First slide

Continuity

Question

Let $f (x) = \{\begin{cases} \frac{a (1 - xsin x) + bcos x + 5}{x^{2}}, x < 0 \\ 3, x = 0 \\ {\{1 + (\frac{P (x)}{x^{2}})\}}^{1 / x}, x > 0 \end{cases}$ where P(x) is a cubic function and f is continuous at x = 0.

Moderate

Question

The range of function g(x) = 3a sin x - b cos x is

A

[-10, 10]
B

[-5, 5]
C

[-12, 12]
D

none of these

Solution

$f (x) = \{\begin{cases} \frac{a (1 - xsin x) + bcos x + 5}{x^{2}}, x < 0 \\ 3, x = 0 \\ {\{1 + (\frac{P (x)}{x})\}}^{1 / x}, x > 0 \end{cases}$
where $P (x) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3}$
f (0) = 3
R.H.L $= lim_{x \to 0^{+}} f (x)$
$\begin{array}{l} = lim_{h \to 0} f (0 + h) \\ = lim_{h \to 0} f (h) \\ = lim_{h \to 0} {\{1 + (\frac{P (h)}{h})\}}^{1 / h} \end{array}$
Since f is continuous at x = 0, R.H.L. exists.
For the existence of R.H.L., $a_{0}, a_{1} = 0$ . Thus,
R.H.L. = $lim_{h \to 0} {(1 + a_{2} h + a_{3} h^{2})}^{1 / h} (1^{\infty} form)$
$= e^{lim_{h \to 0} (1 + a_{2} h + a_{3} h^{2} - 1) (1 / h)} = e^{o_{2}}$
L.H.L = $lim_{x \to 0^{-}} f (x)$
$\begin{array}{l} = lim_{h \to 0} f (0 - h) \\ = lim_{h \to 0} \frac{a (1 - (- h) \sin (- h)) + bcos (- h) + 5}{(- h)^{2}} \\ = lim_{h \to 0} \frac{a (1 - h (h)) + b (1 - \frac{h^{2}}{2!}) + 5}{h^{2}} \end{array}$
For finite value of L.H.L., a + b +5 = 0 and $- a - \frac{b}{2} = 3$
Solving, we get a = -l , b = -4.
Now, S(x) = 3a sin x - b cos x = - 3 sin x + 4 cos x which has range [- 5, 5].
Also, $P (x) = a_{3} x^{3} + (\log_{e} 3) x^{2}$
$\begin{array}{l} P^{''} (x) = 6 a_{3} x + 2 \log_{e} 3 \\ ∴ P^{''} (0) = 2 \log_{e} 3 \end{array}$
Further, $P (x) = b or a_{3} x^{3} + (\log_{e} 3) x^{2} = - 4$ has only one real root, as the graph of $P (x) = a_{3} x^{3} + (\log_{e} 3) x^{2}$ meets y = - 4
only once for negative value of x.

Question

The value of $P^{''} (0)$ is

A

$\log_{e} 9$
B

$\log_{e} 2$
C

2
D

1

Question

If the leading coefficient of P(x) is positive, then the equation P(x) = b has

A

only one real, positive root
B

only one real, negative root
C

three real roots
D

none of these

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App