Answer
Verified
423.9k+ views
Hint: Here we solve the terms in the bracket by converting both numerator and denominator in form of their factors and then cancelling out the same terms to obtain a simpler form. Then we apply the limit to the fraction by breaking the fraction into multiplication of two fractions.
* Limit of a function means the value of the function as it approaches the limit, i.e. \[\mathop {\lim }\limits_{x \to a} f(x) = f(a)\]
* \[\mathop {\lim }\limits_{x \to p} (xy) = \mathop {\lim }\limits_{x \to p} x \times \mathop {\lim }\limits_{x \to p} y\]
* \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1\]
Complete step-by-step answer:
We have to find the value of k such that \[\mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = 5\].
First we write the denominator of the fraction in form of its factors. We can write
\[{x^2} - 4x + 4 = {(x)^2} + {(2)^2} - 2(2)(x)\]
This is of the form \[{a^2} + {b^2} - 2ab\]
By comparing the above equation to \[{(a - b)^2} = {a^2} + {b^2} - 2ab\], we can see that the value of \[a = x,b = 2\].
\[{x^2} - 4x + 4 = {(x - 2)^2}\] … (1)
Now we solve the numerator of the fraction.
We have \[\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} \] as the numerator of the fraction.
Opening the brackets by multiplying the terms we get
\[
\Rightarrow \tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} = \tan (x - 2)\{ {x^2} + kx - 2x - 2k\} \\
\Rightarrow \tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} = \tan (x - 2)\{ {x^2} - 2x + kx - 2k\} \\
\]
Now take x common from the first two terms and k common from the last two terms in the bracket.
\[ \Rightarrow \tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} = \tan (x - 2)\{ x(x - 2) + k(x - 2)\} \]
Pairing the factors we get
\[ \Rightarrow \tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} = \tan (x - 2)\{ (x - 2)(x + k)\} \] … (2)
Substituting equation 1 and 2 in given equation, we get
\[ \Rightarrow \dfrac{{\tan (x - 2)\{ (x - 2)(x + k)\} }}{{{{(x - 2)}^2}}} = \dfrac{{\tan (x - 2)}}{{(x - 2)}} \times \dfrac{{(x - 2)(x + k)}}{{(x - 2)}}\]
Cancel out the same terms from numerator and denominator.
\[ \Rightarrow \dfrac{{\tan (x - 2)\{ (x - 2)(x + k)\} }}{{{{(x - 2)}^2}}} = \dfrac{{\tan (x - 2)}}{{(x - 2)}} \times (x + k)\]
Taking limit on both sides we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)}}{{(x - 2)}} \times (x + k)\]
We know that we can separate the limits as \[\mathop {\lim }\limits_{x \to p} (xy) = \mathop {\lim }\limits_{x \to p} x \times \mathop {\lim }\limits_{x \to p} y\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)}}{{(x - 2)}} \times \mathop {\lim }\limits_{x \to 2} (x + k)\] … (3)
Now we know \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1\]
So, we can apply this property to the first term because \[x \to 2 = x - 2 \to 0\] and as the function is the same as the limit we can apply the property.
\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)}}{{(x - 2)}} = 1\]
Also, \[\mathop {\lim }\limits_{x \to 2} (x + k) = (k + 2)\]
Substituting the values in equation (3) we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = 1 \times (k + 2) = k + 2\]
Now from the statement of the question we know the value of LHS is equal to 5.
\[
\Rightarrow k + 2 = 5 \\
\Rightarrow k = 5 - 2 \\
\Rightarrow k = 3 \\
\]
So, the correct answer is “Option C”.
Note: Students are likely to make mistakes while calculating the limit of the tan function and might substitute the value of x as 2 in the fraction which will give us the answer 0 which is wrong. Keep in mind for fractions like \[\dfrac{{\tan x}}{x}, \dfrac{{\sin x}}{x}\] we always use this way of finding the limit.
* Limit of a function means the value of the function as it approaches the limit, i.e. \[\mathop {\lim }\limits_{x \to a} f(x) = f(a)\]
* \[\mathop {\lim }\limits_{x \to p} (xy) = \mathop {\lim }\limits_{x \to p} x \times \mathop {\lim }\limits_{x \to p} y\]
* \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1\]
Complete step-by-step answer:
We have to find the value of k such that \[\mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = 5\].
First we write the denominator of the fraction in form of its factors. We can write
\[{x^2} - 4x + 4 = {(x)^2} + {(2)^2} - 2(2)(x)\]
This is of the form \[{a^2} + {b^2} - 2ab\]
By comparing the above equation to \[{(a - b)^2} = {a^2} + {b^2} - 2ab\], we can see that the value of \[a = x,b = 2\].
\[{x^2} - 4x + 4 = {(x - 2)^2}\] … (1)
Now we solve the numerator of the fraction.
We have \[\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} \] as the numerator of the fraction.
Opening the brackets by multiplying the terms we get
\[
\Rightarrow \tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} = \tan (x - 2)\{ {x^2} + kx - 2x - 2k\} \\
\Rightarrow \tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} = \tan (x - 2)\{ {x^2} - 2x + kx - 2k\} \\
\]
Now take x common from the first two terms and k common from the last two terms in the bracket.
\[ \Rightarrow \tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} = \tan (x - 2)\{ x(x - 2) + k(x - 2)\} \]
Pairing the factors we get
\[ \Rightarrow \tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} = \tan (x - 2)\{ (x - 2)(x + k)\} \] … (2)
Substituting equation 1 and 2 in given equation, we get
\[ \Rightarrow \dfrac{{\tan (x - 2)\{ (x - 2)(x + k)\} }}{{{{(x - 2)}^2}}} = \dfrac{{\tan (x - 2)}}{{(x - 2)}} \times \dfrac{{(x - 2)(x + k)}}{{(x - 2)}}\]
Cancel out the same terms from numerator and denominator.
\[ \Rightarrow \dfrac{{\tan (x - 2)\{ (x - 2)(x + k)\} }}{{{{(x - 2)}^2}}} = \dfrac{{\tan (x - 2)}}{{(x - 2)}} \times (x + k)\]
Taking limit on both sides we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)}}{{(x - 2)}} \times (x + k)\]
We know that we can separate the limits as \[\mathop {\lim }\limits_{x \to p} (xy) = \mathop {\lim }\limits_{x \to p} x \times \mathop {\lim }\limits_{x \to p} y\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)}}{{(x - 2)}} \times \mathop {\lim }\limits_{x \to 2} (x + k)\] … (3)
Now we know \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1\]
So, we can apply this property to the first term because \[x \to 2 = x - 2 \to 0\] and as the function is the same as the limit we can apply the property.
\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)}}{{(x - 2)}} = 1\]
Also, \[\mathop {\lim }\limits_{x \to 2} (x + k) = (k + 2)\]
Substituting the values in equation (3) we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = 1 \times (k + 2) = k + 2\]
Now from the statement of the question we know the value of LHS is equal to 5.
\[
\Rightarrow k + 2 = 5 \\
\Rightarrow k = 5 - 2 \\
\Rightarrow k = 3 \\
\]
So, the correct answer is “Option C”.
Note: Students are likely to make mistakes while calculating the limit of the tan function and might substitute the value of x as 2 in the fraction which will give us the answer 0 which is wrong. Keep in mind for fractions like \[\dfrac{{\tan x}}{x}, \dfrac{{\sin x}}{x}\] we always use this way of finding the limit.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Why Are Noble Gases NonReactive class 11 chemistry CBSE
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
At which age domestication of animals started A Neolithic class 11 social science CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE