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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.
| 7651. |
Sin60°cos30°+sin30°cos60° |
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Answer» 1 1 And : 1 |
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| 7652. |
log e = ? |
| Answer» Log (e) = 0.4342944819 | |
| 7653. |
Draw the tangent from external point are equal |
| Answer» Given: P is an external point to the circle C(O,r).PQ and PR are two tangents from P to the circle.To Prove: PQ = PRConstruction: Join OPProof:{tex}\\because{/tex}\xa0A tangent to a circle is perpendicular to the radius through the point of contact{tex}\\therefore{/tex}{tex}\\angle{/tex}OQP = 90o = {tex}\\angle{/tex}ORPNow in right triangles POQ and POR,OQ = OR [Each radius r]Hypotenuse. OP = Hypotenuse. OP [common]{tex}\\therefore{/tex}{tex}\\triangle{/tex}POQ\xa0{tex}\\cong{/tex}{tex}\\triangle{/tex}POR\xa0[By RHS rule]{tex}\\therefore{/tex} PQ = PR | |
| 7654. |
What is the meaning of emasculation |
| Answer» The process by which the male part of flowers is cut out to prevent self fertilization... | |
| 7655. |
What do you mean by a.p |
| Answer» Having same common difference | |
| 7656. |
Bpt therom |
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| 7657. |
Sum of n terms of an a. P is -150 and it\'s sum of Next 10terms is -550 . Find its a p |
| Answer» It can be solved by the formula of sumSum of n terms =-150S(n-10)=-550n/2 (2a+(n-1)d)=-150n/2 (2a+(n-1)d)-10/2 (2a+(10-1)d)=-550 | |
| 7658. |
How to find ap |
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Answer» Yes☝☝☝ Tn= a+(n-1)d |
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| 7659. |
Are all quadrilaterals whose diagonals bisect each other parallelogram |
| Answer» Yes | |
| 7660. |
Find a.p of 3n-5n |
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| 7661. |
Find the area of quadrant of a circle whose circumferenceis 22 cm |
| Answer» Let the radius of the circle be r cm.Then, circumference\xa0of the circle = 2{tex}\\pi{/tex}r cmAccording to the question,2{tex}\\pi{/tex}r = 22{tex}\\Rightarrow 2 \\times \\frac { 22 } { 7 } \\times \\mathrm { r } = 22{/tex}{tex}\\Rightarrow \\mathrm { r } = \\frac { 22 \\times 7 } { 2 \\times 22 } \\Rightarrow \\mathrm { r } = \\frac { 7 } { 2 } \\mathrm { cm }{/tex}For a quadrant of a circle,{tex}= \\frac { \\theta } { 360 } \\times \\pi r ^ { 2 }{/tex}{tex}= \\frac { 90 } { 360 } \\times \\frac { 22 } { 7 } \\times \\left( \\frac { 7 } { 2 } \\right) ^ { 2 }{/tex}{tex}= \\frac { 90 } { 360 } \\times \\frac { 22 } { 7 } \\times \\frac { 7 } { 2 } \\times \\frac { 7 } { 2 } = \\frac { 77 } { 8 } \\mathrm { cm } ^ { 2 }{/tex} | |
| 7662. |
What is the method of complete squaring |
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| 7663. |
Will exam from ncert if u sure then tell. |
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| 7664. |
If sec thrita =x+1divide by x then proove that sec thrita×tan thrita=2x or 1divide by x |
| Answer» By the given condition of question\xa0{tex}\\sec \\theta = x + \\frac { 1 } { 4 x }{/tex}{tex}\\therefore \\quad \\tan ^ { 2 } \\theta = \\sec ^ { 2 } \\theta - 1{/tex}{tex}\\Rightarrow \\quad \\tan ^ { 2 } \\theta = \\left( x + \\frac { 1 } { 4 x } \\right) ^ { 2 } - 1 = x ^ { 2 } + \\frac { 1 } { 16 x ^ { 2 } } + \\frac { 1 } { 2 } - 1 = x ^ { 2 } + \\frac { 1 } { 16 x ^ { 2 } } - \\frac { 1 } { 2 } = \\left( x - \\frac { 1 } { 4 x } \\right) ^ { 2 }{/tex}{tex}\\Rightarrow \\quad \\tan \\theta = \\pm \\left( x - \\frac { 1 } { 4 x } \\right){/tex}{tex}\\Rightarrow \\quad \\tan \\theta = \\left( x - \\frac { 1 } { 4 x } \\right) \\text { or, } \\tan \\theta = - \\left( x - \\frac { 1 } { 4 x } \\right){/tex}CASE 1: When\xa0{tex}\\tan \\theta = - \\left( x - \\frac { 1 } { 4 x } \\right) :{/tex}\xa0In this case,{tex}\\sec \\theta + \\tan \\theta = x + \\frac { 1 } { 4 x } + x - \\frac { 1 } { 4 x } = 2 x{/tex}CASE 2: When\xa0{tex}\\theta = - \\left( x - \\frac { 1 } { 4 x } \\right) :{/tex}\xa0In this case,{tex}\\sec \\theta + \\tan \\theta = \\left( x + \\frac { 1 } { 4 x } \\right) - \\left( x - \\frac { 1 } { 4 x } \\right) = \\frac { 2 } { 4 x } = \\frac { 1 } { 2 x }{/tex}Hence,\xa0{tex}\\sec \\theta + \\tan \\theta = 2 x \\text { or } , \\frac { 1 } { 2 x }{/tex} | |
| 7665. |
Who knows parag diwan and goes parag tutorials????? |
| Answer» | |
| 7666. |
State Euclid\'s division lemma.? |
| Answer» a = bq + r, where a= Divided, b= Divisor, q =Quaint and r= Remainder. | |
| 7667. |
Sin a=1 cos=? |
| Answer» 0 | |
| 7668. |
India vs newzealand 1st november pehla T20 dekhna zarrur and speacially my fav. Ro-hit-man sharma |
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| 7669. |
2 + 2 |
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Answer» One more child came?? How do you managed to pass 9th class ?? |
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| 7670. |
SinA ,cosA=2sinA |
| Answer» SinA+cos(90-A) =2sinASinA+SinA=2sinA2SinA =2sinA | |
| 7671. |
prove that all the triangle of equilateral equal |
| Answer» | |
| 7672. |
How to remove fear |
| Answer» If you aren’t ready to face your fears, you probably won’t transcend them.And there’s nothing wrong in that. Everything happens in its own time. | |
| 7673. |
Prove: 3cosx-4sinx=2cosx+sinx |
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| 7674. |
4root3x^2+5x-2root3 |
| Answer» Multipply 4√3 and 2√3U get 24By factorising we get (x+8) (x-3) = 0Therefore x =-8 and x=3 | |
| 7675. |
Find the value of c if one root of the equation 4x2 -kx2+(c+4) = 0 is reciprocal of the other |
| Answer» 8 | |
| 7676. |
Find tb |
| Answer» | |
| 7677. |
How many students use R.D. Sharma for maths and how many R.S. Aggarwal?? |
| Answer» RD SHARMA IS THE BEST...... RD SHARMA | |
| 7678. |
If alpha and bitta are |
| Answer» | |
| 7679. |
Define tangent |
| Answer» A line which is draw from the one point of the circle it is called tangent. If a tangent joint with two point of circle it becomes diameter. So always remember tangent made with a single point | |
| 7680. |
Is 2 a prime no. |
