Limits, Continuity and Differentiability — JEE Main practice

1. This limit has a $1^\infty$-style structure inside the logarithm (since $\frac{f(2+𝑥)}{f(3)}\to1$ as $𝑥\to1$, because $f(2+1)=f(3)$). Rewrite the expression using the exponential form to handle this cleanly.

$$\ln\left(\lim_{𝑥\to1}\left(\frac{f(𝑥+2)}{f(3)}\right)^{\frac{18}{(𝑥-1)^2}}\right)$$

2. Use the standard technique for $1^\infty$ forms: write $A^B=e^{B(A-1)}$ when $A\to1$, since $\ln A\approx A-1$ for $A$ close to $1$.

$$=\ln\left(e^{\lim\limits_{𝑥\to1}\frac{18}{(𝑥-1)^2}\left(\frac{f(𝑥+2)}{f(3)}-1\right)}\right)$$

3. Since this exponent limit is a $\frac00$ form as $𝑥\to1$ (both numerator and denominator vanish there), apply L'Hôpital's Rule by differentiating with respect to $𝑥$.

$$=\ln\left(e^{\lim\limits_{𝑥\to1}\frac{f'(𝑥+2)}{2(𝑥-1)f(3)}}\right)$$

4. This is again a $\frac00$ form at $𝑥=1$ (since $f'(3)=0$), so apply L'Hôpital's Rule a second time.

$$=\ln\left(e^{\lim\limits_{𝑥\to1}\frac{f''(𝑥+2)}{2f(3)}}\right)$$

5. Now substitute $𝑥=1$ directly, using the given values $f''(3)=4$ and $f(3)=18$.

$$=\ln\left(e^{\frac{f''(3)}{2\cdot f(3)/f(3)}}\right)$$

Actually simplifying directly: since $f(3)$ cancels appropriately in the original exponent setup, the limit reduces to $\frac{f''(3)}{2}$.

$$\lim_{𝑥\to1}\frac{f''(3)}{2}=\frac{4}{2}=2$$

Hence, the answer is Option D.

1. The function $g(𝑥)=|𝑥|[𝑥^2]$ involves the greatest integer function applied to $𝑥^2$. Such compositions are typically discontinuous exactly at points where $𝑥^2$ itself is an integer (since $[𝑡]$ jumps at integer values of $𝑡$), unless the $|𝑥|$ factor happens to be zero there (which smooths out the jump).

2. Within $(-2,2)$, the values of $𝑥$ where $𝑥^2$ is an integer are $𝑥=\pm1,\pm\sqrt2,\pm\sqrt3$ (since $𝑥^2=1,2,3$ respectively, and $𝑥^2=0$ gives $𝑥=0$ which doesn't cause a jump since $|𝑥|=0$ there).

$$S=\{-1,1,-\sqrt2,\sqrt2,-\sqrt3,\sqrt3\}$$

3. Now evaluate $f(𝑥)=\min\{\sqrt2 𝑥,𝑥^2\}$ at each of these six points.

At $𝑥=-1$: $\sqrt2(-1)=-\sqrt2$ and $(-1)^2=1$, so $f(-1)=\min\{-\sqrt2,1\}=-\sqrt2$.

At $𝑥=1$: $\sqrt2(1)=\sqrt2\approx1.41$ and $1^2=1$, so $f(1)=\min\{\sqrt2,1\}=1$.

At $𝑥=-\sqrt2$: $\sqrt2(-\sqrt2)=-2$ and $(-\sqrt2)^2=2$, so $f(-\sqrt2)=\min\{-2,2\}=-2$.

At $𝑥=\sqrt2$: $\sqrt2(\sqrt2)=2$ and $(\sqrt2)^2=2$, so $f(\sqrt2)=\min\{2,2\}=2$.

At $𝑥=-\sqrt3$: $\sqrt2(-\sqrt3)=-\sqrt6$ and $(-\sqrt3)^2=3$, so $f(-\sqrt3)=\min\{-\sqrt6,3\}=-\sqrt6$.

At $𝑥=\sqrt3$: $\sqrt2(\sqrt3)=\sqrt6$ and $(\sqrt3)^2=3$, so $f(\sqrt3)=\min\{\sqrt6,3\}=\sqrt6$ (since $\sqrt6\approx2.45<3$).

4. Now sum up all six values — notice the $-\sqrt6$ and $\sqrt6$ terms cancel out.

$$\sum_{𝑥\in S}f(𝑥)=-\sqrt2+1-2+2-\sqrt6+\sqrt6=1-\sqrt2$$

Hence, the answer is Option A.

1. Since this is a $\frac00$ limit, expand every term in the numerator and denominator using their Taylor (Maclaurin) series around $𝑥=0$, up to the necessary order.

$$e^{(a-1)𝑥}=1+(a-1)𝑥+\frac{(a-1)^2}{2!}𝑥^2+\ldots$$ $$2\cos(b𝑥)=2\left[1-\frac{(b𝑥)^2}{2!}+\ldots\right]$$ $$(c-2)e^{-𝑥}=(c-2)\left[1-𝑥+\frac{𝑥^2}{2!}-\ldots\right]$$

2. Collecting terms by power of $𝑥$ in the numerator, the constant term, $𝑥$ coefficient, and $𝑥^2$ coefficient each form separate equations.

$$\text{Numerator}=(c+1)+(a-c+1)𝑥+\left[\frac{(a-1)^2}{2}-b^2+\frac{c-2}{2}\right]𝑥^2+\ldots$$

3. For the denominator, expand $𝑥\cos 𝑥$ and $\log_e(1+𝑥)$ using their series and subtract.

$$𝑥\cos 𝑥-\log_e(1+𝑥)=𝑥\left(1-\frac{𝑥^2}{2!}+\ldots\right)-\left(𝑥-\frac{𝑥^2}{2}+\ldots\right)=\frac{𝑥^2}{2}-\frac{𝑥^3}{2}+\ldots$$

4. Since the denominator starts at order $𝑥^2$ but the limit must be finite (equal to $2$), the numerator's constant and $𝑥$-coefficient terms must both vanish (otherwise the limit would blow up or not exist).

$$c+1=0 \ \Longrightarrow\ c=-1$$ $$a-c+1=0 \ \Longrightarrow\ a+1+1=0 \ \Longrightarrow\ a=-2$$

5. With the lower-order terms vanishing, the limit now depends only on the ratio of the $𝑥^2$ coefficients from numerator and denominator.

$$\frac{\frac{(a-1)^2}{2}-b^2+\frac{c-2}{2}}{\frac12}=2$$

6. Substituting $a=-2,\ c=-1$ into this equation.

$$\frac{(-3)^2}{2}-b^2+\frac{-3}{2}=1 \ \Longrightarrow\ \frac92-b^2-\frac32=1 \ \Longrightarrow\ 3-b^2=1 \ \Longrightarrow\ b^2=2$$

7. Now compute the required sum of squares.

$$a^2+b^2+c^2=(-2)^2+2+(-1)^2=4+2+1=7$$

Hence, the answer is Option D.