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If the sum of the squares of the reciprocals of the roots of alpha and beta of 3x^2+lambdax-1=0 is
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If the sum of the squares of the reciprocals of the roots of alpha and beta of 3x^2+lambdax-1=0 is
The number of real roots of equation : x|x-2|+3|x-3|+1 =0 | JEE MAIN 2025 PYQ Quadratic Equations
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The number of real roots of equation : x|x-2|+3|x-3|+1 =0 | JEE MAIN 2025 PYQ Quadratic Equations
The sum of all real roots of equation (e^2x-4)(6e^2x-5e^x+1)=0| JEE MAIN 2022 MATHS PYQ SOLUTION
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The sum of all real roots of equation (e^2x-4)(6e^2x-5e^x+1)=0| JEE MAIN 2022 MATHS PYQ SOLUTION
The sum of the squares of all the roots of the equation x^2+|2x-3|-4=0 is | JEE MAIN 2025 Maths PYQ
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The sum of the squares of all the roots of the equation x^2+|2x-3|-4=0 is | JEE MAIN 2025 Maths PYQ
Let alpha and beta are roots of x^2+√2x-8=0. If Un=alpha^n+beta^n ,then (U10+√2U9)/2U8 is equal to
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Let alpha and beta are roots of x^2+√2x-8=0. If Un=alpha^n+beta^n ,then (U10+√2U9)/2U8 is equal to
alpha+beta=1, alpha.beta=-1. Pn=alpha^n+beta^n , Pn-1=11, Pn+1=29, then the value of (Pn)^2
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alpha+beta=1, alpha.beta=-1. Pn=alpha^n+beta^n , Pn-1=11, Pn+1=29, then the value of (Pn)^2
let alpha and beta be the roots of the equation 5x^2+6x-2=0 | jee main 2020 Quadratic Equations
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let alpha and beta be the roots of the equation 5x^2+6x-2=0 | jee main 2020 Quadratic Equations
alpha and beta are distinct roots of the equation x^2-(t^2-5t+6)x+1=0, an=alpha^n+beta^n. Then..
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alpha and beta are distinct roots of the equation x^2-(t^2-5t+6)x+1=0, an=alpha^n+beta^n. Then..
alpha and beta be the roots of equation x^2-√2x-√3=0. Pn=alpha^n-beta^n. (11√3-10√2)P10+(11√2+10)..
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alpha and beta be the roots of equation x^2-√2x-√3=0. Pn=alpha^n-beta^n. (11√3-10√2)P10+(11√2+10)..
alpha and beta are roots of x^2+√3x-16=0 ,gamma and delta roots of x^2+3x-1=0.Pn=alpha^n+beta^n, Qn=
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alpha and beta are roots of x^2+√3x-16=0 ,gamma and delta roots of x^2+3x-1=0.Pn=alpha^n+beta^n, Qn=
Newton’s formula: alpha and beta are roots of equation x^2-6x-2=0.an=alpha^n-beta^n,then a10-2a8/2a9
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Newton’s formula: alpha and beta are roots of equation x^2-6x-2=0.an=alpha^n-beta^n,then a10-2a8/2a9
let alpha and beta are roots of equation x^2-6x-2=0. if an=alpha^n-beta^n,then a10-2a8/3a9 | jee
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let alpha and beta are roots of equation x^2-6x-2=0. if an=alpha^n-beta^n,then a10-2a8/3a9 | jee
let alpha and beta are roots of equation x^2-6x-2=0. if an=alpha^n-beta^n,then (a10-2a8)/2a9 | Jee
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let alpha and beta are roots of equation x^2-6x-2=0. if an=alpha^n-beta^n,then (a10-2a8)/2a9 | Jee
the value of 4+1/5+1/4+1/5+1/4+… infinity is | Quadratic Equations JEE MAIN PYQ Simplified
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the value of 4+1/5+1/4+1/5+1/4+… infinity is | Quadratic Equations JEE MAIN PYQ Simplified
the value of 3+1/4+1/3+1/4+1/3+… infinity is | Quadratic Equations JEE MAIN MATHS PYQ Simplified