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Answer» Yes Yes |
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| 7681. |
If √2 is an rational number |
| Answer» Assume √2 as rational√2 = a/b ( where a and b are co- prime) co-prime - no\'s having 1 as common factorSquaring on both sides2 = a^2/b^22b^2 = a^2----12 divides a^2 & a alsoa = 2 cSquaring on both sides a^2 = 4c^2----2From 1 & 2 2b^2 = 4c^2b^2 = 2c^2b^2/2 = c^ 2This shows 2 divides b^2 & b also Hence 2 is a common factor for a and b,this contradicts the fact that a and b are co- prime.so,our assumption is wrong.Therefore, √2 is Irrational. | |
| 7682. |
Find roots of 2x2(square) + 65x -7943 = 0 |
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| 7683. |
Find cot 10 cot 30 cot 80.. |
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| 7684. |
IF SIN ALFA = 1/2 FIND 3 SIN ALFA - 4 SIN3 ALFA... |
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| 7685. |
Area of circle is what ? |
| Answer» πr2 | |
| 7686. |
Find the area of shaded area which the quadrants are intersect to each other in square |
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| 7687. |
Exam of class 10 |
| Answer» Go to CBSE official website and check it | |
| 7688. |
if A+B=90 secA=5÷3 then find the value of cosecB |
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| 7689. |
x-a/x-b+x-b/x-a=a/b+b/a |
| Answer» We have,{tex}\\frac{{x - a}}{{x - b}} + \\frac{{x - b}}{{x - a}} = \\frac{a}{b} + \\frac{b}{a}{/tex}{tex}\\Rightarrow \\frac{{(x - a)(x - a) + (x - b)(x - b)}}{{(x - b)(x - a)}} = \\frac{{{a^2} + {b^2}}}{{ab}}{/tex}{tex} \\Rightarrow \\frac{{{x^2} + {a^2} - 2ax + {x^2} + {b^2} - 2bx}}{{{x^2} - bx - ax + ab}}{/tex}{tex} = \\frac{{{a^2} + {b^2}}}{{ab}}{/tex}{tex}\\Rightarrow \\frac{{2{x^2} - 2ax - 2bx + {a^2} + {b^2}}}{{{x^2} - bx - ax + ab}} = \\frac{{{a^2} + {b^2}}}{{ab}}{/tex}{tex}\\Rightarrow{/tex}{tex} (2x^2 - 2ax - 2bx + a^2 + b^2)ab = (a^2 + b^2)(x^2 - bx - ax + ab){/tex}{tex}\\Rightarrow{/tex}{tex}2abx^2 - 2a^2bx - 2ab^2x + a^3b + ab^3 = a^2x^2 - a^2bx -a^3x + a^3b + b^2x^2 - b^3x - ab^2x + ab^3{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}2abx^2 - a^2x^2 - a^2bx - ab^2x + a^3x + b^3x - b^2x^2 = 0{/tex}{tex}\\Rightarrow{/tex}{tex}(2ab - a^2 - b^2)x^2 + (-a^2b - ab^2 + a^3 + b^3)x = 0{/tex}{tex}\\Rightarrow{/tex} x[(2ab - a2 - b2)x + (a3 + b3 - a2b - ab2)] = 0{tex}\\Rightarrow{/tex} x = 0 or (2ab - a2 - b2)x + a3 + b3 -a2b - ab2 = 0Now,(2ab - a2 - b2)x + a3 + b3 - a2b - ab2 = 0{tex}\\Rightarrow{/tex} (2ab - a2 - b2)x = a2b + ab2 - a3 - b3{tex}\\Rightarrow{/tex} -(a2 - b2 - 2ab)x = a2b - b3 + ab2 - a3{tex}\\Rightarrow{/tex} -(a - b)2x = b(a2 - b2) + a(b2 - a2){tex}\\Rightarrow{/tex} (a - b)2x = -b(a2 - b2) - a(b2 - a2){tex} \\Rightarrow x = \\frac{{ - b({a^2} - {b^2}) + a({a^2} - {b^2})}}{{{{(a - b)}^2}}}{/tex}{tex}\\Rightarrow x = \\frac{{\\left( {{a^2} - {b^2}} \\right)(a - b)}}{{{{(a - b)}^2}}}{/tex}{tex} \\Rightarrow x = \\frac{{{a^2} - {b^2}}}{{a - b}} = \\frac{{(a - b)(a + b)}}{{(a - b)}}{/tex}{tex}\\Rightarrow{/tex} x = a + b{tex}\\therefore{/tex}\xa0x = 0 or x = a + b | |
