NIMCET 2007 — Mathematics PYQ

Step 1: Understand the relationship between second derivative and tangent location

The position of a tangent line relative to a curve depends entirely on the concavity of the function:

• If a curve is concave up (f''(x) > 0), any tangent line drawn to the curve lies below the curve.

• If a curve is concave down (f''(x) < 0), any tangent line drawn to the curve lies above the curve.

Therefore, we need to check the sign of the second derivative $f^{\prime \prime }(x)$ in the given interval $x\in \left(0,\frac{\pi }{2}\right)$.

Step 2: Find the first derivative $f^{\prime }(x)$

The given function is:

$$f(x)=\sin x-\tan x$$

Differentiating with respect to $x$:

$$f^{\prime }(x)=\cos x-{\sec }^{2}x$$

Step 3: Find the second derivative $f^{\prime \prime }(x)$

Now, differentiate $f^{\prime }(x)$ with respect to $x$ again:

$$f^{\prime \prime }(x)=\frac{d}{dx}(\cos x)-\frac{d}{dx}({\sec }^{2}x)$$

Using the chain rule for the second term:

$$f^{\prime \prime }(x)=-\sin x-2\sec x\cdot (\sec x\tan x)$$

$$f^{\prime \prime }(x)=-\sin x-2{\sec }^{2}x\tan x$$

Step 4: Analyze the sign of $f^{\prime \prime }(x)$ in the interval $\left(0,\frac{\pi }{2}\right)$

For the first quadrant where $x\in \left(0,\frac{\pi }{2}\right)$:

• \sin x > 0 \implies -\sin x < 0

• \sec x > 0 \implies 2\sec^2 x > 0

• \tan x > 0

Since all individual trigonometric parts are strictly positive in the first quadrant, the terms preceded by negative signs remain negative:

f''(x) = -(\text{positive}) - (\text{positive}) < 0

Because f''(x) < 0 for all $x\in \left(0,\frac{\pi }{2}\right)$, the function is strictly concave down.

Since the curve bends downwards, any tangent line drawn at any point on this interval will always lie above the curve.