _ { x \rightarrow 0 } \left[ \frac { \log ( 1 + x ) } { e ^ { 2 x } - 1 } \right] |
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| 20. |
OA and ONow using a protractor verify that ZAOB LAOC+and take a ray OC in such a way that ray OC lies between rayshich direction will you face if you start facing north and make(a)รทofa revolution clockwise? (b)-ofarev olution clockwse5. Where will the hand of a clock ston if |
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Answer» (a) East(b) South(c) North |
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| 21. |
1. In the adjoining figure, a circle touches all thefour sides of a quadrilateral ABCD whose sidesare AB-6 cm, BC- 9 cm and CD 8 cm. FindICBSE 20111the length of side AD |
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| 22. |
Find ZAOB in the given figure.30° |
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Answer» The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre. ∠AOB = 180 - 30 = 150° |
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| 23. |
6Find ZAOB in the given figure.30° |
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Answer» We know angle AOB + angle ADB = 180°So, angle AOB = 180° - 30° = 150° |
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| 24. |
22X2. |
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| 25. |
Example 3. Iftan6-2x2x+1)thenfind the values ofsin θ and cos θ |
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Answer» tanA = 2x(x+1) / 2x+1 ---- givensec^2 A = 1+ tan^2 A --- identity. sec^2 A = 1 + [4x^2(x+1)^2] / (2x+1)^2 = [(2x+1)^2 + 4x^2(x+1)^2] / (2x+1)^2= [4x^4 + 8x^3 + 8x^2 + 4x +1 ] / (2x+1)^2= (2x^2 + 2x + 1)^2 / (2x+1)^2 This gives --secA = (2x^2 + 2x + 1) / (2x + 1)So cosA = (2x + 1) / (2x^2 + 2x + 1) Now sin^2 A = (1 - cos^2 A ) = 1 - (2x+1)^2 / (2x^2 + 2x + 1)^2= [(2x^2 + 2x + 1)^2 - (2x+1)^2] / (2x^2 + 2x + 1)^2= (2x^2)*(2x^2 + 4x + 2 ) / (2x^2 + 2x + 1)^2 ---- using [a^2-b^2 =(a+b)*(a-b) ]= (4x^2)*(x^2+2x+1) / (2x^2 + 2x + 1)^2 = [(2x )* (x+1)]^2 / (2x^2 + 2x + 1)^2This gives sinA = (2x )* (x+1) / (2x^2 +2x +1) |
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| 26. |
6Find ZAOB in the given figure.309 |
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| 27. |
In the figure, O' is the centre of the circle.ZAOB = 100find /ADB. /°了100° |
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Answer» Angle AOB=100angle ACB=50 (angle made at centre is double the angle made at any other point) now,ACBD is a cyclic quadrilateral , so angle ADB + angle ACB=180 ( sum of opposite angles of a cyclic quad. is 180) therefore,angle ADB=180-50=130 |
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| 28. |
In the given figure DA丄OP and DB丄OQ and 2ADB = 170° find ZAOB |
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Answer» since DA and DB are _|_ so, angle A and B are = 90° now in Quadrilateral ADBO => 90+90+170° + angle AOB = 360°=> Angle AOB = 360-(90+90+170) = 10° |
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| 29. |
a7¡MAT (Find) lim |
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Answer» please tell if there is any doubt |
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| 30. |
(4x +3)4. If f(x) = (6x-4)**show that fof(x) = x, for all x*-. What is theinverse off? |
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| 31. |
2019] प्रश्न 1. यदि f(x) = 4x +3,x2 तो सिद्ध कीजिए कि सभीx2 के लिए fof (x) = x है। f का प्रतिलोम भी ज्ञातकीजिए। |
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Answer» ghvsvgs to you as soon as possible and Android apps development Android app and Android apps development Android app and wanted to you as soon love to you as well so I can download the to you as soon as possible vvvsv when the watch man Durvesh ko De hai San Francisco and I am to you very good app it to you as a result the |
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| 32. |
के.1 T७ B S l_,%x e e D 78 A |
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Answer» We know that(a^3-b^3)=(a-b)(a^2+ab+b^2)---(1)thena^2+b^2/ab= -1a^2+b^2= -abput this value in equation (1) Answer will be 0 |
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| 33. |
