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Functions question

2025 · 28 Jan · Shift 1 · Q43
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  5. /2025 · 28 Jan · Shift 1 · Q43

Functions question

2025 · 28 Jan · Shift 1 · Q43

JEE MainMathematicsFunctionsMCQ+4 / −1
Let f:R→Rf: \mathbb{R} \rightarrow \mathbb{R}f:R→R be a function defined by f(x)=(2+3a)x2+(a+2a−1)x+b,aeq1f(x)=(2+3 a) x^2+\left(\frac{a+2}{a-1}\right) x+b, a eq 1f(x)=(2+3a)x2+(a−1a+2​)x+b,aeq1. If f(x+y)=f(x)+f(y)+1−27xyf(x+y)=f(x)+f(\mathrm{y})+1-\frac{2}{7} x \mathrm{y}f(x+y)=f(x)+f(y)+1−72​xy, then the value of 28∑i=15∣f(i)∣28 \sum\limits_{i=1}^5|f(i)|28i=1∑5​∣f(i)∣ is
  1. A
    735
  2. B
    675
  3. C
    715
  4. D
    545
View written solutionFree

Correct answer: B

  1. Given

We have

f(x)=(2+3a)x2+(a+2a−1)x+b,a≠1f(x)=(2+3a)x^2+\left(\frac{a+2}{a-1}\right)x+b, \quad a\ne 1f(x)=(2+3a)x2+(a−1a+2​)x+b,a=1

and also

f(x+y)=f(x)+f(y)+1−27xy.f(x+y)=f(x)+f(y)+1-\frac{2}{7}xy.f(x+y)=f(x)+f(y)+1−72​xy.

We need to find

28∑i=15∣f(i)∣.28\sum_{i=1}^5 |f(i)|.28i=1∑5​∣f(i)∣.
  1. Write f(x)f(x)f(x) in standard quadratic form

Let

f(x)=px2+qx+b,f(x)=px^2+qx+b,f(x)=px2+qx+b,

where

p=2+3a,q=a+2a−1.p=2+3a, \qquad q=\frac{a+2}{a-1}.p=2+3a,q=a−1a+2​.

Then

f(x+y)=p(x+y)2+q(x+y)+bf(x+y)=p(x+y)^2+q(x+y)+bf(x+y)=p(x+y)2+q(x+y)+b =px2+2pxy+py2+qx+qy+b.=px^2+2pxy+py^2+qx+qy+b.=px2+2pxy+py2+qx+qy+b.

Also,

f(x)+f(y)+1−27xyf(x)+f(y)+1-\frac{2}{7}xyf(x)+f(y)+1−72​xy =(px2+qx+b)+(py2+qy+b)+1−27xy=(px^2+qx+b)+(py^2+qy+b)+1-\frac{2}{7}xy=(px2+qx+b)+(py2+qy+b)+1−72​xy =px2+py2+qx+qy+2b+1−27xy.=px^2+py^2+qx+qy+2b+1-\frac{2}{7}xy.=px2+py2+qx+qy+2b+1−72​xy.

Since these are equal for all real x,yx,yx,y, compare coefficients.

  1. Compare coefficients

(i) Coefficient of xyxyxy

Left side gives 2p2p2p, right side gives −27-\frac{2}{7}−72​. So,

2p=−27  ⟹  p=−17.2p=-\frac{2}{7} \implies p=-\frac{1}{7}.2p=−72​⟹p=−71​.

But

p=2+3a,p=2+3a,p=2+3a,

so

2+3a=−17.2+3a=-\frac{1}{7}.2+3a=−71​.

Thus,

3a=−17−2=−1573a=-\frac{1}{7}-2=-\frac{15}{7}3a=−71​−2=−715​ a=−1521=−57.a=-\frac{15}{21}=-\frac{5}{7}.a=−2115​=−75​.

(ii) Constant term

Left side constant term is bbb, right side constant term is 2b+12b+12b+1. So,

b=2b+1  ⟹  b=−1.b=2b+1 \implies b=-1.b=2b+1⟹b=−1.
  1. Find qqq

Now

q=a+2a−1.q=\frac{a+2}{a-1}.q=a−1a+2​.

