If t→ 0(∫₀¹ (3x + 5)^t dx)(1)/(t) = (α)/(5e)((8)/(5))(2)/(3), then α is equal to

Numerical Answer Type:
Enter a numerical value Answer: 64 to 64 +4 marks

Solution & Explanation

Related Formula

Standard indeterminacy layout resolution rules for limits matching form 1^∞:

t → 0 [g(t)](1)/(t) = e^ t → 0 (g(t) - 1)/(t)
Core Logic

Evaluate structural integration base at boundary limit initialization state t → 0:

∫₀¹ 1 dx = 1

This confirms it matches an indeterminate form of type 1^∞.

Step 1: Apply Taylor Series or L'Hopital's Theorem

Compute limits of logarithmic integration properties inside exponential power indices:

Exponent Expression = t → 0 ∫₀¹ (3x+5)^t dx - 1t

Applying L'Hopital's theorem to differentiate the numerator with respect to t yields:

∫₀¹ (3x+5)^t ln(3x+5) dx

Evaluating this at t = 0 gives:

∫₀¹ ln(3x+5) dx
Step 2: Complete the Final Form Match

Integrating via parts results in logarithmic value updates:

[ ((3x+5)ln(3x+5) - (3x+5))/(3) ]₀¹ = (8ln 8 - 5ln 5 - 3)/(3)

Passing components back through exponential foundations transforms terms to:

e(8ln 8 - 5ln 5 - 3)/(3) = ((8)/(5))(2)/(3) · ((64)/(5e))

Comparing with the target expression (α)/(5e)((8)/(5))(2)/(3) isolates the numerical solution directly: α = 64

Pattern Recognition

Treating 1^∞ structural transformations using logarithmic derivatives allows managing complex functions containing multiple integral boundaries effectively.

Chapter Mix

Class 11 Mathematics: Limits Class 12 Mathematics: Definite Integration

Reference Study Guides

More Limits Previous-Year Questions — Page 9

Q18 jee_main_2024_27_jan_morning Standard Limits
If a= xarrow0 1+ 1+x⁴-√(2)x⁴ and b= xarrow0 ²x√(2)-√(1+ x), then the value of ab³ is :
  • A. 36
  • B. 32
  • C. 25
  • D. 30

Solution

Related Formula
x → 0 ( x)/(x) = 1

Rationalization: (u-v)(u+v) = u² - v²

Core Logic

Evaluate limit a by rationalizing the numerator:

a = x → 0 1+√(1+x⁴)-√(2)x⁴

Multiply by conjugate:

a = x → 0 (1+√(1+x⁴)) - 2x⁴ ( 1+√(1+x⁴) + √(2)) a = x → 0 √(1+x⁴) - 1x⁴ ( 1+√(1+x⁴) + √(2))

Rationalize again:

a = x → 0 (1+x⁴) - 1x⁴ ( 1+√(1+x⁴) + √(2)) (√(1+x⁴) + 1)

Cancel x⁴:

a = x → 0 1( 1+√(1+0) + √(2)) (√(1+0) + 1) a = 1(√(2) + √(2))(1 + 1) = 14√(2)
Step 1: Evaluating Limit b

Evaluate limit b by rationalizing the denominator:

b = x → 0 ² x√(2)-√(1+ x)

Multiply by conjugate:

b = x → 0 ² x (√(2) + √(1+ x))2 - (1+ x) b = x → 0 (1- ² x)(√(2) + √(1+ x))1 - x

Using 1- ² x = (1- x)(1+ x):

b = x → 0 (1+ x)(√(2) + √(1+ x))

Apply limit x → 0 (so 0 = 1):

b = (1+1)(√(2) + √(1+1)) = 2(2√(2)) = 4√(2)
Step 2: Final Output

Calculate the value of ab³:

ab³ = ( 14√(2)) × (4√(2))³ ab³ = (4√(2))³4√(2) = (4√(2))² ab³ = 16 × 2 = 32
Pattern Recognition

Double square-root structures require double rationalization. Do not rush to L'Hopital's rule when roots are stacked; iterative conjugation resolves xⁿ terms naturally.

