By K. R. Choubey, Chandrakant Choubey & Ravikant Choubey

ISBN-10: 8131758508

ISBN-13: 9788131758502

Direction in arithmetic: A Lecture-wise technique is an entire source that's designed to assist scholars grasp arithmetic for the coveted IIT-JEE, AIEEE, state-level engineering front assessments and all different country senior secondary assessments, as well as the AISSSCE. This meticulously crafted and designed sequence displays the command and authority of the authors at the topic. The sequence adopts a simple step by step method of make studying arithmetic on the senior secondary point a pleased event.

Key gains:

Adopts a well-defined, meticulously deliberate and well dependent studying technique. contains lecture-wise assessments that aid revise every one accomplished lecture. includes pace Accuracy Sheets that enhance the rate and accuracy of scholars and support them revise key options. presents cutting edge guidance and methods which are effortless to use and be mindful. contains solved Topic-Wise query Banks to augment the comprehension and alertness of options.

desk of Contents:

half A Coordinate Geometry Lecture 1 Cartesian Coordinates 1 (Introductions, distance formulation and its software, locus of some extent) Lecture 2 Cartesian Coordinates 2 (Section formulation, region of triangle, zone of quadrilateral) Lecture 2 Cartesian Coordinates 2 (Slope of a line, detailed issues in triangle (centroid, circumcentre centroid, orthocenter, incentre and excentre ) half B directly Line Lecture 1 directly traces 1 (Some very important effects hooked up with one instantly line, point-slope shape, symmetric shape or distance shape, issues shape, intercept shape equation of the immediately traces) Lecture 2 directly traces 2 (Normal shape equation of the immediately line, the final shape equation of the instantly line, relief of the final shape into diversified circumstances, place of issues with admire to the instantly line ax + by way of + c and the perpendicular distance of aspect from the road ax + via + c = zero) Lecture three instantly strains three (Foot of perpendicular, mirrored image element or picture, a few vital effects hooked up with instantly strains, attitude among instantly strains) Lecture four directly traces four (Distance among parallel traces; place of beginning (0, zero) with appreciate to attitude among strains, angular bisectors of 2 given traces, a few details attached with 3 immediately traces) Lecture five immediately traces five (Miscellaneous questions, revision of hetero traces, a few more durable difficulties) half C Pair of hetero strains Lecture 1 Pair of heterosexual traces 1 (Homogeneous equations of moment measure and their numerous varieties) Lecture 2 Pair of hetero strains 2 (Some vital effects hooked up with homogenous pair of hetero line , basic equation of moment measure) half D Circle Lecture 1 Circle 1 D.3 D.14 (Equation of circle in a variety of kinds) Lecture 2 Circle 2 D.15 D.34 (Relative place of element with admire to circle, parametric type of equation of circle, relative place of line and circle) Lecture three Circle three D.35 D.56 (Relative place of circles, pair of tangents and chord of touch draw from an enternal element) half E Conic part Lecture 1 Parabola 1 Lecture 2 Parabola 2 Lecture three Ellipse 1 Lecture four Ellipse 2 (Position of line with appreciate to an ellipse, diameter, tangents and normals, chord of content material) Lecture five Hyperbola attempt Your talents

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**Additional resources for 2D Coordinate Geometry: Course in Mathematics for the IIT-JEE and Other Engineering Entrance Examinations**

**Example text**

Other such progressions are 151, 157, 163; 167, 173, 179. We also know arithmetic progressions of four successive prime numbers with difference 6, for example, 251, 257, 263, 269 and 1741, 1747, 1753, 1759. 14. Simple Theorem of Fermat THEOREM 9. If p is a prime number, then for every integer a the number ap—a is divisible by p. Proof Let p be a given prime number. The theorem is obviously true for a = 1. Now let a be a given natural number and we suppose that the theorem is true for the number a.

It is worth observing that if for a natural number n > 1 (n—1)! + 1 is divisible by n, then n must be a prime number. 3 ... («—1), and so the number {n—1)! + 1, when divided by a, would give the remainder 1, while being divisible by n it must of course be divisible by a. Hence the contradiction, which proves that n must be prime. Thus in order that the natural number n > 1 be prime it is necessary and sufficient that the number (n—1)! + 1 be divisible by n. Theoretically, then, by only one division we can find out whether a number is prime or not.

Since x{ ^ x2 and xx Φ x2, the number xi—x2 is natural and the number yx—y2 is an integer different from zero; so with a suitable sign the number y= ±(yl—y2) is natural and we have x = xl—x2 ^ xx <; m ^y/p and thus x < y/p because the equation x2 — p is not possible, the number p being prime. Similarly we get y < y/p. So the number ax±y9 which with a suitable sign is equal to the number a(xl~x2)— —(yi— y2), is divisible by p. The lemma is thus proved. —this will be a number not divisible by p (being a product of natural numbers less than p and by corollary to Theorem 7).

### 2D Coordinate Geometry: Course in Mathematics for the IIT-JEE and Other Engineering Entrance Examinations by K. R. Choubey, Chandrakant Choubey & Ravikant Choubey

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