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Continuity and Differentiability

Multiple Choice Questions

441.

$If \frac{d}{dx} [(x + 1) (x^{2} + 1) (x^{4} + 1) (x^{8} + 1)] = (15 x^{p} - 16 x^{q} + 1) {(x - 1)}^{- 2}, then (p, q) = ?$

(12, 11)
(15, 14)
(16, 14)
(16, 15)

442.

$\lim_{x \to 0} \frac{\sqrt{1 + x^{2}} - \sqrt{1 - x + x^{2}}}{3^{x} - 1} = ?$

$\frac{1}{\log_{e} (3)}$
$\log_{e} (9)$
$\frac{1}{\log_{e} (9)}$
$\log_{e} (3)$

443.

$If y = \tan^{- 1} (\frac{\sqrt{1 + a^{2} x^{2}} - 1}{ax}), then (1 + a^{2} x^{2}) y'' + 2 a^{2} y' = ?$

- 2a²
a²
2a²
0

444.

$If x^{2} + y^{2} = t + \frac{1}{t} and x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}}, then \frac{dy}{dx} = ?$

$- \frac{x}{y}$
$- \frac{y}{x}$
$\frac{x^{2}}{y^{2}}$
$\frac{y^{2}}{x^{2}}$

445.

$If x = {at}^{2} and y = 2 at, then \frac{d^{2} y}{{dx}^{2}} at t = \frac{1}{2} is$

$- \frac{2}{a}$
$\frac{4}{a}$
$\frac{8}{a}$
$- \frac{4}{a}$

446.

$The equations x - y + 2 z = 4 3 x + y + 4 z = 6 x + y + z = 1 have$

unique solution
infinitely many solutions
no solution
two solutions

447.

The value (s) of x for which the function

$f (x) \{= 1 - x, x < 1 = (1 - x) (2 - x), 1 \leq x \leq 2 3 - x, x > 2\} fails to be continuous is (are)$

1
2
3
all real numbers

448.

$If y = \log_{2} (\log_{2} (x)), then \frac{dy}{dx} = ?$

$\frac{\log_{e} (2)}{{xlog}_{e} (x)}$
$\frac{1}{\log_{e} {(2 x)}^{x}}$
$\frac{1}{({xlog}_{e} (x) \log_{e} (2))}$
$\frac{1}{x {(\log_{2} (x))}^{2}}$

449.
The angle of intersection between the curves y² + x² = a² $\sqrt{2}$ and x² - y² = a² is

$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{6}$

$\frac{π}{12}$

$\frac{π}{4}$

$Given, y^{2} + x^{2} = a^{2} \sqrt{2} . . . (i) and x^{2} - y^{2} = a^{2} . . . (ii) On solving eqs (i) and (ii), we get x = a \sqrt{\frac{\sqrt{2} + 1}{2}}, y = a \sqrt{\frac{\sqrt{2} - 1}{2}} Now, y^{2} + x^{2} = a^{2} \sqrt{2} 2 y {(\frac{dy}{dx})}_{1} + 2 x = 0 \Rightarrow {(\frac{dy}{dx})}_{1} = - \frac{x}{y} ∴ {(\frac{dy}{dx})}_{1} = \frac{- a \sqrt{\frac{\sqrt{2} + 1}{2}}}{a \sqrt{\frac{\sqrt{2} - 1}{2}}} = - \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}} and x^{2} - y^{2} = a^{2} On differentiating$

${(\frac{dy}{dx})}_{2} = \frac{x}{y} = \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}} Then, \tan (θ) = \frac{{(\frac{dy}{dx})}_{1} - {(\frac{dy}{dx})}_{2}}{1 + {(\frac{dy}{dx})}_{1} {(\frac{dy}{dx})}_{2}} = \frac{- \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}} - \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}}}{1 + (- \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}} \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}})} = \frac{- 2 \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}}}{1 - \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}}} = \frac{- 2 \sqrt{(\sqrt{2} + 1) (\sqrt{2} - 1)}}{\sqrt{2} - 1 - \sqrt{2} - 1} = \frac{- 2 \sqrt{2 - 1}}{- 2} = 1 ∴ θ = \tan^{- 1} (1) = \frac{π}{4}$

450.

$If the function f (x) = \{k_{1} {(x - π)}^{2}, x \leq π k_{2} \cos (x), x > π\} is twice differentiable, then the ordered pair (k_{1}, k_{2}) is equal to$

$(\frac{1}{2}, - 1)$
$(\frac{1}{2}, 1)$
(1, 0)
(1, 1)

Previous Year Papers

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Multiple Choice Questions

449. The angle of intersection between the curves y2 + x2 = a22 and x2 - y2 = a2 is π3 π4 π6 π12

449.
The angle of intersection between the curves y² + x² = a² $\sqrt{2}$ and x² - y² = a² is

$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{6}$

$\frac{π}{12}$