Functions — JEE Main practice

1. For $\sin^{-1}(\cdot)$ to be defined, its argument must lie between $-1$ and $1$.

$$-1\le \frac{5-𝑥}{3+2𝑥}\le 1$$

2. Handle the left inequality $\frac{5-𝑥}{3+2𝑥}\ge -1$ first by bringing everything to one side.

$$\frac{2-3𝑥}{3+2𝑥}\le 0 \ \Longrightarrow\ (3𝑥-2)(2𝑥+3)\ge 0$$

Solving this sign condition (roots at $𝑥=\frac23$ and $𝑥=-\frac32$):

$$𝑥\in\left(-\infty,-\frac32\right]\cup\left[\frac23,\infty\right)$$

3. Now handle the right inequality $\frac{5-𝑥}{3+2𝑥}\le 1$ similarly.

$$\frac{8+𝑥}{3+2𝑥}\ge 0 \ \Longrightarrow\ 𝑥\in(-\infty,-8]\cup\left[-\frac32,\infty\right)$$

4. Intersecting the results of steps 2 and 3 gives the domain restriction from the inverse sine term.

$$\left(-\infty,-\frac32\right]\cup\left[\frac23,\infty\right)\ \cap\ \left((-\infty,-8]\cup\left[-\frac32,\infty\right)\right)=(-\infty,-8]\cup\left[\frac23,\infty\right)$$

5. Next, for $\frac{1}{\log_e(10-𝑥)}$ to be defined, we need $10-𝑥>0$ (so the log itself is defined) and $\log_e(10-𝑥)\ne 0$ (so we don't divide by zero). This gives:

$$𝑥<10 \quad \text{and} \quad 10-𝑥\ne 1 \ \Longrightarrow\ 𝑥\ne 9$$

6. Combining this with the result from step 4 gives the overall domain.

$$\big((-\infty,-8]\cup[\tfrac23,\infty)\big)\cap(-\infty,10)-\{9\} = (-\infty,-8]\cup\left[\frac23,10\right)-\{9\}$$

So $\alpha=-8$, $\beta=\frac23$, $\gamma=10$, $\delta=9$.

7. Now compute the required value.

$$\alpha+\beta+\gamma+\delta=-8+\frac23+10+9=\frac{35}{3}$$ $$6(\alpha+\beta+\gamma+\delta)=6\times\frac{35}{3}=70$$

Hence, the answer is Option A.

1. The equation $f(𝑥+𝑦)=f(𝑥)f(𝑦)$ is the defining property of an exponential function, so $f(𝑥)=𝑎^𝑥$ for some constant $𝑎$. Using $f(1)=7$:

$$𝑎^1=7 \ \Longrightarrow\ f(𝑥)=7^𝑥$$

2. For $g$, put $𝑦=1$ in $g(𝑥+𝑦)=g(𝑥𝑦)$.

$$g(𝑥+1)=g(𝑥\cdot 1)=g(𝑥)$$

This means $g$ takes the same value at every consecutive integer, so:

$$g(1)=g(2)=g(3)=\cdots=g(𝑛)=1$$

3. Substitute $f(𝑥)=7^𝑥$ and $g(𝑥)=1$ into the given sum condition.

$$\sum_{𝑥=1}^{𝑛}\frac{7^𝑥}{1}=19607$$

4. This is a geometric series with first term $7$ and common ratio $7$; use the sum formula $\sum = 𝑎\frac{𝑟^𝑛-1}{𝑟-1}$.

$$7\left(\frac{7^𝑛-1}{7-1}\right)=19607$$

5. Simplify to isolate $7^𝑛$.

$$7^𝑛-1=\frac{6}{7}\times 19607=16806$$

6. Solving for $7^𝑛$:

$$7^𝑛=16807=7^5$$

So $𝑛=5$.

Hence, the answer is Option D.

1. Let $\lambda=𝑥^2-10𝑥+85$ to simplify the nested logarithms. For $\log_2\lambda$ to be defined, we need:

$$\lambda>0 \quad \text{...(1)}$$

2. For $\log_5(\cdot)$ to accept a positive argument, we need $7-\log_2\lambda>0$.

$$\lambda<2^7 \quad \text{...(2)}$$

3. For $\log_3(\cdot)$ to accept a positive argument, we need $\log_5(7-\log_2\lambda)>0$, i.e. the inner quantity must exceed $1$.

$$7-\log_2\lambda>1 \ \Longrightarrow\ \lambda<2^6 \quad \text{...(3)}$$

4. Combining conditions (1), (2), and (3), the binding constraint is:

$$0<\lambda<2^6=64$$

5. Substituting back $\lambda=𝑥^2-10𝑥+85$ and solving the resulting quadratic inequality gives:

$$0<𝑥^2-10𝑥+85<64 \ \Longrightarrow\ 𝑥\in(3,7)$$

6. Next, for the $\sin^{-1}(\cdot)$ term, its argument (an absolute value) must lie between $-1$ and $1$, which since it's non-negative really means between $0$ and $1$.

$$-1\le \frac{3𝑥-7}{𝑥-17}\le 1 \ \Longrightarrow\ 𝑥\in[-5,6]$$

7. The domain of $f$ is the intersection of both conditions found in steps 5 and 6.

$$(3,7)\cap[-5,6]=(3,6]$$

So $\alpha=3$ and $\beta=6$.

8. Therefore:

$$\alpha+\beta=3+6=9$$

Hence, the answer is Option A.