Let α and β be the real numbers such that _ x arrow 0 (1)/(x^3)((α)/(2) ∫_0^x (1)/(1-t^2) d t+β x x)=2 . Then the value of α+β is _______

Correct answer: 0 _ x arrow 0 ((α)/(2) ∫_0^x (1)/(1-t^2) d t+β x x)/(x^3) aligned & = _ x arrow 0 ((α)/(2)((1)/(1-x^2))+β x-β x x)/(3 x^2) & = (α)/(2)(1-x^2)^ -1 +β(1-(x^2)/(2!)+(x^4)/(4!) )-β x(x-(x^3)/(3!)+(x^5)/(5!) ) 3 x^2 & =((α)/(2)(1+x^2+x^4 )+β(1-(x^2)/(2!)+(x^4)/(4!) )-β(x^2-(x^4)/(3!) ))/(3 x^2) aligned aligned & = ((α)/(2)+β)+x^2((α)/(2)-(β)/(2)-β)+x^4() 3 x^2 =2( Given ) & (α)/(2)+β=0 and (α-3 β)/(6)=2 & α=-2 β and α=12+3 β & β=-(12)/(5) and α=(24)/(5) aligned α+β=(12)/(5)=2.40

JEE Advanced 2025Paper 1

Exam: JEE Advanced
Year: 2025
Paper: Paper 1
Subject: Mathematics
Chapter: Limits

JEE Advanced 2025 · Paper 1 · Mathematics · Limits

Let α and β be the real numbers such that _ x arrow 0...

Let

\alpha

and

\beta

be the real numbers such that

\lim _{x \rightarrow 0} \frac{1}{x^3}\left(\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} d t+\beta x \cos x\right)=2

. Then the value of

\alpha+\beta

is _______

Correct answer

Step-by-step solution

\lim _{x \rightarrow 0} \frac{\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} d t+\beta x \cos x}{x^3}

\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\frac{\alpha}{2}\left(\frac{1}{1-x^2}\right)+\beta \cos x-\beta x \sin x}{3 x^2} \\ & =\frac{\frac{\alpha}{2}\left(1-x^2\right)^{-1}+\beta\left(1-\frac{x^2}{2!}+\frac{x^4}{4!} \ldots\right)-\beta x\left(x-\frac{x^3}{3!}+\frac{x^5}{5!} \ldots\right)}{3 x^2} \\ & =\frac{\frac{\alpha}{2}\left(1+x^2+x^4 \ldots\right)+\beta\left(1-\frac{x^2}{2!}+\frac{x^4}{4!} \ldots\right)-\beta\left(x^2-\frac{x^4}{3!} \ldots\right)}{3 x^2}\end{aligned}

\begin{aligned} & =\frac{\left(\frac{\alpha}{2}+\beta\right)+\mathrm{x}^2\left(\frac{\alpha}{2}-\frac{\beta}{2}-\beta\right)+\mathrm{x}^4() \ldots}{3 \mathrm{x}^2}=2(\text{ Given }) \\ & \therefore \frac{\alpha}{2}+\beta=0 \text{\text{ and } } \frac{\alpha-3 \beta}{6}=2 \\ & \Rightarrow \alpha=-2 \beta \text{ and } \alpha=12+3 \beta \\ & \Rightarrow \beta=-\frac{12}{5} \text{ and } \alpha=\frac{24}{5}\end{aligned}

\therefore \alpha+\beta=\frac{12}{5}=2.40

Practice Limits on Quantrex Academy →

More from Limits

Let _ x → 2 (( (x-2))(rx^2 + (p-2)x - 2p))/((x-2)^2) = 5 for some r, p ∈ ℝ . If the set… JEE Main 2026 The value of _ x → 0 ((x^2 ^2 x)/(x^2 - ^2 x)) is: JEE Main 2026 Let f(x) = _ y → 0 ((1 - (xy)) (xy))/(y^3) . Then the number of solutions of the equation… JEE Main 2026 The product of all possible values of α , for which _ x → 0 ((1 - (α x) ((α+1)x)… JEE Main 2026 If _ x → 2 (x^3 - 5x^2 + ax + b) ( x-1 - 1) _e(x-1) = m , then a + b + m is equal to : JEE Main 2026 The value of _ x arrow 0 _ e ( (e x) · (e^ 2 x) · … · (e^ 10 x)) e^ 2 -e^ 2 x is equal to JEE Main 2026 If _ x arrow 0 e^ (a-1) x +2 ~b x+(c-2) e^ -x x x- _ e (1+x) =2 , then a^ 2 +b^ 2 +c^ 2… JEE Main 2026 Let [·] denote the greatest integer function and f(x)= _ n arrow ∞ 1 n^ 3 ∑_ k=1 ^ n [ k^… JEE Main 2026 Let f: R arrow(0, ∞) be a twice differentiable function such that f(3)=18, f^ (3)=0 and… JEE Main 2026 Let x_0 be the real number such that e^ x_0 +x_0=0 . For a given real number α , define… JEE Advanced 2025 Given below are two statements : Statement I : _ x arrow 0 ( ^ -1 x+ _e (1+x)/(1-x) -2 x… JEE Main 2025 If the function f(x)=( ( x)- ( x))/( x- x) is continuous at x=0 , then f(0) is equal to… JEE Main 2025 _ x arrow 0^ + (5(x)^ (1)/(3) ) _e(1+3 x^2) ( ^ -1 3 x )^2(e^ 5(x)^ (4)/(3) -1) is equal… JEE Main 2025 If _ x arrow 1^ + ((x-1)(6+λ (x-1))+μ (1-x))/((x-1)^3)=-1 , where λ, μ ∈ ℝ , then λ+μ is… JEE Main 2025 If Lim_ x arrow 0 (( x)/(x))^ (1)/(x^2) =p , then 96 _e p is equal to ______ JEE Main 2025 If _ x arrow 0 ( (2 x)+a (4 x)-b)/(x^4) is finite, then (a+b) is equal to : JEE Main 2025 Full Limits list Limits JEE Maths PYQs JEE Mathematics PYQs All JEE Advanced PYQs