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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.
| 16401. |
Prove that Root 5 irrational |
| Answer» If possible,Let √5 be rational√5=a/b where a and b are integers and co-prime.On squaring both sides,(√5)²=(a/b)²5=a²/b²a²=5b²_(¹)Clearly,5 is a factor of a²Then 5 is also a factor of a.a=5mOn squaring both sides,a²=(5m)²a²=25m²5b²=25m²[From eq(¹)]b²=5m²Clearly,5 is a factor of b²5 is a factor of b.Here,a and b are integers where 5 is a factor of both the integers.So,there is a contradiction arisened 5 is not a co-prime no.Hence,√5 is irrational (By contradiction method) | |
| 16402. |
Question number 8 of Exercise 4.3 |
| Answer» D=360 km Original speed =x km /hIf speed=x+5 km/hATQSpeed=distance÷timeSO, Original time=360 ÷xAnd New time=360÷x+5Now as per ques Original time - 1=new time Or Original time - new time =1So,(360÷x) - (360÷x+5) =1 | |
| 16403. |
Solve the eq using factorisation : X2 -4ax +4a2-b2 =0 |
| Answer» X^2-4ax +4a^2=b^2X2-2ax-2ax+4a2=b2X(x-2a)-2a(x-2a)=b2(x-2a)(x-2a)=b2(X-2a)(x-2a-b2)Hence,these are the factors | |
| 16404. |
1/x-1/1-2 one upon x minus one upon x minus 2 is equal to 3 X is not equal to zero two |
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| 16405. |
4×2 |
| Answer» | |
| 16406. |
Write down the decimal expansion 16/3125 without actual division. |
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Answer» Sorry that emoji is ? * not? 3125=5^5 Prime factorisation of 3125 is in the form of 2^n×5^m Therefore, 16/3125 has terminating decimal expansion? 2^4 ÷ 5^5 |
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| 16407. |
Any five sum related to surface areas |
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| 16408. |
4q-72q-45q-20q=-15+63 |
| Answer» 48/133 | |
| 16409. |
If the points a(x,2) b(-3,-4)c(7,-5) are collinear then find the value of x |
| Answer» Since the points are collinear, then,\xa0Area of triangle = 0{tex}\\frac { 1 } { 2 } \\left[ x _ { 1 } \\left( y _ { 2 } - y _ { 3 } \\right) + x _ { 2 } \\left( y _ { 3 } - y _ { 1 } \\right) + x _ { 3 } \\left( y _ { 1 } - y _ { 2 } \\right) \\right] = 0{/tex}{tex}\\frac { 1 } { 2 } [ x ( - 4 + 5 ) + ( - 3 ) ( - 5 - 2 ) + 7 ( 2 + 4 ) ] = 0{/tex}x + 21 + 42 = 0x = -63 | |
| 16410. |
Given d=5,s9=75 find a and an |
| Answer» d=5 S9=75So 75=9/2(2a+8*5)150=18a+36018a=150-36018a=-210a=-210/18=-11.67an=-11.67+(n-1)*5=-11.67+5n-5an=5n-16.67 | |
| 16411. |
how to do calculation fast in finding mean of ungrouped data |
| Answer» Use formula sigma.fi×xi/sigma.fi | |
| 16412. |
Distance between the point (0,5 )and (- 5,0) |
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Answer» (X^2-x^1)+(y^2-y^1)(-5-0)^2+(5,0)^2(5)^2+(5)^225+2550 Answer is 50 6 |
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| 16413. |
7x square+30+8 |
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Answer» 87 is the answer The answer is simple You have square the number seven and add with thirty and eight(7×7)+30+849+3887 |
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| 16414. |
Write the conditions are set the pair of linear equation has no solution |
