If the function f(x)= cases 2x (k1 + 1)x + (k2 -1)x, & x < 0 4, & x = 0 2x e ( 2 + k1 x2 + k2 x), & x 0 cases is continuous at x = 0, then k1^2 + k2^2 is equal to:

Given: The function is piecewise defined and is continuous at x=0. Find: The value of k1^2+k2^2. For continuity at x=0, the left-hand limit, right-hand limit, and function value must be equal. From the left side, for x<0, x 0^- 2x ( ((k1+1)x)+ ((k2-1)x)) Using x 0 (ax)x=a, =2(k1+1)+2(k2-1) =2k1+2k2 From the right side, for x0, f(x)= 2x e ( 2+k1x2+k2x) Write 2+k1x2+k2x=1+ (k1-k2)x2+k2x Using (1+u) u for small u, x 0^+ 2x e (1+ (k1-k2)x2+k2x) = x 0^+ 2x (k1-k2)x2 =k1-k2 Also, the given value is f(0)=4. So continuity gives 2k1+2k2=4 and k1-k2=4 Hence, k1+k2=2 Solving, 2k1=6 k1=3 k2=2-3=-1 Therefore, k1^2+k2^2=3^2+(-1)^2=10 The correct option is D.

If the function f(x)= cases 2x (k1 + 1)x + (k2 -1)x, & x < 0 4, &… — Solution | JEE Mathematics

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If the function $f(x)=\begin{cases} \frac{2}{x} \left\{ \sin(k_1 + 1)x + \sin(k_2 -1)x \right\}, & x < 0\\ 4, & x = 0\\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x > 0 \end{cases}$ is continuous at $x = 0$ , then $k_1^2 + k_2^2$ is equal to:

A
$8$
B
$20$
C
$5$
D
$10$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The function is piecewise defined and is continuous at $x=0$ .

Find: The value of $k_1^2+k_2^2$ .

For continuity at $x=0$ , the left-hand limit, right-hand limit, and function value must be equal.

From the left side, for $x<0$ ,

\lim_{x\to 0^-} \frac{2}{x}\left(\sin((k_1+1)x)+\sin((k_2-1)x)\right)

Using $\lim_{x\to 0}\frac{\sin(ax)}{x}=a$ ,

=2(k_1+1)+2(k_2-1)

=2k_1+2k_2

Detailed Limit Evaluation

From the right side, for $x>0$ ,

f(x)=\frac{2}{x}\log_e\left(\frac{2+k_1x}{2+k_2x}\right)

Write

\frac{2+k_1x}{2+k_2x}=1+\frac{(k_1-k_2)x}{2+k_2x}

Using $\log(1+u)\approx u$ for small $u$ ,

\lim_{x\to 0^+}\frac{2}{x}\log_e\left(1+\frac{(k_1-k_2)x}{2+k_2x}\right)

=\lim_{x\to 0^+}\frac{2}{x}\cdot \frac{(k_1-k_2)x}{2}

=k_1-k_2

Also, the given value is $f(0)=4$ .

So continuity gives

2k_1+2k_2=4

and

k_1-k_2=4

Hence,

k_1+k_2=2

Solving,

2k_1=6 \Rightarrow k_1=3

k_2=2-3=-1

Therefore,

k_1^2+k_2^2=3^2+(-1)^2=10

The correct option is D.

Common mistakes

Using continuity incorrectly by equating only one-sided limits to each other. For continuity at $x=0$ , both one-sided limits must also equal the given value $f(0)=4$ . Always set LHL = RHL = $f(0)$ .
Applying $\sin(ax)\approx x$ instead of $\sin(ax)\approx ax$ near $x=0$ . The coefficient $a$ must be retained, otherwise the left-hand limit becomes wrong.
Expanding the logarithm incorrectly. For small $u$ , use $\log(1+u)\approx u$ with $u=\frac{(k_1-k_2)x}{2+k_2x}$ , not $u=(k_1-k_2)x$ directly without accounting for the denominator.

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