For a, b\gt 0 , let f(x)= \begincases\frac\tan ((a+1) x)+b \tan xx, & x 0\endcases be a continuous function at x=0 . Then \fracba is equal to :

Limits Continuity and Differentiability PYQ — JEE Main 2024 · 8 Apr · Shift 2 · Q42 with Solution

View written solutionFree

Correct answer: NO UNIQUE ANSWER CAN BE DETERMINED FROM THE STATED CONTINUITY CONDITION ALONE. THE PROBLEM APPEARS MISPRINTED OR INCOMPLETE.

Interpret the piecewise function

The function is

f(x)= \begin{cases} \dfrac{\tan((a+1)x)+b\tan x}{x}, & x\neq 0 \\ a^2+b^2, & x=0 \end{cases}

Since $f$ is continuous at $x=0$ , we must have

\lim_{x\to 0} \frac{\tan((a+1)x)+b\tan x}{x}=a^2+b^2.

Evaluate the limit

Using the standard limit

\lim_{u\to 0}\frac{\tan u}{u}=1,

we get

\tan((a+1)x)\sim (a+1)x, \qquad \tan x\sim x \quad (x\to 0).

Therefore,

\lim_{x\to 0} \frac{\tan((a+1)x)+b\tan x}{x} = \lim_{x\to 0} \left(\frac{\tan((a+1)x)}{x}+b\frac{\tan x}{x}\right).

Now,

\frac{\tan((a+1)x)}{x} = \frac{\tan((a+1)x)}{(a+1)x}\cdot (a+1) \to a+1.

Also,

b\frac{\tan x}{x}\to b.

So the limit is

a+1+b.

Hence continuity at $x=0$ gives

a+b+1=a^2+b^2.

Solve the equation

Rearrange:

a^2-a+b^2-b=1.

Complete squares or factor by grouping:

a(a-1)+b(b-1)=1.

Since $a,b>0$ , let us test values consistent with the options for $\frac ba$ .

Suppose

\frac ba = k \quad \Rightarrow \quad b=ka.

Substitute into

a^2+b^2=a+b+1:

a^2+k^2a^2=a+ka+1.

(1+k^2)a^2-(1+k)a-1=0.

Now check the options.

Option A: $k=4$

17a^2-5a-1=0

Positive root:

a=\frac{5+\sqrt{25+68}}{34}=\frac{5+\sqrt{93}}{34}>0

Then $b=4a>0$ . This gives a valid positive pair.

Option B: $k=5$

26a^2-6a-1=0

Positive root exists, so valid.

Option C: $k=8$

65a^2-9a-1=0

Positive root exists, so valid.

Option D: $k=6$

37a^2-7a-1=0

Positive root exists, so valid.

So from the given continuity equation alone, multiple positive values of $\frac ba$ are possible. Hence the problem as written appears incomplete or misprinted.

Check likely intended form

A common version of this problem is:

f(x)= \begin{cases} \dfrac{\tan((a+1)x)+b\tan x}{x}, & x\ne 0 \\ a^2+b^2, & x=0 \end{cases}

with continuity and perhaps differentiability, or another condition, giving a unique ratio. But with only continuity, $\frac ba$ is not uniquely determined.

Therefore, the stored answer $6$ cannot be concluded from the provided statement.

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: NO UNIQUE ANSWER CAN BE DETERMINED FROM THE STATED CONTINUITY CONDITION ALONE. THE PROBLEM APPEARS MISPRINTED OR INCOMPLETE.

Interpret the piecewise function

The function is

f(x)= \begin{cases} \dfrac{\tan((a+1)x)+b\tan x}{x}, & x\neq 0 \\ a^2+b^2, & x=0 \end{cases}

Since $f$ is continuous at $x=0$ , we must have

\lim_{x\to 0} \frac{\tan((a+1)x)+b\tan x}{x}=a^2+b^2.

Evaluate the limit

Using the standard limit

\lim_{u\to 0}\frac{\tan u}{u}=1,

we get

\tan((a+1)x)\sim (a+1)x, \qquad \tan x\sim x \quad (x\to 0).

Therefore,

\lim_{x\to 0} \frac{\tan((a+1)x)+b\tan x}{x} = \lim_{x\to 0} \left(\frac{\tan((a+1)x)}{x}+b\frac{\tan x}{x}\right).

Now,

\frac{\tan((a+1)x)}{x} = \frac{\tan((a+1)x)}{(a+1)x}\cdot (a+1) \to a+1.

Also,

b\frac{\tan x}{x}\to b.

So the limit is

a+1+b.

Hence continuity at $x=0$ gives

a+b+1=a^2+b^2.

Solve the equation

Rearrange:

a^2-a+b^2-b=1.

Complete squares or factor by grouping:

a(a-1)+b(b-1)=1.

Since $a,b>0$ , let us test values consistent with the options for $\frac ba$ .

Suppose

\frac ba = k \quad \Rightarrow \quad b=ka.

Substitute into

a^2+b^2=a+b+1:

a^2+k^2a^2=a+ka+1.

(1+k^2)a^2-(1+k)a-1=0.

Now check the options.

Option A: $k=4$

17a^2-5a-1=0

Positive root:

a=\frac{5+\sqrt{25+68}}{34}=\frac{5+\sqrt{93}}{34}>0

Then $b=4a>0$ . This gives a valid positive pair.

Option B: $k=5$

26a^2-6a-1=0

Positive root exists, so valid.

Option C: $k=8$

65a^2-9a-1=0

Positive root exists, so valid.

Option D: $k=6$

37a^2-7a-1=0

Positive root exists, so valid.

So from the given continuity equation alone, multiple positive values of $\frac ba$ are possible. Hence the problem as written appears incomplete or misprinted.

Check likely intended form

A common version of this problem is:

f(x)= \begin{cases} \dfrac{\tan((a+1)x)+b\tan x}{x}, & x\ne 0 \\ a^2+b^2, & x=0 \end{cases}

with continuity and perhaps differentiability, or another condition, giving a unique ratio. But with only continuity, $\frac ba$ is not uniquely determined.

Therefore, the stored answer $6$ cannot be concluded from the provided statement.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

Option A: $k=4$

Option B: $k=5$

Option C: $k=8$

Option D: $k=6$

More from Limits Continuity and Differentiability

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

Option A: $k=4$

Option B: $k=5$

Option C: $k=8$

Option D: $k=6$

More from Limits Continuity and Differentiability

Option A: k=4k=4k=4

Option B: k=5k=5k=5

Option C: k=8k=8k=8

Option D: k=6k=6k=6

More from Limits Continuity and Differentiability

Option A: k=4k=4k=4

Option B: k=5k=5k=5

Option C: k=8k=8k=8

Option D: k=6k=6k=6

More from Limits Continuity and Differentiability

Option A: $k=4$

Option B: $k=5$

Option C: $k=8$

Option D: $k=6$

Option A: $k=4$

Option B: $k=5$

Option C: $k=8$

Option D: $k=6$