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

Q29 jee_main_2024_30_jan_morning Differentiability

If the function

f(x) = cases (1)/(|x|) & ,|x| \geq 2 ax² + 2b & ,|x| < 2 cases

is differentiable on

R

, then

48 (a + b)

is equal to

Numerical Answer. Answer: 15 to 15

Solution

Related Formula

Continuity at x=c: x \to c^- f(x) = x \to c^+ f(x)

Differentiability at x=c: x \to c^- f'(x) = x \to c^+ f'(x)

Core Logic

Rewrite the piecewise function without absolute values:

f(x) = cases (1)/(x) & , x \geq 2 ax² + 2b & , -2 < x < 2 -(1)/(x) & , x \leq -2 cases

Step 1: Applying Continuity

For $f(x)$ $f(x)$ to be continuous at $x = 2$ $x = 2$:

x \to 2^- (ax² + 2b) = x \to 2^+ (1)/(x)

a(2)² + 2b = (1)/(2) \Rightarrow 4a + 2b = (1)/(2) (1)

Because the function is even, continuity at $x = -2$ $x = -2$ yields the exact same equation: $4a + 2b = 1/2$ $4a + 2b = 1/2$.

Step 2: Applying Differentiability

Find the derivative $f'(x)$ $f'(x)$ for piecewise sections:

f'(x) = cases -(1)/(x²) & , x > 2 2ax & , -2 < x < 2 (1)/(x²) & , x < -2 cases

For $f(x)$ $f(x)$ to be differentiable at $x = 2$ $x = 2$:

x \to 2^- (2ax) = x \to 2^+ (-(1)/(x²))

4a = -(1)/(4) \Rightarrow a = -(1)/(16)

Step 3: Finding variables and final target

Substitute $a$ $a$ back into equation (1):

4(-(1)/(16)) + 2b = (1)/(2)

-(1)/(4) + 2b = (1)/(2) \Rightarrow 2b = (3)/(4) \Rightarrow b = (3)/(8)

We need to evaluate $48(a + b)$ $48(a + b)$:

48(-(1)/(16) + (3)/(8)) = 48((-1 + 6)/(16)) = 48((5)/(16)) = 3 \times 5 = 15

Pattern Recognition

Piecewise differentiability forces simultaneous linear equations matching function values and their first derivatives at boundary limits.

Chapter Mix

Class 12 Maths: Continuity and Differentiability

Q10 jee_main_2024_31_jan_evening Limits of Functions

Let

f: R \to (0, \infty)

be strictly increasing function such that

x \to \infty (f(7x))/(f(x)) = 1

. Then, the value of

x arrow \infty [ (f(5x))/(f(x)) - 1 ]

is equal to

A. $4$ $4$
B. $0$ $0$
C. $7/5$ $7/5$
D. $1$ $1$

Solution

Related Formula

Sandwich / Squeeze Theorem: If g(x) \leq h(x) \leq k(x) and g(x) = k(x) = L, then h(x) = L

Core Logic

Since $f$ $f$ is a strictly increasing function mapping to $(0,\infty)$ $(0,\infty)$: For $x > 0$ $x > 0$, we have $x < 5x < 7x$ $x < 5x < 7x$. Thus, $f(x) < f(5x) < f(7x)$ $f(x) < f(5x) < f(7x)$. Divide everything by $f(x)$ $f(x)$ (which is strictly positive):

1 < (f(5x))/(f(x)) < (f(7x))/(f(x))

Take the limit as $x \to \infty$ $x \to \infty$:

x\to\infty 1 \leq x\to\infty (f(5x))/(f(x)) \leq x\to\infty (f(7x))/(f(x))

1 \leq x\to\infty (f(5x))/(f(x)) \leq 1

Therefore, $x\to\infty (f(5x))/(f(x)) = 1$ . The required value is:

x arrow \infty [ (f(5x))/(f(x)) - 1 ] = 1 - 1 = 0

Chapter Mix

Class 11 Maths: Limits and Derivatives Class 12 Maths: Continuity and Differentiability

Q14 jee_main_2024_31_jan_evening Differentiability

Consider the function

f:(0,\infty)\to R

defined by

f(x) = e^-| ₑx|

. If

m

and

n

be respectively the number of points at which

f

is not continuous and

f

is not differentiable, then

m + n

is

A. $0$ $0$
B. $3$ $3$
C. $1$ $1$
D. $2$ $2$

Solution

Core Logic

Differentiability diagram for Q14 - JEE Main 2024 Evening

The function is $f(x) = e-|ln x|$ $f(x) = e^{-|\ln x|}$. Rewrite piecewise for $(0, \infty)$ $(0, \infty)$:

f(x) = cases e-(-ln x) & if 0 < x < 1 e-ln x & if x \geq 1 cases

f(x) = cases eln x = x & if 0 < x < 1 1eln x = (1)/(x) & if x \geq 1 cases

Check continuity at $x = 1$ $x = 1$:

x \to 1^- f(x) = x \to 1^- x = 1

x \to 1^+ f(x) = x \to 1^+ (1)/(x) = 1

$f(1) = 1$ $f(1) = 1$. The function is continuous everywhere on $(0, \infty)$ $(0, \infty)$. Thus, $m = 0$ $m = 0$.

