Mathematics Test 37

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Mathematics Test 37

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Weekly Quiz Competition

Question 1
4 / -1

Given P(x) = x⁴ + ax³ +bx² cx + d such that x = 0 is the only real root of Pï (x) = 0. If P(–1) < P(1), then in the interval [–1, 1].

A
P(–1) is the minimum and P(1) is the maximum of P

B
P(–1) is not minimum but P(1) is the maximum of P

C
P(–1) is the minimum but P(1) is not the maximum of P

D
neither P(–1) is the minimum nor P(1) is the maximum of P

Solution

P(x) = x⁴ + ax³ + bx² + cx + d

P'(x) = 4x³ + 3ax² + 2bx + c

As P'(x) = 0 has only root x = 0

⇒ c = 0

So,

P(x) = x⁴ + ax³ + bx² + d

P(-1)<P(1)

⇒ 1 - a + b + d <1 + a + b + d

⇒ a>0

Since, P'(x) =0 only when x=0 and P(x) is differentiable in (-1,1), we should have the maximum and minimum at the points x=-1,0 and 1.

Therefore, the maximum of P(x)=Max{P(0), P(1)} and the minimum of P(x)=Min{P(-1), P(0)}

In the interval [0,1],

P'(x) = 4x³ + 3ax² + 2bx=x(4x² + 3ax + 2b)

Thus, (4x² + 3ax + 2b) has no real roots, therfore

(3a)²-32b<0

b>( 9a² /32 )

Thus, b>0

So, P'(x)>0 for all x lying in the interval [0,1]. Hence, P(x) is increasing in [0,1].

Similarly P'(x)<0 for all x lying in the interval [-1,0]. Hence, P(x) is decreasing in [-1,0].

The minimum is at x=0.

Hence, option B is correct.
Question 2
4 / -1

If 100 time the 100^th term of an AP with an zero common difference equals the 50 times its 50^th term, then the 150^th term of this AP is

A
–150

B
150 times its 50^thterm

C
150

D
zero

Solution

100 (T₁₀₀) = 50 (T₅₀)

⇒ 2[a + 99d] = a + 49 d

⇒ a + 149 d = 0

⇒ T₁₅₀ = 0

Hence, option D is correct.