ISI MS(QMBA) Mock Tests

Let the given equation be $P(x) = x^3 - 6x^2 + 11x - 6 = 0$. The roots are $\alpha, \beta, \gamma$. We need to find an equation whose roots are $2\alpha+1, 2\beta+1, 2\gamma+1$. **Step 1: Define the relationship between the new roots ($y$) and the old roots ($x$).** Let $y$ be a root of the new equation. The relation is $y = 2x+1$. **Step 2: Express $x$ in terms of $y$.** From $y = 2x+1$, we solve for $x$: $2x = y-1$ $x = \frac{y-1}{2}$ **Step 3: Substitute this expression for $x$ into the original equation $P(x) = 0$.** **Step 4: Simplify the equation by clearing denominators.** To eliminate the denominators, multiply the entire equation by $2^3 = 8$: **Step 5: Expand the terms and collect coefficients.** Using the binomial expansion formulas $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ and $(a-b)^2 = a^2 - 2ab + b^2$: * $(y-1)^3 = y^3 - 3y^2(1) + 3y(1)^2 - (1)^3 = y^3 - 3y^2 + 3y - 1$ * $-12(y-1)^2 = -12(y^2 - 2y + 1) = -12y^2 + 24y - 12$ * $44(y-1) = 44y - 44$ * $-48$ Adding these terms together: \begin{align*} & y^3 \quad & \text{(from } y^3 \text{ term)} \\ & (-3 - 12)y^2 = -15y^2 \quad & \text{(from } y^2 \text{ terms)} \\ & (3 + 24 + 44)y = 71y \quad & \text{(from } y \text{ terms)} \\ & (-1 - 12 - 44 - 48) = -105 \quad & \text{(from constant terms)} \end{align*} The new equation is $y^3 - 15y^2 + 71y - 105 = 0$. Replacing $y$ with $x$ as the variable for the new equation (standard practice): $x^3 - 15x^2 + 71x - 105 = 0$. **Why Options are Correct/Incorrect:** * **A. $x^3 - 15x^2 + 71x - 105 = 0$**: This is the correct equation as derived through the systematic substitution method. * **B. $x^3 - 9x^2 + 23x - 15 = 0$**: This option would be obtained if one incorrectly substituted $x = \frac{y+1}{2}$ instead of $x = \frac{y-1}{2}$. This is a common algebraic error in solving for $x$ from $y = 2x+1$. * **C. $x^3 - 12x^2 + 44x - 48 = 0$**: This equation would be obtained if the transformation was only $y=2x$ (roots scaled by 2), completely ignoring the '+1' shift. This represents a partial or incorrect application of the transformation. * **D. $x^3 + 15x^2 + 71x + 105 = 0$**: This option involves multiple sign errors on coefficients, likely from inconsistent application of sign rules during expansion or collecting terms. **Common Mistakes:** 1. Incorrectly solving for $x$ in terms of $y$ from the transformation relation, especially sign errors (e.g., confusing $y = ax+b$ with $x = (y+b)/a$ instead of the correct $x = (y-b)/a$). 2. Arithmetic errors during the expansion of binomial terms like $(y-1)^3$ or $-12(y-1)^2$. 3. Making errors while clearing denominators, such as multiplying by an incorrect power of the scaling factor or distributing the multiplier incorrectly. 4. Failing to collect all like terms correctly or making sign errors when summing the expanded polynomial terms.

