Let - f x - Veft matrix x- 2-frac k x - 2-42- x llendmatrix lright- when x is an int eger otherwise then lim-x rightarrow 2- f x A- exists for every real kB- does not existC- exists for every real k -1D- exists only when k -1

The correct option is A exists for every real k lim_x→2^+f(x)=lim_x→2^+k(x^2 4)/2 x =lim_x→2^+k( 2 x)= 4k lim_x→2^+f(x)=lim_x→2^+k(x^2 4)/2 x=lim_x→2^ k( ...

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Byju's Answer

Standard XI

Mathematics

Relation between Continuity and Differentiability

Let ∼ f x = V...

Question

$L e t f (x) = {\begin{matrix} x^{2} & _{\frac{k (x^{2} - 4)}{2 - x}} \end{matrix} when x is an int eger otherwise then {lim}_{x \to 2} f (x)$

exists only when k=1

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exists for every real k

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exists for every real k =1

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does not exist

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Solution

The correct option is B
exists for every real k

${lim}_{x \to 2^{+}} f (x) = {lim}_{x \to 2^{+}} \frac{k (x^{2} - 4)}{2 - x} = {lim}_{x \to 2^{+}} k (- 2 - x) = - 4 k {lim}_{x \to 2^{+}} f (x) = {lim}_{x \to 2^{+}} \frac{k (x^{2} - 4)}{2 - x} = {lim}_{x \to 2^{-}} k (- 2 - x) = - 4 k$

Hence limits exists for every real value of k.

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Let f(x)={x2k(x2−4)2−xwhen x is an int eger otherwise then limx→2f(x)

The correct option is B exists for every real k limx→2+f(x)=limx→2+k(x2−4)2−x =limx→2+k(−2−x)=−4klimx→2+f(x)=limx→2+k(x2−4)2−x=limx→2−k(−2−x)=−4k Hence limits exists for every real value of k.

$L e t f (x) = {\begin{matrix} x^{2} & _{\frac{k (x^{2} - 4)}{2 - x}} \end{matrix} when x is an int eger otherwise then {lim}_{x \to 2} f (x)$