$2x^2 + kx + 5 = 0$ has no real roots so $D$ &lt; $0$$k^2 - 40$ &lt; $0$$(k - \sqrt{40})(k + \sqrt{40})$ &lt; $0$$k \in (-\sqrt{40}, \sqrt{40})$$x^2 + (k - 5)x + 1 = 0$ has two distinct real roots so $D$ &gt; $0$$(k - 5)^2 - 4$ &gt; $0$$k^2 - 10k + 21$ &gt; $0$$(k - 3)(k - 7)$ &gt; $0$$k \in (-\infty, 3) \cup (7, \infty)$Therefore possible value of k are -6, -5, -4, -3, -2, -1, 0, 1, 2

Let $\frac{x^2 - 6x + 10}{3 - x} = p$$x^2 - 6x + 10 = 3p - px$$x^2 - (6 - p)x + 10 - 3p = 0$Since the equation will have real roots,$(6 - p)^2 - 4 \times (10 - 3p) \ge 0$$p^2 - 12p + 12p + 36 - 40 \ge 0$$p^2 \ge 4$$p \ge 2 ,\ p \le -2$Now, when $p = -2$, $x = 4$. Since it is given that $x &lt; 3$, thus this value will be discarded.Now, $\frac{1}{2}$ and $-\frac{1}{2}$ do not come in the mentioned range.When $p = 2$, $x = 2$Thus, the minimum possible value of p will be 2.Thus, the correct option is C.Alternate explanation:Since $x &lt; 3$,$3 - x &gt; 0$Let $3 - x = y$. So, $y &gt; 0$.Now, $\frac{x^2 - 6x + 10}{3 - x} = \frac{x^2 - 6x + 9 + 1}{3 - x}$$\Rightarrow \frac{(3 - x)^2 + 1}{3 - x}$Since $3 - x = y$, the equation will transform to $\frac{y^2 + 1}{y}$ or $y + \frac{1}{y}$The minimum value of the expression $y + \frac{1}{y}$ for $y &gt; 0$ will be at $y = 1$i.e. Minimum value = $1 + 1 = 2$Thus, the correct option is C.

Let the efficiency of Bob be 3 units/day. So, Alex's efficiency will be 6 units/day, and Cole's will be 2 units/day.Since Bob can finish the job in 40 days, the total work will be 40*3 = 120 units.Since Alex and Bob work on the first day, the total work done = 3 + 6 = 9 units.Similarly, for days 2 and 3, it will be 5 and 8 units, respectively.Thus, in the first 3 days, the total work done = 9 + 5 + 8 = 22 units.The work done in the first 15 days = 22*5 = 110 units.Thus, the work will be finished on the 17th day(since 9 + 5 = 14 units are greater than the remaining work).Since Alex works on two days of every 3 days, he will work for 10 days out of the first 15 days.Then he will also work on the 16th day.The total number of days = 11.

Initially: a glass 500cc milk and a cup 500cc waterStep 1: 150 cc of milk is transferred to the cup from glassAfter step 1: Glass - 350 cc milk, Cup - 150 cc milk and 500 cc waterStep 2: 150 cc of this mixture is transferred from the cup to the glassAfter step 2:Glass - 350 cc milk + 150 cc mixture with milk:water ratio 3:10Cup - 500 cc mixture with milk:water ratio 3:10water in glass : milk in cup = $\frac{10}{13} \times 150 : \frac{3}{13} \times 500 = 1 : 1$

Let the number of students in section A and B be a and b, respectively.It is given, $a = b - 10$$\frac{32a + 60b}{a + b}$ is an integer$\frac{32a + 60(a + 10)}{a + a + 10} = k$$\frac{46a + 300}{a + 5} = k$$k = \frac{46(a + 5)}{a + 5} + \frac{70}{a + 5}$$k = 46 + \frac{70}{a + 5}$a can take values 2, 5, 9, 30, 65Difference 65 - 2 = 63

When x&lt; rf(x) = rf(x) = f(f(x))r = f(r)r= 2r-rr=rWhen x&gt;=rf(x) = 2x-rf(x) = f(f(x))2x-r = f(2x-r)2x-r = 2(2x-r) - r2x-r = 4x-3ror, x=rTherefore x&lt;= r

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/11/7.Screenshot_20221206_172342.png">Area of ABD : Area of BDC = 1:1Therefore, area of ABD = 54Area of ADE : Area of EDB = 1:1Therefore, area of ADE = 27O is the centroid and it divides the medians in the ratio of 2:1Area of BEO : Area of EOD = 2:1Area of EOD = 9

The number of 4-digit numbers possible using 1,1,2, and 4 is $\frac{4!}{2!} = 12$Number of 1's, 2's and 4's in units digits will be in the ratio 2:1:1, i.e. 6 1's, 3 2's and 3 4's.Sum = $6(1) + 3(2) + 3(4) = 24$Similarly, in tens digit, hundreds digit and thousands digit as well.Therefore, sum = $24 + 24(10) + 24(100) + 24(1000) = 24(1111)$Mean = $\frac{24(1111)}{12} = 2222$

It is given,$\frac{5.5 \times 1 \times 8000}{100} + \frac{5.6 \times 1 \times 5000}{100} + \frac{x \times 1 \times 7000}{100} = \frac{5}{100} \times 20000$$440 + 280 + 70x = 1000$x = 4%Interest = $\frac{20000 \times 4 \times 1}{100}$ = Rs 800

Both the cars take 1.5 hrs to meet when they travel towards each other.It is given, speed of slower car is 60 km/hrTherefore, distance covered by slower car before they meet = 60*1.5 = 90 kmThe answer is option C.

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cat 2022 Complete Paper Solution | Slot 3

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<div><p>Suppose k is any integer such that the equation $2 x^2+k x+5=0$ has no real roots and the equation $x^2+(k-5) x+1=0$ has two distinct real roots for x . Then, the number of possible values of k is</p></div>

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<div><p>The minimum possible value of $\frac{x^2-6 x+10}{3-x}$, for x&lt;3, is</p></div>