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JEE Mathematics Solved Question Paper 2016

Multiple Choice Questions

71.

$If \int x^{3} e^{5 x} dx = \frac{e^{5 x}}{5^{4}} [f (x)] + C, then f (x) = ?$

$\frac{x^{3}}{5} - \frac{3 x^{2}}{5^{2}} + \frac{6 x}{5^{3}} - \frac{6}{5^{4}}$
$5 x^{3} - 5^{2} x^{2} + 5^{3} x - 6$
$5^{3} x^{3} - 15 x^{2} + 30 x - 6$
$5^{3} x^{3} - 75 x^{2} + 30 x - 6$

72.

$\int \frac{x}{{(x^{2} + 2 x + 2)}^{2}} dx = ?$

$\frac{x^{2} + 2}{x^{2} + 2 x + 2} - \frac{1}{2} \tan^{- 1} (x - 1) + C$
$\frac{x^{2} - 2}{4 (x^{2} + 2 x + 2)} - \frac{1}{2} \tan^{- 1} (x + 1) + C$
$\frac{x^{2} + 2}{2 (x^{2} + 2 x + 2)} - \frac{1}{2} \tan^{- 1} (x + 1) + C$
$\frac{2 (x - 1)}{x^{2} + 2 x + 2} + \frac{1}{2} \tan^{- 1} (x + 1) + C$

73.

$If \int \log (a^{2} + x^{2}) dx = h (x) + C, then h (x) = ?$

$xlog (a^{2} + x^{2}) + 2 \tan^{- 1} (\frac{x}{a})$
$x^{2} \log (a^{2} + x^{2}) + x + {atan}^{- 1} (\frac{x}{a})$
$xlog (a^{2} + x^{2}) - 2 x + 2 {atan}^{- 1} (\frac{x}{a})$
$x^{2} \log (a^{2} + x^{2}) + 2 x - a^{2} \tan^{- 1} (\frac{x}{a})$

74.

$For x > 0, if \int (\log {(x)}^{5}) dx = ? x [A (\log {(x)}^{5}) + B {(\log (x))}^{4} + C {(\log (x))}^{3} + D {(\log (x))}^{2} + E (\log (x)) + F] + Constant, then A + B + C + D + E + F = ?$

- 44
- 42
- 40
- 36

75.

By the definition of the definite integral, the value of $\lim_{n \to \infty} (\frac{1}{\sqrt{n^{2} - 1}} + \frac{1}{\sqrt{n^{2} - 2^{2}}} + . . . \frac{1}{\sqrt{n^{2} - {(n - 1)}^{2}}})$ is equal to

$π$
$\frac{π}{2}$
$\frac{π}{4}$
$\frac{π}{6}$

76.

$\int_{\frac{π}{4}}^{\frac{π}{4}} (\frac{x + \frac{π}{4}}{2 - \cos (2 x)}) dx$ is equal to

$\frac{8 π \sqrt{3}}{5}$
$\frac{2 π \sqrt{3}}{9}$
$\frac{4 π^{2} \sqrt{3}}{9}$
$\frac{π^{2}}{6 \sqrt{3}}$

77.

The solution of the differential equation $(1 + y^{2}) + (x - e^{\tan^{- 1} (y)}) \frac{dy}{dx} = 0$ , is

${xe}^{\tan^{- 1} (y)} = \tan^{- 1} (y) + C$
${xe}^{2 \tan^{- 1} (y)} = \tan^{- 1} (y) + C$
$2 {xe}^{\tan^{- 1} (y)} = e^{2 \tan^{- 1} (y)} + C$
$x^{2} e^{\tan^{- 1} (y)} = 4 e^{2 \tan^{- 1} (y)} + C$

78.

The solution of the differential equation $(2 x - 4 y + 3) \frac{dy}{dx} + (x - 2 y + 1) = 0$ is

$[\log (2 x - 4 y) + 3] = x - 2 y + C$
$[\log [2 (2 x - 4 y) + 3]] = 2 (x - 2 y) + C$
$\log [2 (x - 2 y) + 5] = 2 (x + y) + C$
$\log [4 (x - 2 y) + 5] = 4 (x + 2 y) + C$

79.

The mid-point of the line segment joining the centroid and the orthocentre of the triangle whose vertices are (a, b),(a, c) and (d, c), is

$(\frac{5 a + d}{6}, \frac{b + 5 c}{6})$
` $(\frac{a + 5 d}{6}, \frac{5 b + c}{6})$
(a, 0)
(0, 0)

80.
The orthocentre of the triangle formed by the lines x + y = 1 and 2y² - xy - 6x² = 0

$(\frac{4}{3}, \frac{4}{3})$

$(\frac{2}{3}, \frac{2}{3})$

$(\frac{2}{3}, \frac{- 2}{3})$

$(\frac{4}{3}, \frac{- 4}{3})$

$(\frac{4}{3}, \frac{4}{3})$

$We have x + y = 1 and 2 y^{2} - xy - 6 x^{2} = 0 \Rightarrow 2 y^{2} - 4 xy + 3 xy - 6 x^{2} = 0 \Rightarrow Equation of sides of ∆ ABC are x + y = 1, 2 y + 3 x = 0 and y - 2 x = 0$

$Solving these equations simultaneously, we get A (0, 0) B (\frac{1}{3}, \frac{2}{3}) and C (- 2, 3) Equation of altitude AD is x - y = 0 . . . (i) Equation of altitude CF is x + 2 y = λ Since, this passes through (- 2, 3) ∴ - 2 + 6 = λ \Rightarrow λ = 4 On solving eqs . (i) and (ii), we get x = \frac{4}{3}, y = \frac{4}{3}$