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Answer for 5 ,6 and 7 questions you can see the explanation in detail here .

5Q) In triangle ABC, the area is \(\frac{1}{2}bc sq.units \). AD is a median to BC. Prove that \(\angle ABC = \frac{1}{2}\angle ADC\).

solution: In a triangle ABC , let the lengths of sides of BC, CA, AB are a, b, c respectively.

Given area of \( \Delta ABC = \frac{1}{2} bc\)

 We know that area of \( \Delta ABC = \frac{1}{2}bc Sin A \)

  equating these two equations we get Sin A =1

  A= \( 90^{\circ}\) , \(\Delta ABC \) is right angle triangle.

  We know that for a right angled triangle , mid point of the hypotenuse is circum center . DA=DB=DC= radius of circumcircle . A circumcircle  passing through the vertices A, B, C with center D .

AC is segment of the circle.

We know that angle made b y segment at the center is twice the angle made at a point on the circle.

 \( \angle ADC = 2 \angle ABC \)

 \( \angle ABC = \frac{1}{2} \angle ADC \).

Hence proved.

 6Q) Solve the inequality, |x-1|+|x+1| < 4.

Solution :

 We know that |x-a| = ± (x-a) 

   |x-1|+|x+1| <4

  x -1 + x + 1 < 4    and  -(x-1) -(x+1) < 4

   2x < 4       and     -2x < 4

   x < 2  and  x > -2

  Therefore the solution is -2 < x < 2 .

7Q) Prove that , if the coefficients of the quadratic equation ax²+bx +c =0 are odd integers , then the roots of the equation cannot be rational numbers.


 if ax² + bx + c = 0 has rational roots then ax² + bx + c must factor across the integers.

for ax² + bx + c to factor across the integers, there must be two integers

that have a product of a∙c and a sum of b.

assuming the above two statements are true, consider a∙c.

since a and c are odd, a∙c is odd. Moreover, the only way a product of two

integers can be odd is if both integers are odd.

now, the sum of any two odd integers is even. Since the product of a∙c is odd, it follows that any two integer factors whose product of a∙c will have a sum that is even. But since b is also odd, this can not happen.

thus, ax² + bx + c can not be factored across the integers if each of a, b, and c is odd. This implies that if each of a, b, and c is odd, ax² + bx + c = 0 will have no rational roots.

  Answer s of the next questions are in next page.

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