L2s log xFind Lim x |
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| 34. |
4. यदि नहुपदीय f(x) = ax?शून्य 2 है, तो 8 का मान है %® -%e %e नहीं a-1)x-1HTH _ |
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Answer» Put in X=2f(X) should be 0a(2)^2-3(a-1)*4-1=4a-3a+12-3=0a+9=0a=-9hence option d |
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| 35. |
\begin { equation } 6 \sin ^{2} \theta-7 \sin \theta+2=0 \end { equation } |
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| 36. |
Find the sum of the first 25 terms of the geometric series16-48 144 -432+... |
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Answer» here a = 16 r = -48/16 = -3 and n = 25 termsso S25 = 16(1-(-3)^25)/1-(-3) = 4(1+3^25) so sum will be 3389154437776. its wrong just now checked its correct think u |
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| 37. |
) {8 X~271 9 e) करन |
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| 38. |
3x +1 3x +22x 2 +1 |
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Answer» (3x+1)/(2x) = (3x+2)/(2x+1) (3x+1)(2x+1) = (3x+1)(2x) 6x*x + 5x + 1 = 6x*x + 2x 3x = -1 x = -1/3 If you find this answer helpful then like it. |
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039. The roots of the quadratic equation 2x2X-6 = 0 are |
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| 40. |
25. Find the number of terms of the AP 18,15¡13,-49and find the sum of all its terms.ă |
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| 41. |
) Iff(x)=14+x+X न जद thenfind {' () |
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| 42. |
find the sum of the first 25 terms of the geometric series 16-48+144-432+.... |
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| 43. |
7. In an AP. ifS S 167 and S235, then find the sum of first 25 terms |
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Answer» Sn = n/2[2a + (n-1)d]S5 + S7 = 1675/2(2a + 4d) + 7/2(2a + 6d) = 16724a + 62d = 33412a + 31d = 167......... (1) S10 = 23510/2(2a + 9d) = 2352a + 9d = 47........... (2) Multiply eq(2) by 612a + 54d = 282......(3) Subtract eq(1) from eq(3)23d = 115d = 115/23 = 5 Put value of d in eq(1)12a + 31*5 = 16712a = 167 - 155a = 12/12 = 1 Then,AP is a, a+d, a+2d, a + 3d......1, 6, 11, 16, 21...... |
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| 44. |
AMPLE Iff (x) is differentiable at xa, find limxma-x-a |
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Answer» since f(x) is differentiable so, f'(x) exists.. now the limit is 0/0 so, using L.H rule lim(x→a) [x²f(a)-a²f(x)]/(x-a) so, = [2xf(a) - a²f'(x)]/1 = 2af(a) - a²f'(a) so, the value is. a[2f(a)-af'(a)] |
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| 45. |
(a) Iff(x) = 2x -1 when x < 0= x2 when x > 0then find $(15). |
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Answer» f(1/2) = (1/2)² = 1/4 |
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| 46. |
4x+3,,2, show that (fof)(x)Iff(x) = 6-4,_3,for all x#2. What is the inverse of3 |
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| 47. |
Q. 11)Iff(x)=sin 5x + afor x>0;4x=x+5-2bfor x<0tor x=0;:find a and b |
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Answer» value of function f(x) at x > 0 is lim x→0 sin5x/4x +a = 5/4 +a at x <0 is 5-2b and at x = 0 is 2 so, 5-2b = 2 => 2b= 5-2 => b = 3/2 also, 5/4+a = 2 => a = 2-5/4 = 3/4. |
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| 48. |
2Check whether (x+ 3)-x'-8 is a quadratic equation ? |
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| 49. |
Check whether (2,-2) is a solution of 2x-y- 6 |
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Answer» If (2,-2) is a solution of 2x - y = 6 Then for x = 2, y = - 2 given equation should be equal to 0 Thus,2*2 - (-2) - 6= 4 + 2 - 6= 6-6= 0 Therefore (2,-2) is solution of given equation |
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| 50. |
Check whether (x-2) is a factor of thepolynomial xx42) |
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Answer» If (x-2) is the factor of the polynomial then x=2 should satisfy the given equation,(2)^3-(2)^2-4= 8-4-4 =0So it's a factor of the polynomial. |
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