Step 1: Calculate the total number of ways to form a committee of 5 members from a group of 7 men and 5 women. Total number of people = $7 + 5 = 12$. Number of ways to choose 5 members from 12 is given by the combination formula $C(n, r) = \frac{n!}{r!(n-r)!}$. Step 2: Calculate the number of favorable ways for the committee to consist of at least 3 men. "At least 3 men" means the committee can have: a) Exactly 3 men and 2 women b) Exactly 4 men and 1 woman c) Exactly 5 men and 0 women Case a) Exactly 3 men and 2 women: Number of ways to choose 3 men from 7: $C(7, 3) = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$. Number of ways to choose 2 women from 5: $C(5, 2) = \frac{5 \times 4}{2 \times 1} = 10$. Ways for Case a) = $C(7, 3) \times C(5, 2) = 35 \times 10 = 350$. Case b) Exactly 4 men and 1 woman: Number of ways to choose 4 men from 7: $C(7, 4) = C(7, 3) = 35$. Number of ways to choose 1 woman from 5: $C(5, 1) = 5$. Ways for Case b) = $C(7, 4) \times C(5, 1) = 35 \times 5 = 175$. Case c) Exactly 5 men and 0 women: Number of ways to choose 5 men from 7: $C(7, 5) = C(7, 2) = \frac{7 \times 6}{2 \times 1} = 21$. Number of ways to choose 0 women from 5: $C(5, 0) = 1$. Ways for Case c) = $C(7, 5) \times C(5, 0) = 21 \times 1 = 21$. Total number of favorable outcomes $n(E) = 350 + 175 + 21 = 546$. Step 3: Calculate the probability. Step 4: Simplify the fraction. Both numerator and denominator are divisible by 2: $\frac{546 \div 2}{792 \div 2} = \frac{273}{396}$ Both are divisible by 3 (sum of digits $2+7+3=12$, $3+9+6=18$): $\frac{273 \div 3}{396 \div 3} = \frac{91}{132}$ Thus, the probability that the committee consists of at least 3 men is $\frac{91}{132}$. **Why other options are incorrect:** * **$\frac{41}{132}$:** This would be the probability of having "at most 2 men" (i.e., 0 men, 5 women; 1 man, 4 women; or 2 men, 3 women). The number of ways for this would be $C(7,0)C(5,5) + C(7,1)C(5,4) + C(7,2)C(5,3) = 1 \times 1 + 7 \times 5 + 21 \times 10 = 1 + 35 + 210 = 246$. So, $P(\text{at most 2 men}) = \frac{246}{792} = \frac{41}{132}$. This is the complement of the correct answer ($1 - \frac{91}{132} = \frac{41}{132}$). * **$\frac{175}{396}$:** This is the probability of having "exactly 3 men and 2 women" ($C(7,3)C(5,2) = 350$) divided by the total number of ways ($792$). So, $\frac{350}{792} = \frac{175}{396}$. This is incorrect because the question asks for "at least 3 men", which includes more cases than just exactly 3 men. * **$\frac{65}{132}$:** This option does not correspond to a straightforward common miscalculation or a different interpretation of the problem. It is likely a distractor without a specific logical derivation from common errors. **Common Mistakes:** 1. **Ignoring all cases for "at least":** Students often calculate only one of the cases (e.g., exactly 3 men) and present it as the final answer, overlooking other valid scenarios (4 men, 5 men). 2. **Confusing "at least" with "at most":** Calculating the probability of the complementary event correctly, but then presenting it as the answer to the original question without subtracting from 1. 3. **Calculation errors in combinations:** Mistakes in calculating $C(n,r)$ values or in summing up the favorable outcomes. 4. **Incorrectly simplifying fractions:** Errors in reducing the final probability to its simplest form.

Step 1: Replace $f(x)$ with $y$. We have $y = \frac{e^x+1}{e^x-1}$. Step 2: Swap $x$ and $y$ to find the inverse relation. Step 3: Solve the new equation for $y$ in terms of $x$. Multiply both sides by $(e^y-1)$: Distribute $x$ on the left side: Gather terms involving $e^y$ on one side and constant terms on the other: Factor out $e^y$ from the left side: Divide by $(x-1)$ to isolate $e^y$: To solve for $y$, take the natural logarithm (base $e$) of both sides: Step 4: Replace $y$ with $f^{-1}(x)$. Let's verify the domain of $f^{-1}(x)$. For the logarithm to be defined, the argument must be positive: $\frac{x+1}{x-1} > 0$. This inequality holds when both numerator and denominator have the same sign, i.e., $x+1 > 0$ and $x-1 > 0$ (implying $x > 1$) or $x+1 < 0$ and $x-1 < 0$ (implying $x < -1$). So, the domain of $f^{-1}(x)$ is $(-\infty, -1) \cup (1, \infty)$. This matches the range of the original function $f(x) = 1 + \frac{2}{e^x-1}$, as $e^x-1$ can take any real value except $-1$ (when $x \to -\infty$, $e^x-1 \to -1$) and $0$ (when $x \to 0$, $e^x-1 \to 0$). As $e^x \in (0, \infty)$, $e^x-1 \in (-1, \infty) \setminus \{0\}$. Therefore, $\frac{2}{e^x-1} \in (-\infty, -2) \cup (0, \infty)$. So $f(x) \in (-\infty, -1) \cup (1, \infty)$. The domain of $f^{-1}(x)$ correctly corresponds to the range of $f(x)$. **Why other options are incorrect:** * **Option (B) $\frac{1}{2}\ln\left(\frac{x+1}{x-1}\right)$:** This extra factor of $\frac{1}{2}$ would appear if the original function was of the form $f(x) = \frac{e^{2x}-1}{e^{2x}+1}$ (hyperbolic tangent) or a similar structure leading to $2y$ before taking logarithm. This indicates an algebraic error in isolating $y$ or a misremembered formula for a different function's inverse. * **Option (C) $\frac{x-1}{x+1}$:** This is simply the reciprocal of the function, or a result of incorrectly swapping terms or variables without properly solving for $y$. An inverse function is not simply the reciprocal of the original function. * **Option (D) $\ln\left(\frac{x-1}{x+1}\right)$:** This results from an algebraic error in the step $e^y = \frac{x+1}{x-1}$, where the numerator and denominator are swapped (i.e., $e^y(x-1) = -(x+1)$ instead of $x+1$). The correct option is $\ln\left(\frac{x+1}{x-1}\right)$. **Common Mistakes:** 1. **Algebraic Errors in Isolating $e^y$:** Students often make mistakes when moving terms around, especially when factoring $e^y$ or dividing by $(x-1)$. Ensure careful distribution and collection of terms. 2. **Incorrect Logarithm Application:** Forgetting to take the natural logarithm (or logarithm of the correct base) to solve for $y$ after isolating the exponential term. 3. **Swapping Numerator and Denominator:** A common error is to incorrectly swap $x+1$ and $x-1$ in the final logarithmic argument, leading to a wrong inverse function. 4. **Ignoring Domain/Range Consistency:** While not explicitly asked to find the domain/range, a quick mental check can sometimes reveal inconsistencies, for example, if the inverse function's domain doesn't match the original function's range, suggesting an error in derivation.

The given series is an Arithmetico-Geometric Progression (AGP). Let the series be $S$. **Method 1: Using the formula for sum to infinity of an AGP** The general term of an AGP is of the form $[a + (n-1)d]r^{n-1}$. In the given series: - The numerators $1, 4, 7, 10, \dots$ form an Arithmetic Progression (AP) with first term $a = 1$ and common difference $d = 4-1 = 3$. - The denominators $1, 5, 25, 125, \dots$ (treating the first term as $1/1$) form a Geometric Progression (GP) with first term $1$ and common ratio $r = \frac{1}{5}$. The sum to infinity of an AGP is given by the formula: Substitute the values $a=1$, $d=3$, and $r=\frac{1}{5}$: To sum these fractions, find a common denominator, which is $16$: **Method 2: Using the 'multiply by $r$ and subtract' method** Let $S = 1 + \frac{4}{5} + \frac{7}{25} + \frac{10}{125} + \dots \quad (1)$ Multiply the series by the common ratio $r = \frac{1}{5}$: Subtract equation (2) from equation (1), shifting the terms to align corresponding denominators: The series on the right-hand side can be written as $1 + 3\left(\frac{1}{5} + \frac{1}{25} + \frac{1}{125} + \dots\right)$. The part in the parenthesis is an infinite Geometric Progression with first term $a' = \frac{1}{5}$ and common ratio $r' = \frac{1}{5}$. The sum of this infinite GP is $S'_{GP} = \frac{a'}{1-r'} = \frac{\frac{1}{5}}{1 - \frac{1}{5}} = \frac{\frac{1}{5}}{\frac{4}{5}} = \frac{1}{4}$. Substitute this back into the equation for $\frac{4}{5}S$: Solve for $S$: Thus, the sum to infinity of the series is $\frac{35}{16}$. **Why other options are incorrect:** * **(A) $\frac{15}{16}$:** This would be obtained if one incorrectly ignores the '1' term in $1 + 3\left(\frac{1}{4}\right)$ during the subtraction method, or if they only calculate the second part of the formula $\frac{dr}{(1-r)^2}$. * **(C) $\frac{5}{4}$:** This would be obtained if one only considers the first part of the formula $\frac{a}{1-r}$ and ignores the second part $\frac{dr}{(1-r)^2}$. * **(D) $\frac{25}{16}$:** This option does not result from a common systematic error but rather likely from a significant arithmetic miscalculation. The final answer is $\boxed{\frac{35}{16}}$ **Common Mistakes:** 1. Forgetting or misapplying the formula for the sum to infinity of an AGP, or incorrectly identifying the 'a', 'd', and 'r' components. 2. Arithmetic errors, especially with fractions and exponents, when calculating the terms of the formula. 3. In the 'multiply by r and subtract' method, forgetting the first term on the right-hand side after subtraction or incorrectly summing the resulting geometric series. 4. Not checking the condition $|r| < 1$ for the sum to infinity to exist (though in this case $r = 1/5$ satisfies it).

