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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.
| 33551. |
Is ncert is enough to pass |
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Answer» If u had do N.C.E.R.T very well and know all about it then u r. to able to do any questions N Of course ncert is best to score good marks |
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| 33552. |
If 3cos theta-4sin theta =2cos theta+sin theta? Find tan theta |
| Answer» {tex} \\Rightarrow 3\\cos \\theta - 4\\sin \\theta = 2\\cos \\theta + \\sin \\theta {/tex}{tex} \\Rightarrow - 4\\sin \\theta - \\sin \\theta = 2\\cos \\theta -3cos \\theta {/tex}{tex} \\Rightarrow - 5\\sin \\theta = - \\cos \\theta {/tex}{tex} \\Rightarrow 5\\sin \\theta = \\cos \\theta {/tex}{tex} \\Rightarrow \\frac{{\\sin \\theta }}{{\\cos \\theta }} = \\frac{1}{5}{/tex}{tex} \\Rightarrow \\tan \\theta = \\frac{1}{5}{/tex} | |
| 33553. |
Find the value of k for which the system of equation have a unique solution x-xy=2 ,3x-3y=7 |
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| 33554. |
Prove that 3+5√2 is irrational. |
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Answer» Let us assume that 3+5√2is rational no. written in the form of p/q where p,q are integer.Suppose p,q have no any common factor .3+2√5=p/q5√2=p/q-3√2 =1/5(p/q-3)Since, 1/5(p/q-3)is rational So, √2 is also a rational .This is contradicts the fact that √2 is irrational.Hence, 3+2√5 is irrational. Let 3+5√2 is a rational numberThen,3+5√2=p/q( rational number always in the form of p/q where q not equal to \'0\')3+5√2=p/q5√2=p/q-3√2=p/q-3/5But we know that √2 is an irrational no° .So , 3+5√2 is an irrational number. |
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| 33555. |
What is zero polynomial |
| Answer» A real number k is called a zero of the polynomial.p(x) ,if p(k). | |
| 33556. |
Solve 35x+23y=209 ; 23x+35y=197 |
| Answer» Solution can be given by cross multiplication method. X=4 and y=3 | |
| 33557. |
If cosA = 4/5,cosB=12/13,3π/2 |
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| 33558. |
Find ten rational number between -1/3 and 1/2 |
| Answer» Multiply both no° by 10 Then we get (-1×10)/(3×10)and (1×10)/2×10)=-10/30 and 10/20Therefore rational no° are-10/30,-11/30............10/20 | |
| 33559. |
Weitage fir every chapter |
| Answer» 11? | |
| 33560. |
If 3 and -5 are the zeroes of a polynomial 6x CUBE + 7y CUBE - ky -3x-4 find k |
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| 33561. |
Discuss the nature of the roots of the equation.4x²-2x+1=0. |
| Answer» D=-12so roots are not real. | |
| 33562. |
Find the product of zeroes of -2x^2+kx+6 |
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Answer» Product of zeros=c/a=6/-2=-3 -3 |
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| 33563. |
Find value of p and q if the polynomial x^4+px+2x^2-3x+q is divisible by (x-1)(x+1) |
| Answer» Np12 | |
| 33564. |
Trignomatry formulas |
| Answer» Get formulae from notes :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 33565. |
If alpha and beta are the zeros of polynomial f(x)=x2+x-2,find the value of 1/alpha-q/beta |
| Answer» Answer=-3/2 | |
| 33566. |
Find the zeros of the polynomial x*x+x-p(p+1). |
| Answer» We have,{tex}f(x) = x^2 + x - p(p + 1){/tex}{tex}= x^2 + (p + 1)x - px - p(p + 1) {/tex}[As x =(p + 1)x - px ]{tex}= x[ x + (p + 1)] - p[ x + (p + 1)]{/tex}{tex}= ( x - p)[x + (p + 1)]{/tex}{tex}{/tex}\xa0Now, f(x) = 0{tex}\\Rightarrow{/tex}\xa0( x - p)[x + (p + 1)] = 0{tex}\\Rightarrow{/tex}\xa0x - p = 0 or x + (p +1) = 0{tex}\\Rightarrow{/tex}\xa0x = p or x = -(p + 1)Thus, the required zeros of f(x) are p and -(p + 1). | |
| 33567. |
Congruence criterion of triangle activity |
