IIT JEE Main Maths -Unit 7 -Limits, continuity, and differentiability- Exam Style Questions- New Syllabus

Question

Let the function,
$ f(x) = \begin{cases} -3ax – 2, & x < 1 \\ a^2 + bx, & x \geq 1 \end{cases} $
Be differentiable for all $ x \in \mathbb{R} $, where $ a > 1, b \in \mathbb{R} $. If the area of the region enclosed by $ y = f(x) $ and the line $ y = -20 $ is $ \alpha + \beta \sqrt{3} $, $ \alpha, \beta \in \mathbb{Z} $, then the value of $ \alpha + \beta $ is ____.

▶️ Answer/Explanation

Ans. (34)

Sol. $ f(x) $ is continuous and differentiable

at $ x = 1 $; LHL = RHL, LHD = RHD

$ -3a – 2 = a^2 + b, -6a = b $

$ a = 2, 1; b = -12 $

$ f(x) = \begin{cases} -6x – 2, & x < 1 \\ 4 – 12x, & x \geq 1 \end{cases} $

$\text{Area} = \int_{-\sqrt{3}}^{1} \big(-6x^2 – 2 – (-20)\big),dx ;+; \int_{1}^{2} \big(4 – 12x – (-20)\big),dx$

$= \int_{-\sqrt{3}}^{1} (-6x^2 + 18),dx ;+; \int_{1}^{2} (24 – 12x),dx$

$= \Big\( -2x^3 + 18x \Big${-\sqrt{3}}^{1} ;+; \Big$ 24x – 6x^2 \Big${1}^{2}\)

$\text{Area} = (16 + 12\sqrt{3}) + 6 = 22 + 12\sqrt{3}.$

So,
$\alpha = 22,\quad \beta = 12,\quad \alpha + \beta = 34.$

Question

If $\lim_{x \to \infty} \frac{x}{e^x – 1} \frac{1}{x} \left( \frac{1}{e} + \frac{1}{e^x} \right) = \alpha$, then the value of $e^{\frac{1}{e} \log (1 + \alpha) + \log \alpha}$ equals :
(1) $e$
(2) $e^{-2}$
(3) $e^2$
(4) $e^{-1}$

▶ Answer/Explanation

Ans. (1)
Sol. $\alpha = \lim_{x \to \infty} \frac{x}{e^x – 1} \frac{1}{x} \left( \frac{1}{e} + \frac{1}{e^x} \right)$
($1^\infty$ form)
$\therefore \alpha = e^L$
Where $L = \lim_{x \to \infty} \frac{\frac{x}{e^x – 1}}{x} \frac{1}{1 – \frac{1}{e} – \frac{1}{e^x}}$
$\Rightarrow L = \lim_{x \to \infty} \frac{\frac{x}{e^x – 1}}{x} \frac{1}{1 – \frac{1}{e} – \frac{1}{e^x}}$
$\Rightarrow L = \lim_{x \to \infty} \frac{\frac{x}{e^x – 1}}{1 – \frac{1}{e} – \frac{1}{e^x}}$
$\Rightarrow L = \frac{e}{1 – \frac{1}{e}}$
$\therefore \alpha = \frac{e}{e – 1}$ $\Rightarrow \log \alpha = \frac{e}{e – 1}$
$\therefore$ Required value = $e^{\frac{1}{e} \log (1 + \alpha) + \log \alpha} = e$

Question

If the function

$
f(x) =
\begin{cases}
\dfrac{2}{x} \left[ \sin{(k_1 + 1)x} + \sin{(k_2 – 1)x} \right], & x < 0 \\(6pt]
4, & x = 0 \\(6pt]
\dfrac{2}{x} \log_e{\left( \dfrac{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

(1) $ 8 $

(2) $ 20 $

(3) $ 5 $

(4) $ 10 $

▶️ Answer/Explanation

$\text{Ans. } (4)$

$\text{Sol. }$

\(
\lim_{x \to 0^-} \frac{2}{x} \left[ \sin{(k_1 + 1)x} + \sin{(k_2 – 1)x} \right] = 4
\(

$\Rightarrow 2(k_1 + 1) + 2(k_2 – 1) = 4$

$\Rightarrow k_1 + k_2 = 2$

\(
\lim_{x \to 0^+} \frac{2}{x} \ln{\left( \frac{2 + k_1 x}{2 + k_2 x} \right)} = 4
\(

$\Rightarrow \lim_{x \to 0} 2 \ln{\left( \frac{1 + \frac{k_1 x}{2}}{1 + \frac{k_2 x}{2}} \right)} \cdot \frac{1}{x} = 4$

Using $\ln(1 + a) \approx a$ for small $a$,

$\Rightarrow 2 \times \frac{1}{x} \times \left( \frac{k_1 x}{2} – \frac{k_2 x}{2} \right) = 4$

$\Rightarrow k_1 – k_2 = 2$

Now, solving the two equations:

\(
\begin{cases}
k_1 + k_2 = 2 \\
k_1 – k_2 = 2
\end{cases}
\Rightarrow
k_1 = 2, \, k_2 = 0
\(

$\therefore k_1^2 + k_2^2 = 2^2 + 0^2 = 4$

However, checking with correct substitution in the logarithmic condition gives
$ k_1 – k_2 = 4 $, not $2$, hence:

\(
\begin{cases}
k_1 + k_2 = 2 \\
k_1 – k_2 = 4
\end{cases}
\Rightarrow
k_1 = 3, \, k_2 = -1
\(

$\therefore k_1^2 + k_2^2 = 3^2 + (-1)^2 = 9 + 1 = 10$

IIT JEE Main Maths -Unit 7 -Limits, continuity, and differentiability- Exam Style Questions- New Syllabus

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