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JEE Mathematics Solved Question Paper 2014

Multiple Choice Questions

71.

The differential equation of the family of parabolas with vertex at (0, - 1) and having axis along the Y-axis is

yy' + 2xy + 1 = 0
xy' + y + 1 = 0
xy' - 2y - 2
xy' - y - 1 = 0

72.

$The solution of x \frac{dy}{dx} = y + e^{\frac{y}{x}} with y (1) = 0 is$

$1 = \log (x) + e^{\frac{y}{x}}$
$\log (x) = e^{- \frac{y}{x}}$
$1 = 2 \log (x) + e^{- \frac{y}{x}}$
$\log (x) + e^{- \frac{y}{x}} = 1$

73.

The solution of cos(y) + (xsin(y) - 1)dy/dx = 0 is,

$xsec (y) = \tan (y) + C$
$\tan (y) - \sec (y) = Cx$
$\tan (y) + \sec (y) = Cx$
$xsec (y) + \tan (y) = C$

74.

Three non-zero non-collinear vectors $\hat{a}, \hat{b} and \hat{c}$ are such that $\hat{a} + 3 \hat{b}$ is collinear with $\hat{c}$ , while $\hat{c}$ is $3 \hat{b} + 2 \hat{c}$ collinear with a. Then $\hat{a} + 3 \hat{b} + 2 \hat{c}$ equals

0
$2 \hat{a}$
$3 \hat{b}$
$4 \hat{c}$

75.

If $\hat{a}, \hat{b} and \hat{c}$ are non-coplanar vectors and if $\hat{d}$ is such that $\hat{d} = \frac{1}{x} (\hat{a} + \hat{b} + \hat{c})$ and $\hat{d} = \frac{1}{y} (\hat{b} + \hat{c} + \hat{d})$ where x and y are non-zero real numbers, then $\frac{1}{xy} (\hat{a} + \hat{b} + \hat{c} + \hat{d})$ equals to

76.

The angle between the lines $\hat{r} = (2 \hat{i} - 3 \hat{j} + \hat{k}) + λ (\hat{i} + 4 \hat{j} + 3 \hat{k})$ and $\hat{r} = (\hat{i} - \hat{j} + 2 \hat{k}) + μ (\hat{i} + 2 \hat{j} - 3 \hat{k})$ is

$\frac{π}{2}$
$\cos^{- 1} (\frac{9}{\sqrt{91}})$
$\cos^{- 1} (\frac{7}{\sqrt{84}})$
$\frac{π}{3}$

77.

If a, b and c are vectors with magnitudes 2, 3 and 4 respectively, then the best upper bound of ${|\hat{a} - \hat{b}|}^{2} + {|\hat{b} - \hat{c}|}^{2} + {|\hat{c} - \hat{a}|}^{2}$ among the given values is

78.

If x, y and z are non-zero real numbers and $\hat{a} = x \hat{i} + 2 \hat{j}$ , $\hat{b} = y \hat{j} + 3 \hat{k}$ and $\hat{c} = x \hat{i} + y \hat{j} + z \hat{k}$ are such that $\hat{a} \times \hat{b} = z \hat{i} - 3 \hat{j} + \hat{k}$ , then $[\hat{a} \hat{b} \hat{c}]$ equals to

79.
If x is real, then the minimum value of y = $\frac{x^{2} - x + 1}{x^{2} + x + 1} is$

3

$\frac{1}{3}$

$\frac{1}{2}$

2

$\frac{1}{3}$

$Let y = \frac{1 - x + x^{2}}{1 + x + x^{2}} On differentiating w . r . t . x, we get \frac{dy}{dx} = \frac{(1 + x + x^{2}) (- 1 + 2 x) - (1 - x + x^{2}) (1 + 2 x)}{{(1 + x + x^{2})}^{2}} = \frac{- 1 + 2 x - x + 2 x^{2} - x^{2} + 2 x^{3} - 1 - 2 x + x + 2 x^{2} - x^{2} - 2 x^{3}}{{(1 + x + x^{2})}^{2}} = \frac{- 2 + 2 x^{2} - 2 x^{2} + 2 x^{2}}{{(1 + x + x^{2})}^{2}} = \frac{2 x^{2} - 2}{{(1 + x + x^{2})}^{2}}$

$Put \frac{dy}{dx} = 0 \Rightarrow x^{2} = 1 \Rightarrow x = \pm 1 Now, \frac{d^{2} y}{{dx}^{2}} = \frac{2 [{(1 + x + x^{2})}^{2} (2 x) - (x^{2} - 1) (2) (1 + x + x^{2}) (1 + 2 x)]}{{(1 + x + x^{2})}^{4}} = \frac{4 [x + x^{2} + x^{3} - x^{2} - 2 x^{3} + 1 + 2 x]}{{(1 + x + x^{2})}^{3}} = \frac{4 (1 + 3 x - x^{3})}{{(1 + x + x^{2})}^{3}}$

$At x = 1, {(\frac{d^{2} y}{{dx}^{2}})}_{x = 1} = \frac{4 [1 + 3 (1) - 1^{3}]}{{(1 + 1 + 1^{2})}^{3}} = \frac{4 (3)}{3^{3}} = \frac{4}{9} > 0 At x = - 1, {(\frac{d^{2} y}{{dx}^{2}})}_{x = - 1} = \frac{4 [1 + 3 (- 1) - {(- 1)}^{3}]}{{[1 + (- 1) + {(- 1)}^{2}]}^{3}} = \frac{4 (1 - 3 + 1)}{{(1 - 1 + 1)}^{3}} = 4 (- 1) = - 4 < 0 ∴ By second derivative test, f is minimum at x = 1 and the minimum value is given by y = \frac{1 - 1 + 1}{1 + 1 + 1} = \frac{1}{3}$