Let D be the domain of the function f(x)=\sin ^-1(\log _3 x(\frac6+2 \log _3 x-5 x)) . If the range of the function g: D arrow \mathbbR defined by g(x)=x-[x],([x] is the greatest integer function), is (\alpha, \beta) , then \alpha^2+\frac5\beta is equal to

Functions PYQ — JEE Main 2023 · 12 Apr · Shift 1 · Q34 with Solution

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Correct answer: D

Given function

We need the domain $D$ of

f(x)=\sin^{-1}\left(\log_{3x}\left(\frac{6+2\log_3 x}{-5x}\right)\right).

Then for $g(x)=x-[x]$ , we need the range of $g$ on $D$ .

Conditions for the logarithm

For

\log_{3x}\left(\frac{6+2\log_3 x}{-5x}\right)

to exist, we need:

Base positive: $3x>0 \Rightarrow x>0$
Base not equal to $1$ : $3x\ne 1 \Rightarrow x\ne \frac13$
Argument positive: $\frac{6+2\log_3 x}{-5x}>0$

Since $x>0$ , denominator $-5x<0$ . Hence numerator must be negative:

6+2\log_3 x<0

3+\log_3 x<0

\log_3 x<-3

x<3^{-3}=\frac1{27}.

So far,

0<x<\frac1{27}.

This already excludes $x=\frac13$ automatically.

Condition for $\sin^{-1}$

y=\log_{3x}\left(\frac{6+2\log_3 x}{-5x}\right),

then for $\sin^{-1}(y)$ to be defined,

-1\le y\le 1.

So we solve

-1\le \log_{3x}\left(\frac{6+2\log_3 x}{-5x}\right)\le 1

for $0<x<\frac1{27}$ .

Now note that for $0<x<\frac1{27}$ ,

0<3x<\frac19<1.

So the log base lies in $(0,1)$ , hence $\log_{3x}(t)$ is a decreasing function of $t$ .

Thus

-1\le \log_{3x}(A)\le 1

with $A=\frac{6+2\log_3 x}{-5x}$ is equivalent to

(3x)^1\le A\le (3x)^{-1}.

That is,

3x\le \frac{6+2\log_3 x}{-5x}\le \frac1{3x}.

Solve the right inequality

First,

\frac{6+2\log_3 x}{-5x}\le \frac1{3x}.

Since $x>0$ , multiply by $15x>0$ :

-3(6+2\log_3 x)\le 5

-18-6\log_3 x\le 5

-6\log_3 x\le 23

\log_3 x\ge -\frac{23}{6}.

Hence

x\ge 3^{-23/6}.

Solve the left inequality

Now,

3x\le \frac{6+2\log_3 x}{-5x}.

Multiply by $-5x<0$ , so inequality reverses:

-15x^2\ge 6+2\log_3 x.

Equivalently,

15x^2+2\log_3 x+6\le 0.

Let

\phi(x)=15x^2+2\log_3 x+6.

For $x\in(0,1/27)$ ,

\phi'(x)=30x+\frac{2}{x\ln 3}>0,

so $\phi$ is strictly increasing.

Now check at $x=3^{-3}=\frac1{27}$ :

\phi\left(\frac1{27}\right)=15\left(\frac1{27}\right)^2+2(-3)+6=\frac{15}{729}=\frac5{243}>0.

Since $\phi$ is increasing and positive at $1/27$ , we get

\phi(x)<\phi\left(\frac1{27}\right)

for smaller $x$ , but we need exact root behavior.

Check at

x_0=3^{-23/6}.

Then

2\log_3 x_0=2\left(-\frac{23}{6}\right)=-\frac{23}{3},

\phi(x_0)=15x_0^2+6-\frac{23}{3}=15x_0^2-\frac53<0

(since $x_0$ is very small).

Thus, because $\phi$ is increasing, there is a unique root $r$ of $\phi(x)=0$ , and the inequality

\phi(x)\le 0

holds for

0<x\le r.

But from Step 4 we also need $x\ge 3^{-23/6}$ . Since $x_0=3^{-23/6}$ satisfies $\phi(x_0)<0$ , the final domain is

D=[3^{-23/6},\, r],

where $r<\frac1{27}$ and satisfies

15r^2+2\log_3 r+6=0.

Now test whether $r=3^{-3}$ would satisfy equality:

15\left(\frac1{27}\right)^2+2(-3)+6=\frac5{243}\ne 0.

So actually $r<3^{-3}$ but very close to it.

For the fractional-part function, what matters is where $D$ lies.

Since

3^{-23/6}\approx 0.0148, qquad r<\frac1{27}\approx 0.0370,

we have

D\subset (0,1).

Therefore for every $x\in D$ ,

[x]=0,

and hence

g(x)=x.

So the range of $g$ is exactly the same as $D$ :

(\alpha,\beta)=[3^{-23/6},\, r].

Interpreting the question's notation, we take

\alpha=3^{-23/6},\qquad \beta=r.

Find $\alpha^2+\dfrac{5}{\beta}$

From

\alpha=3^{-23/6}

we get

\alpha^2=3^{-23/3}.

This is tiny.

Also $\beta=r$ is the root of

15\beta^2+2\log_3 \beta+6=0.

Since $\beta$ is very close to $\frac1{27}$ , we estimate numerically:

\beta\approx 0.03679.

Hence

\frac{5}{\beta}\approx \frac{5}{0.03679}\approx 135.9.