| Answer» a1/a2=b1/b2 but not equal to c1 by c2 | |
| 16415. |
What is sino |
| Answer» 0 | |
| 16416. |
Express the trigonometry ratioes sin a,tan a,in terms of cot a |
| Answer» We have to express the trigonometric ratios sin A, sec A\xa0and tan A\xa0in terms of cot A.For sin A,By using identity {tex}cosec ^ { 2 } A - \\cot ^ { 2 } A = 1{/tex}{tex}\\Rightarrow cosec ^ { 2 } A = 1 + \\cot ^ { 2 } A{/tex}{tex}\\Rightarrow \\frac { 1 } { \\sin ^ { 2 } A } = 1 + \\cot ^ { 2 } A{/tex}{tex}\\Rightarrow \\sin ^ { 2 } A = \\frac { 1 } { 1 + \\cot ^ { 2 } A }{/tex}{tex} \\Rightarrow \\quad \\sin A = \\frac { 1 } { \\sqrt { 1 + \\cot ^ { 2 } A } }{/tex}For tan A,{tex} \\tan A = \\frac { 1 } { \\cot A }{/tex} | |
| 16417. |
Write a rationsl and irrational number between root 2 and root 3 |
| Answer» 1.5 --rational1.50550555055550..... irrational | |
| 16418. |
2^x+3^y=17and 2^x+2-3^y+1=5 then find the value of x and y |
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Answer» Here; 2^x+3^y=17 (i) 2^x+2-3^y+1=5 2^x-3^y+3=5 2^x-3^y=5-3 2^x-3^y=2 (ii)Adding (i) and (ii) to get : 2^x+3^y+2^x-3^y = 17+2 4^x=19 x=19/4Putting x=19/4 in (i) : 2×19/4 +3^y=17 19/2+3^y=17 3^y=17-19/2 3^y=15/2 y=5/2 Sorry, but you have to wait tonight i will answer you tommorow at 1 pm Give me amswer |
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| 16419. |
If a=sinA and y= b tanA then prove that a^2y^2-b^2x^2= x^2y^2 |
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| 16420. |
Important questions of ch-1 |
| Answer» All | |
| 16421. |
If tanA + secA =l prove that secA= l^2+1/2l |
| Answer» tanA + secA =l\xa0or secA+tanA= I ---------------------(1)We know\xa0that\xa0{tex}\\sec^2\\mathrm A-\\tan^2\\mathrm A=\\mathrm 1-------(2){/tex}Dividing (2) by (1) we get{tex}\\begin{array}{l}\\frac{\\sec^2\\mathrm A-\\tan^2\\mathrm A}{\\mathrm{secA}+\\mathrm{tanA}}=\\frac1{\\mathrm I}\\\\\\mathrm{secA}-\\mathrm{tanA}=\\frac1{\\mathrm I}-----------(3)\\end{array}{/tex}Adding (1) and (3) we get{tex}\\begin{array}{l}2\\mathrm{secA}=\\mathrm I+\\frac1{\\mathrm I}\\\\2\\mathrm{secA}=\\frac{\\mathrm I^2+1}{\\mathrm I}\\\\\\mathrm{secA}=\\frac{\\mathrm I^2+1}{2\\mathrm I}\\\\\\mathrm{Hence}\\;\\mathrm{proved}\\end{array}{/tex} | |
| 16422. |
prove that tan cube theta -1 / tan theta - 1 = secant square theta + tan theta |
| Answer» | |
| 16423. |
Ex_6.5 1,2,3,4,5,6,7,8,9,10 sum |
| Answer» You can check NCERT Solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 16424. |
Chapter -6 integers class6 |
| Answer» Check NCERT Solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 16425. |
Why in trignometry the angle should only be in 90 degree? |
| Answer» Ratios can be found on principle of pythagoras theorem | |
| 16426. |
Is the important questions which are available in this app are really Important? |
| Answer» Yes | |
| 16427. |
Find the zeroes of the polynomial 6x2-7x-3 |
| Answer» =6x²-7x-3=6x²+2x-9x-3=2x(3x+1)-3(3x+1)=(2x-3)(3x+1)⇒2x-3=0 ⇒3x+1=0⇒x=3/2 ⇒x= - 1/3α=3/2 ,β= - 1/3 | |
| 16428. |
Explain chapter =1exercise =1.1 in Hindi |
| Answer» Check NCERT Solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 16429. |
The latgest number whivh divides 615 and 963 leaving reminder 6 in each case |
| Answer» 615-6=609. 963-6= 957. 615=3×7×29. 963= 3×11×29. HCF= 3×29=87. | |
| 16430. |
I have questions regarding how to make such word problems equations in chapter 3 |
| Answer» Let the no of item of any substance be xAnd other yThen put the value in equation | |
| 16431. |
If α & β are zeroes of ax^2+bx+c then find polynomial whose zeroes are 1/α and 1/β |
| Answer» a/c | |
| 16432. |
If a=pq×q and b=pppq then find the lcm of a and b |
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Answer» The LCM is p^3q^2 P^3q^2 |
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| 16433. |
What is the Syllabus of Math for Mid Term Examination . |
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Answer» Ch 1 to 8 Chapter 1,2,3,4,5,6,7 and 14,15 in murlidhar D. A. V Ch123681116 |
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| 16434. |
13.1 8th question |
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| 16435. |
I have doubts in forming the equation in quadratic equation can anyone explain |
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Answer» What is doubt First read given statement carefully and then make your equation in the form of (axsquare+bx+c) The standard form is ax square+bx+c first check what is in a questionchoose a question and now yes |
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| 16436. |
If the mean of 5 observations x,x+2,x+4,x+6 and x+8 is 11 , find the value of x. |
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Answer» X=7 x=9 |
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| 16437. |
Important formula |
| Answer» Of what | |
| 16438. |
Obtain all zeroes of 3x⁴-15x³+13x²+25x-30 if two of it\'s zero\'s are √5/3 and -√5/3 |
| Answer» We have the polynomial f(x) = (3x4 - 15x3 + 13x2 + 25x - 30).Since\xa0{tex}\\sqrt { \\frac { 5 } { 3 } } \\text { and } - \\sqrt { \\frac { 5 } { 3 } }{/tex}\xa0are the zeros of f(x),i.e.,\xa0{tex}x=-\\sqrt{\\frac53}{/tex}\xa0and {tex}x=\\sqrt{\\frac53}{/tex}i.e.,\xa0{tex}x+\\sqrt{\\frac53}=0\\;and\\;x-\\sqrt{\\frac53}=0{/tex} so it follows that eachone of\xa0{tex}\\left( x - \\sqrt { \\frac { 5 } { 3 } } \\right) \\operatorname { and } \\left( x + \\sqrt { \\frac { 5 } { 3 } } \\right){/tex}\xa0is a factor of f(x).so using\xa0{tex}\u200b\u200b\\left(a+b\\right)\\left(a-b\\right)=a^2-b^2{/tex}, we get{tex}\\therefore \\quad \\left( x - \\frac { \\sqrt { 5 } } { \\sqrt { 3 } } \\right) \\left( x + \\frac { \\sqrt { 5 } } { \\sqrt { 3 } } \\right) = 0 \\Rightarrow \\left( x ^ { 2 } - \\frac { 5 } { 3 } \\right)= 0 \\Rightarrow \\frac { \\left( 3 x ^ { 2 } - 5 \\right) } { 3 }= 0{/tex}\xa0is also a factor of f(x).Consequently, (3x2\xa0- 5) is a factor of f{x).On dividing the polynomial by (3x2 - 5), we get{tex}\\therefore{/tex}\xa0f(x) = 3x4 - 15x3 + 13x2 + 25x - 30= (3x2\xa0- 5 )(x2\xa0- 5x + 6).By middle term factorisation. We get,={tex}\\left(3x^2-5\\right)\\left(x^2-2x-3x+6\\right){/tex}=({tex}\\lbrack(\\sqrt{3x})^2-\\left(\\sqrt5\\right)^2\\rbrack{/tex}{tex}\\left[x\\left(x-2\\right)-3\\left(x-2\\right)\\right]{/tex}By