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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.
| 28601. |
Prove that:cos1°cos2°cos 3°...........cos180°=0 |
| Answer» If it is series ,and are in multiplicationThere cos90°will also be presentAs cos90°=0 therefore whole ans convert into 0 Because any number mutiplied by 0 is 0? | |
| 28602. |
What is value of cos 67 - sin 23 |
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Answer» 0 0 0 Sin(90 - 67 )- sin 23 ==== sin 23 - sin 23 ====0 |
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| 28603. |
Find hcf of 245 & 285 |
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Answer» 245 = 5 × 7 × 7285 = 3 × 5 × 19Therefore, HCF = 5 HCF(245,285)=5 |
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| 28604. |
P is a rational number prove |
| Answer» | |
| 28605. |
If n is odd integer then show that n^2-1 is divisible by 8 |
| Answer» Let n = 4q + 1 (an odd integer){tex}\\therefore \\quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \\quad \\text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}{tex}= 16{q^2} + 8q{/tex}{tex}= 8 \\left( 2 q ^ { 2 } + q \\right){/tex}= 8m, which is divisible by 8. | |
| 28606. |
Prove that (√5) is an irrational number |
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Answer» Let √5be a rational number say p/q where q is not equal to zero and p and q are coprime.√5=p/qSquaring both side5=p²/q²5q²=p²----------(1)5q²is divisible by 5p² is divisible by 5p is divisible by 5let p=5aputting the value in equation 15q²=p²5q²=(5a)²5q²=25a²q²=5a²5a² is divisible by 5q² is divisible by 5q is divisible by 5 Therefore both p&q have common factor 5We supposed p&q are coprime.Our supposition is wrong √5 is irrational. let root 5 be rationalthen it must in the form of p/q [q is not equal to 0][p and q are co-prime]root 5=p/q=> root 5 ×q = psquaring on both sides=> 5×q×q = p×p ------> 1p×p is divisible by 5p is divisible by 5p = 5c [c is a positive integer] [squaring on both sides ]p×p = 25c×c --------- > 2sub p×p in 15×q×q = 25×c×cq×q = 5×c×c=> q is divisble by 5thus q and p have a common factor 5there is a contradictionas our assumsion p &q are co prime but it has a common factorso\xa0√5 is an irrational |
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| 28607. |
4+7×9=16 |
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Answer» 16......❌ ........67.....✔️ Wrong .... Its answer is 67 |
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| 28608. |
an=4 ,d=2, sn=-14 find \'n\' and \'a\' . |
| Answer» an=4a+(n-1)d=4a+2n-2=4a+2n=6a=6-2n-------(1)Sn=-14n/2(2a+(n-1)d)=-14n/2(2a+(n-1)2)=-14n(a+n-1)=-14n(6-2n+n-1)=-14-----from (1)n(5-n)=-145n-n²=-14n²-5n-14=0n²-(7-2)n-14=0n²-7n+2n-14=0n(n-7)+2(n-7)=0(n+2)(n-7)=0Therefore, n=7Putting the value of n in equation (1)a=6-2×7a=6-14a=-8 | |
| 28609. |
if sinA +cos A /sinA - cos A = 5/3 then find the value of 7tan A+ 2 / 2tan A + 7 |
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Answer» Cross multiplyUll get3sinA +3cosA=5SinA -5cosATranspose3cosa + 5cosa=5sina -3sina8cosa=2sina8/2=sina/cosa4=tan AThe answer woul be 28 +2/8+730/9 plz friends answer this question |
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| 28610. |
A man save 18% of his monthly income. If he save ₹3780 per month. What is his monthly income |
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Answer» ₹21000 I understood.Let his salary be x, So, 18/100.x=3780X=3780.100/18=21000rupee How? ₹21000 |
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| 28611. |
prove that the tangests drawn at the end of a circel are parallel |
| Answer» See ncerr | |
| 28612. |
1/a +1/b 1/x = 1/a+b+x |
