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Limits and Derivatives

Multiple Choice Questions

11.

$\lim_{x \to \infty} {(\frac{x + 5}{x + 2})}^{x + 3}$ equals

e
e²
e³
e⁵

e³

$\lim_{x \to \infty} {(\frac{x + 5}{x + 2})}^{x + 3} = \lim_{x \to \infty} {(1 + \frac{3}{x + 2})}^{x + 3} = \lim_{x \to \infty} {[{(1 + \frac{3}{x + 2})}^{\frac{x + 2}{3}}]}^{\frac{3 (x + 3)}{x + 2}} = e^{\lim_{x \to \infty} (\frac{1 + \frac{3}{x}}{1 + \frac{2}{x}})} = e^{3}$

12.

The differential equation which represents the family of curves y=c₁e^c₂xe, where c₁ and c₂ are arbitrary constants, is

y' =y²
y″ = y′ y
yy″ = y′
yy″ = y′

yy″ = y′

y c₁e^c₂x = …..(i)
y' = c₁c₂e^c₂x
y' = c₂y.....(from (i) ....(ii)
y" = c₂y' ....... (iii)
from (ii) & (iii)
$fraction numerator straight y apostrophe over denominator straight y end fraction space equals space fraction numerator straight y " over denominator straight y apostrophe end fraction rightwards double arrow space yy " space equals space left parenthesis straight y apostrophe right parenthesis squared$

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13.

The value of $\lim_{x \to 2} \frac{5}{\sqrt{2} - \sqrt{x}}$ is

10 $\sqrt{2}$
+ $\infty$
- $\infty$
does not exist

does not exist

$\lim_{x \to 2} \frac{5}{\sqrt{2} - \sqrt{x}} LHL = \lim_{x \to 2^{-}} \frac{5}{\sqrt{2} - \sqrt{x}} = \lim_{h \to 0} \frac{5}{(\sqrt{2} - \sqrt{2 - h})} \times \frac{(\sqrt{2} + \sqrt{2 - h})}{(\sqrt{2} + \sqrt{2 - h})} = \lim_{h \to 0} \frac{5 \sqrt{2} + \sqrt{2 + h}}{2 - 2 + h} = \infty$

$RHL = \lim_{x \to 2^{+}} \frac{5}{\sqrt{2} - \sqrt{x}} = \lim_{h \to 0} \frac{5}{(\sqrt{2} - \sqrt{2 - h})} \times \frac{(\sqrt{2} + \sqrt{2 + h})}{(\sqrt{2} + \sqrt{2 + h})} = \lim_{h \to 0} \frac{5 \sqrt{2} + \sqrt{2 + h}}{2 - 2 - h} = - \infty ∵ LHL \neq RHL$

Thus, limit does not exist.

14.

If y = $\tan^{- 1} \sqrt{\frac{1 - \sin (x)}{1 + \sin (x)}}$ , then the value of $\frac{dy}{dx}$ at x = $\frac{π}{6}$ is

- $\frac{1}{2}$
$\frac{1}{2}$
1
- 1

- $\frac{1}{2}$

$Given, y = \tan^{- 1} \sqrt{\frac{1 - \sin (x)}{1 + \sin (x)}} = \sqrt{\frac{1 - \cos (\frac{π}{2} - x)}{1 + \cos (\frac{π}{2} - x)}} = \tan^{- 1} |\tan (\frac{π}{4} - \frac{x}{2})| = \frac{π}{4} - \frac{x}{2} [∵ x = \frac{π}{6}] \Rightarrow \frac{dy}{dx} = - \frac{1}{2}$

15.

The value of $\lim_{x \to 2} \frac{e^{3 x - 6} - 1}{\sin (2 - x)}$

$\frac{3}{2}$
3
- 3
- 1

- 3

$\lim_{x \to 2} \frac{e^{3 x - 6} - 1}{\sin (2 - x)} = \underset{x \to 2}{\lim -} \frac{e^{3 x - 6} (3)}{\cos (2 - x)} [∵ using L' Hospital' s rule)] = - \frac{3 e^{0}}{\cos (0)} = - 3$

16.

The value of the limit $\lim_{x \to 1} \frac{\sin (e^{x - 1} - 1)}{\log (x)}$

0
e
$\frac{1}{e}$
1

$\lim_{x \to 1} \frac{\sin (e^{x - 1} - 1)}{\log (x)} = \lim_{h \to 0} \frac{\sin (e^{h} - 1)}{\log (1 + h)} = \lim_{h \to 0} \frac{\sin (e^{h} - 1)}{(e^{h} - 1)} \times \frac{(e^{h} - 1)}{\log (1 + h)} = 1 \times \lim_{h \to 0} \frac{(h + \frac{h^{2}}{2!} + . . .)}{(h - \frac{h^{2}}{2!} + . . . \infty)} = 1 \times 1 = 1$

17.

The value of $\lim_{n \to \infty} [\frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + . . . + \frac{n}{n^{2} + n^{2}}]$ is

$\frac{π}{4}$
$\log (2)$
0
1

$\frac{π}{4}$

$\lim_{n \to \infty} [\frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + . . . + \frac{n}{n^{2} + n^{2}}] = \lim_{n \to \infty} \sum_{r = 1}^{n} \frac{n}{n^{2} + r^{2}} = \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} \frac{1}{1 + {(\frac{r}{n})}^{2}}$

= $\int_{0}^{1} \frac{d x}{1 + x^{2}} = {[\tan^{- 1} (x)]}_{0}^{1} = \frac{π}{4}$

18.

Let f(x) = $\frac{\sqrt{x + 3}}{x + 1}$ , then the value of $\lim_{x \to - 3 - 0} f (x)$ is

0
does not exist
$\frac{1}{2}$
$- \frac{1}{2}$

does not exist

$\lim_{x \to - 3 - 0} \frac{\sqrt{x + 3}}{x + 1}$

But $\sqrt{x + 3}$ is not defined on left hand side of - 3.

Hence, function is not defined.

19.

The equation of the tangent to the curve y = x +4/x², that is parallel to the x-axis, is

y= 0
y= 1
y= 2
y= 2

y= 2

We have,
$straight y space equals space straight x space plus space 4 over straight x squared$
On differentiating w.r.t x, we get

$dy over dx space equals space 1 space minus space 8 over straight x cubed$
since the tangent is parallel to X- axis, therefore
dy/dx = 0
⇒ x³ = 8

⇒ x = 2 abd y =3

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20.

If g(x) is a polynomial satisfying g(x) g(y) = g(x) + g(y) + g(xy) - 2 for all real x and y and g(2) = 5, then $\lim_{x \to 3}$ g(x) is

Since, g(x) g(y) = g(x) + g(y) + g(xy) - 2

Now, at x = 0, y = 2, we get

g(0) g(2) = g(0) + g(2) + g(0) - 2

$[∵ g (2) = 5]$

$\Rightarrow 5 g (0) = 5 + 2 g (0) - 2 \Rightarrow 3 g (0) = 3 \Rightarrow g (0) = 1$

g(x) is given in a polynomial and by the relation given g(x) cannot be linear.

Let g(x) = x² + k $[∵ g (0) = 1]$

$∴ From Eq . (i) (x^{2} + 1) (y^{2} + 1) = x^{2} + 1 + y^{2} + 1 + x^{2} y^{2} + 1 - 2 ∴ \lim_{x \to 3} g (x) = g (3) = 3^{2} + 1 = 10$