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Let f(x) = y. Then,
{tex}\frac{{3x - 4}}{5} = y{/tex}
{tex}\; \Rightarrow x = \frac{{5y + 4}}{3}{/tex}
{tex} \Rightarrow {f^{ - 1}}(y) = \frac{{5y + 4}}{3}{/tex}
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NCERT Exemplars are basically practice-books that include extra questions of a higher level and are meant for aiding in-depth learning. They are used especially for JEE mains and JEE advanced exams. The NCERT exemplar books contain conceptual sums, which cover CBSE board exams and competitive exams (JEE main and JEE advanced). CBSE board exams sometimes contain in-depth sums. So, it is beneficial if you go through these books.
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The second derivative tells us a lot about the qualitative behaviour of the graph. If the second derivative is positive at a point, the graph is concave up. If the second derivative is positive at a critical point, then the critical point is a local minimum. The second derivative will be zero at an inflection point.
In physics, jerk is the rate of change of acceleration; that is, the time derivative of acceleration, and as such the second derivative of velocity, or the third time derivative of position.
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Check last year papers here : <a href="https://mycbseguide.com/cbse-question-papers.html">https://mycbseguide.com/cbse-question-papers.html</a>
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We have, y2 = 8x and x = 2
{tex}\therefore{/tex} Area of shaded region, {tex}A = 2\int_0^2 {\sqrt {8x} dx = 2.2\sqrt 2 \int_0^2 {{x^{1/2}}dx} } {/tex}
{tex}= 4.\sqrt 2 .\left[ {2.\frac{{{x^{3/2}}}}{3}} \right]_0^2 = 4\sqrt 2 \left[ {\frac{2}{3}.2\sqrt 2 - 0} \right]{/tex}
{tex}= \frac{{32}}{2}{/tex} sq units
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