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JEE MATHEMATICS MCQ: CONTINUITY

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Let f:R→Rf:\mathbb{R} \to \mathbb{R} be defined as
f(x)=[[ex],x<0aex+[x−1],b+[sin⁡(πx)],0⩽x<11⩽x<2[e−x]−c,x⩾2f\left( x \right) = \left[ {\begin{array}{cccccccccccccccccccc} {\left[ {{e^x}} \right],}&{x < 0} \\ {\begin{array}{cccccccccccccccccccc} {a{e^x} + \left[ {x - 1} \right],} \\ {b + \left[ {\sin \left( {\pi x} \right)} \right],} \end{array}}&{\begin{array}{cccccccccccccccccccc} {0 \leqslant x < 1} \\ {1 \leqslant x < 2} \end{array}} \\ {\left[ {{e^{ - x}}} \right] - c,}&{x \geqslant 2} \end{array}} \right.
Where a,b,c∈Ra,b,c \in \mathbb{R} and [t]\left[ t \right] denotes greatest integer less than or equal to t.t. Then, which of the following statements is true?
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Correct option: C
If ff is discontinuous at exactly one point, then a+b+c≠1a + b + c \ne 1
The correct answer is 'If ff is discontinuous at exactly one point, then a+b+c≠1a + b + c \ne 1' because continuity conditions lead to constraints on a, b, and c, and the function cannot be continuous at all points.

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If the function
f(x)={{1+∣sin⁡x∣}a∣sin⁡x∣,−π6<x<0b,x=0etan⁡2xtan⁡3x,0<x<π6f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{{\left\{ {1 + \left| {\sin x} \right|} \right\}}^{\frac{a}{{\left| {\sin x} \right|}}}}, - \frac{\pi }{6} < x < 0}\\ {b,x = 0}\\ {{e^{\frac{{\tan 2x}}{{\tan 3x}}}},0 < x < \frac{\pi }{6}} \end{array}} \right.
Is continuous at x=0x = 0 f(x)={1+cos⁡−1{cot⁡x},x<π2π[x]+1,x≥π2f(x)=\left\{\begin{array}{rr}1+\cos ^{-1}\{\cot x\}, & x<\frac{\pi}{2} \\ \pi[x]+1, & x \geq \frac{\pi}{2}\end{array}\right.
Where [] denotes greatest integer and {}\{ \} denotes fractional part function, Then value of jump of discontinuity is: If f:R→Rf:R \to R is defined by f(x)={cos⁡3x−cos⁡xx2,for  x≠0λ,for  x=0f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\cos 3x - \cos x}}{{{x^2}}},\mathrm{for}\,\,x \ne 0}\\ {\lambda ,\mathrm{for}\,\,x = 0} \end{array}} \right. and if ff is continuous at x=0x = 0, then λ\lambda is equal to Let ff be a continuous function satisfying f(x)f(y)=f(x) f(y)= f(x)+f(y)+f(xy)−2f(x)+f(y)+f(x y)-2 for all x,y∈Rx, y \in {R} and f(2)=5f(2)=5 then lim⁡x→4f(x)\lim \limits_{{x \rightarrow 4}} f(x) is If f(x)={x2+3x−10x2+2x−15, when x≠−5a, when x=−5\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}\frac{x^2+3 x-10}{x^2+2 x-15} \text {, when } x \neq-5 \\ a, \quad \text { when } x=-5\end{array}\right. is continuous at x=−5x=-5, then the value of 'a' will be If f(x)={xsin⁡1x,x≠0k,x=0f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {x\sin \frac{1}{x},x \ne 0}\\ {k,x = 0} \end{array}} \right. is continuous at x=0x = 0, then the value of kk is
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