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| 33568. |
Prove that sin theta-coos theta+1/$in theta+cos theta-1=1/$ec theta-tan theta |
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| 33569. |
，difine division lemma theorem |
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| 33570. |
define division algorithm theorem |
| Answer» The no is uniquely divide and have no excutee no is show common factors | |
| 33571. |
Is ncert sufficient for 10Th board exams |
| Answer» Yes NCERT is sufficient for 10th board exam. if you want more practice may purchase NCERT examplears .Make notes and start revising .make accurate notes revis as many times you can till examination . | |
| 33572. |
Sin48.sec42+cos48.cosec42=o prove that |
| Answer» Sin48 = cos42 Using sinA = cos (90-A)Cos48 = sin42 Using cosA= sin(90-A)Now back to question cos42.sec42 + sine42.cosec42Cos42.cos42+sine42.sine42 ANS= 1 | |
| 33573. |
Ex 4,3 Ka 9th question |
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| 33574. |
Find a quadratic polynomial whose zeroes exceed the zeroes of ×2+5×+6 by a number 2 |
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| 33575. |
How tan=sin/cos |
| Answer» Ncert book ki page -175 me he | |
| 33576. |
Prove that root a +rootB is an irrational number if root a b is rational |
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| 33577. |
Find value of the k of the equation three x square +k root 3+4=0. |
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| 33578. |
Express the following in the form of p/q (1) 37.09158 (2) 423.04567 |
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| 33579. |
(3x+1) (x-4) |
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Answer» 3x2 + x - 16 3x^2-11x-4 3x(x-4)+1(x-4)3x^2-12x+x-43x^2-11x-4 Multiple by 1st equation to 2 equation 3x²-11x-4 Solutions |
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| 33580. |
Find the area of quadrilateral ABCD whose vertices are A (1,1)B (7,-3)C(12,2)and D(7,21). |
| Answer» Area of quadrilateral ABCD = ar {tex}\\triangle{/tex}ABD + ar {tex}\\triangle{/tex}BCDar {tex}\\triangle{/tex}ABD =\xa0{tex}\\frac 12{/tex}[ 1(- 3 - 21) + 7(21 - 1) + 7(1 + 3)]=\xa0{tex} \\frac { 1 } { 2 } [ - 24 + 7 \\times 20 + 7 \\times 4 ]{/tex}{tex}= \\frac { 1 } { 2 } {/tex}[-24 + 140 + 28]{tex}= \\frac { 1 } { 2 } \\times 144{/tex}\xa0= 72 sq. unitsar {tex}\\triangle{/tex}BCD=\xa0{tex}\\frac { 1 } { 2 }{/tex}[7(2 - 21) + 12(21 + 3) + 7(-3 - 2)]=\xa0{tex}\\frac { 1 } { 2 } [ 7 \\times - 19 + 12 \\times 24 + 7 \\times - 5 ]{/tex}=\xa0{tex}\\frac12{/tex}[-133 + 288 - 35]=\xa0{tex}\\frac 12{/tex}[288 - 168]{tex}= \\frac { 1 } { 2 } \\times 120{/tex}= 60 sq. unitsHence, Area Quadrilateral ABCD = 72 + 60 = 132 sq units | |
| 33581. |
√log(sin(x square-3/1) |
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| 33582. |
What is the full form of AP |
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Answer» Arithmetic progress Arithmetic Progression |
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| 33583. |
Find the largest number of four digit exactly divisible by 12,15,18 and 27 |
| Answer» 9720. You can find the answer by taking LCM of all the numbers then divide 9999 by the LCM and subtract the remainder from 9999. | |
| 33584. |
Derive pth theoram |
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| 33585. |
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60\' |
| Answer» Radius of the sector = 6 cmAngle of the sector (A) = 60°Area of the sector =( A/360)× πr^2 sq.cm= (60/360)× π×6×6 square.cm\xa0=6π square.cm.\xa0 | |
| 33586. |
SinA(1+tanA)+cosA(1+cotA)=secA +cosecA...how to prove |
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| 33587. |
If p,q are prime ,then find H.C.F of (p,q) |