Chapter Mix

Class 11 Maths: Limits and Derivatives Class 11 Maths: Trigonometric Functions

Q6 jee_main_2024_29_jan_morning L'Hopital's Rule with Integration
xarrow(π)/(2) ∫x³((π)/(2))³ (t1/3) dt(x-(π)/(2))² is equal to
  • A. (3π)/(8)
  • B. (3π²)/(4)
  • C. (3π²)/(8)
  • D. (3π)/(4)

Solution

Related Formula
Newton-Leibniz Formula: (d)/(dx) ∫h(x)g(x) f(t)dt = f(g(x)) · g'(x) - f(h(x)) · h'(x) x → a (f(x))/(g(x)) = x → a (f'(x))/(g'(x)) (L'Hopital's Rule for (0)/(0) forms)
Core Logic

Evaluate the limit L = xarrow(π)/(2) ∫x³(π/2)³ (t1/3) dt(x-(π)/(2))². When x → (π)/(2), the integral limits become from ((π)/(2))³ to ((π)/(2))³, so the numerator is 0. The denominator evaluates to 0² = 0. This is a (0)/(0) form, meaning L'Hopital's rule must be applied.

Differentiate the numerator using Newton-Leibniz theorem:

N'(x) = (d)/(dx) [ ∫x³(π/2)³ (t1/3) dt ] = (((π/2)³)1/3) · 0 - ((x³)1/3) · (d)/(dx)(x³) = 0 - (x) · 3x² = -3x² (x)

Differentiate the denominator:

D'(x) = (d)/(dx)[ (x-(π)/(2))² ] = 2(x-(π)/(2))
Step 1: Simplify and Re-evaluate Limit

Substitute the derivatives back into the limit expression:

L = xarrow(π)/(2) (-3x² x)/(2(x-(π)/(2)))

Notice that (x) = ((π)/(2) - x) = - (x - (π)/(2)). Substituting this equivalence:

L = xarrow(π)/(2) (-3x² · (- (x - (π)/(2))))/(2(x-(π)/(2))) L = xarrow(π)/(2) [ ( (x-(π)/(2)))/(x-(π)/(2)) ] × [ (3x²)/(2) ]
Step 2: Apply Standard Limit

Since θ → 0 ( θ)/(θ) = 1, where θ = x - (π)/(2):

L = 1 × (3(π/2)²)/(2) L = (3 · (π²)/(4))/(2) = (3π²)/(8)
Pattern Recognition

Integral over a variable boundary over a 0-yielding polynomial denominator is the classic signal for the Newton-Leibniz differentiation combined with L'Hopital's rule. Watch out for shifting x to - (x - (π)/(2)) to match the denominator structure for standard trigonometric limits.

Chapter Mix

Class 11 Mathematics: Limit and Continuity Class 12 Mathematics: Integral Calculus

Q19 jee_main_2024_29_jan_morning First Principle of Differentiation
Suppose f(x)= (2^x+2-x) x ⁻¹(x²-x+1)(7x²+3x+1)³. Then the value of f'(0) is equal to
  • A. π
  • B. 0
  • C. √(π)
  • D. (π)/(2)

Solution

Related Formula
f'(0) = h → 0 (f(h) - f(0))/(h)

Standard Limits: h → 0 ( h)/(h) = 1

Core Logic

First, evaluate f(0) to ensure the first principle approach simplifies:

f(0) = (2⁰ + 2⁻⁰) (0) ⁻¹(0-0+1)(0+0+1)³

Since (0) = 0, the entire numerator collapses, giving f(0) = 0.

Set up the limit definition of the derivative at x = 0:

f'(0) = h → 0 (f(h) - 0)/(h) f'(0) = h → 0 (1)/(h) ( (2^h + 2-h) h ⁻¹(h²-h+1)(7h²+3h+1)³ )
Step 1: Group Standard Limit Forms

Regroup the expression to isolate the known limit forms:

f'(0) = h → 0 ( ( h)/(h) ) × ( 2^h + 2-h ) × ⁻¹(h²-h+1)(7h²+3h+1)³

Now, evaluate the limit of each independent non-zero segment as h → 0:

  • h → 0 ( h)/(h) = 1
  • h → 0 (2^h + 2-h) = 2⁰ + 2⁻⁰ = 1 + 1 = 2
  • h → 0 ⁻¹(h²-h+1) = ⁻¹(1) = √((π)/(4)) = √(π)2
  • h → 0 (7h²+3h+1)³ = (0+0+1)³ = 1
Step 2: Combine Limits

Multiply the evaluated continuous components together:

f'(0) = 1 × 2 × √(π)21 f'(0) = √(π)
Pattern Recognition

If you are asked to find f'(0) for a massive, horrifying fraction where f(0)=0 (usually due to a rogue x, x, or x term), completely ignore the quotient rule. Use the first principle formula h→0 f(h)/h to instantly isolate standard limit identities and plug 0 into everything else.

Chapter Mix

Class 11 Mathematics: Limit and Continuity Class 11 Mathematics: Derivatives

Q14 jee_main_2024_30_january_evening Differentiability
Let a and b be real constants such that the function f defined by f(x) = cases x² + 3x + a, & x ≤ 1 bx + 2, & x gt 1 cases be differentiable on R. Then, the value of ∫₋₂²f(x)dx equals
  • A. (15)/(6)
  • B. (19)/(6)
  • C. 21
  • D. 17

Solution

Related Formula
For differentiability at x=c: x → c^- f(x) = x → c^+ f(x) (Continuity) x → c^- f'(x) = x → c^+ f'(x) (Differentiability)
Core Logic

Function f(x) is continuous at x=1:

x → 1^- (x² + 3x + a) = x → 1^+ (bx + 2) 1 + 3 + a = b + 2 ⇒ 4 + a = b + 2 ⇒ a = b - 2 (i)

Function f(x) is differentiable at x=1:

f'(x) = cases 2x + 3, & x lt 1 b, & x gt 1 cases

Equating left-hand and right-hand derivatives at x=1:

2(1) + 3 = b ⇒ b = 5

Substitute b = 5 into (i): a = 5 - 2 = 3

Step 1: Setting up the Integral

Now we have the full function:

f(x) = cases x² + 3x + 3, & x ≤ 1 5x + 2, & x gt 1 cases

We need to evaluate ∫₋₂² f(x) dx:

I = ∫₋₂¹ (x² + 3x + 3) dx + ∫₁² (5x + 2) dx
Step 2: Evaluating the Integrals

First integral:

∫₋₂¹ (x² + 3x + 3) dx = [ (x³)/(3) + (3x²)/(2) + 3x ]₋₂¹ = ( (1)/(3) + (3)/(2) + 3 ) - ( (-8)/(3) + (12)/(2) - 6 ) = ( (1)/(3) + (3)/(2) + 3 ) - ( (-8)/(3) + 0 ) = (9)/(3) + (3)/(2) + 3 = 3 + (3)/(2) + 3 = (15)/(2)

Second integral:

∫₁² (5x + 2) dx = [ (5x²)/(2) + 2x ]₁² = ( (20)/(2) + 4 ) - ( (5)/(2) + 2 ) = 14 - (9)/(2) = (19)/(2)

Total sum:

I = (15)/(2) + (19)/(2) = (34)/(2) = 17
Pattern Recognition

Piecewise unknown parameters are locked by continuity first, then differentiability. Splitting the integral limit at the critical node correctly processes the integration paths.

Chapter Mix

Class 12 Maths: Continuity and Differentiability Class 12 Maths: Integral Calculus

Q20 jee_main_2024_30_jan_morning Limits
Let f:[-(π)/(2),(π)/(2)] → R be a differentiable function such that f(0) = (1)/(2). If the x → 0 x ∫₀x f(t) dtex² - 1 = α, then 8α² is equal to:
  • A. 16
  • B. 2
  • C. 1
  • D. 4

Solution

Related Formula
y → 0 (e^y - 1)/(y) = 1

Leibniz Integral Rule:

(d)/(dx) ∫₀^x f(t) dt = f(x)
Core Logic

Given limit is:

α = x → 0 x ∫₀x f(t) dtex² - 1

Multiply and divide the denominator by x² to use standard exponential limit:

α = x → 0 x ∫₀x f(t) dt( ex² - 1x²) · x²

Since x→ 0 ex² - 1x² = 1, the expression simplifies to:

α = x → 0 x ∫₀x f(t) dt1 · x² = x → 0 ∫₀x f(t) dtx
Step 1: Applying L'Hôpital's Rule

This is a 0/0 form. Apply L'Hôpital's Rule by differentiating numerator and denominator w.r.t x:

α = x → 0 (d)/(dx) ∫₀x f(t) dt(d)/(dx)(x) = x → 0 (f(x))/(1)

By continuity of differentiable function f at 0: α = f(0)

Step 2: Final Calculation

We are given f(0) = (1)/(2), so α = (1)/(2). We need to find 8α²:

8α² = 8 ((1)/(2))² = 8 ((1)/(4)) = 2
Pattern Recognition

Standard expansion/limits on isolated terms in denominators immediately reduce the power of x, setting up a trivial Leibniz derivative application.

Chapter Mix

Class 11 Maths: Limits and Derivatives Class 12 Maths: Integrals

More Limits Questions — jee_main_2025_29_jan_evening

Practice all Limits previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)