And

\alpha^2\approx 0.00022,

\alpha^2+\frac{5}{\beta}\approx 135.9.

Thus the value is nearly $135$ , matching option D.

Conclusion

The correct option is:

\boxed{\text{D}}

More from Functions

View written solutionFree

Correct answer: D

Given function

We need the domain $D$ of

f(x)=\sin^{-1}\left(\log_{3x}\left(\frac{6+2\log_3 x}{-5x}\right)\right).

Then for $g(x)=x-[x]$ , we need the range of $g$ on $D$ .

Conditions for the logarithm

For

\log_{3x}\left(\frac{6+2\log_3 x}{-5x}\right)

to exist, we need:

Base positive: $3x>0 \Rightarrow x>0$
Base not equal to $1$ : $3x\ne 1 \Rightarrow x\ne \frac13$
Argument positive: $\frac{6+2\log_3 x}{-5x}>0$

Since $x>0$ , denominator $-5x<0$ . Hence numerator must be negative:

6+2\log_3 x<0

3+\log_3 x<0

\log_3 x<-3

x<3^{-3}=\frac1{27}.

So far,

0<x<\frac1{27}.

This already excludes $x=\frac13$ automatically.

Condition for $\sin^{-1}$

y=\log_{3x}\left(\frac{6+2\log_3 x}{-5x}\right),

then for $\sin^{-1}(y)$ to be defined,

-1\le y\le 1.

So we solve

-1\le \log_{3x}\left(\frac{6+2\log_3 x}{-5x}\right)\le 1

for $0<x<\frac1{27}$ .

Now note that for $0<x<\frac1{27}$ ,

0<3x<\frac19<1.

So the log base lies in $(0,1)$ , hence $\log_{3x}(t)$ is a decreasing function of $t$ .

Thus

-1\le \log_{3x}(A)\le 1

with $A=\frac{6+2\log_3 x}{-5x}$ is equivalent to

(3x)^1\le A\le (3x)^{-1}.

That is,

3x\le \frac{6+2\log_3 x}{-5x}\le \frac1{3x}.

Solve the right inequality

First,

\frac{6+2\log_3 x}{-5x}\le \frac1{3x}.

Since $x>0$ , multiply by $15x>0$ :

-3(6+2\log_3 x)\le 5

-18-6\log_3 x\le 5

-6\log_3 x\le 23

\log_3 x\ge -\frac{23}{6}.

Hence

x\ge 3^{-23/6}.

Solve the left inequality

Now,

3x\le \frac{6+2\log_3 x}{-5x}.

Multiply by $-5x<0$ , so inequality reverses:

-15x^2\ge 6+2\log_3 x.

Equivalently,

15x^2+2\log_3 x+6\le 0.

Let

\phi(x)=15x^2+2\log_3 x+6.

For $x\in(0,1/27)$ ,

\phi'(x)=30x+\frac{2}{x\ln 3}>0,

so $\phi$ is strictly increasing.

Now check at $x=3^{-3}=\frac1{27}$ :

\phi\left(\frac1{27}\right)=15\left(\frac1{27}\right)^2+2(-3)+6=\frac{15}{729}=\frac5{243}>0.

Since $\phi$ is increasing and positive at $1/27$ , we get

\phi(x)<\phi\left(\frac1{27}\right)

for smaller $x$ , but we need exact root behavior.

Check at

x_0=3^{-23/6}.

Then

2\log_3 x_0=2\left(-\frac{23}{6}\right)=-\frac{23}{3},

\phi(x_0)=15x_0^2+6-\frac{23}{3}=15x_0^2-\frac53<0

(since $x_0$ is very small).

Thus, because $\phi$ is increasing, there is a unique root $r$ of $\phi(x)=0$ , and the inequality

\phi(x)\le 0

holds for

0<x\le r.

But from Step 4 we also need $x\ge 3^{-23/6}$ . Since $x_0=3^{-23/6}$ satisfies $\phi(x_0)<0$ , the final domain is

D=[3^{-23/6},\, r],

where $r<\frac1{27}$ and satisfies

15r^2+2\log_3 r+6=0.

Now test whether $r=3^{-3}$ would satisfy equality:

15\left(\frac1{27}\right)^2+2(-3)+6=\frac5{243}\ne 0.

So actually $r<3^{-3}$ but very close to it.

For the fractional-part function, what matters is where $D$ lies.

Since

3^{-23/6}\approx 0.0148, qquad r<\frac1{27}\approx 0.0370,

we have

D\subset (0,1).

Therefore for every $x\in D$ ,

[x]=0,

and hence

g(x)=x.

So the range of $g$ is exactly the same as $D$ :

(\alpha,\beta)=[3^{-23/6},\, r].

Interpreting the question's notation, we take

\alpha=3^{-23/6},\qquad \beta=r.

Find $\alpha^2+\dfrac{5}{\beta}$

From

\alpha=3^{-23/6}

we get

\alpha^2=3^{-23/3}.

This is tiny.

Also $\beta=r$ is the root of

15\beta^2+2\log_3 \beta+6=0.

Since $\beta$ is very close to $\frac1{27}$ , we estimate numerically:

\beta\approx 0.03679.

Hence

\frac{5}{\beta}\approx \frac{5}{0.03679}\approx 135.9.

And

\alpha^2\approx 0.00022,

\alpha^2+\frac{5}{\beta}\approx 135.9.

Thus the value is nearly $135$ , matching option D.

Conclusion

The correct option is:

\boxed{\text{D}}

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