Advanced Math MCQ Quiz in मल्याळम - Objective Question with Answer for Advanced Math - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

For the absolute value equation $|5x-1|=14$, we consider the two cases for the absolute value:

1. $5x-1=14$

2. $5x-1=-14$

For the first equation, $5x-1=14$:

Add $1$ to both sides:

$5x=15$

Divide by $5$:

$x=3$

For the second equation, $5x-1=-14$:

Add $1$ to both sides:

$5x=-13$

Divide by $5$:

$x=-2.6$

Thus, the solutions are $x=3$ and $x=-2.6$, which correspond to option 2.

The starting height corresponds to $t=0$. Substituting $t=0$ into the equation yields $h=-4.5(0{)}^2+9(0)+12$. This simplifies to $h=12$. Hence, the starting height is 12 meters. Incorrect options arise from mistakes in evaluating the equation at $t=0$.

First, rewrite the equation $x^2-6x=8$ as $x^2-6x-8=0$. To solve this quadratic equation, look for two numbers that multiply to $-8$ and add to $-6$. These numbers are $-8$ and $1$. Therefore, factor the equation as $(x-8)(x+1)=0$. Using the zero-product property, set each factor to zero: $x-8=0$ or $x+1=0$. Solving these gives $x=8$ and $x=-1$. The smallest solution is $-1$.

To find the number of gadgets that minimizes the production cost, we look for the vertex of the quadratic function $C=4x^2-12x+20$. The vertex $x$-coordinate is found using $x=-\frac{b}{2a}$, where $a=4$ and $b=-12$. Thus, $x=-\frac{-12}{2(4)}=\frac{12}{8}=1.5$. Therefore, the production cost is minimized at $x=1.5$, or $1.5$ gadgets. Option $3$ is the correct choice. Options $0$, $1$, and $3$ do not yield the minimum cost as they do not correspond to the vertex of the parabola.

The height of the ball when it reaches the ground is 0. Therefore, we set $-4.9t^2+20t+1.5=0$ and solve for $t$. This is a quadratic equation in standard form, $a{t}^2+bt+c=0$, where $a=-4.9$, $b=20$, and $c=1.5$. Solving this using the quadratic formula $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we have: $t=\frac{-20\pm \sqrt{{20}^2-4(-4.9)(1.5)}}{2(-4.9)}=\frac{-20\pm \sqrt{400+29.4}}{-9.8}$. Simplifying, $t=\frac{-20\pm \sqrt{429.4}}{-9.8}$. The positive solution is approximately $t=4.1$. Therefore, the correct option is 3. Option 1 (1.5) and Option 2 (3.5) are too early for the ball to reach the ground, and Option 4 (4.5) is slightly beyond the correct time.

A quadratic function $a{x}^2+bx+c$ has only one real root when its discriminant $b^2-4ac$ is zero. For $f(x)=x^2-6x+k$, the discriminant is $(-6{)}^2-4(1)(k)=36-4k$. Set the discriminant to zero for one real root: $36-4k=0$. Solving for $k$, we have $4k=36$ and $k=9$. Therefore, the value of $k$ is 9.

To simplify $4m^2-12m^2+3m^2$, combine the coefficients of the like terms, $m^2$. This gives us $(4-12+3)m^2=-5m^2$. Therefore, option 4 is the correct answer. Option 1 incorrectly adds the exponents. Option 2 incorrectly combines coefficients. Option 3 incorrectly calculates the result.

The function $g(x)$ halves in value for each increase in $x$ by $1$. Therefore, the function can be expressed as $g(x)=a(0.5{)}^x$. Given $g(1)=10$, we can substitute $10$ for $g(1)$ to find $a$. Hence, $g(x)=10(0.5{)}^{x-1}$ represents the function correctly. Option 1 correctly uses the initial condition at $x=1$. Option 2 and Option 3 misrepresent the decay process, while Option 4 uses incorrect scaling.

The function $h(x)$ decreases by $30\mathrm{\%}$ with every unit increase in $x$. This indicates an exponential decay model $h(x)=a(1-0.3{)}^x$. Given $h(0)=20$, $a=20$. Therefore, $h(x)=20(0.7{)}^x$. Option 1 is correct. Option 2 suggests growth rather than decay. Option 3 incorrectly models the decay rate, and Option 4 misrepresents the exponential factor.

To find $V(100)$, substitute $x=100$ into the function: $V(100)=1.05(100{)}^2+20$. First, calculate $(100{)}^2=10000$. Then $1.05(10000)=10500$. Add $20$: $10500+20=10520$. Thus, $V(100)=10520$. Option 3 is correct, as it reflects the correct computation of the investment value.