To form distinct 4-digit even numbers using the digits 0, 1, 2, 3, 4, 5 without repetition, we must consider two main restrictions: 1. The number must be even, meaning its unit's digit must be 0, 2, or 4. 2. The thousand's digit (first digit) cannot be 0. We will use case analysis based on the unit's digit: **Case 1: The unit's digit is 0.** Let the 4-digit number be represented by $d_1 d_2 d_3 d_4$. * **$d_4$ (Unit's digit):** Must be 0. There is 1 choice (0). * **$d_1$ (Thousand's digit):** Since 0 is already used for $d_4$, and $d_1$ cannot be 0, we can choose from the remaining 5 digits {1, 2, 3, 4, 5}. So, there are 5 choices for $d_1$. * **$d_2$ (Hundred's digit):** Two digits have been used (one for $d_4$, one for $d_1$). From the remaining 4 digits, there are 4 choices for $d_2$. * **$d_3$ (Ten's digit):** Three digits have been used. From the remaining 3 digits, there are 3 choices for $d_3$. Number of ways for Case 1 = $5 \times 4 \times 3 \times 1 = 60$. **Case 2: The unit's digit is 2 or 4.** Let's consider these two possibilities together because the restriction on $d_1$ (not being 0) interacts similarly with both. * **$d_4$ (Unit's digit):** Can be 2 or 4. There are 2 choices. * **$d_1$ (Thousand's digit):** This is the tricky part. One non-zero digit (2 or 4) has been used for $d_4$. The remaining digits include 0. $d_1$ cannot be 0. So, from the initial 6 digits, one non-zero digit is used, and 0 is not allowed for $d_1$. This leaves $6 - 2 = 4$ choices for $d_1$ (e.g., if $d_4=2$, the choices for $d_1$ are {1, 3, 4, 5}). * **$d_2$ (Hundred's digit):** Two digits have been used ($d_4$ and $d_1$). From the remaining 4 digits (which now includes 0), there are 4 choices for $d_2$. * **$d_3$ (Ten's digit):** Three digits have been used. From the remaining 3 digits, there are 3 choices for $d_3$. Number of ways for Case 2 = (Choices for $d_1$) $\times$ (Choices for $d_2$) $\times$ (Choices for $d_3$) $\times$ (Choices for $d_4$) Number of ways for Case 2 = $4 \times 4 \times 3 \times 2 = 96$. **Total number of distinct 4-digit even numbers:** Summing the ways from both cases: Total = Case 1 + Case 2 = $60 + 96 = 156$. Therefore, 156 distinct 4-digit even numbers can be formed. **Why other options are incorrect:** * **(A) $120$:** This might be obtained by incorrect handling of cases or a calculation error, possibly by treating the leading digit restriction less carefully. For instance, if one assumed 5 choices for the first digit (excluding 0) and then simply chose 3 positions, without proper case analysis for the last digit. Or, it could be $P(5,3) \times 1 = 60$ for each of the 2,4 options if 0 was not an issue in the $d_1$ choice or if cases were missed. ($P(5,3) \times 2 = 120$ if 0 was allowed at start). This isn't how it works with the leading zero and even number constraints combined. * **(C) $180$:** This would be the answer if the leading digit could be 0, but only for numbers ending in 2 or 4. If we incorrectly calculated $P(5,3) \times 3$ (for the 3 even digits 0, 2, 4), treating all positions similarly, without accounting for the leading zero restriction: $5 \times 4 \times 3 \times 3 = 180$. This overlooks the first digit cannot be zero when the last digit is non-zero (2 or 4). Specifically, it would be $5 \times 4 \times 3 \times 1 = 60$ (for 0) and $5 \times 4 \times 3 \times 1 = 60$ (for 2) and $5 \times 4 \times 3 \times 1 = 60$ (for 4), leading to 180, if the leading zero constraint was completely ignored or mishandled. * **(D) $192$:** This value does not correspond to any standard miscalculation for this type of problem and is likely a result of several compounding errors. The final answer is $\boxed{156}$. **Common Mistakes:** 1. **Forgetting the leading zero restriction:** Many students forget that the first digit of a number cannot be 0, especially when 0 is part of the available digits. This leads to overcounting. 2. **Not using case analysis when restrictions interact:** The 'even number' restriction (last digit) interacts with the 'no leading zero' restriction. If the last digit is 0, the first digit has more options than if the last digit is 2 or 4 (because 0 is no longer available to cause a leading zero issue). Failing to separate these cases leads to incorrect counts. 3. **Incorrectly managing available choices:** After placing digits in restricted positions, students sometimes miscount the remaining available digits for subsequent positions, often forgetting to include 0 when it becomes available for non-leading positions. 4. **Confusing permutations with repetition vs. without repetition:** The problem explicitly states "without repetition", which means choices decrease for each subsequent position. Forgetting this would lead to $n^r$ type calculations.