Let f : R → R be defined as f x = 2 sin - π x 2 , if x 1 If f x is continuous on R , then a + b equals :

f x is continuous on R ⇒ f 1 - = f 1 = f 1 + a + 1 + b = lim x → 1 + sin πx a + 1 + b = 0 ⇒ a + b = - 1 ⇒ Also f - 1 - = f - 1 = f - 1 + lim x → - 1 2 sin - π x 2 = a - 1 + b a - 1 + b = 2 Either a - 1 + b = 2 or a - 1 + b = - 2 a + b = 3 . . . 2 or a + b = - 1 . . . 3 from 1 and 2 ⇒ a + b = 3 = - 1 (reject) from 1 and 3 ⇒ a + b = - 1

Mathematics - Continuity - Differentiability Question with Solution | TestHub

MathematicsContinuity - DifferentiabilityContinuity- MiscellaneousEasy2 minPYQ_2021

MathematicsEasysingle choice

Let $f : R \to R$ be defined as $f (x) = \{\begin{cases} 2 \sin (- \frac{π x}{2}), & if x < - 1 \\ |a x^{2} + x + b|, & if - 1 \leq x \leq 1 \\ \sin (π x), & if x > 1 \end{cases}$

If $f (x)$ is continuous on $R$ , then $a + b$ equals :

Options:

Answer:

Solution:

$f (x)$ is continuous on $R$

$\Rightarrow f (1^{-}) = f (1) = f (1^{+})$

$|a + 1 + b| = \lim_{x \to 1^{+}} \sin (πx)$

$|a + 1 + b| = 0 \Rightarrow a + b = - 1$

$\Rightarrow$ Also

$f (- 1^{-}) = f (- 1) = f (- 1^{+})$

$\lim_{x \to - 1} 2 \sin (\frac{- π x}{2}) = |a - 1 + b|$

$|a - 1 + b| = 2$

Either $a - 1 + b = 2$ or $a - 1 + b = - 2$

$a + b = 3 . . . (2)$ or $a + b = - 1 . . . (3)$

from $(1)$ and $(2)$ $\Rightarrow a + b = 3 = - 1$ (reject)

from $(1)$ and $(3)$ $\Rightarrow a + b = - 1$

Stream:JEESubject:MathematicsTopic:Continuity - DifferentiabilitySubtopic:Continuity- Miscellaneous

⏱ 2mℹ️ Source: PYQ_2021

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