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5. Continuity and Differentiation
hard
ધારો કે $f$ અને $g$ એ $(-2,2)$ પરનાં એવા દ્વિ વિકલનીય ચુગ્મ વિધેયો છે કે જેથી $f\left(\frac{1}{4}\right)=0, f\left(\frac{1}{2}\right)=0, f(1)=1$ અને $g\left(\frac{3}{4}\right)=0, g(1)=2 .$ ,તો $(-2,2)$ માં, $f(x) g^{\prime \prime}(x)+f^{\prime}(x) g^{\prime}(x)=0$ ના ઉકેલોની ન્યૂનતમ સંખ્યા $\dots\dots$છે.
A
$0$
B
$2$
C
$4$
D
$6$
(JEE MAIN-2022)
Solution
Let $h(x)=f(x) g^{\prime}(x) \rightarrow 5$ roots
$\because f ( x )$ is even $\Rightarrow$
$f \left(\frac{1}{4}\right)= f \left(\frac{1}{2}\right)= f \left(-\frac{1}{2}\right)= f \left(\frac{1}{4}\right)=0$
$g ( x )$ is even $\Rightarrow g \left(\frac{3}{4}\right)= g \left(-\frac{3}{4}\right)=0$
$g ^{\prime}( x )=0$ has minimum one root
$h^{\prime}( x )$ has at last $4$ roots
Standard 12
Mathematics
