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Continuity of a composite of two piecewise functions

This type of question, asked 5 times: 2021 Paper 6, 25 Jun 2022, 28 Jun 2022, 26 Jul 2022, 6 Apr 2026 S2. All 5 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 6

    85
    Let f:R→Rf:R \rightarrow R and g:R→Rg:R \rightarrow R be defined as f(x)={x+a,x<0∣x−1∣,x≥0f(x) =\left\{ \begin{matrix} x+a, & x< 0 \\ |x- 1|, & x\geq 0 \end{matrix} \right. and
    g(x)={x+1,x<0(x−1)2+b,x≥0g(x) =\left\{ \begin{matrix} x+ 1, & x< 0 \\ (x- 1)^{2}+b, & x\geq 0 \end{matrix} \right.
    where a,ba,b are non-negative real numbers. If (gof)(x) is continuous for all x∈Rx\in R, then a+ba+b is equal to
  2. Question 2

    25 Jun 2022

    172
    Let f(x)=[2x2+1]f(x) =\left\lbrack 2x^{2}+ 1 \right\rbrack and g(x)={2x−3,x<02x+3,x≥0g(x) =\left\{ \begin{matrix} 2x - 3, & x < 0 \\ 2x + 3, & x \geq 0 \end{matrix} \right., where [t]\lbrack t\rbrack is the greatest integer ≤t\leq t. Then, in the open interval (−1,1)( - 1,1), the number of points where fog is discontinuous is equal to
  3. Question 3

    28 Jun 2022

    156
    Let f,g:R→Rf,g:R \rightarrow R be functions defined by f(x)={[x],x<0∣1−x∣,x≥0f(x) =\left\{ \begin{matrix} \lbrack x\rbrack, & x < 0 \\ |1 - x|, & x \geq 0 \end{matrix} \right. and
    g(x)={ex−xx<0(x−1)2−1,x≥0g(x) =\left\{ \begin{matrix} e^{x}- x & x < 0 \\ (x - 1)^{2}- 1, & x \geq 0 \end{matrix} \right.
    where [x]\lbrack x\rbrack denote the greatest integer less than or equal to xx. Then, the function fog is discontinuous at exactly:

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  4. Question 4

    26 Jul 2022

    68
    If f(x)={x+a,x≤0∣x−4∣,x>0f(x) =\left\{ \begin{matrix} x + a, & x \leq 0 \\ |x - 4|, & x > 0 \end{matrix} \right. and
    g(x)={x+1,x<0(x−4)2+b,x≥0g(x) =\left\{ \begin{matrix} x + 1 & ,x < 0 \\ (x - 4)^{2}+ b, & x \geq 0 \end{matrix} \right.
    are continuous on R, then (gof) (2) + (fog) (-2) is equal to:

    Tap an option to check your answer.

  5. Question 5

    6 Apr 2026 S2

    75
    Let f(x)={x3+8x<0x2−4x≥0f(x) = \left\{ \begin{matrix} x^{3} + 8 & x < 0 \\ x^{2} - 4 & x \geq 0 \end{matrix} \right. and g(x)={(x−8)1/3;x<0(x+4)1/2;x≥0g(x) = \left\{ \begin{matrix} (x - 8)^{1/3}; & x < 0 \\ (x + 4)^{1/2}; & x \geq 0 \end{matrix} \right. Then the number of points, where the function gof is discontinuous, is ____\_\_\_\_ .