Let f ( x ) is a twice differentiable function, such that f (1-2 x )= f (1+2 x ) ∀ x ∈ R, then minimum number of roots of equation (f prime prime(x))2+f prime(x) ⋅ f prime prime prime(x)=0 in x ∈(-5,10) is (given that f(2)=f(5)=f(10) ) is

Question

Let f ( x ) is a twice differentiable function, such that f (1-2 x )= f (1+2 x ) ∀ x ∈ R, then minimum number of roots of equation (f prime prime(x))2+f prime(x) ⋅ f prime prime prime(x)=0 in x ∈(-5,10) is (given that f(2)=f(5)=f(10) ) is (A) 2 (B) 4 (C) 6 (D) 8

Tardigrade Admin · Accepted Answer

f (2)= f (0) and f(5)=f(-3) and f (10)= f (-4) ∴ f prime( x )=0 arrow min .4 times and f prime prime( x )=0 arrow min .3 times ∴ f prime( x ) ⋅ f prime prime( x )=0 arrow min .7 times.

Q. Let $f (x)$ is a twice differentiable function, such that $f (1 - 2 x) = f (1 + 2 x) \forall x \in R$ , then minimum number of roots of equation $(f^{''} (x))^{2} + f^{'} (x) \cdot f^{'''} (x) = 0$ in $x \in (- 5, 10)$ is (given that $f (2) = f (5) = f (10)$ ) is

2

4

6

8

$f (2) = f (0)$
and $f (5) = f (- 3)$
and $f (10) = f (- 4)$
$∴ f^{'} (x) = 0 \to min .4$ times
and $f^{''} (x) = 0 \to min .3$ times
$∴ f^{'} (x) \cdot f^{''} (x) = 0 \to min .7$ times.

Q. Let f(x) is a twice differentiable function, such that f(1−2x)=f(1+2x)∀x∈R, then minimum number of roots of equation (f′′(x))2+f′(x)⋅f′′′(x)=0 in x∈(−5,10) is (given that f(2)=f(5)=f(10) ) is

2

4

6

8

f(2)=f(0) and f(5)=f(−3) and f(10)=f(−4) ∴f′(x)=0→min.4 times and f′′(x)=0→min.3 times ∴f′(x)⋅f′′(x)=0→min.7 times.

Q. Let $f (x)$ is a twice differentiable function, such that $f (1 - 2 x) = f (1 + 2 x) \forall x \in R$ , then minimum number of roots of equation $(f^{''} (x))^{2} + f^{'} (x) \cdot f^{'''} (x) = 0$ in $x \in (- 5, 10)$ is (given that $f (2) = f (5) = f (10)$ ) is

$f (2) = f (0)$
and $f (5) = f (- 3)$
and $f (10) = f (- 4)$
$∴ f^{'} (x) = 0 \to min .4$ times
and $f^{''} (x) = 0 \to min .3$ times
$∴ f^{'} (x) \cdot f^{''} (x) = 0 \to min .7$ times.