This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 67501. |
1.Find the principal values of the following:(0 (a) sin-112in(P.B. 2017) |
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Answer» Explanation : sin30 = 1/2 sin^(-1)(1/2) = 30° Solution : 30° If you find this answer helpful then like it |
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| 67502. |
18-6-7 में से 8-4r+3x-x' को घटा |
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Answer» 14x^4-6x^2+3x^3+15 is your answer |
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| 67503. |
Teacher asked the students, to find tan0 if 7 sin 2e+3Arnav answered it as Is the answer correct? Justify it.V3 |
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| 67504. |
( 8. If 11, 109,999 is divided by 1111, then what is theremainder? |
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Answer» Its clearly see if i add 1 in this number so, it is became a multiple of 1111. => 11,109,999 + 1 = 11,110,000 ; Its clearly see that this is multiple of 1,111. If we add +1 in that number, so we need to subtract -1 from 1,111. Then we will get the perfect remainder. 1,111 -1 = 1110 So, The remainder is 1,110. |
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| 67505. |
16 Subtract 2r-3y +11 from the sum of 2x-3y +7 and -5x -7. |
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| 67506. |
1111प्रश्न चिह्न (?) के स्थान पर दया आसेगाप्रश्न चिह्न (2) के स्थान पर क्या आये |
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Answer» 4 is the right answer 4 is the correct answer |
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| 67507. |
है| ( 005 32 + 005 3 - cos 2x =0बे. क ~— N |
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Answer» cos 3x + cos x - cos 2x = 0[cos 3x + cos x ] - cos 2x = 02 cos [(3x + x)/2] .cos [(3x-x)/2] - cos 2x = 02 cos 2x . cos x - cos 2x = 0cos 2x [ 2 cos x - 1 ] = 0cos 2x = 0 or cos x = 1/2 = cosπ/32x = (2n +1 )π/2 or x = 2mπ +-π/3x = (2n + 1 )π/4 or x = 2mπ +-π/3 ; where n , m∈ I |
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| 67508. |
A hot water system (Geyser) consists of a cylindrical pipe of length 14 m anddiameter 5 cm. Find the total radiating surface of hot water system. |
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| 67509. |
stance of 29 cm from the centre of a circde of radiusA paint d the length of the tangent drawn from P to the dircle.20 cm. FindC8SE 2017) |
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| 67510. |
In fig tangents PO and PR are drawn to a cirele such thattangent PQ. Find RQSRPo-30P. A chord RS is drawn parallel to the030 |
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Answer» PLEASE LIKE IT, IF YOU FIND THE ANSWER HELPFUL. |
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| 67511. |
Similar Triangles213Prove that if the areas of two similar triangles are equal, then they are congrueple-8. Prov |
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| 67512. |
MathematicsDefe |
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Answer» Make both equal to 15 xy, and therefore equal to each other. 20x + 30y = 24x + 27y 3y = 4x you can then substitute this in the first equation 4x+8x = x (4x) make this into a quadratic 4x squared - 12x = 0, the only real integer is 3 (the other is zero). x = 3, and y must therefore be 4. Like my answer if you find it useful! |
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| 67513. |
हा. ८00 + 56८9.८0५६००1+005 0 |
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Answer» Like if you find it useful |
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| 67514. |
6. 1111 e S Dx>0 | - 005 2 |
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Answer» tnx alka |
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| 67515. |
\frac { x } { y } = \frac { a + 2 } { a - 2 } , \text { let us show that } \frac { x ^ { 2 } - y ^ { 2 } } { x ^ { 2 } + y ^ { 2 } } = \frac { 4 a } { a ^ { 2 } + 4 } |
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| 67516. |
बे नर,. Ifcos O + 500 0 न /2 cos 8, prov |
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| 67517. |
Prov '(la sin[)rcÂŤâssb)2 _1-cosd"V N\ TS sinb+cos8 ) 1% coml |
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Answer» LHS = [(1+sinA-cosA )/(1+sinA+cosA)]² =[ (1+sin²A+cos²A+2sinA-2sinAcosA-2cosA)/(1+sin²A+cos²A+2sinA+2sinAcosA+2cosA) ]² = [( 1+1+2sinA-2sinAcosA-2cosA)/(1+1+2sinA+2sinAcosA+2cosA)]² = { ( 2(1+sinA) -2cosA(sinA+1) / [ 2(1+sinA) + 2cosA(sinA +1 ) ]}² = { (2(1+sinA)[1 - cosA] /[2(1+sinA )(1 + cosA )] }² = [ ( 1 - cosA ) / ( 1 + cosA ) ]² = RHS |
