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) = begincases frac1|x| & ,|x| geq 2 \\ ax^2 + 2b & ,|x| < 2 endcases

is differentiable on

mathbbR

, then

48 (a + b)

is equal to

Numerical Answer. Answer: 15 to 15

Solution

### Related Formula

textContinuity at x=c: lim_x to c^- f(x) = lim_x to c^+ f(x)

textDifferentiability at x=c: lim_x to c^- f'(x) = lim_x to c^+ f'(x)

### Core Logic Rewrite the piecewise function without absolute values:

f(x) = begincases frac1x & , x geq 2 \\ ax^2 + 2b & , -2 < x < 2 \\ -frac1x & , x leq -2 endcases

### Step 1: Applying Continuity For

f(x)

to be continuous at

x = 2

:

lim_x to 2^- (ax^2 + 2b) = lim_x to 2^+ frac1x

a(2)^2 + 2b = frac12 Rightarrow 4a + 2b = frac12 quad dots (1)

Because the function is even, continuity at

x = -2

yields the exact same equation:

4a + 2b = 1/2

. ### Step 2: Applying Differentiability Find the derivative

f'(x)

for piecewise sections:

f'(x) = begincases -frac1x^2 & , x > 2 \\ 2ax & , -2 < x < 2 \\ frac1x^2 & , x < -2 endcases

For

f(x)

to be differentiable at

x = 2

:

lim_x to 2^- (2ax) = lim_x to 2^+ left(-frac1x^2right)

4a = -frac14 Rightarrow a = -frac116

### Step 3: Finding variables and final target Substitute

a

back into equation (1):

4left(-frac116right) + 2b = frac12

-frac14 + 2b = frac12 Rightarrow 2b = frac34 Rightarrow b = frac38

We need to evaluate

48(a + b)

:

48left(-frac116 + frac38right) = 48left(frac-1 + 616right) = 48left(frac516right) = 3 times 5 = 15

### Pattern Recognition Piecewise differentiability forces simultaneous linear equations matching function values and their first derivatives at boundary limits. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Continuity and Differentiability

Q10 jee_main_2024_31_jan_evening Limits of Functions

Let

f: mathbbR to (0, infty)

be strictly increasing function such that

lim_x to infty fracf(7x)f(x) = 1

. Then, the value of

lim_x rightarrow infty left[ fracf(5x)f(x) - 1 right]

is equal to

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

Solution

### Related Formula

textSandwich / Squeeze Theorem: If g(x) le h(x) le k(x) text and lim g(x) = lim k(x) = L, text then lim h(x) = L

### Core Logic Since

f

is a strictly increasing function mapping to

(0,infty)

: For

x > 0

, we have

x < 5x < 7x

. Thus,

f(x) < f(5x) < f(7x)

. Divide everything by

f(x)

(which is strictly positive):

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

Take the limit as

x to infty

:

lim_xtoinfty 1 le lim_xtoinfty fracf(5x)f(x) le lim_xtoinfty fracf(7x)f(x)

1 le lim_xtoinfty fracf(5x)f(x) le 1

Therefore,

lim_xtoinfty fracf(5x)f(x) = 1

. The required value is:

lim_x rightarrow infty left[ fracf(5x)f(x) - 1 right] = 1 - 1 = 0

### Evaluation Rubric / Model Answer null ### 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 mathbbR

defined by

f(x) = e^-|log_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|

. Rewrite piecewise for

(0, infty)

:

f(x) = begincases e^-(-ln x) & textif 0 < x < 1 \\ e^-ln x & textif x ge 1 endcases

f(x) = begincases e^ln x = x & textif 0 < x < 1 \\ frac1e^ln x = frac1x & textif x ge 1 endcases

Check continuity at

x = 1

:

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

lim_x to 1^+ f(x) = lim_x to 1^+ frac1x = 1

f(1) = 1

. The function is continuous everywhere on

(0, infty)

. Thus,

m = 0

. Check differentiability at

x = 1

:

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

RHD = lim_x to 1^+ f'(x) = -frac1x^2Big|_x=1 = -1

Since

LHD neq RHD

, the function is not differentiable at

x = 1

. Thus,

n = 1

. Finally,

m + n = 0 + 1 = 1

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Continuity and Differentiability

Q27 jee_main_2024_31_jan_evening Maclaurin Series / L'Hopital

If

lim_x to 0 fracax^2e^x - b log_e(1 + x) + cxe^-xx^2sin x = 1

, then

16(a^2 + b^2 + c^2)

is equal to

Numerical Answer. Answer: 81 to 81

Solution

### Related Formula

e^x = 1 + x + fracx^22! + dots

ln(1+x) = x - fracx^22 + fracx^33 - dots

sin x approx x implies x^2sin x approx x^3

### Core Logic Expand the numerator terms using Maclaurin series around

x=0

:

ax^2 left(1 + x + fracx^22 + dotsright) - b left(x - fracx^22 + fracx^33 - dotsright) + cx left(1 - x + fracx^22 - fracx^36 + dotsright)

Denominator behavior is

x^3

. Group by powers of

x

: Coefficient of

x

:

-b + c = 0 implies c = b

Coefficient of

x^2

:

a + fracb2 - c = 0 implies a = c - fracb2 = fracb2

Coefficient of

x^3

:

a - fracb3 + fracc2 = 1

Substitute

a = b/2

and

c = b

into the

x^3

equation:

fracb2 - fracb3 + fracb2 = 1

b - fracb3 = 1 implies frac2b3 = 1 implies b = frac32

This gives

c = frac32

and

a = frac34

. Calculate the required value:

16(a^2 + b^2 + c^2) = 16left(frac916 + frac94 + frac94right)

= 9 + 36 + 36 = 81

### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives

Q7 jee_main_2024_31_jan_morning Exponential Limits

lim_xto 0frace^2|sin x| - 2|sin x| - 1x^2

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

Solution

### Core Logic Evaluate

lim_x to 0 frace^2|sin x| - 2|sin x| - 1x^2

. Multiply and divide by

|sin x|^2

:

= lim_x to 0 frace^2|sin x| - 2|sin x| - 1|sin x|^2 times fracsin^2 xx^2

### Step 1: Substitution and L'Hôpital Let

|sin x| = t

. As

x to 0, t to 0

.

lim_t to 0 frace^2t - 2t - 1t^2 times lim_x to 0 fracsin^2 xx^2

The second limit evaluates to

1

. Using L'Hôpital's Rule on the first limit:

= lim_t to 0 frac2e^2t - 22t = lim_t to 0 frac4e^2t2 = 2

### Step 2: Final Result

2 times 1 = 2

### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives

If $lim_tto 0left(int_0^1 (3x + 5)^t dxright)^frac1t = fracalpha5eleft(frac85right)^frac23$ , then $alpha$ $\alpha$ is equal to

Solution & Explanation

Reference Study Guides

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If lim_tto 0left(int_0^1 (3x + 5)^t dxright)^frac1t = fracalpha5eleft(frac85right)^frac23$\lim_{t\to 0}\left(\int_0^1 (3x + 5)^t dx\right)^{\frac{1}{t}} = \frac{\alpha}{5e}\left(\frac{8}{5}\right)^{\frac{2}{3}}$, then alpha$\alpha$ is equal to

Solution & Explanation

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More Limits Previous-Year Questions — Page 8

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If $lim_tto 0left(int_0^1 (3x + 5)^t dxright)^frac1t = fracalpha5eleft(frac85right)^frac23$ , then $alpha$ $\alpha$ is equal to