JEE Main 2023:Let x = 2 be a root of the equation x²+px+q=0 and f(x) = { 1-cos(x²-4px+q²+8q+16)

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Let x=2 be a root of equation x^2+px+q=0 &f(x)=1-cos(x2-4px+q2+8q+16)/(x-2p)^2,x#2p &0at x=2p then Let x=2 be a root of equation x^2+px+q=0 &f(x)=1-cos(x2-4px+q2+8q+16)/(x-2p)^2,x#2p &0at x=2p then
Let x = 2 be a root of the equation x^4 +px +q = 0 and f(x) = { 1- cos(x^2 - 4px +q^2 +8q +16) Let x = 2 be a root of the equation x^4 +px +q = 0 and f(x) = { 1- cos(x^2 - 4px +q^2 +8q +16)
Let x = 2 be a root of the equation x2+px+q=0 and f(x)=1−cos⁡x2−4px+q2+8q+16(x−2p)4,x≠2p0,x=2pThen Let x = 2 be a root of the equation x2+px+q=0 and f(x)=1−cos⁡x2−4px+q2+8q+16(x−2p)4,x≠2p0,x=2pThen
Let x=2 be root of the equation x^2+px+q=0 and f(x)=(1-cos(x^2-4px+q^2+8q+16)/(x-2p)^4, x\neq 0.. Let x=2 be root of the equation x^2+px+q=0 and f(x)=(1-cos(x^2-4px+q^2+8q+16)/(x-2p)^4, x\neq 0..
Q) Let 𝑓:𝑅→𝑅 be defined by 𝑓(𝑥)=𝑥/(1+𝑥^2 ), 𝑥∈𝑅. Then the range of 𝑓 i #MATHS #jee #jee2025 #jeemain Q) Let 𝑓:𝑅→𝑅 be defined by 𝑓(𝑥)=𝑥/(1+𝑥^2 ), 𝑥∈𝑅. Then the range of 𝑓 i #MATHS #jee #jee2025 #jeemain
Q) Let 𝑦=𝑦(𝑥) be the solution of the differential equation (𝑥𝑦−5𝑥^2 √(1+𝑥^2 ))𝑑𝑥+(1+𝑥^2 ) Q) Let 𝑦=𝑦(𝑥) be the solution of the differential equation (𝑥𝑦−5𝑥^2 √(1+𝑥^2 ))𝑑𝑥+(1+𝑥^2 )
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