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| 67518. |
e that tangent drawn at any point of a circleProvs perpendicular to the radius through the point ofcontact.[4 |
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| 67519. |
i)If sinh x =-, findProv4cosh (2x) and sinh (2x)tan |
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Answer» tq |
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| 67520. |
थञ! 0 - 0059 + 1 ?11 () + 005 9 - | 3 srove that |
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| 67521. |
8. A volume of a cuboidal box is 250 cm3. If its length is 20 cm and height is 5 cm, find its breadth9. A pit 8 m Ă 5 m x 2 m. is dug. The soil dug out is spread evenly on a piece of land which is |
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Answer» If you find this solution helpful, Please like it. |
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| 67522. |
the defeminat S |
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Answer» Determinant is (-5)(4)-(3)(-4)=-20+12=-8(answer) |
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| 67523. |
Fld ul&. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is5 cm. Find the outer curved surface area of the bowl. |
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| 67524. |
10. SĂmplify each of the following products:(m+n)'(m-7)IT(i) (2a-3b)(3b|a)(4a2+9b2)(ii) |
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| 67525. |
A Volume of a cuboidal box is 250 cm3 If it's length is 20 cm and height is 5 cm find its breadth |
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| 67526. |
-3(a + b) +4(2a-3b)-(2a-b) |
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Answer» -3(a+b) + 4 ( 2a - 3b) - (2a-b)-3a - 3b + 8a - 12b - 2a + b-3a + 8a - 2a - 3b - 12b + b 8a - 5a - 15b + b3a - 14b TQ |
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| 67527. |
5. गुणनखन कीजिए।।।।। । ।(ii)?। ।6. निम्नलिखित धनों का प्रसारित रूप में लिखा:(ii) (2 3 )।2।।। (ii) (102)7. उपयुक्त सर्वसमिकाएँ प्रयोग करके निम्नलिखित के मान ज्ञात कीजिए:(iii) (998)(।) (0)8. निम्नलिखित में से प्रत्येक का गुणनखंडन कीजिए:(ii) 8' ' 12 + 6ab(iv) (54727144ub+1012/y (Nah(1) 8/1.35 - 225 ।(101) 27 125(५) 27212} {"{(11))3सत्यापित कीजिए: (1) । ( ।) (10. निम्नलिखित में से प्रत्येक का गुणनखंडन कीजिए:9.(11) (04/ 345)(1) 2712-125संकेत: देखिए प्रश्न !11. गुणनखंडन कीजिए: 27 ।। ११ ।1) + (-२)। । । । :) (12. सत्यापित कॉजिए: ।। - ३४५८- () हा, ता दिखाइए कि । 1 + = 3.81 है।13. यदि ४।14. वास्तव में धनों का परिकलन किए बिना निम्नलिखित में से प्रत्येक काम(ii) (28) + (15) । (13)(i) (12) (7) (5)रातों जिनमें उनके क्षेत्रफल दिए गए हैं, में से प्रत्येक के |
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Answer» 3znx sms+₹ }|÷|×#×|(. |××| |
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| 67528. |
स्व डे- कि. % 20-3b+4C 7gL P उस! |
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Answer» P= 2 will be the answer |
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| 67529. |
1 सें 20 तक के प्रशतों1. |] ए+ 04 |
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| 67530. |
abIf 20 3b - 6 then prove that c+ b |
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Answer» thanks |
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| 67531. |
1 3x+4y_3x 4y, let us show that yevX(1) If_3u +4v3u-4y!! |
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Answer» 3x+4y/3u+4v = 3x-4y/3u-4v; (3x+4y)(3u-4v)=(3x-4y)(3u+4v); (9xu-12xv+12yu-16vy)=(9ux-12xv-12uy-16vy)=-24xu+24th = 24xu=24yv= xu=uy; x/y=u/v |
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| 67532. |
HTTRĺŻ3x + 4y 26+ 4y-26 |
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| 67533. |
cosd 50 # 0 #+ 005 /I-tanA 1-04 |
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Answer» Please hit the like button |
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| 67534. |
(A)x+ 3y(C) x -4y(B) x+ 4y(D) 2x -3y |
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Answer» option c is the answer |
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| 67535. |
7x+4y-12=0 ,3x+4y+4=0 |
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| 67536. |
ST T T S MR et(X) sin3A +cos2B=2 20T cos 2A +sin 3B G शान कह १ _53k |
