Let f:[0,1] arrow R be a twice differentiable function in (0,1) such that f(0)=3 and f(1)=5. If the line y=2 x+3 intersects the graph of f at only two distinct points in (0,1), then the least number of points x ∈(0,1), at which f prime prime( x )=0, is

Question

Let f:[0,1] arrow R be a twice differentiable function in (0,1) such that f(0)=3 and f(1)=5. If the line y=2 x+3 intersects the graph of f at only two distinct points in (0,1), then the least number of points x ∈(0,1), at which f prime prime( x )=0, is  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/741a114b20d0360931dd40844b7d0971-.png" /> f prime(a)=f prime(b)=f prime(c)=2 ⇒ f prime prime(x) text is zero for atleast x1 ∈(a, b) x2 ∈(b, c)

Q. Let $f : [0, 1] \to R$ be a twice differentiable function in $(0, 1)$ such that $f (0) = 3$ and $f (1) = 5$ . If the line $y = 2 x + 3$ intersects the graph of $f$ at only two distinct points in $(0, 1)$ , then the least number of points $x \in (0, 1)$ , at which $f^{''} (x) = 0$ , is ___

$f^{'} (a) = f^{'} (b) = f^{'} (c) = 2$
$\Rightarrow f^{''} (x) is zero$
for atleast $x_{1} \in (a, b) & x_{2} \in (b, c)$

Q. Let f:[0,1]→R be a twice differentiable function in (0,1) such that f(0)=3 and f(1)=5. If the line y=2x+3 intersects the graph of f at only two distinct points in (0,1), then the least number of points x∈(0,1), at which f′′(x)=0, is ___

f′(a)=f′(b)=f′(c)=2 ⇒f′′(x) is zero for atleast x1​∈(a,b)&x2​∈(b,c)

Q. Let $f : [0, 1] \to R$ be a twice differentiable function in $(0, 1)$ such that $f (0) = 3$ and $f (1) = 5$ . If the line $y = 2 x + 3$ intersects the graph of $f$ at only two distinct points in $(0, 1)$ , then the least number of points $x \in (0, 1)$ , at which $f^{''} (x) = 0$ , is ___

$f^{'} (a) = f^{'} (b) = f^{'} (c) = 2$
$\Rightarrow f^{''} (x) is zero$
for atleast $x_{1} \in (a, b) & x_{2} \in (b, c)$