For alpha, beta, gamma R, if 0 x2 alpha x + (gamma - 1) ex2 2x - beta x = 3, then beta + gamma

Q18 jee_main_2024_27_jan_morning Standard Limits

If

a=lim_xrightarrow0fracsqrt1+sqrt1+x^4-sqrt2x^4

and

b=lim_xrightarrow0fracsin^2xsqrt2-sqrt1+cos x

, then the value of

ab^3

is :

A. $36$ $36$
B. $32$ $32$
C. $25$ $25$
D. $30$ $30$

Solution

### Related Formula

lim_x to 0 fracsin xx = 1

Rationalization:

(u-v)(u+v) = u^2 - v^2

### Core Logic Evaluate limit

a

by rationalizing the numerator:

a = lim_x to 0 fracsqrt1+sqrt1+x^4-sqrt2x^4

Multiply by conjugate:

a = lim_x to 0 frac(1+sqrt1+x^4) - 2x^4 (sqrt1+sqrt1+x^4 + sqrt2)

a = lim_x to 0 fracsqrt1+x^4 - 1x^4 (sqrt1+sqrt1+x^4 + sqrt2)

Rationalize again:

a = lim_x to 0 frac(1+x^4) - 1x^4 (sqrt1+sqrt1+x^4 + sqrt2) (sqrt1+x^4 + 1)

Cancel

x^4

:

a = lim_x to 0 frac1(sqrt1+sqrt1+0 + sqrt2) (sqrt1+0 + 1)

a = frac1(sqrt2 + sqrt2)(1 + 1) = frac14sqrt2

### Step 1: Evaluating Limit b Evaluate limit

b

by rationalizing the denominator:

b = lim_x to 0 fracsin^2 xsqrt2-sqrt1+cos x

Multiply by conjugate:

b = lim_x to 0 fracsin^2 x (sqrt2 + sqrt1+cos x)2 - (1+cos x)

b = lim_x to 0 frac(1-cos^2 x)(sqrt2 + sqrt1+cos x)1 - cos x

Using

1-cos^2 x = (1-cos x)(1+cos x)

:

b = lim_x to 0 (1+cos x)(sqrt2 + sqrt1+cos x)

Apply limit

x to 0

(so

cos 0 = 1

):

b = (1+1)(sqrt2 + sqrt1+1) = 2(2sqrt2) = 4sqrt2

### Step 2: Final Output Calculate the value of

ab^3

:

ab^3 = left(frac14sqrt2right) times (4sqrt2)^3

ab^3 = frac(4sqrt2)^34sqrt2 = (4sqrt2)^2

ab^3 = 16 times 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

x^n

terms naturally. ### Evaluation Rubric / Model Answer null ### 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

lim_xrightarrowfracpi2 fracint_x^3^(fracpi2)^3 cos(t^1/3) dt(x-fracpi2)^2

is equal to

A. $frac3pi8$ $\frac{3\pi}{8}$
B. $frac3pi^24$ $\frac{3\pi^2}{4}$
C. $frac3pi^28$ $\frac{3\pi^2}{8}$
D. $frac3pi4$ $\frac{3\pi}{4}$

Solution

### Related Formula

textNewton-Leibniz Formula: fracddx int_h(x)^g(x) f(t)dt = f(g(x)) cdot g'(x) - f(h(x)) cdot h'(x)

lim_x to a fracf(x)g(x) = lim_x to a fracf'(x)g'(x) quad text(L'Hopital's Rule for frac00 text forms)

### Core Logic Evaluate the limit

L = lim_xrightarrowfracpi2 fracint_x^3^(pi/2)^3 cos(t^1/3) dt(x-fracpi2)^2

. When

x to fracpi2

, the integral limits become from

(fracpi2)^3

to

(fracpi2)^3

, so the numerator is 0. The denominator evaluates to

0^2 = 0

. This is a

frac00

form, meaning L'Hopital's rule must be applied. Differentiate the numerator using Newton-Leibniz theorem:

N'(x) = fracddx left[ int_x^3^(pi/2)^3 cos(t^1/3) dt right]

= cosleft(left((pi/2)^3right)^1/3right) cdot 0 - cosleft((x^3)^1/3right) cdot fracddx(x^3)

= 0 - cos(x) cdot 3x^2 = -3x^2 cos(x)