| Answer» See its quite simple1/ a + 1/b + 1/x = 1/ a + b + xDo one thing bring 1/x to right side with 1/a+b+xand take the l.c.m of both the sides Then cross multiply it and u will get ur value... of A and BI hope u got ur answer...? | |
| 28613. |
How to find the value of any trigonometric value geometrically? |
| Answer» Its there in rd sharma | |
| 28614. |
2sinA2A=√3 |
| Answer» A=30° | |
| 28615. |
Tan4A+tan2A=sec4A-sec2A |
| Answer» Explanation: Use one of the Pythagorean identity namely,sec2A=1+tan2A\xa0sec4A−sec2A=(sec2A)2−sec2A=(1+tan2A)2−(1+tan2A)=1+2tan2A+tan4A−1−tan2A=tan4A+tan2A | |
| 28616. |
For what value of k the equation x^2+(k+1)+k+4)=0 has equal roots? |
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Answer» K=3 and k=-1 For real and equal roots d=b sq. -4 ac nikal lena usse k ki value nikal jayegi Give me complete solve answer K+1 ke baad x lgta hain reh gaya , isiliye answer k=5,-3 value put krke dekh lena ho skta 1 value satisfy na ho |
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| 28617. |
Find the nth term of the -10-15-20-25 |
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Answer» Nth term is 4 N term is -5 (n+5) n term -5 n-5 Nth term= -10-5n |
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| 28618. |
What is blue print of science n maths can any one send me |
| Answer» You can check the pattern here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 28619. |
If tanA= N tanBAnd sinA= M sinBThen, prove that cos^2A= (M^2-1)/(N^2-1) |
| Answer» Given,\xa0tan A = n tan B{tex} \\Rightarrow{/tex} tanB = {tex}\\frac{1}{n}{/tex}tan A{tex}\\Rightarrow{/tex}\xa0cotB =\xa0{tex}\\frac { n } { \\tan A }{/tex}..........(1)Also given,\xa0sin A = m sin B{tex}\\Rightarrow{/tex}\xa0sin B =\xa0{tex}\\frac{1}{m}{/tex}sin A{tex}\\Rightarrow{/tex}\xa0cosec B =\xa0{tex}\\frac { m } { \\sin A }{/tex}.....(2)We know that, cosec2B - cot2B = 1, hence from (1) & (2) :-{tex} \\quad \\frac { m ^ { 2 } } { \\sin ^ { 2 } A } - \\frac { n ^ { 2 } } { \\tan ^ { 2 } A } = 1{/tex}{tex}\\Rightarrow \\quad \\frac { m ^ { 2 } } { \\sin ^ { 2 } A } - \\frac { n ^ { 2 } \\cos ^ { 2 } A } { \\sin ^ { 2 } A } = 1{/tex}{tex}\\Rightarrow \\quad \\frac { m ^ { 2 } - n ^ { 2 } \\cos ^ { 2 } A } { \\sin ^ { 2 } A } = 1{/tex}{tex}\\Rightarrow{/tex}\xa0m2 - n2cos2A = sin2A{tex}\\Rightarrow{/tex}\xa0m2 - n2cos2A = 1 - cos2A{tex}\\Rightarrow{/tex}\xa0m2 - 1 = n2cos2A - cos2A{tex}\\Rightarrow{/tex}\xa0m2 - 1 = (n2 - 1) cos2A{tex}\\Rightarrow \\quad \\frac { m ^ { 2 } - 1 } { n ^ { 2 } - 1 } ={/tex}\xa0cos2A | |
| 28620. |
if secA =x +1/4x , then prove that secA + tan A = 2x |
| Answer» | |
| 28621. |
Prove that underroot 5 irrational no by contradiction method |
| Answer» Let √5 be a rational number say,p/q where q is not equal to zero and p&q are co prime.√5=p/qSquaring both side5=p²/q²5q²=p²--------------(1)5q² is divisible by 5q²is divisible by 5q is divisible by 5Let q=5aPutting the value in (1)5q²=(5a)²5q²=25a²q²=5a²5a² is divisible by 5q²is divisible by 5q is divisible by 5p&q has common factor 5But we assumed they are co prime.It\'s a contradiction. Our supposition is wrong. Therefore √5 is irrational. | |
| 28622. |
How to upload a pic of any solution for any ques. |
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Answer» Many members uploaded photo of their answers....by the way its not thr on google See on YouTube or Google |
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| 28623. |
Ex 10.2 question no.5 |
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Answer» See in this app only Given: a circle c(o,r), |
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| 28624. |
Find the zoroes of the polynomial x^2 -3 and verify the relationship between the zeroes. |
