Question

Let f(x) be a cubic polynomial with leading coefficient unity such that f(a) = b
and f'(a)=f''(a)=0. Suppose g(x)=f(x)-f(a)
+(a-x)f'(x)+3(x-a)^ 2 for which conclusion of Rolle’s theorem in [a, b] holds at x = 2, where x ∈(a,b)
1. The value of f''(2) , is
A) a prime number B) a composite number
C) an even number D) an odd number
52. The value of definite integral ∫ (a to b) f(x) is p/q
(where p & q are relatively prime numbers), then :
A) p+q=385 B)p-5q=51 C)p+q=6 D)p-5q=5

Answer 1

Answer:

1. A

Step-by-step explanation:

i dont know the else left one

Answer 2

given that,

$f'(a)-f''(a)=0$ , $x=a$ is equal to two times the root of $f(x)$ then $f'(x)=p(x-a)^{2}$

therefore $p=1$ because the leading coefficient $f'(x)=1$

$f'(x)=x^{2} -2ax+a^{2}+b---- > eq1$

$f'(x)=\frac{x^{3} }{3} -ax^{2} +a^{2} x$

as we know,

$f'(x)=f'(a)=b$

∴$b=\frac{a^{3} }{3} -a^{3} +a^{3}$

$b=-\frac{a^{3} }{3}$, let's sub this value in equation1

$f'(x)=x^{2} -2ax+a^{2}-\frac{a^{3} }{3}$

$g(x)=f(x)-f(a)+a-x(f'(x))+3(a-x)^{2}$

$g'(x)=f'(x)+(a-x)f''(x)-f'(x)+6(x-a)$

let's sub as $x=2$ then, $g'(2)=0$ and the equation will be

$0=f'(2)+(a-2)f''(2)-f'(2)+6(2-a)$

∴$f''(2)=\frac{6(a-2)}{(a-2)}$

           $=6$

therefore, 6 is an even number and the correct option is (c) and the correct value of definite integral ∫ (a to b) f(x) is p/q is an option (c)p+q=6

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