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 6

Q75 jee_main_2025_04_april_morning Differentiability of Maximum Functions
Let m and n be the number of points at which the function f(x) = x, x³, x⁵, , x²¹ for x in R is not differentiable and not continuous, respectively. Then m + n is equal to
Numerical Answer. Answer: 3 to 3

Solution

Related Formula

A function is non-differentiable at sharp corner transition points where left-hand and right-hand derivatives do not match.

Core Logic

Analyze the behavior of powers of x across significant transition domains: For x < -1: x is the largest because higher odd powers of negative fractions decrease rapidly (x > x³ > x⁵...). For -1 ≤ x < 0: x²¹ is largest (closest to zero from below). For 0 ≤ x < 1: x is largest. For x ≥ 1: x²¹ is largest.

f(x) = cases x, & x < -1 x²¹, & -1 ≤ x < 0 x, & 0 ≤ x < 1 x²¹, & x ≥ 1 cases
Step 1: Continuity and Differentiability Checks

At critical intersection boundaries x = -1, 0, 1, f(x) matches continuous values perfectly, so n = 0. Now check derivative transitions f'(x):

f'(x) = cases 1, & x < -1 21x²⁰, & -1 < x < 0 1, & 0 < x < 1 21x²⁰, & x > 1 cases

At x = -1: LHD = 1, RHD = 21(-1)²⁰ = 21 Non-differentiable. At x = 0: LHD = 0, RHD = 1 Non-differentiable. At x = 1: LHD = 1, RHD = 21(1)²⁰ = 21 Non-differentiable.

Step 2: Conclusion

Thus, the function is non-differentiable at exactly 3 points (x = -1, 0, 1), so m = 3. Since n = 0:

m + n = 3 + 0 = 3
Pattern Recognition

Maximum boundary tracking curves for standard power elements always form continuous shapes but introduce non-differentiable sharp corners at every intersection crossover point.

Chapter Mix

Class 12 Mathematics: Limits, Continuity and Differentiability

Q58 jee_main_2025_07_april_evening Polynomial Limits and Extrema
Let f: R → R be a polynomial function of degree four having extreme values at x = 4 and x = 5. If x→ 0 f(x)x² = 5, then f(2) is equal to :
  • A. 12
  • B. 10
  • C. 8
  • D. 14

Solution

Related Formula

For a finite limit x→ 0 (f(x))/(xⁿ) = L, the lowest powers of x below degree n in the polynomial f(x) must vanish.

Core Logic

Let the 4th-degree polynomial be:

f(x) = ax⁴ + bx³ + cx² + dx + e

Given:

xarrow 0 (ax⁴ + bx³ + cx² + dx + e)/(x²) = 5

For the limit to exist and equal 5, the terms dx and e must be 0, and the coefficient of x² must be equal to 5:

c = 5, d = 0, e = 0

Thus, the polynomial simplifies to:

f(x) = ax⁴ + bx³ + 5x²
Step 1: Use Extrema Conditions

Differentiating f(x) with respect to x:

f'(x) = 4ax³ + 3bx² + 10x = x(4ax² + 3bx + 10)

Since f(x) has extreme values at x=4 and x=5, f'(4) = 0 and f'(5) = 0. This means 4 and 5 are roots of the quadratic factor 4ax² + 3bx + 10 = 0.

Step 2: Solve Coefficients

Using properties of roots for 4ax² + 3bx + 10 = 0:

Product of roots = 4 · 5 = 20 = (10)/(4a) 4a = (10)/(20) = (1)/(2) a = (1)/(8) Sum of roots = 4 + 5 = 9 = -(3b)/(4a)

Substituting 4a = (1)/(2):

9 = -(3b)/(1/2) = -6b b = -(9)/(6) = -(3)/(2)

Our full polynomial is:

f(x) = (1)/(8)x⁴ - (3)/(2)x³ + 5x²
Step 3: Calculate f(2)

Evaluate at x = 2:

f(2) = (1)/(8)(2⁴) - (3)/(2)(2³) + 5(2²) = (16)/(8) - (24)/(2) + 20 = 2 - 12 + 20 = 10
Pattern Recognition

Whenever a limit explicitly matches a denominator power xⁿ, it directly yields both the lower-order coefficients as zeroes and the xⁿ coefficient as the limit value.

Chapter Mix

Class 12 Mathematics: Limits, Continuity and Differentiability

Q71 jee_main_2025_07_april_evening Continuity of Functions
If the function f(x) = ( ( x) - ( x))/( x - x) is continuous at x = 0, then f(0) is equal to ________.
Numerical Answer. Answer: 2 to 2

Solution

Related Formula

For continuity at x=0, f(0) = x → 0 f(x).

Core Logic

We need to evaluate the limit:

x arrow 0 ( ( x) - ( x))/( x - x)

Adding and subtracting x inside the numerator:

x arrow 0 (( ( x) - x) + ( x - x) + ( x - ( x)))/( x - x)

Divide individual parts by x³ across standard series layouts directly yields the combined fractional evaluation equal to 2.

Step 1: Final Resolution

The limit evaluates cleanly to 2. Therefore, for continuity, f(0) = 2.

Pattern Recognition

Expansion of expansion functions like ( x) simplifies smoothly when paired strategically with basic structural Taylor series expansions.

Chapter Mix

Class 12 Mathematics: Limits, Continuity and Differentiability

Q73 jee_main_2025_07_april_evening Limits of Roots and Functions
For t > -1, let αₜ and βₜ be the roots of the equation ((t + 2) ^ (1)/(7) - 1) x ^ 2 + ((t + 2) ^ (1)/(6) - 1) x + ((t + 2) ^ (1)/(2 1) - 1) = 0. If t arrow - 1 ⁺ α_ t = a and t arrow - 1 ⁺ β_ t = b, then 72 (a + b) ^ 2 is equal to
Numerical Answer. Answer: 98 to 98

Solution

Related Formula

Sum of roots for a quadratic equation Ax² + Bx + C = 0 satisfies:

α + β = -(B)/(A)
Core Logic

We need to find t → -1 (αₜ + βₜ) = a + b:

a + b = t → -1 - (t+2)1/6 - 1(t+2)1/7 - 1

Let y = t+2. As t → -1, y → 1.

a + b = y → 1 - y1/6 - 1y1/7 - 1
Step 1: Evaluate Limit

Applying L'Hopital's Rule or standard limit templates:

a + b = -((1)/(6))/((1)/(7)) = -(7)/(6)

Squaring the sum alignment:

(a + b)² = (49)/(36) 72(a + b)² = 72 · (49)/(36) = 98
Pattern Recognition

Treating (α + β) collectively allows direct evaluation via standard root identities without solving for individual root entities.

Chapter Mix

Class 12 Mathematics: Limits, Continuity and Differentiability Class 11 Mathematics: Quadratic Equations

Q57 jee_main_2025_24_jan_evening Continuity and Differentiability of Composite Functions
Let [x] denote the greatest integer function, and let m and n respectively be the numbers of the points, where the function f(x)=[x]+|x-2|, -2
  • A. 6
  • B. 9
  • C. 8
  • D. 7

Solution

Related Formula

The greatest integer function [x] is discontinuous at all integer points. The absolute value function |x-x₀| is continuous everywhere but non-differentiable at its corner tip x = x₀.

Core Logic

Break down the function f(x) = [x] + |x-2| in the open domain (-2, 3) across sub-intervals between integers:

f(x) = cases -2 - (x-2) = -x & -2 < x < -1 -1 - (x-2) = -x+1 & -1 ≤ x < 0 0 - (x-2) = -x+2 & 0 ≤ x < 1 1 - (x-2) = -x+3 & 1 ≤ x < 2 2 + (x-2) = x & 2 ≤ x < 3 cases
Step 1: Count Discontinuity Points (m)

Evaluate the limits at internal integers -1, 0, 1, 2:

  • At x = -1: LHL = 1, RHL = 2 ⇒ Discontinuous.
  • At x = 0: LHL = 1, RHL = 2 ⇒ Discontinuous.
  • At x = 1: LHL = 1, RHL = 2 ⇒ Discontinuous.
  • At x = 2: LHL = 1, RHL = 2 ⇒ Discontinuous.
  • Thus, f(x) is discontinuous at exactly 4 integer locations , meaning m = 4.

Step 2: Count Non-Differentiability Points (n)

Since discontinuity automatically implies non-differentiability, the points -1, 0, 1, 2 are non-differentiable. Let's check if there are other sharp corners. The modulus part |x-2| turns sharp at x=2, which is already covered in our discontinuity list. Hence, there are no additional non-differentiable points.

Thus, n = 4.

Step 3: Total Evaluation

Calculate the Σ requested :

m + n = 4 + 4 = 8
Pattern Recognition

For expressions containing [x], the discontinuity at integers usually drives the overall non-differentiability tally, making any coincidental sharp points from continuous elements redundant if they happen at the exact same integers.

Chapter Mix

Class 12 Mathematics: Limits, Continuity and Differentiability

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