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Answer» Its solution is on page no. 30 of ncert book Zeros are root 3 and - root3 |
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| 28625. |
Write the condition for similartrily of triangles |
| Answer» ●All angle of traingle should be similar.●coressponding side of traingles are proportional to each other.For better ans you should refer to Ncert | |
| 28626. |
Sign of congurence |
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Answer» ~= ~= |
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| 28627. |
Real no all Answers |
| Answer» Check it on this app or net. | |
| 28628. |
If sec A = x + 1/x then prove that secA .tan A = 2x or 1/2x |
| Answer» Given,\xa0sec\xa0{tex}\\theta{/tex}\xa0= x +\xa0{tex}\\frac{1}{4 x}{/tex}We know that, tan\xa0{tex}\\theta{/tex}\xa0=\xa0{tex}\\pm \\sqrt{\\sec ^{2} \\theta-1}{/tex}{tex}=\\pm \\sqrt{\\left(x+\\frac{1}{4 x}\\right)^{2}-1}{/tex}{tex}=\\pm \\sqrt{x^2 + \\frac{1}{(4x)^2}+2.x. \\frac{1}{(4x)}-1}{/tex}{tex}=\\pm \\sqrt{x^2 + \\frac{1}{(4x)^2}+ \\frac{1}{(2)}-1}{/tex}{tex}=\\pm \\sqrt{x^2 + \\frac{1}{(4x)^2}- \\frac{1}{(2)}}{/tex}{tex}=\\pm \\sqrt {\\left(x-\\frac{1}{4 x}\\right)^2}{/tex}{tex}=\\pm\\left(x-\\frac{1}{4 x}\\right){/tex}Now, sec\xa0{tex}\\theta{/tex}\xa0+ tan\xa0{tex}\\theta{/tex}\xa0=\xa0{tex}\\left(x+\\frac{1}{4 x}\\right) \\pm\\left(x-\\frac{1}{4 x}\\right){/tex}i)\xa0sec\xa0{tex}\\theta{/tex}\xa0+ tan\xa0{tex}\\theta{/tex}\xa0=\xa0{tex}\\left(x+\\frac{1}{4 x}\\right) +\\left(x-\\frac{1}{4 x}\\right){/tex}{tex}=2x{/tex}ii)\xa0sec\xa0{tex}\\theta{/tex}\xa0+ tan\xa0{tex}\\theta{/tex}\xa0=\xa0{tex}\\left(x+\\frac{1}{4 x}\\right) -\\left(x-\\frac{1}{4 x}\\right){/tex}{tex}=\\frac{1}{4x}-(-\\frac{1}{4x}) \\\\=\\frac{1}{4x}+\\frac{1}{4x}\\\\=\\frac{2}{4x}{/tex}{tex}=\\frac{1}{2 x}{/tex} | |
| 28629. |
Sec+ tan=p then the value of cosec |
| Answer» Secθ+tanθ=p ----------------------(1)∵, sec²θ-tan²θ=1or, (secθ+tanθ)(secθ-tanθ)=1or, secθ-tanθ=1/p ----------------(2)Adding (1) and (2) we get,2secθ=p+1/por, secθ=(p²+1)/2p∴, cosθ=1/secθ=2p/(p²+1)∴, sinθ=√(1-cos²θ)=√[1-{2p/(p²+1)}²]=√[1-4p²/(p²+1)²]=√[{(p²+1)²-4p²}/(p²+1)²]=√[(p⁴+2p²+1-4p²)/(p²+1)²]=√(p⁴-2p²+1)/(p²+1)=√(p²-1)²/(p²+1)=(p²-1)/(p²+1)∴, cosecθ=1/sinθ=1/[(p²-1)/(p²+1)]=(p²+1)/(p²-1)\xa0 | |
| 28630. |
To prove sin2A+cos2A=1 |
| Answer» How to prove here its too long, plz check it on good, hope yu can understand ? | |
| 28631. |
Is there any chapter wise blueprint |
| Answer» Ya, yu can check it in Google........... | |
| 28632. |
Prove that √2 id irrational |
| Answer» Let √2 be a rational number So it is written in a form of p/q and a co prime numberSo: √2 =p/q √2/q=p And we see that √2 is a irrational number so √2/q is also a irrational number and it is equal to p which is an rational so our assumption is wrong it is a irrational number.I hope this Answer will help you ? | |
| 28633. |
An equilT |
| Answer» | |
| 28634. |
Find the value of (-1)^n+(-1)^2n+(-1)^2n+1+(-1)4n+2 , where n its a positive odd integer |
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Answer» Pls give the steps 0 |
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| 28635. |
No. Not divisible by 8, at random between 1 and 100 . Find the probability |
| Answer» Here n=98 Number divisible by 8 are(8,16....96)M=12 p(A)=m/n=12/98 | |
| 28636. |
Prove that n raise to power 3 is divisible by 6 if n is any +ve integer. |
| Answer» n3\xa0- n = n (n2\xa0- 1) = n (n - 1) (n + 1)\xa0Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.⇒ n (n – 1) (n + 1) is divisible by 3.\xa0Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q, then n is divisible by 2.If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.⇒ n (n – 1) (n + 1) is divisible by 2.Since, n (n – 1) (n + 1) is divisible by 2 and 3.∴ n (n-1) (n+1) = n3\xa0- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)\xa0\xa0 | |
