Nonlinear Functions MCQ Quiz in मल्याळम - Objective Question with Answer for Nonlinear Functions - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The function $p(n)=300,000(1.12{)}^n$ provides the profit $n$ years after the product launch. $p(7)\approx 545,616$ implies that after 7 years, the profit is approximately $545,616$ dollars.

Option 1 is correct, as it directly reflects the profit at $n=7$. 

The function $R(t)=350\times (1.05{)}^{\frac{3}{5}t}$ describes revenue growth. To find the percentage increase every 10 months, calculate how many years 10 months is: $\frac{10}{12}=\frac{5}{6}$ of a year. Substitute $t=\frac{5}{6}$ into the function: $R(\frac{5}{6})=350\times (1.05{)}^{\frac{3}{5}\times \frac{5}{6}}=350\times (1.05{)}^{\frac{1}{2}}$. This results in approximately 1.03, indicating a 3% increase. Therefore, the revenue increases by 3% every 10 months.

Substitute $t=5$ into $h(t)=-4.9t^2+50t+5$. Compute $-4.9(5{)}^2+50(5)+5$. $5^2=25$, so $-4.9\times 25=-122.5$. Then, $50\times 5=250$. $-122.5+250+5=132.5$. Therefore, $h(5)=132.5$. The correct answer is 132.5.

Evaluate $s(x)$ for each $x$ value and subtract 6. For $x=-5$, $s(-5)=(-5+1)(-5-4)(-5+5)=-4\times -9\times 0=0$, so $y=0-6=-6$. For $x=-1$, $s(-1)=(-1+1)(-1-4)(-1+5)=0\times -5\times 4=0$, hence $y=0-6=-6$. For $x=4$, $s(4)=(4+1)(4-4)(4+5)=5\times 0\times 9=0$, thus $y=0-6=-6$. The correct table is therefore Option 4, which shows $y=-6$ for each $x$.

The function $g(x)=200(1.05{)}^x$ models exponential growth of an investment. $g(6)$ represents the value of the investment 6 years after the initial amount was invested. Calculating $g(6)$, we get $g(6)=200(1.05{)}^6$, which is approximately 268. This means that after 6 years, the value of the investment is approximately 268 dollars. Therefore, option 1 is correct. Option 2 is incorrect because it suggests the value increased by 268 dollars, not that it became 268 dollars. Option 3 is incorrect because it implies multiplication by the original value rather than the current value. Option 4 is incorrect because it refers to a percentage increase over a single year, not over 6 years.

The function $h(t)=9t^2$ describes the height of a ball as a function of time $t$. Thus, $h(3)=81$ implies that after $3$ seconds, the ball reaches a height of $81$ feet. Option 1 correctly interprets this situation, while the other options present incorrect relationships between time and height or misstate the initial or maximum conditions.

The function $h(t)=1200(0.95{)}^t$ models exponential decay, meaning the population decreases each year by a certain percentage. $h(3)$ represents the population 3 years after the project began. Calculating $h(3)=1200(0.95{)}^3$, we get $h(3)$ which is approximately 1083. Thus, option 1 is correct because it interprets $h(3)$ as the fish population being approximately 1083 after 3 years. Option 2 is incorrect because it implies a decrease of 1083 rather than a value of 1083. Option 3 is incorrect as it misinterprets the multiplication factor. Option 4 is incorrect as it misrepresents the yearly decrease as applying to only one year.

To find the equation of the parabola, we use the vertex form $y=a(x-h{)}^2+k$, where $(h,k)$ is the vertex. Given $(h,k)=(3,5)$, the equation becomes $y=a(x-3{)}^2+5$. The parabola passes through $(5,1)$, so substitute $x=5$ and $y=1$ into the equation: $1=a(5-3{)}^2+5$. This simplifies to $1=4a+5$, giving $-4=4a$, so $a=-1$. Therefore, the equation is $y=-1(x-3{)}^2+5$. Option 3 is incorrect because $a=-0.5$ does not satisfy the equation with the given point, while Option 4 represents an upward-opening parabola.

The function $p(x)=50000(1.02{)}^x$ describes exponential growth of a city's population. $p(5)$ estimates the population 5 years after 2015. Calculating $p(5)$, we find $p(5)=50000(1.02{)}^5$, approximately 55204. This means that in the year 2020, the population is about 55204, making option 1 correct. Option 2 is incorrect as it mistakenly suggests a multiplication by 55204. Option 3 is incorrect as it implies an addition rather than a final value. Option 4 is incorrect because it correctly states the growth rate but does not interpret the specific output of the function.

To find the annual decrease rate, we start with the fact that the equipment value halves every 6 years, meaning the decay factor over 6 years is 0.5. We need the annual rate using:

$(1-r_{annual}{)}^6=0.5$

Taking the 6th root gives:

$1-r_{annual}={0.5}^{1/6}$

Calculating this gives

$1-r_{annual}\approx 0.882$

Therefore,

$r_{annual}\approx 1-0.882=0.118$

or 11.8%.

Thus, the correct answer is option 1.