JEE Advanced · Mathematics · 12. Circle

The straight line $2 x-3 y=1$ divides the circular region $x^2+y^2 \leq 6$ into two parts. If $S=\left\{\left(2, \frac{3}{4}\right),\left(\frac{5}{2}, \frac{3}{4}\right),\left(\frac{1}{4},-\frac{1}{4}\right),\left(\frac{1}{8}, \frac{1}{4}\right)\right\}$, then the number of point(s) in $S$ lying inside the smaller part is

Verified Solution

Answer & Solution

Correct Answer

(B) 2

Step-by-step Solution

Detailed explanation

$x^2+y^2 \leq 6$ and $2 x-3 y=1$ is shown as, For the point to lie in the shaded part, origin and the point lie on opposite side of straight line $L$.

$\therefore$ For any point in shaded part $L>0$ and for any point inside the circle $S < 0$.
Now, for $\left(2, \frac{3}{4}\right) L: 2 x-3 y-1$
\[
L: 4-\frac{9}{4}-1=\frac{3}{4}>0
\]
and $S: x^2+y^2-6$,
\[
S: 4+\frac{9}{16}-6 < 0
\]
$\Rightarrow\left(2 \frac{3}{4}\right)$ lies in shaded part.
For $\left(\frac{5}{2}, \frac{3}{4}\right) L: 5-9-1 < 0 \quad$ [neglect]
For $\left.\left(\frac{1}{4},-\frac{1}{4}\right) L: \frac{1}{2}+\frac{3}{4}-1>0\right\}$
$\therefore\left(\frac{1}{4},-\frac{1}{4}\right)$ lies in the shaded part. For $\left(\frac{1}{8}, \frac{1}{4}\right) L: \frac{1}{4}-\frac{3}{4}-1 < 0$ [neglect] $\Rightarrow$ Only 2 points lie in the shaded part.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Let

f (x) = x + \log_{e} x - x \log_{e} x, x \in (0, \infty) .

•

Column 1 contains information about zeros of

f (x), f^{'} (x)

and

f^{''} (x) .

•

Column 2 contains information about the limiting behaviour of

f (x), f^{'} (x)

and

f^{''} (x)

at infinity.

•

Column 3 contains information about increasing-decreasing nature of

f (x)

and

f^{'} (x) .

Column 1	Column 2	Column 3
(I) $f (x) = 0$ for some $x \in (1, e^{2})$	(i) $\lim_{x \to \infty} f (x) = 0$	(P) $f$ is increasing in (0, 1)
(II) $f^{'} (x) = 0$ for some $x i n (1, e)$	(ii) $\lim_{x \to \infty} f (x) = - \infty$	(Q) $f$ is decreasing in $(e, e^{2})$
(III) $f^{'} (x) = 0$ for some $x \in (0, 1)$	(iii) $\lim_{x \to \infty} f^{'} (x) = - \infty$	(R) $f^{'}$ is increasing in (0, 1)
(IV) $f^{''} (x) = 0$ for some $x \in (1, e)$	(iv) $\lim_{x \to \infty} f^{''} (x) = 0$	(S) $f^{'}$ is decreasing in $(e, e^{2})$

Which of the following options is the only Incorrect combination?

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