Substitute a=−57a=-\frac{5}{7}a=−75​:

a+2=−57+2=97,a+2=-\frac{5}{7}+2=\frac{9}{7},a+2=−75​+2=79​, a−1=−57−1=−127.a-1=-\frac{5}{7}-1=-\frac{12}{7}.a−1=−75​−1=−712​.

Hence,

q=9/7−12/7=−34.q=\frac{9/7}{-12/7}=-\frac{3}{4}.q=−12/79/7​=−43​.

Therefore,

f(x)=−17x2−34x−1.f(x)=-\frac{1}{7}x^2-\frac{3}{4}x-1.f(x)=−71​x2−43​x−1.
  1. Compute f(1),f(2),f(3),f(4),f(5)f(1),f(2),f(3),f(4),f(5)f(1),f(2),f(3),f(4),f(5)

It is convenient to write

f(x)=−(17x2+34x+1),f(x)=-\left(\frac{1}{7}x^2+\frac{3}{4}x+1\right),f(x)=−(71​x2+43​x+1),

so all these values are negative, hence

∣f(i)∣=−f(i).|f(i)|=-f(i).∣f(i)∣=−f(i).

For x=1x=1x=1

f(1)=−17−34−1f(1)=-\frac{1}{7}-\frac{3}{4}-1f(1)=−71​−43​−1 =−4+21+2828=−5328.=-\frac{4+21+28}{28}=-\frac{53}{28}.=−284+21+28​=−2853​.

So,

∣f(1)∣=5328.|f(1)|=\frac{53}{28}.∣f(1)∣=2853​.

For x=2x=2x=2

f(2)=−47−32−1f(2)=-\frac{4}{7}-\frac{3}{2}-1f(2)=−74​−23​−1 =−8+21+1414=−4314.=-\frac{8+21+14}{14}=-\frac{43}{14}.=−148+21+14​=−1443​.

So,

∣f(2)∣=4314.|f(2)|=\frac{43}{14}.∣f(2)∣=1443​.

For x=3x=3x=3

f(3)=−97−94−1f(3)=-\frac{9}{7}-\frac{9}{4}-1f(3)=−79​−49​−1 =−36+63+2828=−12728.=-\frac{36+63+28}{28}=-\frac{127}{28}.=−2836+63+28​=−28127​.

So,

∣f(3)∣=12728.|f(3)|=\frac{127}{28}.∣f(3)∣=28127​.

For x=4x=4x=4

f(4)=−167−3−1=−167−4=−447.f(4)=-\frac{16}{7}-3-1=-\frac{16}{7}-4=-\frac{44}{7}.f(4)=−716​−3−1=−716​−4=−744​.

So,

∣f(4)∣=447.|f(4)|=\frac{44}{7}.∣f(4)∣=744​.

For x=5x=5x=5

f(5)=−257−154−1f(5)=-\frac{25}{7}-\frac{15}{4}-1f(5)=−725​−415​−1 =−100+105+2828=−23328.=-\frac{100+105+28}{28}=-\frac{233}{28}.=−28100+105+28​=−28233​.

So,

∣f(5)∣=23328.|f(5)|=\frac{233}{28}.∣f(5)∣=28233​.
  1. Sum the absolute values

Convert all to denominator 282828:

∣f(1)∣=5328,|f(1)|=\frac{53}{28},∣f(1)∣=2853​, ∣f(2)∣=8628,|f(2)|=\frac{86}{28},∣f(2)∣=2886​, ∣f(3)∣=12728,|f(3)|=\frac{127}{28},∣f(3)∣=28127​, ∣f(4)∣=17628,|f(4)|=\frac{176}{28},∣f(4)∣=28176​, ∣f(5)∣=23328.|f(5)|=\frac{233}{28}.∣f(5)∣=28233​.

Thus,

\sum_{i=1}^5 |f(i)|=\frac{53+86+127+176+233}{28}= rac{675}{28}.

Therefore,

28∑i=15∣f(i)∣=675.28\sum_{i=1}^5 |f(i)|=675.28i=1∑5​∣f(i)∣=675.
  1. Option check

The value is

675.\boxed{675}.675​.

So the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.

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