| Answer» 1 | |
| 33588. |
Find the zeroes of x^2+√15 |
| Answer» It will have imaginary root. The root is 15 ( ¡ = √-1). | |
| 33589. |
Show that the square of any odd positive integer is of the form 8m+1 for some integer m. |
| Answer» Since, any odd positive integer n is of the form 4m\xa0+ 1 or 4m + 3.if n = 4m + 1n2 = (4m + 1)2= 16m2 + 8m + 1= 8(2m2 + m) + 1So n2 = 8q + 1 ........... (i)) (where q = 2m2 + m is a positive integer)If n = (4m + 3)n2 = (4m + 3)2= 16m2 + 24m + 9= 8(2m2 + 3m + 1) + 1So n2 = 8q + 1 ...... (ii) (where q = 2m2 + 3m + 1 is a positive integer)From (i) and (ii) we conclude that the square of an odd positive integer is of the form 8q + 1, for some integer q. | |
| 33590. |
comperment paper |
| Answer» Check Papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 33591. |
What is the L.C.M. of 196 and 38220 |
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Answer» 38220 3822p |
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| 33592. |
What are the zeros of the polynomial |
| Answer» Zeros or roots of the polynomial means a number, a decimal or a fraction when substituted in the place of x in the polynomial the final answer comes 0. | |
| 33593. |
sin²30cos²45+4tan²30+1/2sin²90-2cos²+1/24cos²0 please simply |
| Answer» We know that, Sin30°=(1/2), Cos45°=(1/√2), Sin90°=1, Cos90°=0, Cos0°=1 & tan30°=(1/√3), putting these values in the given expression, we get :-{tex}{\\sin ^2}30^\\circ {\\cos ^2}45^\\circ + 4{\\tan ^2}30^\\circ + \\;\\frac{1}{2}{\\sin ^2}90^\\circ - 2\\cos^2 90^\\circ + \\frac{1}{{24}}{\\cos ^2}0^\\circ {/tex}{tex} = {\\left( {\\frac{1}{2}} \\right)^2} \\times {\\left( {\\frac{1}{{\\sqrt 2 }}} \\right)^2} + 4{\\left( {\\frac{1}{{\\sqrt 3 }}} \\right)^2} + \\frac{1}{2} \\times {\\left( 1 \\right)^2} - 2 \\times {\\left( 0 \\right)^2} + \\frac{1}{{24}}{\\left( 1 \\right)^2}{/tex}{tex} = \\frac{1}{4} \\times \\frac{1}{2} + 4 \\times \\frac{1}{3} + \\frac{1}{2} \\times 1 - 2 \\times 0 + \\frac{1}{{24}} \\times 1{/tex}{tex} = \\frac{1}{8} + \\frac{4}{3} + \\frac{1}{2} - 0 + \\frac{1}{{24}}{/tex}{tex} = \\frac{{3 + 32 + 12 - 0 + 1}}{{24}}{/tex}{tex} = \\frac{{48}}{{24}} = 2{/tex} | |
| 33594. |
If a b are zero polynomial |
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| 33595. |
find th pair of equations which have unique solution x=1,x=5 |
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| 33596. |
Can i get previous years question papers 10th stander all subjects |
| Answer» In this app | |
| 33597. |
Please give some tips for compartment exam |
| Answer» Study trigonometry ,real number and algebra for passing | |
| 33598. |
nsquared-n is even for every positive integer.. How to prove |
| Answer» Any positive integer is of the form 2q or 2q + 1, for some integer q.{tex}\\therefore{/tex} When n = 2q{tex}\\style{font-family:Arial}{n^2\\;-\\;n\\;=\\;n(n\\;-\\;1)\\;=\\;2q(2q\\;-\\;1)=\\;2m,}{/tex}where m = q(2q - 1) ( m is any integer)This is divisible by 2When n = 2q + 1{tex}\\style{font-family:Arial}{\\begin{array}{l}n^2\\;-\\;n\\;=\\;n(n\\;-\\;1)\\;=\\;(2q\\;+\\;1)(2q+1-1)\\\\=2q(2q+1)\\end{array}}{/tex}= 2m, when m = q(2q + 1) ( m is any integer)which is divisible by 2.Hence, n2 - n is divisible by 2 for every positive integer n. | |
| 33599. |
1plus 1 |
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Answer» How 99 2 2 99 |
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| 33600. |
explain in decimal form 15/ 1600 |
| Answer» the decimal expansion of given rational number is:{tex}\\frac { 15 } { 1600 } = \\frac { 15 } { 2 ^ { 4 } \\times 10 ^ { 2 } } = \\frac { 15 \\times 5 ^ { 4 } } { 2 ^ { 4 } \\times 5 ^ { 4 } \\times 10 ^ { 2 } } = \\frac { 9375 } { 10 ^ { 6 } } = .009375{/tex} | |