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    Anyone get Question 4 in the maths HL paper? SStudent

    Let f (x) = 3x + 5, for x ∈ ℝ.

    (a) Find the value of f (7).

    (b) Write f (k ) in terms of k.

    (c) Using your answer to part (b), or otherwise, find the value of k for which f (k ) = k.

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      For part a put 7 in for x and solve it: f(7)=3(7) + 5


      I think i got b and c wrong :] It seemed too easy lol.

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      i got a but not b and c

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      f(x) is a function. To keep things simple, you sub the 7 into the slot of X. Therefore;

      A) f(7) = 3(7) + 5, f(7) = 26

      B) Using what I said above, you slot K into the position of X. Therefore, 3x + 5 becomes 3k + 5

      C) Because we know what f(k) is, 3k + 5, we can leave that equal to k. Therefore; 3k + 5 = k, moving things around it becomes 3k - k = -5. In the end K works out to be -5/2 or -2.5.

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      i but down for (B) : f(k)=3k + 5

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      ugh looking back i could have probably figured that out if it wasn't an exam...ugh

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