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Answer» Answer:1/2Step-by-step explanation:Here, sin 3A + cos 2B = 2 Let sin 3A = cos 2B = 1We know that, sin 90 = 1 Here, sin 3A = 1 ∴ 3A = 90 A = 90 ÷ 3 = 30 We get A = 30 here. Let's find B. We know that, cos 0 = 1 Here, cos 2B = 1 ∴ 2B = 0 B = 0 ÷ 2 = 0 We get B = 0. Now, cos 2A + sin 3B = cos 2 × 30 + sin 3 × 0 = cos 60 + sin 0 = 1/2 + 0 = 1/2 |
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| 67537. |
) (3b + 2a)(3b + 2a) |
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Answer» (3b+2a)(3b+2a)(3b+2a)² (a+b)²=a²+b²+2ab (3b+2a)²=9b²+4a²+12ab |
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| 67538. |
B= 203 Sin 3B + 2005 (28+5)ocas 38 - sus (9B-100) |
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| 67539. |
(3b + 2a)(3b + 2a |
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Answer» PLEASE LIKE IT, IF YOU FIND THE ANSWER HELPFUL. |
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| 67540. |
A Basket ball leam has lost otGames and has won 13 Grame104 the |
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Answer» Game lost= 7Game won=13Total game played=13+7=20∴Percentage of game did they win=(13/20)×100=65% |
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| 67541. |
6 (2a + 3b)? â 8(2a + 3b) |
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| 67542. |
3x (a - 3b) - y (a-3b) |
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Answer» taking (a - 3b ) common(3x - y ) ( a - 3b) |
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| 67543. |
(2x+ 3y) (2a -3b) + (3r-4y) (2a -3b) |
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Answer» Take (2a-3b) as common:- =(2a-3b){(2x+3y)+(3x-4y)}=(2a-3b){2x +3y+3x-4y}=(2a-3b)(5x-y) |
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| 67544. |
to te oluameter ot baller |
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| 67545. |
of I m long tube if the density of the metal be 7.8 gm/cm6. The radius & height of a cylinder are in the rati5:7 & its volume is 550 cm3. Find its radius.(Ď-22/7)nd the cost of plastering the inner curved surface at Rs 3 per square meter. A glass cylindetr& İts external radius is 9cm.. How many cubic meters of earth must be dugout to sink a well 22.5 m deep & of diameter 7m?with diameter 20cm has water to a height of 9 cm.1. |
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| 67546. |
A tree breaks due to storm and the brouches t. The distancthe foot of the tree to the point whettouches the ground is 8 m. Find thehebends so that the top of the tree tomaking an angle 30° with it |
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| 67547. |
sin (A + 3B)+sin (3A + B) 4 -sin 2A +sin 2B(a) 2 cos (A + B) (b) 2 sin (A-B)(c) 2 sin (A + B) (d) 2 cos (A-B) |
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Answer» Answer:a)2cos (A+B) Explanation :{sin(A +3B) + sin(3A +B)}/{sin2A +sin2B} use formula,sinA + sinB = 2sin(A+B)/2.cos(A-B)/2 sin2A = 2sinA.cosA cos(-a) = cosa now, {2sin(A + 3B + 3A + B)/2.cos(A +3B-3A-B)/2 }/{2sin(A + B).cos(A -B)} => sin(2A +2B).cos(B - A)/sin(A + B).cos(A -B) =>2sin(A +B).cos(A + B).cos(B -A)/sin(A + B).cos(A - B) => 2cos( A + B) option a is the right answer |
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| 67548. |
cos A0. Im and sin BdCOSAthen, prove that (m2 + n2)cos Bn then, prove that (7722 + n 2) cos2 B = n2 s=cos B |
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Answer» Like my answer if you find it useful! |
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| 67549. |
27.Acylindricalglass of radius 7 cm is filled with mango juice upto a height of 17 em. Whenrefully observed, the bottom of the glass is covered by a transparent cone of height 6m ndadius 7 cm.What is the actual volume of the juice ? How much juice the seller is saving?In which category, you put the seller(a) dishonest(b) cheat(c) not fair trade practice(d) all of the above |
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Answer» Given:cylinder:radius=7 cmheight=17 cm cone:height= 6 cmradius= 7 cm solution:volume of cylinder=22/7 *r^2 *h=22/7 *7*7* 17=2618 cm^3volume of cone= 1/3 * 22/7 * r^2 *h= 1/3 *22/7 * 7*7*6= 308 cm^3total volume of juice in the glass = 2618-308=2310 cm^3= 2.31 litresin each glass the seller will save 0.31 liters of juice. Like my answer if you find it useful! thanks |
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| 67550. |
15. Prove that2兀+ cos2 7π =2.8cos*-+ cos"-+cos2兀3x (兀8 |
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