| 7690. |
the literacy rate of 40 states is given below if mean is 63.5 find x and y\xa0 |
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| 7691. |
Example15 of ch_8 trignometry |
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| 7692. |
Sum of the roots of the eqn 2^33x-2 + 2^11x+1 = 2^22x+1 + 1 |
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| 7693. |
(cosecA-cotA)²=1 |
| Answer» Its an identity | |
| 7694. |
In AP what is important |
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Answer» In u ya for u Every thing is important in me???? |
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| 7695. |
What is cos |
| Answer» The cosine function, along with sine and tangent, is one of the three most common trigonometric functions. In any right triangle, the cosine of an angle is the length of the adjacent side (A) divided by the length of the hypotenuse (H). In a formula, it is written simply as \'cos\'. | |
| 7696. |
When are board exams for 2018 class 10 |
| Answer» New Delhi: Central Board of Secondary Education (CBSE) will conduct the class 10 board examination in the month of March, contrary to the ongoing rumours about exam being held in February. Students, who will be appearing for the examination, next year must start their preparation, now onwards. The Board has released sample question papers and marking scheme for all the main subjects for which the board exam will be held. CBSE class 10 sample question paper and marking scheme for exam 2017-2018 will be available at the official website of the Board at cbse.nic.in or cbseacademic.in. | |
| 7697. |
To find the middle term of an AP 1 ,8,15,------ 505 |
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| 7698. |
If opposite vertices of square are 3,2 and 4,1.find other two vertices |
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| 7699. |
What is complete the square method |
| Answer» The given equation is:7x2\xa0+ 3x - 4 = 0Multiply each term by 7, we obtain:\xa049x2\xa0+ 21x - 28 = 0{tex}\\Rightarrow{/tex}\xa049x2\xa0+ 21x = -28On adding\xa0{tex}\\left( \\frac { 3 } { 2 } \\right) ^ { 2 }{/tex} on both sides, we get(7x)2\xa0+ 2\xa0{tex}\\times{/tex}\xa07x\xa0{tex}\\times \\frac { 3 } { 2 } + \\left( \\frac { 3 } { 2 } \\right) ^ { 2 } = 28 + \\left( \\frac { 3 } { 2 } \\right) ^ { 2 }{/tex}\xa0{tex}\\Rightarrow \\left( 7 x + \\frac { 3 } { 2 } \\right) ^ { 2 } = 28 + \\frac { 9 } { 4 }{/tex}{tex}\\Rightarrow \\left( 7 x + \\frac { 3 } { 2 } \\right) ^ { 2 } = \\frac { 121 } { 4 }{/tex}{tex}\\Rightarrow 7 x + \\frac { 3 } { 2 } = \\pm \\frac { 11 } { 2 }{/tex}Therefore, either 7x =\xa0{tex}- \\frac { 3 } { 2 } - \\frac { 11 } { 2 }{/tex}\xa0or 7x =\xa0{tex}- \\frac { 3 } { 2 } + \\frac { 11 } { 2 }{/tex}{tex}\\Rightarrow{/tex}\xa07x = -7 or 7x = 4{tex}\\Rightarrow{/tex}\xa0x = -1 or x =\xa0{tex}\\frac { 4 } { 7 }{/tex}Hence, the roots of given equation are {tex}\\frac { 4 } { 7 }{/tex} and\xa0-1. | |
| 7700. |
4 cosβ = 11 sinβ |
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