Check differentiability at $x = 1$ $x = 1$:

LHD = x \to 1^- f'(x) = 1

RHD = x \to 1^+ f'(x) = -(1)/(x²)|x=1 = -1

Since $LHD \neq RHD$ $LHD \neq RHD$, the function is not differentiable at $x = 1$ $x = 1$. Thus, $n = 1$ $n = 1$.

Finally, $m + n = 0 + 1 = 1$ $m + n = 0 + 1 = 1$.

Chapter Mix

Class 12 Maths: Continuity and Differentiability

Q27 jee_main_2024_31_jan_evening Maclaurin Series / L'Hopital

If

x \to 0 ax²e^x - b ₑ(1 + x) + cxe-xx² x = 1

, then

16(a² + b² + c²)

is equal to

Numerical Answer. Answer: 81 to 81

Solution

Related Formula

e^x = 1 + x + (x²)/(2!) +

ln(1+x) = x - (x²)/(2) + (x³)/(3) -

x \approx x x² x \approx x³

Core Logic

Expand the numerator terms using Maclaurin series around $x=0$ $x=0$:

ax² (1 + x + (x²)/(2) + ) - b (x - (x²)/(2) + (x³)/(3) - ) + cx (1 - x + (x²)/(2) - (x³)/(6) + )

Denominator behavior is $x³$ $x^3$. Group by powers of $x$ $x$: Coefficient of $x$ $x$: $-b + c = 0 c = b$ $-b + c = 0 \implies c = b$ Coefficient of $x²$ $x^2$: $a + (b)/(2) - c = 0 a = c - (b)/(2) = (b)/(2)$ Coefficient of $x³$ $x^3$: $a - (b)/(3) + (c)/(2) = 1$ $a - \frac{b}{3} + \frac{c}{2} = 1$

Substitute $a = b/2$ $a = b/2$ and $c = b$ $c = b$ into the $x³$ $x^3$ equation:

(b)/(2) - (b)/(3) + (b)/(2) = 1

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

This gives $c = (3)/(2)$ $c = \frac{3}{2}$ and $a = (3)/(4)$ $a = \frac{3}{4}$.

Calculate the required value:

16(a² + b² + c²) = 16((9)/(16) + (9)/(4) + (9)/(4))

= 9 + 36 + 36 = 81

Chapter Mix

Class 11 Maths: Limits and Derivatives

Q7 jee_main_2024_31_jan_morning Exponential Limits

x\to 0 e2| x| - 2| x| - 1x²

A. $is equal to -1$ $\text{is equal to } -1$
B. $does not exist$ $\text{does not exist}$
C. $is equal to 1$ $\text{is equal to } 1$
D. $is equal to 2$ $\text{is equal to } 2$

Solution

Core Logic

Evaluate $x \to 0 e2| x| - 2| x| - 1x²$ . Multiply and divide by $| x|²$ $|\sin x|^2$:

= x \to 0 e2| x| - 2| x| - 1| x|² \times ( ² x)/(x²)

Step 1: Substitution and L'Hôpital

Let $| x| = t$ $|\sin x| = t$. As $x \to 0, t \to 0$ $x \to 0, t \to 0$.

t \to 0 e2t - 2t - 1t² \times x \to 0 ( ² x)/(x²)

The second limit evaluates to $1$ $1$. Using L'Hôpital's Rule on the first limit:

= t \to 0 2e2t - 22t = t \to 0 4e2t2 = 2

Step 2: Final Result

$2 \times 1 = 2$ $2 \times 1 = 2$

Chapter Mix

Class 11 Maths: Limits and Derivatives

If t→ 0(∫₀¹ (3x + 5)^t dx)(1)/(t) = (α)/(5e)((8)/(5))(2)/(3)$\lim_{t\to 0}\left(\int_0^1 (3x + 5)^t dx\right)^{\frac{1}{t}} = \frac{\alpha}{5e}\left(\frac{8}{5}\right)^{\frac{2}{3}}$, then α$\alpha$ is equal to

Solution & Explanation

Related Formula

Core Logic

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

Step 2: Complete the Final Form Match

Pattern Recognition

Chapter Mix

Reference Study Guides

More Limits Previous-Year Questions — Page 10

Solution

Related Formula

Core Logic

Step 1: Applying Continuity

Step 2: Applying Differentiability

Step 3: Finding variables and final target

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Substitution and L'Hôpital

Step 2: Final Result

Chapter Mix

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If $t\to 0(\int₀¹ (3x + 5)^t dx)(1)/(t) = (α)/(5e)((8)/(5))(2)/(3)$ , then $α$ $\alpha$ is equal to