Coachify CAT Club
Master CAT Algebra Questions with practice questions and detailed solutions.
Stocks A, B and C are priced at rupees 120, 90 and 150 per share, respectively. A trader holds a portfolio consisting of 10 shares of stock A, and 20 shares of stocks B and C put together. If the total value of her portfolio is rupees 3300, then the number of shares of stock B that she holds, is:
Text Explanation:
Let $3 \leq x \leq 6$ and $[x^2] = [x]^2$, where $[x]$ is the greatest integer not exceeding $x$. If set $S$ represents all feasible values of $x$, then a possible subset of $S$ is .
The number of distinct pairs of integers $(x, y)$ satisfying the inequalities $x$ >$ y \geq 3$ and $x + y $<$ 14$ is .
It is given, $x$>$y\geq3$ and $x+y<14$
Now for $y=3$, the values of $x$ can be 4,5,6,7,8,9,10 (7 cases)
Then for $y=4$, the values of $x$ can be 5,6,7,8,9 (5 cases)
Then for $y=5$, the values of $x$ can be 6,7,8 (3 cases)
Then for $y=6$, the values of $x$ will be 7 (1 case)
So, total number of cases =1+3+5+7=16 cases
A value of $c$ for which the minimum value of $f(x) = x^2 - 4cx + 8c$ is greater than the maximum value of $g(x) = -x^2 + 3cx - 2c$, is .
First function $f\left(x\right)=x^2-4cx+8$
For this function $a>0$, so minimum value will occur at $x=-\dfrac{b}{2a}=-\left(-\dfrac{4c}{2}\right)=2c$
So, the minimum value of the function is = $2c^2-4c\left(2c\right)+8c=-4c^2+8c$
Second function $g(x)=-x^{2}+3cx-2c$
For this function $a<0$, so maximum value will occur at $x=-\dfrac{b}{2a}=-\dfrac{\left(-3c\right)}{2}=\dfrac{3c}{2}$
So, the maximum value of the function is = $-\left(\dfrac{3c}{2}\right)^2+3c\left(\dfrac{3c}{2}\right)-2c=\dfrac{9c^2}{4}-2c$
So, as per the given condition,
$\dfrac{9c^2}{4}-2c$<$-4c^2+8c$
or, $\dfrac{9c^2}{4}+4c^2$<$8c+2c$
or, $\dfrac{25c^2}{4}$<$10c$
or, $\dfrac{5c^2}{4}$<$2c$
or, $5c^2$<$8c$
or, $5c^2-8c$<$0$
or, $c\left(c-\dfrac{8}{5}\right)$<$0$
or, $0$
So, the value of $c$ which lies in this range is $\dfrac{1}{2}$
Let $f(x)=\frac{x}{(2x-1)}$ and $g(x)=\frac{x}{(x-1)}$. Then, the domain of the function $h(x)=f(g(x))+g(f(x))$ is all real numbers except
The set of all real values of x for which $(x^{2}-|x+9|+x)$ > 0, is
We are asked to solve
$x^2 $- |x+9| + x > 0
Split into two cases based on the absolute value.
Case 1: $x+9 \ge 0 \Rightarrow x \ge -9$
$|x+9| = x+9$
Inequality becomes:
$x^2 - (x+9) + x $> 0$ \implies x^2 - 9$ >$ 0 \implies (x-3)(x+3)$ > 0
So x < -3$ \text{ or } x $> 3. Combined with $x \ge -9$, we get
$x \in [-9,-3) \cup (3,\infty)$
Case 2: x+9 < 0$ \Rightarrow x $< -9
$|x+9| = -(x+9) = -x - 9$
$x^2 $- (-x-9) + x > $0 \implies x^2 + 2x + 9$ > 0
The quadratic $x^2 + 2x + 9$ has discriminant $(4 - 36 = -32 < 0)$, so always positive. But in this case $(x < -9)$, so inequality is satisfied. Thus $(x < -9)$ also works.
x < -3$ \text{ or } x $> 3
So the solution set is
$(-\infty,-3) \cup (3,\infty)$
If $9^{x^{2}+2x-3}-4(3)^{x^{2}+2x-2}+27=0$ then the product of all possible values of x is
Let's assume that $x^{2}+2x-3 = t$
$9^{x^{2}+2x-3}-4(3^{x^{2}+2x-2})+27=0$ can be written as $9^t-4(3^{t+1})+27=0$
$3^{2t}-12(3^t)+27=0$
Let's assume that $3^t = y$
$y^2-12y+27=0$
$(y-9)(y-3)=0$
y = 3 or 9 $\Rightarrow$ t = 1 or 2
Let's solve when t = 1
$x^{2} + 2x - 3 = 1 \Rightarrow x^{2} + 2x - 4 = 0$
$b^2-4ac\ =\ 4+16\ =\ 20$
Positive, so the equation has real roots.
