If \alpha=\lim_x arrow 0^+(\frace^\sqrt\tan x-e^\sqrtx\sqrt\tan x-\sqrtx) and \beta=\lim_x arrow 0(1+\sin x)^\frac12 \cot x are the roots of the quadratic equation a x^2+b x-\sqrte=0 , then 12 \log _e(a+b) is equal to \underline\hspace2cm .

Limits Continuity and Differentiability PYQ — JEE Main 2024 · 8 Apr · Shift 2 · Q58 with Solution

View written solutionFree

Correct answer: 6

Find $\alpha$

We need

\alpha=\lim_{x\to 0^+}\frac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}.

This is of the form

\frac{f(u)-f(v)}{u-v}

with $f(t)=e^t$ , $u=\sqrt{\tan x}$ , $v=\sqrt{x}$ .

As $x\to 0^+$ ,

\sqrt{\tan x}\to 0,\qquad \sqrt{x}\to 0.

Hence, by the standard limit / Mean Value Theorem idea,

\alpha=f'(0)=e^0=1.

So,

\boxed{\alpha=1}.

Find $\beta$

We need

\beta=\lim_{x\to 0}(1+\sin x)^{\frac12\cot x}.

Take logarithm:

\ln \beta=\lim_{x\to 0}\frac12\cot x\cdot \ln(1+\sin x).

Now use standard small-angle limits:

\sin x\sim x,\qquad \ln(1+u)\sim u \text{ as } u\to 0,

\ln(1+\sin x)\sim \sin x.

Therefore,

\ln \beta=\frac12\lim_{x\to 0}\cot x\,\sin x n=\frac12\lim_{x\to 0}\frac{\cos x}{\sin x}\sin x =\frac12\lim_{x\to 0}\cos x =\frac12.

Hence,

\beta=e^{1/2}=\sqrt e.

So,

\boxed{\beta=\sqrt e}.

Use roots of quadratic

Given $\alpha$ and $\beta$ are the roots of

ax^2+bx-\sqrt e=0.

For a quadratic with roots $\alpha,\beta$ :

\alpha+\beta=-\frac{b}{a},\qquad \alpha\beta=\frac{-\sqrt e}{a}.

Using $\alpha=1$ , $\beta=\sqrt e$ :

\alpha\beta=\sqrt e.

So,

\frac{-\sqrt e}{a}=\sqrt e \quad\Rightarrow\quad a=-1.

Now,

\alpha+\beta=1+\sqrt e=-\frac{b}{a}.

Since $a=-1$ ,

1+\sqrt e=b.

Thus,

a+b=-1+(1+\sqrt e)=\sqrt e.

Compute required value

We need

12\ln(a+b)=12\ln(\sqrt e).

Now,

\ln(\sqrt e)=\ln(e^{1/2})=\frac12.

Therefore,

12\ln(a+b)=12\cdot \frac12=6.

So the required integer is

\boxed{6}.

Comparison with stored answer

Stored correct answer = $6$ .

Our derived answer also equals $6$ , so it agrees.

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: 6

Find $\alpha$

We need

\alpha=\lim_{x\to 0^+}\frac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}.

This is of the form

\frac{f(u)-f(v)}{u-v}

with $f(t)=e^t$ , $u=\sqrt{\tan x}$ , $v=\sqrt{x}$ .

As $x\to 0^+$ ,

\sqrt{\tan x}\to 0,\qquad \sqrt{x}\to 0.

Hence, by the standard limit / Mean Value Theorem idea,

\alpha=f'(0)=e^0=1.

So,

\boxed{\alpha=1}.

Find $\beta$

We need

\beta=\lim_{x\to 0}(1+\sin x)^{\frac12\cot x}.

Take logarithm:

\ln \beta=\lim_{x\to 0}\frac12\cot x\cdot \ln(1+\sin x).

Now use standard small-angle limits:

\sin x\sim x,\qquad \ln(1+u)\sim u \text{ as } u\to 0,

\ln(1+\sin x)\sim \sin x.

Therefore,

\ln \beta=\frac12\lim_{x\to 0}\cot x\,\sin x n=\frac12\lim_{x\to 0}\frac{\cos x}{\sin x}\sin x =\frac12\lim_{x\to 0}\cos x =\frac12.

Hence,

\beta=e^{1/2}=\sqrt e.

So,

\boxed{\beta=\sqrt e}.

Use roots of quadratic

Given $\alpha$ and $\beta$ are the roots of

ax^2+bx-\sqrt e=0.

For a quadratic with roots $\alpha,\beta$ :

\alpha+\beta=-\frac{b}{a},\qquad \alpha\beta=\frac{-\sqrt e}{a}.

Using $\alpha=1$ , $\beta=\sqrt e$ :

\alpha\beta=\sqrt e.

So,

\frac{-\sqrt e}{a}=\sqrt e \quad\Rightarrow\quad a=-1.

Now,

\alpha+\beta=1+\sqrt e=-\frac{b}{a}.

Since $a=-1$ ,

1+\sqrt e=b.

Thus,

a+b=-1+(1+\sqrt e)=\sqrt e.

Compute required value

We need

12\ln(a+b)=12\ln(\sqrt e).

Now,

\ln(\sqrt e)=\ln(e^{1/2})=\frac12.

Therefore,

12\ln(a+b)=12\cdot \frac12=6.

So the required integer is

\boxed{6}.

Comparison with stored answer

Stored correct answer = $6$ .

Our derived answer also equals $6$ , so it agrees.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability