JEE Main 2025 2 April Evening Shift
Exam JEE Main
Year 2025
Date 2 April
Shift Evening Shift
Subject Mathematics
Chapter Limits JEE Main 2025 · 2 April · Evening Shift · Mathematics · Limits
If _ x arrow 0 ( (2 x)+a (4 x)-b)/(x^4) is finite, then (a+b) is...
If
lim x → 0 cos ( 2 x ) + a cos ( 4 x ) − b x 4 \lim _{x \rightarrow 0} \frac{\cos (2 x)+a \cos (4 x)-b}{x^4} lim x → 0 x 4 c o s ( 2 x ) + a c o s ( 4 x ) − b is finite, then
( a + b ) (a+b) ( a + b ) is equal to :
Options A 1 2 \frac{1}{2} 2 1 B 0 0 0 C 3 4 \frac{3}{4} 4 3 D − 1 -1 − 1 Correct answer A. 1 2 \frac{1}{2} 2 1
Step-by-step solution
lim x → 0 cos 2 x + a cos 4 x − b x 4 = \lim _{x \rightarrow 0} \frac{\cos 2 x+a \cos 4 x-b}{x^4}= lim x → 0 x 4 c o s 2 x + a c o s 4 x − b = finite
L = { 1 − ( 2 x ) 2 2 ‾ + ( 2 x ) 4 4 ‾ ⋯ } } + a { 1 − ( 4 x ) 2 2 ‾ + ( 4 x ) 4 ⌊ 4 … ⋅ } − b x 4 L = ( 1 + a − b ) − x 2 ( 2 + 8 a ) + x 4 ( 2 3 + 32 3 a ) + x 6 ( ) … x 4 \begin{aligned} & L=\frac{\left\{1-\frac{(2 x)^2}{\underline{2}}+\frac{(2 x)^4}{\underline{4} \cdots\}\}+a\left\{1-\frac{(4 x)^2}{\underline{2}}+\frac{(4 x)^4}{\lfloor 4} \ldots \cdot\right\}-b}\right.}{x^4} \\ & L=\frac{(1+a-b)-x^2(2+8 a)+x^4\left(\frac{2}{3}+\frac{32}{3} a\right)+x^6() \ldots}{x^4}\end{aligned} L = x 4 { 1 − 2 ( 2 x ) 2 + 4 ⋯ }} + a { 1 − 2 ( 4 x ) 2 + ⌊ 4 ( 4 x ) 4 … ⋅ } − b ( 2 x ) 4 L = x 4 ( 1 + a − b ) − x 2 ( 2 + 8 a ) + x 4 ( 3 2 + 3 32 a ) + x 6 ( ) … ∴ 1 + a − b = 0 and 2 + 8 a = 0 ⇒ a = − 1 4 b = a + 1 = − 1 4 + 1 = 3 4 ∴ a + b = − 1 4 + 3 4 = 1 2 \begin{aligned} & \therefore 1+\mathrm{a}-\mathrm{b}=0 \text{ and } 2+8 \mathrm{a}=0 \Rightarrow \mathrm{a}=-\frac{1}{4} \\ & \mathrm{ b}=\mathrm{a}+1 \\ & =-\frac{1}{4}+1=\frac{3}{4} \\ & \therefore \mathrm{a}+\mathrm{b}=-\frac{1}{4}+\frac{3}{4}=\frac{1}{2}\end{aligned} ∴ 1 + a − b = 0 and 2 + 8 a = 0 ⇒ a = − 4 1 b = a + 1 = − 4 1 + 1 = 4 3 ∴ a + b = − 4 1 + 4 3 = 2 1 $
Practice Limits on Quantrex Academy →
← Previous question If Lim_ x arrow 0 (( x)/(x))^ (1)/(x^2) =p , then 96 _e p is equal to ______Next question → For α, β, γ, ∈ R , if _ x arrow 0 x^2 α x+(γ-1) e^ x^2 2 x-β x =3 , then β+γ-α is equal…
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 α and β be the real numbers such that _ x arrow 0 (1)/(x^3)((α)/(2) ∫_0^x (1)/(1-t^2)… JEE Advanced 2025 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
Full Limits list
Limits JEE Maths PYQs JEE Mathematics PYQs
All JEE Main PYQs
