Subject

Mathematics

Class

JEE Class 12

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JEE Mathematics Solved Question Paper 2017

Multiple Choice Questions

The number of all numbers having 5 digits, with distinct digits is

99999
$9 \times {}^{9}P_{4}$
${}^{10}P_{5}$
${}^{9}P_{4}$

The greatest integer which divides (p + 1) (p + 2) (p + 3) .... (p + q) for all p E N and fixed q $\in$ N is

Let ${(1 + x + x^{2})}^{9} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{18} x^{18}$ . Then,

$a_{0} + a_{2} + . . . + a_{18} = a_{0} + a_{3} + . . . + a_{17}$
$a_{0} + a_{2} + . . . + a_{18}$ is even
$a_{0} + a_{2} + . . . + a_{18}$ is divisible by 9
$a_{0} + a_{2} + . . . + a_{18}$ is divisible by 3 but not by 9

Let f : R ➔ R be such that f is injective and f(x)f(y) = f(x + y) for $⟇$ x, y $\in$ R. If f(x), f(y), f(z) are in G.P., then x, y, z are in

AP always
GP always
AP depending on the value of x, y, z
GP depending on the value of x, y, z

The probability that a non-leap year selected at random will have 53 Sunday is,

6.
The equation sin x(sin(x) + cos(x)) = k has real solutions, where k is a real number. Then,

$0 \leq k \leq \frac{1 + \sqrt{2}}{2}$

$2 - \sqrt{3} \leq k \leq 2 + \sqrt{3}$

$0 \leq k \leq 2 - \sqrt{3}$

$\frac{1 - \sqrt{2}}{2} \leq k \leq \frac{1 + \sqrt{2}}{2}$

$\frac{1 - \sqrt{2}}{2} \leq k \leq \frac{1 + \sqrt{2}}{2}$

Let, f(x) = sin(x) (sin(x) + cos(x))

= $\sin^{2} (x) + \sin (x) \cos (x)$

= $\frac{1 - \cos (2 x)}{2} + \frac{2 \sin (x) \cos (x)}{2}$

= $\frac{1}{2} - \frac{1}{2} \cos (2 x) + \frac{1}{2} \sin (2 x)$

= $\frac{1}{2} + \frac{1}{2} \sin (2 x - \cos (2 x))$

Now, $- \sqrt{{(1)}^{2} + {(1)}^{2}} \leq \sin (2 x) - \cos (2 x) \leq \sqrt{{(1)}^{2} + {(1)}^{2}}$

$[∵ - \sqrt{a^{2} + b^{2}} \leq asin (x) + bcos (x) \leq \sqrt{a^{2} + b^{2}}]$

$\Rightarrow - \sqrt{2} \leq \sin (2 x) - \cos (2 x) \leq \sqrt{2}$

$\Rightarrow - \frac{\sqrt{2}}{2} \leq \frac{\sin (2 x) - \cos (2 x)}{2} \leq \frac{\sqrt{2}}{2}$

$\Rightarrow \frac{1}{2} - \frac{\sqrt{2}}{2} \leq \frac{1}{2} + \frac{\sin (2 x) - \cos (2 x)}{2} \leq \frac{1}{2} + \frac{\sqrt{2}}{2}$

$\Rightarrow \frac{1 - \sqrt{2}}{2} \leq f (x) \leq \frac{1 + \sqrt{2}}{2}$

$\Rightarrow \frac{1 - \sqrt{2}}{2} \leq k \leq \frac{1 + \sqrt{2}}{2}$

Transforming to parallel axes through a point (p, q), the equation

2x² + 3xy + 4y² + x + 18y + 25 = 0 becomes 2x² + 3xy + 4y²= 1. Then,

p = - 2, q = 3
p = 2, q = - 3
p = 3, q = - 4
p = - 4, q = 3

Let A(2,- 3) and B(- 2,1) be two angular points of AABC. If the centroid of the triangle moves on the line 2x + 3y = 1, then the locus of the angular point C is given by

2x + 3y = 9
2x - 3y = 9
3x + 2y = 5
3x + 2y = 3

The point P(3, 6) is first reflected on the line y =x and then the image point Q is again reflected on the line y = - x to get the image point Q'. Then, the circumcentre of the APQQ' is

(6, 3)
(6,- 3)
(3,- 6)
(0, 0)

10.

Let d₁and d₂ be the lengths of the perpendiculars drawn from any point of the line 7x - 9y + 10 = 0 upon the lines 3x + 4y = 5 and 12x +5y = 7, respectively. Then,