JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 20)
Let f : R $$ \to $$ R and g : R $$ \to $$ R be defined as
$$f(x) = \left\{ {\matrix{ {x + a,} & {x < 0} \cr {|x - 1|,} & {x \ge 0} \cr } } \right.$$ and
$$g(x) = \left\{ {\matrix{ {x + 1,} & {x < 0} \cr {{{(x - 1)}^2} + b,} & {x \ge 0} \cr } } \right.$$,
where a, b are non-negative real numbers. If (gof) (x) is continuous for all x $$\in$$ R, then a + b is equal to ____________.
$$f(x) = \left\{ {\matrix{ {x + a,} & {x < 0} \cr {|x - 1|,} & {x \ge 0} \cr } } \right.$$ and
$$g(x) = \left\{ {\matrix{ {x + 1,} & {x < 0} \cr {{{(x - 1)}^2} + b,} & {x \ge 0} \cr } } \right.$$,
where a, b are non-negative real numbers. If (gof) (x) is continuous for all x $$\in$$ R, then a + b is equal to ____________.
Answer
1
Explanation
$$g[f(x)] = \left[ {\matrix{
{f(x) + 1} & {f(x) < 0} \cr
{{{(f(x) - 1)}^2} + b} & {f(x) \ge 0} \cr
} } \right.$$
$$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x + a < 0\& x < 0} \cr {|x - 1| + 1} & {|x - 1| < 0\& x \ge 0} \cr {{{(x + a - 1)}^2} + b} & {x + a \ge 0\& x < 0} \cr {{{(|x - 1| - 1)}^2} + b} & {|x - 1| \ge 0\& x \ge 0} \cr } } \right.$$
$$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x \in ( - \infty , - a)\& x \in ( - \infty ,0)} \cr {|x - 1| + 1} & {x \in \phi } \cr {{{(x + a - 1)}^2} + b} & {x \in [ - a,\infty )\& x \in [0,\infty )} \cr {{{(|x - 1| - 1)}^2} + b} & {x \in R\& x \in [0,\infty )} \cr } } \right.$$
$$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x \in ( - \infty , - a)} \cr {{{(x + a - 1)}^2} + b} & {x \in [ - a,0)} \cr {{{(|x - 1| - 1)}^2} + b} & {x \in [0,\infty )} \cr } } \right.$$
g(f(x)) is continuous.
At x = $$-$$a
-a + a + 1 = (-a + a - 1)2 + b
$$ \Rightarrow $$ 1 = b + 1
$$ \Rightarrow $$ b = 0
at x = 0
(a $$-$$1)2 + b = (|0 - 1| - 1)2 + b
$$ \Rightarrow $$ (a $$-$$1)2 + b = b
$$ \Rightarrow $$ a = 1
$$ \Rightarrow $$ a + b = 1
$$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x + a < 0\& x < 0} \cr {|x - 1| + 1} & {|x - 1| < 0\& x \ge 0} \cr {{{(x + a - 1)}^2} + b} & {x + a \ge 0\& x < 0} \cr {{{(|x - 1| - 1)}^2} + b} & {|x - 1| \ge 0\& x \ge 0} \cr } } \right.$$
$$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x \in ( - \infty , - a)\& x \in ( - \infty ,0)} \cr {|x - 1| + 1} & {x \in \phi } \cr {{{(x + a - 1)}^2} + b} & {x \in [ - a,\infty )\& x \in [0,\infty )} \cr {{{(|x - 1| - 1)}^2} + b} & {x \in R\& x \in [0,\infty )} \cr } } \right.$$
$$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x \in ( - \infty , - a)} \cr {{{(x + a - 1)}^2} + b} & {x \in [ - a,0)} \cr {{{(|x - 1| - 1)}^2} + b} & {x \in [0,\infty )} \cr } } \right.$$
g(f(x)) is continuous.
At x = $$-$$a
-a + a + 1 = (-a + a - 1)2 + b
$$ \Rightarrow $$ 1 = b + 1
$$ \Rightarrow $$ b = 0
at x = 0
(a $$-$$1)2 + b = (|0 - 1| - 1)2 + b
$$ \Rightarrow $$ (a $$-$$1)2 + b = b
$$ \Rightarrow $$ a = 1
$$ \Rightarrow $$ a + b = 1
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