Answer
Verified
387.9k+ views
Hint: We should know that the formula of newton’s forward method is used as follows: If\[f(x)=f(a)+m\left( f(b)-f(a) \right)\] where a and b are two integers, \[a < x < b\], \[m=\dfrac{x-a}{h}\] where \[h=b-a\]. By using this formula, we can find the value of \[f(1.3)\].
Complete step by step answer:
Before solving the question, we should know how the newton’s forward formula is used. If\[f(x)=f(a)+m\left( f(b)-f(a) \right)\] where a and b are two integers, \[a\le x\le b\], \[m=\dfrac{x-a}{h}\] where \[h=b-a\].
From the question, it was given that \[f(1)=10,f(2)=14\]. Now we should find the value of \[f(1.3)\] using newton’s forward formula.
Now let us compare \[f(1)=10,f(2)=14\] with \[f(a),f(b)\]. Then the value of a is equal to 1 and the value of b is equal to 2.
We know that \[h=b-a\].
Let us consider
\[h=b-a....(3)\]
Now let us substitute equation (1) and equation (2) in equation (3), then we get
\[\begin{align}
& \Rightarrow h=2-1 \\
& \Rightarrow h=1....(4) \\
\end{align}\]
From the question, it is clear that we should find the value of \[f(1.3)\].
Now we will let us assume \[f(x)\] with \[f(1.3)\].
Then, we get
\[\Rightarrow x=1.3....(5)\]
We know that \[m=\dfrac{x-a}{h}\].
Let us consider
\[m=\dfrac{x-a}{h}......(6)\]
So, let us substitute equation (5), equation (1) and equation (4) in equation (6), then we get
\[\begin{align}
& \Rightarrow m=\dfrac{1.3-1}{1} \\
& \Rightarrow m=0.3......(7) \\
\end{align}\]
We know that \[f(x)=f(a)+m\left( f(b)-f(a) \right)\] where a and b are two integers, \[a\le x\le b\], \[m=\dfrac{x-a}{h}\] where \[h=b-a\].
Let us consider
\[f(x)=f(a)+m\left( f(b)-f(a) \right).....(8)\]
We know that \[f(1)=10,f(2)=14\].
So, let us consider
\[\begin{align}
& f(1)=10.....(9) \\
& f(2)=14.....(10) \\
\end{align}\]
Now we will substitute equation (3), equation (4), equation (5), equation (6), equation (7), equation (9) and equation (10) in equation (8), then we get
\[\begin{align}
& \Rightarrow f(1.3)=10+\left( 0.3 \right)\left( 14-10 \right) \\
& \Rightarrow f(1.3)=10+\left( 0.3 \right)\left( 4 \right) \\
& \Rightarrow f(1.3)=10+1.2 \\
& \Rightarrow f(1.3)=11.2......(11) \\
\end{align}\]
Now from equation (11), it is clear that the value of \[f(1.3)\] is equal to 11.2.
So, the correct answer is “Option B”.
Note: Students may have a misconception that If\[f(x)=f(a)+m\left( f(a)-f(b) \right)\] where a and b are two integers, \[a < x < b\], \[m=\dfrac{x-a}{h}\] where \[h=b-a\]. If this misconception is followed then the whole problem may go wrong. So, this misconception should be avoided. Students should not make any calculation mistakes while solving the problem to get a correct solution.
Complete step by step answer:
Before solving the question, we should know how the newton’s forward formula is used. If\[f(x)=f(a)+m\left( f(b)-f(a) \right)\] where a and b are two integers, \[a\le x\le b\], \[m=\dfrac{x-a}{h}\] where \[h=b-a\].
From the question, it was given that \[f(1)=10,f(2)=14\]. Now we should find the value of \[f(1.3)\] using newton’s forward formula.
Now let us compare \[f(1)=10,f(2)=14\] with \[f(a),f(b)\]. Then the value of a is equal to 1 and the value of b is equal to 2.
We know that \[h=b-a\].
Let us consider
\[h=b-a....(3)\]
Now let us substitute equation (1) and equation (2) in equation (3), then we get
\[\begin{align}
& \Rightarrow h=2-1 \\
& \Rightarrow h=1....(4) \\
\end{align}\]
From the question, it is clear that we should find the value of \[f(1.3)\].
Now we will let us assume \[f(x)\] with \[f(1.3)\].
Then, we get
\[\Rightarrow x=1.3....(5)\]
We know that \[m=\dfrac{x-a}{h}\].
Let us consider
\[m=\dfrac{x-a}{h}......(6)\]
So, let us substitute equation (5), equation (1) and equation (4) in equation (6), then we get
\[\begin{align}
& \Rightarrow m=\dfrac{1.3-1}{1} \\
& \Rightarrow m=0.3......(7) \\
\end{align}\]
We know that \[f(x)=f(a)+m\left( f(b)-f(a) \right)\] where a and b are two integers, \[a\le x\le b\], \[m=\dfrac{x-a}{h}\] where \[h=b-a\].
Let us consider
\[f(x)=f(a)+m\left( f(b)-f(a) \right).....(8)\]
We know that \[f(1)=10,f(2)=14\].
So, let us consider
\[\begin{align}
& f(1)=10.....(9) \\
& f(2)=14.....(10) \\
\end{align}\]
Now we will substitute equation (3), equation (4), equation (5), equation (6), equation (7), equation (9) and equation (10) in equation (8), then we get
\[\begin{align}
& \Rightarrow f(1.3)=10+\left( 0.3 \right)\left( 14-10 \right) \\
& \Rightarrow f(1.3)=10+\left( 0.3 \right)\left( 4 \right) \\
& \Rightarrow f(1.3)=10+1.2 \\
& \Rightarrow f(1.3)=11.2......(11) \\
\end{align}\]
Now from equation (11), it is clear that the value of \[f(1.3)\] is equal to 11.2.
So, the correct answer is “Option B”.
Note: Students may have a misconception that If\[f(x)=f(a)+m\left( f(a)-f(b) \right)\] where a and b are two integers, \[a < x < b\], \[m=\dfrac{x-a}{h}\] where \[h=b-a\]. If this misconception is followed then the whole problem may go wrong. So, this misconception should be avoided. Students should not make any calculation mistakes while solving the problem to get a correct solution.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE