Let f: R → R be defined as f(x) = \ array ll a - bcos2x x ^ 2 ; x 1...

At $x = 1, f\left(x\right)$ is continuous,

$⇒ f\left({1}^{-}\right) = f\left(1\right) = f\left({1}^{+}\right)$

$⇒ f\left(1\right) = 3 + c . . . \left(i\right)$

$⇒ f\left({1}^{+}\right) = \lim_{h \to 0}2\left(1 + h\right) + 1$

$⇒ f\left({1}^{+}\right) = \lim_{h \to 0}3 + 2h = 3 . . . \left(ii\right)$

Using \left(i\right)\text{and}\left(ii\right)$,</p><p>$⇒ c = 0$</p><p>At$x = 0, f\left(x\right)$is continuous,</p><p>$⇒ f\left({0}^{-}\right) = f\left(0\right) = f\left({0}^{+}\right) . . . \left(iii\right)$</p><p>$⇒ f\left(0\right) = f\left({0}^{+}\right) = 2 . . . \left(iv\right)$</p><p>So,$f\left({0}^{-}\right)$has to be equal to$2$.</p><p>$⇒ \lim_{h \to 0}\dfrac{a - bcos\left(2h\right)}{{h}^{2}}$</p><p>$⇒ \lim_{h \to 0}\dfrac{a - b\left\{1 - \dfrac{4{h}^{2}}{2 !} + \dfrac{16{h}^{4}}{4 !} + . ..\right\}}{{h}^{2}}$</p><p>$⇒ \lim_{h \to 0}\dfrac{a - b + b\left\{2{h}^{2} - \dfrac{2}{3}{h}^{4} . ..\right\}}{{h}^{2}}$</p><p>For limit to exist$a - b = 0$and limit is$2b$$. . . \left(v\right)$

Using, \left(iii\right), \left(iv\right)\text{and}\left(v\right)$</p><p>$⇒ a = b = 1$</p><p>Checking differentiability at$x = 0$.</p><p>$⇒ LHD: \lim_{h \to 0}\dfrac{\dfrac{1 - \cos 2h}{{h}^{2}} - 2}{- h}$</p><p>$⇒ \lim_{h \to 0}\dfrac{1 - \left(1 - \dfrac{4{h}^{2}}{2 !} + \dfrac{16{h}^{4}}{4 !} - . ..\right) - 2{h}^{2}}{- {h}^{3}} = 0$</p><p>$⇒ RHD: \lim_{h \to 0}\dfrac{{\left(0 + h\right)}^{2} + 2 - 2}{h} = 0$</p><p>Function is differentiable at every point in its domain</p><p>$∴ m = 0$</p><p>$⇒ m + a + b + c = 0 + 1 + 1 + 0$</p><p>$⇒ m + a + b + c = 2$