K - - x -0end cases If f x is continuous for all x - R- then A- a - b -0B- f -0-0C- k -2D- f -0 does not exist

The correct options are A a b=0 B f'(0) =0 C k = 2Given : nbsp;f(x) = lim_n →∞e^x^2 1+[(a+b)x (a b)]x^2nx^2n+1 + cos x 1, amp; x ∈ R {0} nbsp; k, a ...

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Byju's Answer

Standard XII

Mathematics

Differentiability

k , & x =0end...

Question

Let $f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} lim n \to \infty \frac{e^{x^{2}} - 1 + [(a + b) x - (a - b)] x^{2 n}}{x^{2 n + 1} + cos x - 1}, & x \in R - {0} k, & x = 0 \end{matrix}$
If $f (x)$ is continuous for all $x \in R$ , then

a - b = 0

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f^{'} (0) = 0

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k = - 2

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f^{'} (0)

does not exist

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Solution

The correct options are
A $a - b = 0$
B $f^{'} (0) = 0$
C $k = - 2$
Given : $f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} lim n \to \infty \frac{e^{x^{2}} - 1 + [(a + b) x - (a - b)] x^{2 n}}{x^{2 n + 1} + cos x - 1}, & x \in R - {0} k, & x = 0 \end{matrix}$
$\Rightarrow f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{e^{x^{2}} - 1}{cos x - 1}, & 0 < x^{2} < 1 k, & x = 0 \frac{(a + b) x - (a - b)}{x}, & x^{2} > 1 \end{matrix}$

Checking the continuity at $x = 1$ , we get
$lim x \to 1^{-} \frac{e^{x^{2}} - 1}{cos x - 1} = \frac{e - 1}{cos 1 - 1} lim x \to 1^{+} \frac{(a + b) x - (a - b)}{x} = 2 b \Rightarrow b = \frac{e - 1}{2 (cos 1 - 1)}$

Checking the continuity at $x = - 1$ , we get
$lim x \to - 1^{+} \frac{e^{x^{2}} - 1}{cos x - 1} = \frac{e - 1}{cos 1 - 1} lim x \to - 1^{-} \frac{(a + b) x - (a - b)}{x} = \frac{- (a + b) - a + b}{- 1} \Rightarrow 2 a = \frac{e - 1}{(cos 1 - 1)} ∴ a = b$

Now, as $f (x)$ is continuous for all $x \in R$ , so
$f (0) = lim x \to 0 \frac{e^{x^{2}} - 1}{cos x - 1}$
Using L'Hospital's Rule, we get
$\Rightarrow k = lim x \to 0 \frac{e^{x^{2}} \times 2 x}{- sin x} = - 2$

$f^{'} (0^{+}) = lim h \to 0 \frac{f (h) - f (0)}{h} \Rightarrow f^{'} (0^{+}) = lim h \to 0 \frac{\frac{e^{h^{2}} - 1}{cos h - 1} + 2}{h} \Rightarrow f^{'} (0^{+}) = lim h \to 0 \frac{3 - e^{h^{2}} - 2 cos h}{2 h {sin}^{2} \frac{h}{2}} \Rightarrow f^{'} (0^{+}) = lim h \to 0 \frac{2 (3 - e^{h^{2}} - 2 cos h)}{h^{3}}$
Using L'Hospital's Rule, we get
$\Rightarrow f^{'} (0^{+}) = lim h \to 0 \frac{2 (- e^{h^{2}} 2 h + 2 sin h)}{3 h^{2}} \Rightarrow f^{'} (0^{+}) = lim h \to 0 \frac{2 (- e^{h^{2}} 4 h^{2} - 2 e^{h^{2}} + 2 cos h)}{6 h} \Rightarrow f^{'} (0^{+}) = lim h \to 0 \frac{4 (- e^{h^{2}} + cos h)}{6 h} \Rightarrow f^{'} (0^{+}) = lim h \to 0 \frac{4 (- e^{h^{2}} 2 h - sin h)}{6} \Rightarrow f^{'} (0^{+}) = 0$
Similarly,
$f^{'} (0^{-}) = 0 ∴ f^{'} (0) = 0$

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