The given equation x³ – ax² + bx – a = 0 can be rewritten as:

x (x² + b) – a(x² + 1) = 0

In case b = 1, then the equation becomes

(x – a)( x² + 1) = 0

Now here x² + 1 = 0 will give imaginary roots.

Hence, if the given equation has three real roots, then b ≠ 1.

Answers: (d)

Let y = n³ – 7n² + 11n – 5

Now n = 1 is a root of the above equation.

Hence, it can be written as

n³ – 7n² + 11n – 5 = (n – 1)(n² – 6n + 5) = (n – 1)²(n – 5)

Now, (n – 1)² is always positive. So the whole expression will be positive if n – 5 > 0 i.e. n > 5.

Since n is an integer so its least value will be n = 6.

Now n ≥ m and m is also an integer.

Hence, the least value of m is 6.

 Answers: (c)

Let the total pages be ‘n’. So we have  {n(n+1)}/2 = 100

Since one page was added twice, so 1000 is not the actual sum but it is an increased sum.

We have n(n + 1) = 2000.

Now by hit and trial we can say that the value of n = 44 i.e. initially there were 44 pages and their sum was (44 x 45)/2 = 990

Since the given sum is 1000, so we can say that the page number 10 was added twice.

Answers: (a)

Since Raman made a mistake in constant term, so his product of roots will be wrong but his sum of roots will be correct.

Hence, the sum of roots is 4 + 3 = 7.

Manoj made a mistake in coefficient of ‘x’, so his sum of roots will be wrong but product of roots will be correct.

Hence, the product of roots is 3 × 2 = 6.

Hence, the required equation is x² – 7x + 6 = 0.

The roots of this equations are 6 and 1.

Answers: (d)

We have the equation x²– (α – 2)x– α –1 = 0. its roots are p and q.

So, we have sum of roots = p+q = a–2 and the product of roots, pq = –a –1

Now p²+q² = (p+q)² – 2pq = (a–2)² + 2(a+1)= a² +4 – 4a + 2a + 2 = (a – 1)² + 5

Since (α – 1)² is a perfect square, so its minimum value will be 0, when α = 1.

In that case, the minimum value of p² + q²will be 5.