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MATHMAA

14th KVJ Math Olympiad question paper and solutions are below. It helps the students who appear for KVS Math Olympiad  and other competitive exams.  Here are the questions and answers for them.

1Q) Show that for any natural number n, the fraction $$\frac{21n+4}{14n+3}$$ is in its lowest term.

Solution:  To prove that for all natural number n , the fraction $$\frac{21n+4}{14n+3}$$ is in its lowest term, its enough to show that HCF of (21n+4, 14n+3)=1 .

21n+4= 1(14n+3)+(7n+1)

14n+3 =2(7n+1)+1

7n+1 = 1(7n+1) + 0

(21n+4, 14n+3) = 1

Therefore fraction $$\frac{21n+4}{14n+3}$$ is in its lowest term.

2Q) a) Factorise : $$x^{6}+ 5x^{3}+8$$

b) Prove that $$3a^{4} - 4a^{3}b+b^{4}\geq 0$$ for all real numbers a and b.

Solution:

a) Factorise : $$x^{6}+ 5x^{3}+8$$

$$x^{6}+ 5x^{3}+8$$=$$(x^{2})^{3} + 2^{3} + (-x)^{3} -3x^{2}*2*(-x)$$

We know the formula that

a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)

Let a = x^2 , b=2, c= -x Then plug in to the formula.

$$x^{6}+ 5x^{3}+8$$ = $$(x^{2})^3+2^{3}-x^{3}+6x^{3}$$

=$$(x^{2})^{3} + 2^{3} + (-x)^{3} -3x^{2}*2*(-x)$$

=$$(x^{2}+2-x)((x^{2})^{2}+2^{2}+(-x)^{2}-2*x^{2}-2*(-x)-x^{2}*(-x))$$

=$$(x^{2}+2-x)(x^{4}+4+x^{2}-2x^{2}+2x+x^3)$$

=$$(x^{2}-x+2)(x^{4}+x^{3}-x^{2}+2x+4)$$

b) $$3a^{4} - 4a^{3}b+b^{4}\geq 0$$ for all real numbers a and b .

=$$a^{4}+2a^{4} +b^{4}- 4a^{3}b$$

=$$a^{4}+b^{4} +2a^{4} - 4a^{3}b$$

=$$(a^{2})^{2} + (b^{2})^2 + 2a^{4} -4a^{3}b$$

= $$(a^{2})^{2} + (b^{2})^{2} -2a^{2}b^{2} + 2a^{2}b^{2} +2a^{4} - 4a^{3}b$$

= $$(a^{2}-b^{2})^{2} + 2a^{2}b^{2} + 2a^{4}-4a^{3}b$$

= $$(a^{2}-b^{2})^{2}+2a^{2}(b^{2} + a^{2} -2ab)$$

= $$(a^{2}-b^{2})^{2} +2a^{2}(a-b)^{2} \geq 0$$ for all a, b

As square is always positive and sum of the squares also positive.

Hence proved.

4Q) Solve the inequality , |x-1| + |x+1| $$\leq 4$$.