using\xa0{tex}a^2-b^2=\\left(a+b\\right)\\left(a-b\\right){/tex}we get,{tex}= ( \\sqrt { 3 } x + \\sqrt { 5 } ) ( \\sqrt { 3 } x - \\sqrt { 5 } ) ( x - 2 ) ( x - 3 ){/tex}{tex}\\therefore{/tex}\xa0f(x) = 0 , so either factors can equated to zero to get the roots\xa0{tex}\\Rightarrow ( \\sqrt { 3 } x + \\sqrt { 5 } ) = 0 \\text { or } ( \\sqrt { 3 } x - \\sqrt { 5 } ) = 0{/tex}{tex}\\Rightarrow{/tex}(x - 2) = 0 or (x - 3) = 0{tex}\\Rightarrow x = - \\sqrt { \\frac { 5 } { 3 } } \\text { or } x = \\sqrt { \\frac { 5 } { 3 } }{/tex}\xa0or x = 2 or x = 3Hence, we get all zeros of f(x) are\xa0{tex}\\sqrt { \\frac { 5 } { 3 } } , - \\sqrt { \\frac { 5 } { 3 } }{/tex}, 2 and 3. | |
| 16439. |
Why the values of trigonometry can\'t changed??? |
| Answer» | |
| 16440. |
RAT=42 and CAT=57 then LATE= |
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Answer» Can u please upload the solution please 72 70 |
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| 16441. |
i want blueprint of cbse exam |
| Answer» | |
| 16442. |
Which chapter come in final |
| Answer» Whole chapters of all subjects except Social Science | |
| 16443. |
Number of terms |
| Answer» an=a(n-1)d | |
| 16444. |
a1x +b1y +c1=0a2x +b2y +c1=0 |
| Answer» The question itself is wronga1x +b1y +c1=0 & a2x +b2y +c2 /_____/ | |
| 16445. |
What is the formula for acute angle triangle of one side square is equal to? |
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| 16446. |
I want a set of questions of class 11 relations and functions |
| Answer» | |
| 16447. |
Sum and products of zeroes of quadratic polynomial are -1/4 & 1/2 respectively. Find the polynomial. |
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Answer» Sum= -1/4 product= 1/2 use formula make quadratic polynomial formula is x² - (sum of zeroes)x +(product of zeroes) putting the value x²- (-1/4)x + (1/2) ans ayega x² + 1/4+1/2. Fir hum multiply krege 4 se jisse divide ki value chali jaye like 4x²+ 4× 1/4 + 4×1/2 to humare pss equation ayegi 4x²+ x +2 ye humara quadratic polynomial banega (a+b)=-1/4(ab)=1/2Putting the value of (a+b),(ab).P(x)=x2+( a +b)x-(ab) x2-1/4x-1/24x2-1x-2hence the polynomial is 4x2-1x-2 |
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| 16448. |
All the formulas of trignometry |
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| 16449. |
What is polynomial |
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Answer» A polynomial looks like this:\texample of a polynomial this one has 3 terms\tPolynomial\xa0comes from\xa0poly-\xa0(meaning "many") and\xa0-nomial\xa0(in this case meaning "term") ... so it says "many terms"A polynomial can have:\tconstants\xa0(like\xa03,\xa0−20, or\xa0½)variables\xa0(like\xa0x\xa0and\xa0y)exponents\xa0(like the 2 in y2), but only\xa00, 1, 2, 3, ...\xa0etc are allowed\t Combination of constant and variable with +,-,×,÷ sign are polynomials |
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| 16450. |
tan(A-B)=tanA-tanB /1+tanA*tanB |
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Answer» Isko prof krna h ya value nikalni hai 1 |
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