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the value of 3+1/4+1/3+1/4+1/3+… infinity is | Quadratic Equations JEE MAIN MATHS PYQ Simplified
If alpha, beta are roots of quadratic equation 2x^2+6x-5=0 then find alpha+beta, alpha-beta
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If alpha, beta are roots of quadratic equation 2x^2+6x-5=0 then find alpha+beta, alpha-beta
alpha, beta are roots of quadratic equation y^2-2y-7=0 find alpha^2 + beta^2 | easy method
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alpha, beta are roots of quadratic equation y^2-2y-7=0 find alpha^2 + beta^2 | easy method
alpha, beta are roots of quadratic equation y^2-2y-7=0 find alpha^3 + beta^3 | easy method
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alpha, beta are roots of quadratic equation y^2-2y-7=0 find alpha^3 + beta^3 | easy method
Sum of roots of Quadratic equation is double their product. find k if equation is x^2-4kx+k+3=0
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Sum of roots of Quadratic equation is double their product. find k if equation is x^2-4kx+k+3=0
Alpha and beta are roots of the Quadratic equation x^2-5x+6=0 then find alpha^2 + beta^2
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Alpha and beta are roots of the Quadratic equation x^2-5x+6=0 then find alpha^2 + beta^2
Alpha and beta are roots of quadratic equation x^2-5x+6=0 find alpha/beta + beta/alpha
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Alpha and beta are roots of quadratic equation x^2-5x+6=0 find alpha/beta + beta/alpha
1+2+3+...+1000 = | Sum of first n natural numbers | 1+2+3+...+50 = | 1+2+3+...+2000=
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1+2+3+...+1000 = | Sum of first n natural numbers | 1+2+3+...+50 = | 1+2+3+...+2000=
z1=2-i , z2=1+i find z1+z2+1/z1-z2+i | modulus of complex numbers
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z1=2-i , z2=1+i find z1+z2+1/z1-z2+i | modulus of complex numbers
1^3+2^3+3^3+...+n^3 formula | 1^3+2^3+3^3+...+10^3 = |1^3+2^3+3^3+...+20^3 = |1^3+2^3+3^3+...+30^3
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1^3+2^3+3^3+...+n^3 formula | 1^3+2^3+3^3+...+10^3 = |1^3+2^3+3^3+...+20^3 = |1^3+2^3+3^3+...+30^3
find the derivative of x = cos(xy) with respect to x | x = cos(xy) , find dy/dx | Differentiation
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find the derivative of x = cos(xy) with respect to x | x = cos(xy) , find dy/dx | Differentiation
y= cos^-1(e^x) find dy/dx | y= cos inverse (e^x) find dy/dx | 11th 12th Differentiation jee cbse
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y= cos^-1(e^x) find dy/dx | y= cos inverse (e^x) find dy/dx | 11th 12th Differentiation jee cbse
y=cos√x find dy/dx | y=cos sqrt x find dy/dx| y= cos root x find dy/dx | 12th Differentiation
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y=cos√x find dy/dx | y=cos sqrt x find dy/dx| y= cos root x find dy/dx | 12th Differentiation
y=e^sin^-1x find dy/dx | y=e^sin^inverse x find dy/dx |inverse trigonometry differentiation 12th
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y=e^sin^-1x find dy/dx | y=e^sin^inverse x find dy/dx |inverse trigonometry differentiation 12th
y=e^√x√x√x... then d^2y/dx^2 at x=log3 |differentiation class 12th solutions jee main CBSE maths
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y=e^√x√x√x... then d^2y/dx^2 at x=log3 |differentiation class 12th solutions jee main CBSE maths
x^2+y^2=t+1/t and x^4+y^4=t^2+1/t^2 then dy/dx |Differentiation 12th JEE CBSE Engineering CETs
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x^2+y^2=t+1/t and x^4+y^4=t^2+1/t^2 then dy/dx |Differentiation 12th JEE CBSE Engineering CETs