| 28637. |
Prove the volume of frustrum and its surface area too?? |
| Answer» 1/3πh(R2-r2+r+R) | |
| 28638. |
9th term of an AP is-32 and the sum of its 11th and 13th term is-94 .find the AP |
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Answer» Sorry I didn\'t saw the negative sign that\'s why my ap is wrong A=-7. D=-15/8 AP:-8,-3,2,7.... T9=a+8d=-32a=-32-8d-----------(1)T11+T13=a+10d+a+12d=-942a+22d=-94a+11d=-47-32-8d+11d=-473d=-15d=-5now from (1)a=-32+11(-5)=-32-8(-5=-32+40=8so AP is8,3,-2,-7,-12,-17........\xa0\xa0 Let a be the first term and d be the common difference of the AP. Then,Hence, the common difference of the AP is -5. |
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| 28639. |
If 7 times the 7th term of an AP is equal to 11 times its 11th term , then find its 18th term |
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Answer» 0 let a = first term and d=common difference.According to question,7a7\xa0= 11 a11=>7(a+6d) = 11(a+10d)=> 7a-11a = 110d-42d=> -4a = 68d=>a = -17d ......(1)Now , a18\xa0= a\xa0+17d\xa0but a = -17 d ..by (1)so\xa0put this\xa0value , we get,a18=-17d +17d =0\xa0Hence proved. |
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| 28640. |
The following table gives production yield per hectare of wheat of 100 farms of a village. |
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Answer» \xa0The following table gives production yield per hectare of wheat of 100 farms of a village.\tProduction yield50 - 5555 - 6060 - 6565 - 7070 - 7575 - 80Number of farms2812243816\t We can obtain cumulative frequency distribution of more than type as following:\xa0\tProduction yield(lower class limits)Cumulative frequencyMore than or equal to 50100More than or equal to 55100 - 2 = 98More than or equal to 6098 - 8 = 90More than or equal to 6590 - 12 = 78More than or equal to 7078 - 24 = 54More than or equal to 7554 - 38 = 16\t\xa0Now, taking lower class limits on\xa0x-axis and their respective cumulative frequencies on\xa0y-axis, we can obtain the ogive as follows. |
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| 28641. |
1/x+1+1/2x-1=11/7x+9 |
| Answer» | |
| 28642. |
Why only class x chapter- real number have RD sharma solution |
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| 28643. |
Sin2 - cos2= |
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Answer» -1 -1 -1 -1 -1 |
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| 28644. |
Cos2 20+cos2 70/sec2 50 - cot2 40+2cosec 2 58-2cot 58 *tan 32 - 4tan 13*tan37*tan45*tan53*tan77 |
| Answer» Please write this question correctly | |
| 28645. |
If p,q,r are in A.p,then find the value of p3 + r3-8q3 |
| Answer» | |
| 28646. |
What is meant by sn and an generally |
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Answer» Sn =sum of nth termAn=nth term Sn =sum. An=term |
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| 28647. |
HCF OF 20 AND 149 |
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Answer» 1 How 1 |
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| 28648. |
Prove that sum of zeros |
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Answer» -b/a Complete yur question....... |
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| 28649. |
but how k= 0 |
| Answer» fkflp | |
| 28650. |
If x,y,z are in AP then find the value of (x+y-z)(y+z-x) |
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Answer» Let y be first term of AP so X = a-d Y = aZ = a+d X + y - z = aY+z-x = 2a +2d So answer is y*2z |
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