Differentiate the denominator:

D'(x) = fracddxleft[ (x-fracpi2)^2 right] = 2(x-fracpi2)

### Step 1: Simplify and Re-evaluate Limit Substitute the derivatives back into the limit expression:

L = lim_xrightarrowfracpi2 frac-3x^2 cos x2(x-fracpi2)

Notice that

cos(x) = sin(fracpi2 - x) = -sin(x - fracpi2)

. Substituting this equivalence:

L = lim_xrightarrowfracpi2 frac-3x^2 cdot (-sin(x - fracpi2))2(x-fracpi2)

L = lim_xrightarrowfracpi2 left[ fracsin(x-fracpi2)x-fracpi2 right] times left[ frac3x^22 right]

### Step 2: Apply Standard Limit Since

lim_theta to 0 fracsin thetatheta = 1

, where

theta = x - fracpi2

:

L = 1 times frac3(pi/2)^22

L = frac3 cdot fracpi^242 = frac3pi^28

### 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

cos x

to

-sin(x - fracpi2)

to match the denominator structure for standard trigonometric limits. ### Evaluation Rubric / Model Answer null ### 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)=frac(2^x+2^-x)tan xsqrttan^-1(x^2-x+1)(7x^2+3x+1)^3

. Then the value of

f'(0)

is equal to

A. $pi$ $\pi$
B. $0$ $0$
C. $sqrtpi$ $\sqrt{\pi}$
D. $fracpi2$ $\frac{\pi}{2}$

Solution

### Related Formula

f'(0) = lim_h to 0 fracf(h) - f(0)h

Standard Limits:

lim_h to 0 fractan hh = 1

### Core Logic First, evaluate

f(0)

to ensure the first principle approach simplifies:

f(0) = frac(2^0 + 2^-0)tan(0)sqrttan^-1(0-0+1)(0+0+1)^3

Since

tan(0) = 0

, the entire numerator collapses, giving

f(0) = 0

. Set up the limit definition of the derivative at

x = 0

:

f'(0) = lim_h to 0 fracf(h) - 0h

f'(0) = lim_h to 0 frac1h left( frac(2^h + 2^-h)tan hsqrttan^-1(h^2-h+1)(7h^2+3h+1)^3 right)

### Step 1: Group Standard Limit Forms Regroup the expression to isolate the known limit forms:

f'(0) = lim_h to 0 left( fractan hh right) times left( 2^h + 2^-h right) times fracsqrttan^-1(h^2-h+1)(7h^2+3h+1)^3

Now, evaluate the limit of each independent non-zero segment as

h to 0

: 1.

lim_h to 0 fractan hh = 1

2.

lim_h to 0 (2^h + 2^-h) = 2^0 + 2^-0 = 1 + 1 = 2

3.

lim_h to 0 sqrttan^-1(h^2-h+1) = sqrttan^-1(1) = sqrtfracpi4 = fracsqrtpi2

4.

lim_h to 0 (7h^2+3h+1)^3 = (0+0+1)^3 = 1

### Step 2: Combine Limits Multiply the evaluated continuous components together:

f'(0) = 1 times 2 times fracfracsqrtpi21

f'(0) = sqrtpi

### Pattern Recognition If you are asked to find

f'(0)

for a massive, horrifying fraction where

f(0)=0

(usually due to a rogue

sin x

,

tan x

, or

x

term), completely ignore the quotient rule. Use the first principle formula

lim_hto0 f(h)/h

to instantly isolate standard limit identities and plug 0 into everything else. ### Evaluation Rubric / Model Answer null ### 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) = begincases x^2 + 3x + a, & x le 1 \\ bx + 2, & x gt 1 endcases

be differentiable on

mathbbR

. Then, the value of

int_-2^2f(x)dx

equals

A. $frac156$ $\frac{15}{6}$
B. $frac196$ $\frac{19}{6}$
C. $21$ $21$
D. $17$ $17$

Solution

### Related Formula

textFor differentiability at x=c:\\ lim_x to c^- f(x) = lim_x to c^+ f(x) quad text(Continuity)\\ lim_x to c^- f'(x) = lim_x to c^+ f'(x) quad text(Differentiability)

### Core Logic Function

f(x)

is continuous at

x=1

:

lim_x to 1^- (x^2 + 3x + a) = lim_x to 1^+ (bx + 2)

1 + 3 + a = b + 2 Rightarrow 4 + a = b + 2 Rightarrow a = b - 2 quad dots(i)

Function

f(x)

is differentiable at

x=1

:

f'(x) = begincases 2x + 3, & x lt 1 \\ b, & x gt 1 endcases

Equating left-hand and right-hand derivatives at

x=1

:

2(1) + 3 = b Rightarrow 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) = begincases x^2 + 3x + 3, & x le 1 \\ 5x + 2, & x gt 1 endcases

We need to evaluate

int_-2^2 f(x) dx

:

I = int_-2^1 (x^2 + 3x + 3) dx + int_1^2 (5x + 2) dx

### Step 2: Evaluating the Integrals First integral:

int_-2^1 (x^2 + 3x + 3) dx = left[ fracx^33 + frac3x^22 + 3x right]_-2^1

= left( frac13 + frac32 + 3 right) - left( frac-83 + frac122 - 6 right)

= left( frac13 + frac32 + 3 right) - left( frac-83 + 0 right) = frac93 + frac32 + 3 = 3 + frac32 + 3 = frac152

Second integral:

int_1^2 (5x + 2) dx = left[ frac5x^22 + 2x right]_1^2

= left( frac202 + 4 right) - left( frac52 + 2 right) = 14 - frac92 = frac192

Total sum:

I = frac152 + frac192 = frac342 = 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Continuity and Differentiability Class 12 Maths: Integral Calculus

Q20 jee_main_2024_30_jan_morning Limits

Let

f:left[-fracpi2,fracpi2right] to mathbbR

be a differentiable function such that

f(0) = frac12

. If the

lim_x to 0 fracx int_0^x f(t) dte^x^2 - 1 = alpha

, then

8alpha^2

is equal to:

A. $16$ $16$
B. $2$ $2$
C. $1$ $1$
D. $4$ $4$

Solution

### Related Formula

lim_y to 0 frace^y - 1y = 1

Leibniz Integral Rule:

fracddx int_0^x f(t) dt = f(x)

### Core Logic Given limit is:

alpha = lim_x to 0 fracx int_0^x f(t) dte^x^2 - 1

Multiply and divide the denominator by

x^2

to use standard exponential limit:

alpha = lim_x to 0 fracx int_0^x f(t) dtleft(frace^x^2 - 1x^2right) cdot x^2

Since

lim_xto 0 frace^x^2 - 1x^2 = 1

, the expression simplifies to:

alpha = lim_x to 0 fracx int_0^x f(t) dt1 cdot x^2 = lim_x to 0 fracint_0^x 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

:

alpha = lim_x to 0 fracfracddx int_0^x f(t) dtfracddx(x) = lim_x to 0 fracf(x)1

By continuity of differentiable function

f

at

0

:

alpha = f(0)

### Step 2: Final Calculation We are given

f(0) = frac12

, so

alpha = frac12

. We need to find

8alpha^2

:

8alpha^2 = 8 left(frac12right)^2 = 8 left(frac14right) = 2

### Pattern Recognition Standard expansion/limits on isolated terms in denominators immediately reduce the power of

x

, setting up a trivial Leibniz derivative application. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives Class 12 Maths: Integrals

For $alpha, beta, gamma in mathbbR$ $\alpha, \beta, \gamma \in \mathbb{R}$, if $lim_x to 0 fracx^2 sin alpha x + (gamma - 1) e^x^2sin 2x - beta x = 3$ , then $beta + gamma - alpha$ $\beta + \gamma - \alpha$ is equal to:

Solution & Explanation

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For alpha, beta, gamma in mathbbR$\alpha, \beta, \gamma \in \mathbb{R}$, if lim_x to 0 fracx^2 sin alpha x + (gamma - 1) e^x^2sin 2x - beta x = 3$\lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1) e^{x^2}}{\sin 2x - \beta x} = 3$, then beta + gamma - alpha$\beta + \gamma - \alpha$ is equal to:

Solution & Explanation

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For $alpha, beta, gamma in mathbbR$ $\alpha, \beta, \gamma \in \mathbb{R}$, if $lim_x to 0 fracx^2 sin alpha x + (gamma - 1) e^x^2sin 2x - beta x = 3$ , then $beta + gamma - alpha$ $\beta + \gamma - \alpha$ is equal to: