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Functions and Graphs questions for IPMAT, with solutions

10 free Functions and Graphs practice questions for IPMAT Indore, Rohtak and JIPMAT, 4 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Functions and Graphs on Learn.

  1. Question 1Multiple choiceAsked in IPMAT Indore 2020Functions

    Given f(x)=x2+log⁡3xf(x)=x^{2}+\log_{3}x and g(y)=2y+f(y)g(y)=2y+f(y), then the value of g(3)g(3) equals

    1. A.1616
    2. B.1515
    3. C.2525
    4. D.2626
    Show the answer and solution

    Answer: A. 1616

    First evaluate f(3)f(3): f(3)=32+log⁡33=9+1=10f(3)=3^2+\log_33=9+1=10. Then g(3)=2⋅3+f(3)=6+10=16g(3)=2\cdot3+f(3)=6+10=16. The common slip is forgetting that log⁡33=1\log_33=1, which gives 1515.

  2. Question 2Multiple choiceWritten by IPMATwayFunctions

    If ff is an odd function with f(3)=5f(3)=5, then f(−3)+f(0)f(-3)+f(0) equals

    1. A.00
    2. B.−4-4
    3. C.−5-5
    4. D.55
    Show the answer and solution

    Answer: C. −5-5

    An odd function satisfies f(−x)=−f(x)f(-x)=-f(x). So f(−3)=−f(3)=−5f(-3)=-f(3)=-5. Putting x=0x=0: f(0)=−f(0)f(0)=-f(0), so f(0)=0f(0)=0. Hence f(−3)+f(0)=−5+0=−5f(-3)+f(0)=-5+0=-5.

  3. Question 3Type the answerWritten by IPMATwayFunctions

    For real numbers aa and bb, define a@b=a2+b2a@b=a^{2}+b^{2} and a#b=a−ba\#b=a-b. Then (3@4)#(2@1)(3@4)\#(2@1) equals

    Show the answer and solution

    Answer: 20

    Evaluate each bracket with @@: 3@4=32+42=253@4=3^2+4^2=25 and 2@1=22+12=52@1=2^2+1^2=5. Then apply #\#: 25#5=25−5=2025\#5=25-5=20.

  4. Question 4Multiple choiceAsked in IPMAT Indore 2025Graphs and Behaviour

    If y=a+blog⁡exy=a+b\log_{e}x, then

    1. A.y−ay-a is proportional to xbx^{b}
    2. B.log⁡ey\log_{e}y is proportional to xx
    3. C.1y−a\dfrac{1}{y-a} is proportional to xbx^{b}
    4. D.eye^{y} is proportional to xbx^{b}
    Show the answer and solution

    Answer: D. eye^{y} is proportional to xbx^{b}

    Exponentiate both sides: ey=ea+blog⁡ex=ea⋅eblog⁡exe^y=e^{a+b\log_ex}=e^a\cdot e^{b\log_ex}. Since eblog⁡ex=(elog⁡ex)b=xbe^{b\log_ex}=\left(e^{\log_ex}\right)^b=x^b, we get ey=ea xbe^y=e^a\,x^b. As eae^a is a constant, eye^y is proportional to xbx^b.

  5. Question 5Multiple choiceAsked in IPMAT Rohtak 2019Graphs and Behaviour

    If the minimum value of f(x)=x2+2bx+2c2f(x)=x^{2}+2bx+2c^{2} is greater than the maximum value of g(x)=−x2−2cx+b2g(x)=-x^{2}-2cx+b^{2} for real xx, then

    1. A.∣c∣>2∣b∣|c|>\sqrt{2}|b|
    2. B.2∣c∣>b\sqrt{2}|c|>b
    3. C.0<c<2b0<c<\sqrt{2}b
    4. D.no real values of bb and cc
    Show the answer and solution

    Answer: A. ∣c∣>2∣b∣|c|>\sqrt{2}|b|

    Complete the squares: f(x)=(x+b)2+2c2−b2f(x)=(x+b)^2+2c^2-b^2, so min⁡f=2c2−b2\min f=2c^2-b^2. And g(x)=−(x+c)2+b2+c2g(x)=-(x+c)^2+b^2+c^2, so max⁡g=b2+c2\max g=b^2+c^2. The condition 2c2−b2>b2+c22c^2-b^2>b^2+c^2 simplifies to c2>2b2c^2>2b^2, i.e. ∣c∣>2 ∣b∣|c|>\sqrt2\,|b|.

  6. Question 6Type the answerWritten by IPMATwayRelations

    The number of onto functions from a set with 44 elements to a set with 22 elements is

    Show the answer and solution

    Answer: 14

    There are 24=162^4=16 functions from a 44-element set to a 22-element set. A function fails to be onto only if it misses one of the two values, i.e. it is constant. There are 22 constant functions. So the number of onto functions is 16−2=1416-2=14.

  7. Question 7Type the answerWritten by IPMATwayRelations

    The maximum number of equivalence relations on the set A={1,2,3}A=\{1,2,3\} is

    Show the answer and solution

    Answer: 5

    Each equivalence relation on AA corresponds to a partition of AA into classes. The partitions of {1,2,3}\{1,2,3\} are: {1},{2},{3}\{1\},\{2\},\{3\}; {1,2},{3}\{1,2\},\{3\}; {1,3},{2}\{1,3\},\{2\}; {2,3},{1}\{2,3\},\{1\}; and {1,2,3}\{1,2,3\}. That is 55 partitions, so there are 55 equivalence relations.

  8. Question 8Multiple choiceAsked in IPMAT Indore 2021Functions

    Suppose that a real-valued function f(x)f(x) of real numbers satisfies f(x+xy)=f(x)+f(xy)f(x+xy)=f(x)+f(xy) for all real xx, yy, and that f(2020)=1f(2020)=1. Compute f(2021)f(2021).

    1. A.20212020\dfrac{2021}{2020}
    2. B.20202019\dfrac{2020}{2019}
    3. C.11
    4. D.20202021\dfrac{2020}{2021}
    Show the answer and solution

    Answer: A. 20212020\dfrac{2021}{2020}

    Additivity gives f(nx)=nf(x)f(nx)=nf(x) for integers and f(pqx)=pqf(x)f\left(\tfrac{p}{q}x\right)=\tfrac{p}{q}f(x). With r=20212020r=\tfrac{2021}{2020} and x=2020x=2020: f(2021)=20212020f(2021)=\tfrac{2021}{2020}.

  9. Question 9Multiple choiceWritten by IPMATwayGraphs and Behaviour

    How many real values of xx satisfy ∣x2−4x∣=4|x^2-4x|=4?

    1. A.44
    2. B.22
    3. C.11
    4. D.33
    Show the answer and solution

    Answer: D. 33

    ∣x2−4x∣=4|x^2-4x|=4 means x2−4x=4x^2-4x=4 or x2−4x=−4x^2-4x=-4. First: x2−4x−4=0x^2-4x-4=0 has discriminant 16+16=32>016+16=32>0, roots x=2±22x=2\pm2\sqrt2, two distinct values. Second: x2−4x+4=0x^2-4x+4=0, i.e. (x−2)2=0(x-2)^2=0, one repeated root x=2x=2. This differs from 2±222\pm2\sqrt2. In the graph, the line y=4y=4 cuts the W-shaped curve at two outer points and touches the middle peak at (2,4)(2,4). Total of 33 solutions.

  10. Question 10Multiple choiceWritten by IPMATwayRelations

    On the set {1,2,3,4}\{1,2,3,4\}, how many relations are reflexive and symmetric but not transitive?

    1. A.4949
    2. B.6464
    3. C.1515
    4. D.6363
    Show the answer and solution

    Answer: A. 4949

    A reflexive relation must contain (i,i)(i,i) for all four elements. A symmetric relation includes (i,j)(i,j) together with (j,i)(j,i), so each of the (42)=6\binom42=6 unordered pairs is in or out: 26=642^6=64 relations. Those that are also transitive are exactly the equivalence relations, which correspond to partitions of a 4-element set: Bell number B4=15B_4=15. Hence reflexive, symmetric, not transitive: 64−15=4964-15=49.