Question

Statement-I : The equation $\frac{x^3}{4}-\sin \pi x+3=2\frac{1}{2}$ has at least one solution in $[2,2]$
Because
Statement-II : If $f:[a,b]$ $\to R$ be a function & let $c$ be a number such that $f\left(a\right)<c<f\left(b\right)$, then there is at least one number $n$ $\in$ $(a,b)$ such that $f\left(n\right)=c$

• A. Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I
• B. Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I
• C. Statement-I is true, Statement-II is false
• D. Statement-I is false, Statement-II is true

Answer 1

The correct option is C Statement-I is true, Statement-II is false

$f\left(x\right)=\frac{x^3}{4}-\sin \pi x+\frac{1}{2}$ has at least two solution in the interval $[-2,2]$.

$f^1\left(x\right)=\frac{3x^2}{4}-\pi \cos \pi x$

$f\left(2\right)=\frac{2^3}{4}-\sin 2\pi +\frac{1}{2}=\frac{5}{2}>0$

$f\left(0\right)=\frac{0^3}{4}-\sin \pi 0+\frac{1}{2}=\frac{1}{2}>0$

$f(-2)=\frac{(-2{)}^3}{4}-\sin (-2)\pi +\frac{1}{2}=\frac{-3}{2}<0$

$f\left(\frac{1}{2}\right)<0$

As we can clearly see that we have at least two roots.

One root in between $(-2,0)$ and other between $(\frac{1}{2},2)$

Statement-I is true, Statement-II is false