Product of possible value of x = -4
Let's solve for t = 2
$x^{2}+2x-3 = 2 \Rightarrow x^{2}+2x-5 = 0$
$b^2-4ac\ =\ 4+20\ =\ 24$
Product of possible value of x = -5
The product of all values = 20
Suppose a, b, c are three distinct natural numbers, such that $3ac = 8(a+b)$. Then, the smallest possible value of $3a+2b+c$ is
The equations $3x^{2}-5x+p=0$ and $2x^{2}-2x+q=0$ have one common root. The sum of the other roots of these two equations is
If m and n are integers such that $(m+2n)(2m+n)=27$, then the maximum possible value of $2m-3n$ is
If $f(x)=(x^{2}+3x)(x^{2}+3x+2)$, then the sum of all real roots of the equation $\sqrt{f(x)+1}=9701,$ is
In a school with 1500 students, each student chooses any one of the streams out of science, arts, and commerce, by paying a fee of Rs 1100, Rs 1000, and Rs 800, respectively. The total fee paid by all the students is Rs 15,50,000. If the number of science students is not more than the number of arts students, then the maximum possible number of science students in the school is
For real values of x, the range of the function $f(x)=\frac{2x-3}{2x^{2}+4x-6}$ is
Let $y = \dfrac{2x-3}{2x^{2}+4x-6}$
$\Rightarrow 2yx^2 + 4yx - 6y = 2x - 3$
$\Rightarrow 2yx^2 + (4y-2)x - 6y + 3 = 0$
Equation (1) is a quadratic in $x$, where $x$ is real. Therefore, the discriminant of the quadratic has to be greater than or equal to zero.
$(4y-2)^2 + 4\times 2y\times (6y-3) \geq 0$
$16y^2 + 4 - 16y - 24y + 48y^2 \geq 0$
$64y^2 - 40y + 4 \geq 0$
$16y^2 - 10y + 1 \geq 0$
The roots of the quadratic above will be $\dfrac{10\pm 6}{32} = \dfrac{1}{2}$ or $\dfrac{1}{8}$
Since the coefficient of $y^2$ is positive, the quadratic will be less than zero in the range $\left(\dfrac{1}{8},\dfrac{1}{2}\right)$
The quadratic will be greater than or equal to zero otherwise.
Therefore, the domain the quadratic, or possible values of $y$, which is the range of $f(x)$, will be,
$\left(-\infty, \dfrac{1}{8}\right] \cup \left[\dfrac{1}{2}, \infty \right)$
Option C is the correct answer.
If $(x^{2}+\frac{1}{x^{2}})=25$ and $x>0$ then the value of $(x^{7}+\frac{1}{x^{7}})$ is
$\left(x+\dfrac{1}{x}\right)^2 = x^2+\dfrac{1}{x^2} + 2 = 25+2 = 27$
Therefore, $\left(x+\dfrac{1}{x}\right) = \sqrt{27} = 3\sqrt{3}$
Also, $\left(x+\dfrac{1}{x}\right)^3 = x^3 + \dfrac{1}{x^3} + 3\left(x+\dfrac{1}{x}\right)$
Therefore, $x^3 + \dfrac{1}{x^3} = (3\sqrt{3})^3 - 9\sqrt{3} = 72\sqrt{3}$
Also, $\left(x^2+\dfrac{1}{x^2}\right)^2 = x^4 + \dfrac{1}{x^4} + 2$
Therefore, $x^4 + \dfrac{1}{x^4} = (25)^2 - 2 = 623$
Lastly, $\left(x^4+\dfrac{1}{x^4}\right)\left(x^3+\dfrac{1}{x^3}\right) = x^7+\dfrac{1}{x^7} + x + \dfrac{1}{x}$
Therefore, $x^7+\dfrac{1}{x^7} $= $623*( 72\sqrt{3})$ - $3\sqrt{3}$ = $44853\sqrt{3}$
Option A is the correct answer.
