JEE Advance - Mathematics (2018 - Paper 1 Offline - No. 4)
For every twice differentiable function $$f:R \to [ - 2,2]$$ with $${(f(0))^2} + {(f'(0))^2} = 85$$, which of the following statement(s) is(are) TRUE?
There exist r, s $$ \in $$ R, where r < s, such that f is one-one on the open interval (r, s)
There exists x0 $$ \in $$ ($$-$$4, 0) such that |f'(x0)| $$ \le $$ 1
$$\mathop {\lim }\limits_{x \to \infty } f(x) = 1$$
There exists $$\alpha $$$$ \in $$($$-$$4, 4) such that f($$\alpha $$) + f"($$\alpha $$) = 0 and f'($$\alpha $$) $$ \ne $$ 0
Explanation
We have,
$${(f(0))^2} + {(f'(0))^2} = 85$$
and $$f:R \to [ - 2,2]$$
(a) Since, f is twice differentiable function, so f is continuous function.
$$ \therefore $$ This is true for every continuous function.
Hence, we can always find x $$ \in $$ (r, s), where f(x) is one-one.
$$ \therefore $$ This statement is true.
(b) By L.M.V.T.
$$f'(c) = {{f(b) - f(a)} \over {b - a}}$$
$$ \Rightarrow |f'(c)| = \left| {{{f(b) - f(a)} \over {b - a}}} \right|$$
$$ \Rightarrow |f'({x_0})| = \left| {{{f(0) - f( - 4)} \over {0 + 4}}} \right|$$
$$ = \left| {{{f(0) - f( - 4)} \over 4}} \right|$$
Range of f is [$$-$$2, 2]
$$ \therefore $$ $$ - 4 \le f(0) - f( - 4) \le 4$$
$$ \Rightarrow 0 \le \left| {{{f(0) - f( - 4)} \over 4}} \right| \le 1$$
Hence, |f'(x0)| = 1.
Hence, statement is true.
(c) As no function is given, then we assume
$$f(x) = 2\sin \left( {{{\sqrt {85} x} \over 2}} \right)$$
$$ \therefore $$ $$f'(x) = \sqrt {85} \cos \left( {{{\sqrt {85} x} \over 2}} \right)$$
Now, $${(f(0))^2} + {(f'(0))^2} = {(2\sin 0)^2} + {(\sqrt {85} \cos 0)^2}$$
$${(f(0))^2} + {(f'(0))^2} = 85$$
and $$\mathop {\lim }\limits_{x \to \infty } f(x)$$ does not exists.
Hence, statement is false.
(d) From option b, $$|f'({x_0})|\, \le 1$$
$${x_0} \in ( - 4,0)$$
$$ \therefore $$ $${(f'({x_0}))^2}\, \le 1$$
Hence, $$g({x_0}) = {(f({x_0}))^2} + {(f'({x_0}))^2}\, \le 4 + 1$$
[$$ \because $$ f(x0) $$ \in $$ [$$-$$2, 2]
$$ \Rightarrow g({x_0})\, \le 5$$
Now, let p $$ \in $$ ($$-$$4, 0) for which g(p) = 5
Similarly, let q be smallest positive number q $$ \in $$ (0, 4)
such that g(q) = 5
Hence, by Rolle's theorem is (p, q)
g'(c) = 0 for $$\alpha $$ $$ \in $$ ($$-$$4, 4) and since g(x) is greater than 5 as we move from x = p to x = q
and f(x))2 $$ \le $$ 4
$$ \Rightarrow $$ (f'(x))2 $$ \ge $$ 1 in (p, q)
Thus, g'(c) = 0
$$ \Rightarrow $$ f'f + f'f" = 0
So, f($$\alpha $$) + f"($$\alpha $$) = 0 and f'($$\alpha $$) $$ \ne $$ 0
Hence, statement is true.
$${(f(0))^2} + {(f'(0))^2} = 85$$
and $$f:R \to [ - 2,2]$$
(a) Since, f is twice differentiable function, so f is continuous function.
$$ \therefore $$ This is true for every continuous function.
Hence, we can always find x $$ \in $$ (r, s), where f(x) is one-one.
$$ \therefore $$ This statement is true.
(b) By L.M.V.T.
$$f'(c) = {{f(b) - f(a)} \over {b - a}}$$
$$ \Rightarrow |f'(c)| = \left| {{{f(b) - f(a)} \over {b - a}}} \right|$$
$$ \Rightarrow |f'({x_0})| = \left| {{{f(0) - f( - 4)} \over {0 + 4}}} \right|$$
$$ = \left| {{{f(0) - f( - 4)} \over 4}} \right|$$
Range of f is [$$-$$2, 2]
$$ \therefore $$ $$ - 4 \le f(0) - f( - 4) \le 4$$
$$ \Rightarrow 0 \le \left| {{{f(0) - f( - 4)} \over 4}} \right| \le 1$$
Hence, |f'(x0)| = 1.
Hence, statement is true.
(c) As no function is given, then we assume
$$f(x) = 2\sin \left( {{{\sqrt {85} x} \over 2}} \right)$$
$$ \therefore $$ $$f'(x) = \sqrt {85} \cos \left( {{{\sqrt {85} x} \over 2}} \right)$$
Now, $${(f(0))^2} + {(f'(0))^2} = {(2\sin 0)^2} + {(\sqrt {85} \cos 0)^2}$$
$${(f(0))^2} + {(f'(0))^2} = 85$$
and $$\mathop {\lim }\limits_{x \to \infty } f(x)$$ does not exists.
Hence, statement is false.
(d) From option b, $$|f'({x_0})|\, \le 1$$
$${x_0} \in ( - 4,0)$$
$$ \therefore $$ $${(f'({x_0}))^2}\, \le 1$$
Hence, $$g({x_0}) = {(f({x_0}))^2} + {(f'({x_0}))^2}\, \le 4 + 1$$
[$$ \because $$ f(x0) $$ \in $$ [$$-$$2, 2]
$$ \Rightarrow g({x_0})\, \le 5$$
Now, let p $$ \in $$ ($$-$$4, 0) for which g(p) = 5
Similarly, let q be smallest positive number q $$ \in $$ (0, 4)
such that g(q) = 5
Hence, by Rolle's theorem is (p, q)
g'(c) = 0 for $$\alpha $$ $$ \in $$ ($$-$$4, 4) and since g(x) is greater than 5 as we move from x = p to x = q
and f(x))2 $$ \le $$ 4
$$ \Rightarrow $$ (f'(x))2 $$ \ge $$ 1 in (p, q)
Thus, g'(c) = 0
$$ \Rightarrow $$ f'f + f'f" = 0
So, f($$\alpha $$) + f"($$\alpha $$) = 0 and f'($$\alpha $$) $$ \ne $$ 0
Hence, statement is true.
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