MHT CET 2020 Math Solved Paper

We have f(x)=4x3−7,x∈R.
fis one-one. Let x1,x2∈R and f(x1)=f(x2).
⇒4x13−7=4x23−7⇒4x13=4x2‌3
⇒‌‌x13=x23⇒x13−x23=0
⇒(x1−x2)(x12+x1x2+x22)=0
⇒(x1−x2)[(x1+‌

x2

2

)2+‌

3x22

4

]=0
⇒x1−x2=0, because the other factor is nonzero.
⇒x1=x2 ∴f is one-one.
f is onto. Let k∈R any real number.
f(x)=k⇒4x3−7=k⇒x=(‌

k+7

4

)1∕3
Now(‌

k+7

4

)1∕3∈R, because k∈R and f[(‌

k+7

4

)1∕3]=4[(‌

k+7

4

)1∕3]3−7
=4(‌

k+7

4

)−7=k
∴‌‌k is the image of (‌

k+7

4

)1∕3
∴‌‌f is onto. ∴‌‌f is a bijective function.