NIMCET 2007 — Mathematics PYQ

Step 1: Understand the geometric meaning of the derivative

The first derivative of a function, $f^{\prime }(x)$, represents the slope of the tangent line to the curve at any point $x$. The slope is also given by $\tan (\theta )$, where $\theta$ is the angle made by the tangent with the positive direction of the $x$-axis.

• At $x=1$, the tangent makes an angle of $\pi /6$:

  $$f^{\prime }(1)=\tan \left(\frac{\pi }{6}\right)=\frac{1}{\sqrt{3}}$$

• At $x=4$, the tangent makes an angle of $\pi /4$:

  $$f^{\prime }(4)=\tan \left(\frac{\pi }{4}\right)=1$$

Step 2: Solve the definite integral using substitution

We need to evaluate the integral:

$$I={\int }_1^4{f}^{\prime }(x)f^{\prime \prime }(x)dx$$

Let us use the method of substitution. Let:

$$t=f^{\prime }(x)$$

Differentiating both sides with respect to $x$:

$$dt=f^{\prime \prime }(x)dx$$

Step 3: Change the limits of integration

Since we changed our variable from $x$ to $t$, we must also update the upper and lower integration limits:

• When $x=1$ (Lower limit):

  $$t=f^{\prime }(1)=\frac{1}{\sqrt{3}}$$

• When $x=4$ (Upper limit):

  $$t=f^{\prime }(4)=1$$

Step 4: Compute the integration value

Substitute $t$ and $dt$ along with the new limits into the integral:

$$I={\int }_{1/\sqrt{3}}^{1}tdt$$

Now evaluate the standard integral $\int tdt=\frac{t^2}{2}$:

$$I={\left[\frac{t^2}{2}\right]}_{1/\sqrt{3}}^1$$

Apply the upper and lower limits:

$$I=\frac{1}{2}\left[(1{)}^2-{\left(\frac{1}{\sqrt{3}}\right)}^2\right]$$

$$I=\frac{1}{2}\left[1-\frac{1}{3}\right]$$

$$I=\frac{1}{2}\left[\frac{2}{3}\right]$$

$$I=\frac{1}{3}$$

Thus, the value of the given integral is $1/3$.