A cubic polynomial $f(x) = a{x^3} + b{x^2} + cx + d$ has a graph which is tangent to the x- axis at $2$ has another x- intercept at $ - 1$ and has y- intercept at $ - 2$ as shown. The values of $a + b + c + d$ is
A. $ - 2$
B. $ - 1$
C. $0$
D. $1$
Answer
545.1k+ views
Hint: In this question, we are given a function with unknown constants and their graph. We have to find the values of these constants.
So, we will find the values of the function at different values of $x$ , then solve them to get the values of $a,b,c,d$ .
The point to remember here is, since, it has a tangent at the x-axis at the point $2$ , so, the derivative of this function at the point $2$ will be zero.
Complete step-by-step answer:
We are given a function with unknown constants $f(x) = a{x^3} + b{x^2} + cx + d$ and its graph.
To find the values of $a$ , $b$ , $c$ and $d$ .
First, we will differentiate the given function and find its value at $x = 2$ .
On differentiating, we get, $f'(x) = 3a{x^2} + 2bx + c$ , now, putting $x = 2$ in it, we get, $f'(x) = 3a{(2)^2} + 2b(2) + c$ , i.e., $f'(2) = 8a + 4b + c$ .
From the graph, it is clear that, $f'(2) = 0$ , i.e., $8a + 4b + c = 0$ … (1)
Also, from the graph, $f(2) = 0$ , $f( - 1) = 0$ and $f(0) = - 2$ , so, first putting, $x = 2$ in the given function $f(2) = 8a + 4b + 2c + d = 0$ , then, putting, $x = - 1$ , we get, $f( - 1) = - a + b - c + d = 0$ and finally, putting $x = 0$ , we get, $f(0) = d = - 2$ .
Now, since we have the value of $d$ , so we will put it in all equations, then the equation becomes,
$f( - 1) = - a + b - c = 2$
$f(2) = 8a + 4b + 2c = 2$
$f'(2) = 8a + 4b + c = 0$
On solving these equations simultaneously, we get, $a = - \dfrac{1}{2}$ , $b = \dfrac{3}{2}$ , $c = 0$ and $d = - 2$ .
Hence, the value of $a + b + c + d = - \dfrac{1}{2} + \dfrac{3}{2} + 0 - 2$ , least common multiple of the denominators is $2$ , so we get, $a + b + c + d = \dfrac{{ - 1 + 3 + 0 - 4}}{2}$ , on simplifying, we get, $a + b + c + d = \dfrac{{ - 2}}{2} = - 1$ .
So, the correct answer is “Option B”.
Note: If a function has a tangent at a point $x = a$ , then, it means, it is the critical point of that function and the derivative of the function is zero at that point.
Perform the calculations carefully, to avoid silly mistakes.
One should know how to read a graph to solve such questions.
So, we will find the values of the function at different values of $x$ , then solve them to get the values of $a,b,c,d$ .
The point to remember here is, since, it has a tangent at the x-axis at the point $2$ , so, the derivative of this function at the point $2$ will be zero.
Complete step-by-step answer:
We are given a function with unknown constants $f(x) = a{x^3} + b{x^2} + cx + d$ and its graph.
To find the values of $a$ , $b$ , $c$ and $d$ .
First, we will differentiate the given function and find its value at $x = 2$ .
On differentiating, we get, $f'(x) = 3a{x^2} + 2bx + c$ , now, putting $x = 2$ in it, we get, $f'(x) = 3a{(2)^2} + 2b(2) + c$ , i.e., $f'(2) = 8a + 4b + c$ .
From the graph, it is clear that, $f'(2) = 0$ , i.e., $8a + 4b + c = 0$ … (1)
Also, from the graph, $f(2) = 0$ , $f( - 1) = 0$ and $f(0) = - 2$ , so, first putting, $x = 2$ in the given function $f(2) = 8a + 4b + 2c + d = 0$ , then, putting, $x = - 1$ , we get, $f( - 1) = - a + b - c + d = 0$ and finally, putting $x = 0$ , we get, $f(0) = d = - 2$ .
Now, since we have the value of $d$ , so we will put it in all equations, then the equation becomes,
$f( - 1) = - a + b - c = 2$
$f(2) = 8a + 4b + 2c = 2$
$f'(2) = 8a + 4b + c = 0$
On solving these equations simultaneously, we get, $a = - \dfrac{1}{2}$ , $b = \dfrac{3}{2}$ , $c = 0$ and $d = - 2$ .
Hence, the value of $a + b + c + d = - \dfrac{1}{2} + \dfrac{3}{2} + 0 - 2$ , least common multiple of the denominators is $2$ , so we get, $a + b + c + d = \dfrac{{ - 1 + 3 + 0 - 4}}{2}$ , on simplifying, we get, $a + b + c + d = \dfrac{{ - 2}}{2} = - 1$ .
So, the correct answer is “Option B”.
Note: If a function has a tangent at a point $x = a$ , then, it means, it is the critical point of that function and the derivative of the function is zero at that point.
Perform the calculations carefully, to avoid silly mistakes.
One should know how to read a graph to solve such questions.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

What is the maximum height of a projectile class 11 physics CBSE

Describe various types of epithelial tissues with the class 11 biology CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

Potato is a stem and sweet potato is a root Justify class 11 biology CBSE

Simon Commission came to India in A 1927 B 1928 C 1929 class 11 social science CBSE

