g :R→R be two real valued functions defined as
\(f(x) = \begin{cases} -|x + 3| & x < 0 \\ e^x, & x \geq 0 \end{cases}\)
and
\(g(x) = \begin{cases} x^2 + k_1x ,& x < 0 \\ 4x + k_2 ,& x \geq 0 \end{cases}\)
where k1 and k2 are real constants. If (goƒ) is differentiable at x = 0, then (goƒ) (–4) + (goƒ) (4) is
equal to:
g :R→R be two real valued functions defined as
\(f(x) = \begin{cases} -|x + 3| & x < 0 \\ e^x, & x \geq 0 \end{cases}\)
and
\(g(x) = \begin{cases} x^2 + k_1x ,& x < 0 \\ 4x + k_2 ,& x \geq 0 \end{cases}\)
where k1 and k2 are real constants. If (goƒ) is differentiable at x = 0, then (goƒ) (–4) + (goƒ) (4) is
equal to:
4(e4 + 1)
2(2e4 + 1)
4e4
2(2e4 – 1)
The Correct Option is D
Solution and Explanation
The correct answer is (D) : 2(2e4 – 1)
∵ goƒ is differentiable at x = 0
So R.H.D = L.H.D
\(\frac{d}{dx}(4e^x + k_2) = \frac{d}{dx}\left((-|x + 3|)^2 - k_1|x + 3|\right)\)
⇒ 4 = 6 –k1 ⇒ k1 = 2
Also g(ƒ(0+)) = g(ƒ(0–))
⇒ 4 + k2 = 9 – 3k1⇒k2 = –1
Now g(ƒ(–4)) + g(ƒ(4))
= g(–1) + g(e4) = (1 – k1) + (4e4 + k2)
= 4e4 – 2
= 2(2e4 – 1)
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Concepts Used:
Functions
A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.
Kinds of Functions
The different types of functions are -
One to One Function: When elements of set A have a separate component of set B, we can determine that it is a one-to-one function. Besides, you can also call it injective.
Many to One Function: As the name suggests, here more than two elements in set A are mapped with one element in set B.
Moreover, if it happens that all the elements in set B have pre-images in set A, it is called an onto function or surjective function.
Also, if a function is both one-to-one and onto function, it is known as a bijective. This means, that all the elements of A are mapped with separate elements in B, and A holds a pre-image of elements of B.
Read More: Relations and Functions